Mahesh

Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B Definite Integrals Important Questions Very Short Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2B Definite Integrals Important Questions Very Short Answer Type

Question 1.
Evaluate \(\int_0^a\left(a^2 x-x^3\right) d x\).
Solution:

Question 2.
Evaluate \(\int_0^4 \frac{x^2}{1+x} d x\). [(TS) May ’15]
Solution:

Question 3.
Evaluate \(\int_0^\pi \sqrt{2+2 \cos \theta} d \theta\). [(AP) May ’16, Mar. ’18. ’16]
Solution:

Question 4.
Evaluate \(\int_2^3 \frac{2 x}{1+x^2} d x\). [(TS) Mar. ’20. ’16; Mar. ’17 (AP): Mar. ’12]
Solution:

Question 5.
Evaluate \(\int_0^1 \frac{x^2}{x^2+1} d x\). [(TS) May ’18, ’10]
Solution:

Question 6.
Evaluate \(\int_0^3 \frac{x}{\sqrt{x^2+16}} d x\). [Mar. ’17 (TS)]
Solution:
Put x² + 16 = t² then 2x dx = 2t dt
x dx = t dt
Lower limit: x = 0
⇒ t² = 16
⇒ t = 4
Upper limit: x = 3
⇒ t² = 25
⇒ t = 5

Question 7.
Evaluate \(\int_0^a \frac{d x}{x^2+a^2}\). [(TS) May ’19; (AP) ’15]
Solution:

Question 8.
Find \(\int_0^2 \sqrt{4-x^2} d x\). [Mar. ’07, May ’03]
Solution:

Question 9.
Evaluate \(\int_0^a \sqrt{a^2-x^2} d x\). [(TS) Mar. ’16]
Solution:
Put x = a sin θ then dx = a cos θ dθ
Lower limit: x = 0 ⇒ θ = 0
Upper limit: x = a ⇒ θ = \(\frac{\pi}{2}\)

Question 10.
Evaluate \(\int_0^1 \frac{d x}{\sqrt{3-2 x}}\). [Mar. ’19 (AP)]
Solution:
Put 3 – 2x = t² then -2 dx = 2t dt
dx = -t dt
Lower limit: x = 0
⇒ t² = 3
⇒t = √3
Upper limit: x = 1
⇒ t² = 1
⇒ t = 1

Question 11.
Evaluate \(\int_1^5 \frac{d x}{\sqrt{2 x-1}}\). [(TS) Mar. ’15]
Solution:
Put 2x – 1 = t² then 2dx = 2t dt
dx = t dt
Lower limit: x = 1
⇒ t² = 1
⇒ t = 1
Upper limit: x = 5
⇒ t² = 9
⇒ t = 3

Question 12.
Evaluate \(\int_0^{\pi / 2} \frac{\sin ^5 x}{\sin ^5 x+\cos ^5 x} d x\). [(AP) May ’19, Mar. ’17; Mar. ’14]
Solution:
Let I = \(\int_0^{\pi / 2} \frac{\sin ^5 x}{\sin ^5 x+\cos ^5 x} d x\)

Question 13.
Evaluate \(\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\). [(TS) Mar. ’20, ’14]
Solution:

Question 14.
Evaluate \(\int_0^{\pi / 2} \frac{\cos ^{\frac{5}{2}} x}{\sin ^{\frac{5}{2}} x+\cos ^{\frac{5}{2}} x} d x\). [(AP) May ’15]
Solution:

Question 15.
Evaluate \(\int_0^{\pi / 2} \frac{\sin ^2 x-\cos ^2 x}{\sin ^3 x+\cos ^3 x} d x\). [May ’12]
Solution:

Question 16.
Evaluate \(\int_0^2|1-x| d x\). [(AP) May ’19); Mar. ’15; (TS) Mar. ’18; May ’16]
Solution:

Question 17.
Evaluate \(\int_0^4|2-x| d x\). [(AP) May ’17; ’13]
Solution:

Question 18.
Find \(\int_0^{\pi / 2} \sin ^{10} x d x\)
Solution:

Question 19.
Find \(\int_0^{\pi / 2} \sin ^4 x d x\). [May ’06, ’02]
Solution:

Question 20.
Find \(\int_0^{\pi / 2} \cos ^{11} x d x\). [(TS) Mar. ’19]
Solution:

Question 21.
Find \(\int_0^{\pi / 2} \sin ^4 x \cdot \cos ^5 x d x\). [Mar. ’10]
Solution:

Question 22.
Find \(\int_0^{\pi / 2} \sin ^5 x \cdot \cos ^4 x d x\). [(AP) Mar. ’15]
Solution:

Question 23.
Find \(\int_0^{\pi / 2} \sin ^6 x \cdot \cos ^4 x d x\). [(AP) Mar. ’19; May ’16]
Solution:

Question 24.
Find \(\int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^4 x d x\). [(AP) Mar. ’20, ’18, ’16; May ’16 (TS); Mar. ’13]
Solution:
Let f(x) = sin²x cos⁴x
f(-x) = sin²(-x) cos⁴(-x)
= (-sin x)² (cos x)⁴
= sin²x cos⁴x
= f(x)
∴ f(x) is an even function.
We know that f(x) is even then

Question 25.
Find \(\int_{-\pi / 2}^{\pi / 2} \sin ^3 \theta \cos ^3 \theta d \theta\). [May ’14]
Solution:
Let, f(θ) = sin³θ . cos³θ
New, f(-θ) = sin³(-θ) cos³(-θ)
= -sin³θ . cos³θ
= -f(θ)
∴ f(θ) is an odd function.
∴ \(\int_{-\pi / 2}^{\pi / 2} \sin ^3 \theta \cos ^3 \theta d \theta\) = 0

Question 26.
Find \(\int_0^{2 \pi} \sin ^2 x \cos ^4 x d x\). [(AP) May ’18; (TS) Mar. ’15; ’14]
Solution:
Let f(x) = sin²x cos⁴x
Now f(2π – x) = sin²(2π – x) cos⁴(2π – x) = sin²x cos⁴x = f(x)
Also t(π – x) = sin²(π – x) cos⁴(π – x) = sin²x cos⁴x = f(x)
∴ \(\int_0^{2 \pi} \sin ^2 x \cos ^4 x d x=2 \int_0^\pi \sin ^2 x \cos ^4 x d x\)

Question 27.
Find \(\int_0^\pi \sin ^3 x \cos ^3 x d x\). [(TS) May ’15]
Solution:
Let f(x) = sin³x . cos³x
Now f(π – x) = sin³(π – x) cos³(π – x)
= -sin³x cos³x
= -f(x)
∴ \(\int_0^\pi \sin ^3 x \cos ^3 x d x\) = 0

Question 28.
Find the area bounded by the parabola y = x², the X-axis, and the lines x = -1, x = 2. [(TS) May ’18, ’16; (AP) ’15]
Solution:
Given y = x², X-axis i.e., y = 0
x = -1; x = 2
y = x²

Question 29.
Find the area of the region bounded by y = x³ + 3, X-axis and x = -1, x = 2. [(TS) Mar. ’20; May ’17; Mar. ’12]
Solution:
Given y = x³ + 3, y = 0, x = -1, x = 2

Question 30.
Find the area enclosed between the curves x = 4 – y², x = 0. [Mar. ’11, ’10]
Solution:
Given, x = 4 – y² and x = 0
Solving, 4 – y² = 0 then y = ±2
∴ y = 2 and y = -2

Question 31.
Find the area bounded between the curves y² – 1 = 2x and x = 0.
Solution:
Given y² – 1 = 2x and x = 0 ……..(2)
⇒ x = \(\frac{\mathrm{y}^2-1}{2}\) ……(1)

Solving (1) & (2)
o = \(\frac{\mathrm{y}^2-1}{2}\)
⇒ y² – 1 = 0
⇒ y² = 1
⇒ y = ±1

Question 32.
Find the area enclosed between the curve x² = 4y, x = 2, y = 0.
Solution:
Given x² = 4y, x = 2, y = 0
Solving, x² = 4y = 4(0) = 0

Question 33.
Find the area enclosed between the curves y = x², y = 2x. [May ’13]
Solution:
Given y = x² and y = 2x

Question 34.
Find the area enclosed between the curves, y = x², y = x³. [(TS) May ’19]
Solution:
Given y = x² and y = x³

Question 35.
Evaluate \({Lim}_{n \rightarrow \infty}\left[\frac{1}{n+1}+\frac{1}{n+2}+\ldots \ldots . .+\frac{1}{6 n}\right]\)
Solution:

Question 36.
Evaluate \({Lim}_{n \rightarrow \infty} \frac{1+2^4+3^4+\ldots \ldots \ldots+n^4}{n^5}\). [(AP) Mar. ’20]
Solution:

Question 37.
Evaluate \({Lt}_{n \rightarrow \infty} \sum_{i=1}^n \frac{i}{\mathbf{n}^2+i^2}\)
Solution:

Question 38.
Find the area bounded by y = sin x, x-axis, x = 0, and x = π.
Solution:
Given curve is y = sin x, the x-axis
x = 0 and x = π

Question 39.
Find the area under the curve f(x) = sin x in [0, 2π]. [May ’09]
Solution:
Let y = f(x) = sin x

Question 40.
Find the area of one of the curvilinear triangles bounded by y = sin x, y = cos x, and X-axis. [Mar. ’19 (AP)]
Solution:
Given curves are
y = sin x ……..(1)
y = cos x …….(2)

Solving (1) and (2)
sin x = cos x
⇒ \(\frac{\sin x}{\cos x}\) = 1
⇒ tan x = 1
⇒ x = \(\frac{\pi}{4}\)
OAB is one of the curvilinear triangles bounded by y = sin x, y = cos x and XY-axes
Required area A = \(\int_0^{\pi / 4} \sin x d x+\int_{\pi / 4}^{\pi / 2} \cos x d x\)

Question 41.
Find the area bounded between curves y = x², y = √x. [Mar. ’18 (TS)]
Solution:
Given curves are
y = x² …….(1)
y = √x ………(2)
From (1) and (2)
√x = x²
⇒ x = x⁴
⇒ x(1 – x³) = 0
⇒ x = 0, x³ = 1
⇒ x = 1

Question 42.
Evaluate \({Lt}_{n \rightarrow \infty} \frac{2^k+4^k+6^k+\ldots \ldots+(2 n)^k}{n^{k+1}}\) by using the methods of finding the definite integral of the limit of a sum. [(AP) May ’18]
Solution:

Question 43.
Evaluate \(\int_0^{\pi / 4} \sec ^4 \theta d \theta\). [May ’14]
Solution:

Question 44.
Evaluate \(\int_0^{\pi / 2} \sin ^2 \mathrm{x} d \mathbf{x}\). [May ’95]
Solution:

Question 45.
Evaluate \(\int_{-1}^1 \frac{1}{1+x^2} d x\). [May ’94, Mar. ’92]
Solution:

Question 46.
Evaluate \(\int_0^1 \sin ^{-1} x \mathrm{dx}\). [Mar. ’99]
Solution:

Question 47.
Evaluate \(\int_1^2 \log x d x\). [May ’98, ’94]
Solution:

Question 48.
Evaluate \(\int_0^1 \frac{d x}{e^x+e^{-x}}\). [Mar. ’06]
Solution:

Question 49.
Evaluate \(\int_{\mathbf{a}}^{\mathbf{b}} \frac{|\mathbf{x}|}{\mathbf{x}} \mathbf{d x}\). [May ’03]
Solution:
\(\int_{\mathbf{a}}^{\mathbf{b}} \frac{|\mathrm{x}|}{\mathrm{x}} \mathrm{dx}=[|\mathrm{x}|]_{\mathrm{a}}^{\mathbf{b}}\) = |b| – |a|

Question 50.
Find \(\int_0^{\pi / 2} \cos ^8 x \mathbf{d x}\)
Solution:

Question 51.
Find \(\int_0^{\pi / 2} \sin ^7 x d x\). [Mar. ’17 (AP)]
Solution:

Question 52.
Find \(\int_0^{2 \pi} \sin ^4 x \cdot \cos ^6 x d x\). [Mar. ’19 (TS); (AP) May ’17]
Solution:

Question 53.
Find \(\int_0^\pi \sin ^3 x \cos ^6 x d x\)
Solution:
Given, Let, f(x) = sin³x cos⁶x
Now, f(π – x) = sin³(π – x) cos⁶(π – x)
= sin³x . cos⁶x
= f(x)

Question 54.
Evaluate \(\int_0^\pi \cos ^3 x \cdot \sin ^4 x d x\). [Mar. ’00]
Solution:
Let f(x) = cos³x sin⁴x
Then f(π – x) = cos³(π – x) sin⁴(π – x)
= -cos³x sin⁴x
= -f(x)
∴ f(x) is an odd function
∴ \(\int_0^\pi \cos ^3 x \cdot \sin ^4 x d x\) = 0

Question 55.
Evaluate \(\int_0^{\pi / 2} \frac{1}{1+\tan x} d x\). [Mar. ’02, May ’99]
Solution:

Question 56.
Find the area under the curve f(x) = cos x in [0, 2π].
Solution:
Let Y = f(x) = cos x; y = f(x) = cos x

Question 57.
Find the area enclosed between the curves y = e^x, y = x; x = 0, x = 1.
Solution:

Question 58.
Find the area enclosed between the curves x = 2 – 5y – 3y², x = 0.
Solution:
Given x = 2 – 5y – 3y², x = 0
x = 2 – 5y – 3y²

Solving, 2 – 5y – 3y² = 0
⇒ 3y² + 5y – 2 = 0
⇒ 3y² + 6y – y – 2 = 0
⇒ 3y(y + 2) – 1(y + 2) = 0
⇒ (y + 2)(3y – 1) = 0
∴ y = -2 and y = \(\frac{1}{3}\)

Question 59.
Find the area enclosed between the curve y² = 3x, x = 3.
Solution:
Given y² = 3x and x = 3
Solving, y² = 3(3)
y = ±3
∴ y = 3 and y = -3

Question 60.
Find the area of the right-angled triangle with base b and altitude h, using the fundamental theorem of integral calculus.
Solution:
Let OAB be a right-angled triangle and ∠B = 90°.
Choose ‘O’ as the origin and OB as the +ve x-axis.
If OB = b, and AB = h then A = (b, h).
So, the equation of OA is y = (\(\frac{h}{b}\))x

Question 61.
Find the area bounded between the curves y = x² + 1, y = 2x – 2, x = -1, x = 2. [(AP) May ’16]
Solution:
Given curves are y = x² + 1, y = 2x – 2
x = -1 and x = 2

Question 62.
Find the area cut off between the line y = 0 and the parabola y = x² – 4x + 3,
Solution:
Given, y = x² – 4x + 3 ……(1)
and y = o ……..(2)
Solving (1) and (2)
x² – 4x + 3 = 0
⇒ x² – 3x – x + 3 = 0
⇒ x(x – 3) – 1(2 – 3) = 0
⇒ (x – 1)(x – 3) = 0
⇒ x = 1, 3

Question 63.
Evaluate \(\int_{-\pi / 2}^{\pi / 2} \sin |\mathbf{x}| \mathbf{d x}\). [Mar. ’17 (TS)]
Solution:

Question 64.
Show that \(\int_0^{\pi / 2} \sin ^n x d x=\int_0^{\pi / 2} \cos ^n x \mathbf{d x}\)
Solution:

Question 65.
Evaluate \(\int_0^{\pi / 2} x \sin x d x\). [Mar. ’18 (TS)]
Solution:

Question 66.
Evaluate \(\int_0^a(\sqrt{a}-\sqrt{x})^2 d x\). [Mar. ’19 (TS)]
Solution:

Question 67.
Evaluate \(\int_0^1 x e^{-x^2} d x\)
Solution:
Put -x² = t then -2x dx = dt
x dx = \(\frac{-\mathrm{dt}}{2}\)
Lower limit: x = 0 ⇒ t = 0
Upper limit: x = 1 ⇒ t = -1

Question 68.
Evaluate \(\int_{-1}^2 \frac{x^2}{x^2+2} d x\)
Solution:

Question 69.
Evaluate \(\int_0^{\pi / 2} x^2 \sin x d x\)
Solution:

Question 70.
Evaluate \({Lt}_{n \rightarrow \infty} \frac{\sqrt{n+1}+\sqrt{n+2}+\ldots \ldots \ldots+\sqrt{n+n}}{n \sqrt{n}}\)
Solution:

Question 71.
Evaluate \(\int_0^a \mathbf{x}(\mathbf{a}-\mathbf{x})^{\mathbf{n}} \mathbf{d x}\)
Solution:

Question 72.
Evaluate \(\int_0^{\pi / 2} \tan ^5 x \cos ^8 x d x\)
Solution:

Question 73.
Find the area bounded by the curves y = sin x and y = cos x between any two consecutive points of intersection. [Mar. ’18 (AP)]
Solution:

Question 74.
Evaluate \(\int_0^\pi \sin x d x\)
Solution:
\(\int_0^\pi \sin \mathrm{x} d \mathrm{x}=[-\cos \mathrm{x}]_0^\pi\)
= -[cos π – cos 0]
= -[-1 – 1]
= -(-2)
= 2

Question 75.
Evaluate \(\int_2^3 \frac{2 x}{1+x^2} d x\)
Solution:
Put 1 + x² = t, then 2x dx = dt
L.L: x = 2 ⇒ t = 5
U.L: x = 3 ⇒ t = 10

Question 76.
Evaluate \(\int_1^4 x \sqrt{x^2-1} d x\)
Solution:
Put x² – 1 = t²
⇒ 2x dx = 2t dt
⇒ x dx = t dt
L.L: x = 1 ⇒ t = 0
U.L: x = 4 ⇒ t = √15

Question 77.
Evaluate \(\int_0^\pi(1+\cos x)^3 d x\)
Solution:

Question 78.
Evaluate \(\int_0^5(x+1)\) dx as a limit of a sum.
Solution:

Question 79.
Evaluate \(\int_0^4 x^2 d x\) as a limit of a sum.
Solution:

Question 80.
Find the area in sq. units bounded by the x-axis, part of the curve y = 1 + \(\frac{8}{x^2}\), and the ordinates x = 2 and x = 4.
Solution:

Question 81.
Find the area bounded by the curve y = log x, the x-axis, and the straight line x = e.
Solution:
Given that y = log x ………(1)
x-axis, i.e., y = 0 ……….(2)
x = e
Solve (1) & (2)
0 = log e^x
⇒ x = e⁰
⇒ x = 1

= (e log e – e) – (log 1 – 1)
= (e – e) – (0 – 1)
= 0 – (-1)
= 1 sq. units

Question 82.
Find the area enclosed within the curve |x| + |y| = 1.
Solution:
Given curve is |x| + |y| = 1
⇒ ±x ± y = 1
Given curve represents 4 straight lines
Take x + y = 1 ⇒ y = 1 – x

Telangana TSBIE TS Inter 2nd Year Accountancy Study Material 2nd Lesson Consignment Accounts Textbook Questions and Answers.

TS Inter 2nd Year Accountancy Study Material 2nd Lesson Consignment Accounts

Short Answer Questions

Question 1.
Distinguish the differences between the consignment and sales.
Answer:
Consignment Sending of goods from the owner to his agents for sale on a commission basis is called consignment. The agent only sells the goods on behalf of the owner for the sake of commission at the risk of the owner.

Question 2.
How the unsold stock is valued in consignment?
Answer:
At the end of the accounting period, the unsold goods should be valued properly. Other wise, the true profit cannot be ascertained. The value of unsold stock includes the purchasing price of unsold stock and proportionate direct expenses or nonrecurring ex¬penses incurred by the consignor and consignee.
Valuation of unsold stock.

Question 3.
How the abnormal loss is valued in consignment?
Answer:
I. Introduction:
1) Abnormal loss is a loss which arises due to fire, accidents, riot, flood, theft, pilfer¬age, sabotage, negligence, inefficiency etc.

2) It is unnatural and unexpected. This loss is beyond the control of human being. This loss can be insured.

3) Sometimes, insurance company admits the claim in part or full. The value of abnor¬mal loss is ascertained and it is taken on the credit side of consignment account.

4) The method of calculation of abnormal loss is to the method of calculation of clos¬ing stock.
Abnormal loss = Cost of goods sent on consignment + Expenses of consignor x

II. Entries:
1. When there is no insurance claim: –

2. When insurance claim is admitted by insurance company:

Question 4.
List out some of the recurring and non-recurring expenses of the consignor and consignee.
The following table gives details of recurring expenses in curred by consignor and consignee:
Recuring Expenses:

Non-Recuring Expenses:

Question 5.
Proforma of consignment accounts in the books of consignor.
Answer:
Proforma of consignment account in the books of consignor:

Very Short Answer Questions

Question 1.
What is consignment?
Answer:
The process of sending goods by consignor to a consignee for the purpose of sale on commission basis is called consignment.

Question 2.
Parties in consignment.
Answer:
There are two parties in consignment They are: A) consignor B) consignee.
A) Consignor: The person who send the goods on consignment is called consignor (or) principal.
B) Consignee: The person to whom goods are sent is called consignee or agent.

Question 3.
Proforma Invoice.
Answer:

1. The statement which is prepared by the consignor and send to the consignee stating that the details of the goods consigned is called proforma Invoice.
2. It contains quantity, weight, expenses incurrred, mode of transportation and minimum price at which the goods are to be sold.

Question 4.
Account sales.
Answer:

1. Consignee has to inform its consignor about the amount of sales and expenses in¬curred by him for making sales. For this purpose, the consignee has to prepare a state¬ment which is known as account Sales.
2. Account sales is a statement showing that the amount of sales, other expenses in¬curred, advance remitted by him and amount payable by him to the consignor.

Question 5.
Delecredere commission.
Answer:

1. It is the extra commission than the ordinary commission which is allowed by the consignor to the consignee for bearing the risk of bad debts arising out of credit sales made by him.
2. When this commission is allowed to the consignee, the consignee has to bear the loss due to the bad debts.

Question 6.
Over-riding commission.
Answer:

1. It is an additional commission paid by the consignor to the consignee to enhance the sales or for selling goods at prices higher than prices fixed by the consignor or boost up the sales of new products.
2. This commission is also calculated on gross proceeds of sales.

Question 7.
Invoice price.
Answer:

1. The consignor, instead of sending the goods on consignment at cost price, may send it at a higher price than the cost price. This price is known as Invoice price or selling price.
2. The intention of the consignor is that to hide or conceal the actual profit earned by him from the consignee.

Question 8.
Loading.
Answer:

1. The difference between the cost price and the invoice price of goods is known as “Loading”.
2. It also called as “Excess price over the Loading = Invoice price – Cost price
3. Loading = Invoice price – Cost price

Question 9.
Normal loss.
Answer:

1. Normal loss is a loss that arises due to evaporation, leakage and breaking the bulk material into small pieces.
2. It is a natural, unavoidable and inherent in the nature of goods sent on consignment.

Question 10.
Abnormal Loss.
Answer:

1. Abnormal loss is a loss which arises due to fire, accidents, riot, flood, theft, pilferage, sabotage, negligence, inefficiency etc.
2. It is unnatural and unexpected. This loss is beyond the control of human being. This loss can be insured.
3. Sometimes, insurance company admits the claim in part or full. The value of abnor¬mal loss is ascertained and it is taken on the credit side of consignment account.
4. The method of calculation of abnormal loss is to the method of calculation of closing stock.
   Abnormal loss = Cost of goods sent on consignment + Expenses of consignor x
   (Damaged or lost quantity/Total quantity of goods consigned)

Practical Problems

A) Short Problems:

1. The cost of price of goods is 20,000. But the invoice price of the goods is 24,000. Find the loading?
Answer:
Given:
Cost of price of goods = 20,000
The invoice price of goods = 24,000
Rs. Loading = Invoice Price – Cost price
Loading = 24,000 – 20,000 = 4,000.

Question 2.
The cost of the goods sent on consignment is Rs. 20,000 but the invoice price i sl0% more than the cost price, what will be the invoice price of the goods?
Answer:
Method (1):
Given:
Cost of goods consign = 20,000
Invoice price = 10% more than the cost price
= cost price + 10% on cost price (i.e., loading)
= 20,000 + [20,000 x 10/100]
= 20,000 + 2000
Rs. Invoice price = 22,000

Method (2)
Given:
Cost of goods consign = 20,000

Question 3.
Jeevan Surya sent goods on consignment of invoice price Rs. 15,000. This price is 20% more than the cost price. What will be cost price of the goods?
Answer:
Given that:
Invoice price = 15,000

Cost of goods = Invoice price – Loading
= 15,000 – 2,500
= 12,500

Question 4.
500 cases of goods were consigned at a cost of Rs. 150 each. The consignor paid Rs. 2,000 towards insurance and freight. The consignee paid t 2,000 for carriage and Rs. 1,000 for salaries Consignee sold only 400 cases of goods. Find the value of unsold goods?
Answer:
Total cases of goods consigned = 500 cases
(-) Consignee sold = 400 cases
unsold goods = 100 cases

Valuation of unsold stock of consignment:
Cost of unsold goods (100 cases x 150/-)= 15,000
Add roportionate non recurring expenses of consignor = 400

Proportionate non recurring expenses of consigner = 400

value of unsold goods = 15,800

Question 5.
Hardhikpatil consigned goods costing Rs. 50,000 to Vidhvanpatilwhose recurring and non-recurring expenses on the same amounted to Rs. 5,000 and t 2,000 respectively. Vidhvanpatil sold 3/4th of the goods. Find the value of unsold goods?
Answer:
Cost of goods consigned = 50,000
Given that vidhvanpatil sold ¾th goods
unsold goods = ¼th goods = 50,000 x ¼ = 12,500

Value of unsold goods :

Question 6.
Nidhi consigned 100 bales of cloth to Srikari at Rs. 1,000 per bale. Nidhi incurred the expenses Rs. 5,000. Srikari incurred Rs. 6,000 for packing and Rs. 2,000 for rent. Srikari sold 80 bales of cloth at the rate Rs. 1,500 per bale. Ascertain the value of consignment stock?
Answer:
Total goods consiged = 100 bales of cloth
(-) Sold bales of cloth = 80 bales
.*. Unsold bales of cloth = 20 bales

valuation of consignment stock:

Question 7.
Ascertain the value of closing stock from the following information.
No. of Units consigned : 5,000
Cost of each unit Rs. 10
Expenses of consignor Rs. 4.000
Expenses consingee Rs. 1,000.
No. of units sold by the consignee : 4,000
Answer:
Total goods consigned = 5000
(-) Sold goods = 4000
.-. Unsold goods = 1000

Value of consignment stock :

Question 8.
Anisa consigned goods consting Rs. 20,000 to Prasanna on consignment basis. Anisa paid Rs. 5,000 for carriage and cartage. On the way l/5th of the consignment was lost by fire. Ascertain the amount of abnormal loss.
Ans Total cost of goods consigned = 20,000
Cost of goods loss by fire = 1/5 on 20,000
The amount of Abnormal loss = = 20,000 x 1/5 = 4,000

The abnormal loss value is Rs. 5,000.

Question 9.
100 tons of coal is consigned at the rate of Rs. 900 per ton. Freight charges paid by the consignor amounted to Rs. 5,000. Loss due to loading and unloading of coal is estimated at 5 tones. 75 tons of coal is sold at the rate of Rs. 1,200. Find the value of unsold stock.
Answer:
Completion of unsold stock :

Value of unsold stock

Unsold stock value = 20,000

B) Long Problems

Question 1.
Manasa consigned goods value of Rs. 60,000 to Suhas. Manasa paid transport charges Rs. 4,000. Suhas sent the account sales of consignment stating that the entire stock was sold for Rs. 75,000. He paid Rs. 2,000 for cartage. His is entitled comission of Rs. 3,000. He sent bank draft for the balance. Write journal entries in the books of both the parties.
I. In the books of Manasa (consignor) :

Ledger Accounts in the books of consignor Manasa
Journal Entries

II. In the books of Suhas (Consignee) :
Journal Entries

Ledger Accounts :

Question 2.
Hari of Hyderabad consigned goods value Rs. 30,000 to Sumaiatha of Surath. Ilari paid for forwarding charges Rs. 1,500 and drew a bill of two months on sumaiatha for Rs. 15,000. The bill was discounted for Rs. 14,500. Sumaiatha sent account sales of consignment stating that the entire stock was sold for Rs. 38,000, agent commission Rs. 2,000 and sent bank draft for the balance. Write journal entries in the books of both the parties.
Answer:
In the books of Hari consignor :
Journal Entries

Ledger Accounts

II. In the books of Sumalatha consignee :
Journal Entries


Ledger Accounts

Question 3.
Arjun of Kamareddy and Vittal of Karimnagar are in consignment business. Vittal sent goods to Arjun Rs. 20,000. Vittal paid freight Rs. 800, insurance Rs. 700. Arjun met sales expenses rs. 750. Arjun sold the entire stock for Rs. 30,000 and he is entitled to a commission of 5% on sales. Prepare necessary ledger accounts in the books of Vittal and Arjun.
Answer:
1. in the books of Vittal (consignor) :

Question 4.
Shivabhiram of Secunderabad consigned to Anil of vijavawada 100 boxes of medi¬cines at Rs. 500 each. Shivabhiram paid freight and insurance amounting to Rs. 2,500. Anil sent Rs. 20,000 as an advance payment against consignment. Shivabhuiram received account sales from Anil containing the following particulars :
a. Gross sale proceeds are Rs. 74,000.
b. Transportaion and warehousing charges Rs. 2,500.
c. Commission charged at 10% on gross proceeds.
Prepare necessary ledger accounts in the books of consignor and consignee.
Answer:
I. In the books of Shivabhiram (consignor) :


In the books of Anil (Consignee) :

Given : Total stock = 80,000
Sold = 80%
remaining unsold stock cost = 20% i.e., = 80,000 x 20/100 = 16,000

Question 5.
Karthik of Karimnagar sent sports material worth Rs. 80,000 to Supreeth on a consign¬ment basis. Karthik spent Rs. 2,000 for insurance. Supreeth while taking the goods spent Rs. 1,000 for transport and Rs. 1,500 for Godown rent. 80% of stock was sold out. Calculate the value of unsold stock.
Answer:
Valuation of Unsold stock

Question 6.
On 1st January, 2020 Srinu of Srinivasanagar consigned goods value Rs. 30,000 to Arun of Alwal. Srinu paid cartage and other expenses Rs. 2,000. On 31st March, 2020 Arun sent an account sales with the following information :
a. 50 of the goods sold for Rs. 5,000.
b. Arun incurred expenses Rs. 1,750.
c. Arun entitled to receive commission @ 6% on sales.
Bank draft was enclosed for the balance due. Pass necessary journal entries in the books of both the parties.
Answer:
II. In the books of Srinu (Consignee)
Journal Entries

Ledger accounts :

Working Note :
Valuation of Unsold stock

II. In the books of Aran (Consignee) :
Journal Entries


Ledger accounts:

Question 7.
On 1-1-2020 Bajaji of Hyderabad consigned goods valued at Rs. 50,000 to Shivaji of Sholapur. Balaji paid cartage and other expenses Rs. 2,400. On 31-3-2020 sent the account sales with the following information.
a. 3/4th of the goods sold for Rs. 48,000
b. Shivaji incurred expenses amounting to Rs. 1,200.
c. Shivaji is entitled to receive commission @ 5% on sales.
Bank draft was enclosed for the balance. Prepare necessary ledger accounts in the books of Balaji.
Answer:
Ledger Accounts in the books of Balaji

Working Note :
Valuation of Consignment stock (unsold stock)

Question 8.
Amith & Co of New Delhi consigned 50 cell phones to Teja & Co. of Hyderabad. The cost of each cell phone was Rs. 2,500. Amit & Co paid Insurance Rs. 500, Freight Rs. 2,500/ am account sale was received from Teja & Co showing the sales of 40 cell phones a Rs. 3.000 each. The following expenses were deducted by them :
Selling expenses : Rs. 1.600. Commission : Rs. 3,000. Amith & Co received a bank draft for the balance due. Prepare important ledger account in the books of Amith & Co and Teja & Co.
Answer:
Ledger accounts in the books of Amith & Co (consignor) :


II. In the books of Teja and Co consignee :

Working Note :
Valuation of closing stock

Question 9.
Naveen of Agra sends 50 IV sets each costing Rs. 15,000 to Sravan of Warangal to be on a consignment basis. He incured the following expenses ; freight Rs. 2,000 and insurance X 5,000. Sravan sold 45 TVs for Rs. 7,50,000 and paid X 10,000 as shop rent, which is borne by Naveen as per terms and conditions of the consignment. Consignee is entitled for a commission of X 200 per TV sold. Assuming that, Sravan settled the account by sending the bank draft to Naveen. Prepare necessary ledger accounts in the books of Naveen and Sravan.
Answer:
Ledger accounts in the books of Naveen (consignor) :


Ledger account in the books of Sravan : (consignee) :

Working Note :
Valuation of unsold stock

Question 10.
Akhil of Hyderabad consigned goods worth Rs. 20,000 to his agent Anil of Kamareddy on consignment. Akhil spend Rs. 1,000 on transport, Rs. 500 on insurance. Anil sent Rs. 5,000 as advance. After two months, Akhil received the account sales as follows:
a. Half of the goods were sold f 24,000, of which Rs. 4,000 sold credit.
b. Selling expenses were Rs. 1,200.
c. Ordinary commission 8% and delecredere commission 2% on sales.
d. Bad debts to beamounted to Rs. 400.
Give ledger accounts in the books of both the parties.
Answer:
Ledger accounts in the books of Akhil (consignor) :


Working Note :
Valuation of consignment stock

Question 11.
Manaswi of Medak consigned 50 boxes of medicine to Vedhagna of yellareddy. The cost of each box is X 800 and invoice price is X 1,000 each box. Manaswi paid Rs. 2,500 for forwarding goods. Vedhagna advanced X 25,000 to Vedhagna and sent ac¬count sales with the details sales 40 boxes at the rate of X 1,000 each. Selling expenses are X 1,000 and his commission is X 1,500. Pass necessary journal entries in the books of Manaswi.
Answer:
Journal Entries




Working Note :
Valuation of consignment stock

Question 12.
Ragamai of Kochin consigned 200 cases of ayurvedic medicines to Sricharani of Hyderabad. The cost of each case is Rs. 400 but the invoice price is Rs. 500 per case. Ragamai paid Rs. 1,200 towards packing sand carriage. Sricharani informed through account sales that 180 cases are sold at Rs. 520 each. Expenses paid by Sricharani were freight Rs. 800. Commission has to be calculated at 5% on sales. Prepare Consignment account and consignee’s account in the boooks of Ragami.
Answer:
Ledger accounts in the books of Ragamai

Working Note :
Valuation of unsold stock on consignment

Question 13.
Rajani of Nagpur consigned goods to Praveen of Patna. The value of Rs. 40,000 in¬voiced at 25% above the cost. Rajani paid Rs. 1,500 for carriage and Rs. 500 for insur¬ance. Rajau drew a bill on Praveen for Rs. 20,000 payable after 3 months and received Praveen’s acceptance. Later, Rajani received account sales from Praveen stating the following : Total goods sold for Rs. 60,000 and the sales expenses Rs. 1,000 and his commission 5% on sales.
Praveen deducted all the above and sent a cheque for remaining balance. Prepare necessary accounts in the books Rajani and Praveen.
Answer:


Working Note :
Given cost of the goods = 40,000
Invoice price = cost + loading
= 40,000 + 25% on cost
= 40,000 + [ 40,000 x 25/100 ]
= 40,000 + 10,000 = Invoice price = 50,000

Question 14.
Reepthika Reddy sent goods to Rishika Reddy of Adilabad on consignment worth Rs. 30,000 at invoice price of cost plus 20%. Reepthika Reddy paid transport and insurance charges of Rs. 800. Rishika Reddy sent an account sales by showing 3\4th goods are sold for Rs. 28,000 and sales expenses Rs. 700. She sent a draft for an amount due after deducting 5% commission. Give the necessary accounts in the books of both the parties.
Answer:
Ledger accounts in the books of Reepthika Reddy :

In the books of Rishika Reddy a/c:

Working Note :
1) Invoice price = Cost price + 20% on cos
= 30,000 + [30,000 x 20/100]
= 30,000 + 6000 = 36,000
2) Valuation of consignment stock

3) Stock reserve = 6,000 x 1/4 = 1,500

Question 15.
Srinivas of Srinagar consigned 10,000 cases of tinned fruits costintg Rs. 75,000 to Kiran of Kanpur on 1-1-2020 charging him a pro forma invoice price to show a profit of 25% on sales. Srinivas paid Rs. 6,000 towsards freight. During the half year ending June. 30, 2020 Kiran incurred Rs. 4,000 in expenditure and reported sales of 8,000 cases for Rs. 82,000. Kiran was entitled to a commission of 5% on sales. He has sent a demand draft for the balance due to Srinivas. Prepare the necessary ledger accounts in the books of Srinivas.
Answer:
Ledger accounts in the books of Srinivas :



Working Note :
1) Invoice price = cost + profit [ie., 25% on sales]

2) Valuation of unsold stock :

Question 16.
Arun sent 100 sewing machines on consignment basis to Tarim. The cost of each machine was Rs. 300, but the consignor prepared the proforma invoice at 25% abobe the cost. The company spent Rs. 800 on various expenses.
After receiving the machines, Tarun had to spend Rs. 950 for freight and Rs. 1,100 for godown rent. By the end of the year Tarun sold 80 Machines @ 1410 per machine. He was entitled to a commission of 5% on sales. Prepare necessary accounts in the books of Arun,
Answer:
In the books of Arun :

Working Note :
1) Invoice price = cost + 25% on cash
= 30,000 + 30,000 x 25/100
= 30,000 + 7,500 = 37,500.
2) Valuation of unsold stock of consignment

3) Stock Reserve = 20 x 75 = 1,500

Question 17.
500 sanitizer bottles were consigned by Anuhya and co. Hyderabad to Nithya and co. Secunderabad costing Rs. 500 each. Expenses incurred by Anuhya and co. amounted to Rs. 5,000. On the way 5 sanitizer bottles were completely damaged due to bad han¬dling and insurance company admitted the calim to the extent of Rs. 2,500.
Nithya and co, took delivery of the remaining sanitizer bottles and incurred ex¬penses Rs. 12,000. Nithya and co. sold 495 sanitizer bottles for Rs. 40,000, It is entitled to a commission of 5% on sales. Prepare Consignment account, Nithya and co. a/c in the books of Anuhya and co. assuming that Nithya and co. remitted amount due by them.
Answer:
In the books of Anuhya and Co :

Working Note :
Abnormal loss :

Question 18.
Raja oil mills, kamareddy consigned 500 kgs of ghee to the Vijay dealers of Yellareddy. Cost of each Kg. is 800. Raja oil mills paid Rs. 5,000 for various expenses. During the transit, 50 kgs of ghee were accidentally destroyed.
Vijay received remaining stock of ghee and reported that entire ghee is sold for Rs. 5,00,000 and his expenses are 12,000. He is entitled to a 5% commission on sales. Prepare necessary accounts in the books of Raja oil mills assuming that Vijay remitted amount due by them.
Answer:

Working Note:

Question 19.
Pavan Kumar of Hyderabad consigns to Kiran Kumar of Nizamabad 50 eases of goods at a cost of Rs. 500 per case, Pavan Kumar incurred the expenses Rs. 500. Kiran Kumar paid Rs. 400.4 cases were destroyed in transit. t
Kiran Kumar received remaining stock and sold entire stock for Rs. 30,000. He is entitled for a commission of 5% on sales and sent the bank draft for full settlement of account. Show necessary accounts in the books of Pavan Kumar.
Answer:

Working Note:

= 2,040

Question 20.
On 1 January, 2020 Bharat coal company Ltd. Consigned to Vijay dealers, vijayawada. 400 tons of coal cost of the coal being Rs. 180 per ton. The company had paid Rs. 6,000 towards freight and insurance.
Vijay took delivery of the goods consigned on 10th January, 2020. On 31st March, 2020 the consignee reported that :
1. There was a shortage of 10 tons due to loading and unloading of the coal, (Treated as normal loss)
2. 380 tons were sold @ Rs. 250 per ton.
3. He had paid Rs. 3,000 towards godown rent and selling expenses.
Vijay dealers were entitled to a commission of Rs. 4,000. Show the necessary accounts in the books of Bharath Coal Company Ltd.
Answer:
In the books of Bharath Ltd. coal company :


Working Note :
1) Valuation of unsold stock :
Unsold coal tons = total – Normal loss – sold .
= 400 tons – 10 tons – 380 tons unsold =10 tons
= 2,000

consignment stock = 2,000

Telangana TSBIE TS Inter 2nd Year Chemistry Study Material 9th Lesson Biomolecules Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material 9th Lesson Biomolecules

Very Short Answer Questions (2 Marks)

Question 1.
Define Carbohydrates.
Answer:
Carbohydrates are optically active polyhydroxy aldehydes or ketones or molecules which provide such units on hydrolysis.

Question 2.
Name the different types of carbohydrates on the basis of their hydrolysis. Give one example for each.
Answer:
Carbohydrates are classified into three types on the basis of their hydrolysis.

1. Monosaccharides Ex: glucose
2. Oligosaccharides Ex: sucrose
3. Polysaccharides Ex: starch

Question 3.
Why are sugars classified as reducing and non – reducing sugars ?
Answer:
In monosaccharides, whether they are aldoses or ketoses, the functional groups are free and reduce Fehling’s solution etc. In some disaccharides such as sucrose, the reducing groups of the monosaccharides are bonded. They do not show reducing properties. On the other hand, in sugars such as maltose and lactose, the functional groups are free and reduce Fehling’s solution. Hence, sugars are classified into reducing and non – reducing sugars.

Question 4.
What do you understand from the names
a) aldopentose and
b) ketoheptose ?
Answer:
a) Aldopentose means a monosaccharide having an aldehyde group and contains five carbon atoms in its molecule.
b) Ketoheptose means a monosaccharide having a ketogroup and contains seven carbon atoms in its molecule.

Question 5.
Write two methods of preparation of glucose. [IPE 14]
Answer:
Preparation of glucose:
1) From sucrose (Cane sugar): Sucrose on boiling with dilute HCl or H₂SO₄ in alcoholic solution undergoes hydrolysis and gives equal amounts of glucose and fructose. On cooling the resulting solution, glucose being much less soluble than fructose, separates out.

2) From starch : Commercially glucose is made by the hydrolysis of starch by heat¬ing with very dilute sulphuric acid at 120°C under pressure.

Question 6.
Glucose reacts with bromine water to give gluconic acid. What information do you get from this reaction about the structure of glucose ?
Answer:
Glucose is oxidised by a mild oxidising agent like bromine water to give gluconic acid, a monocarboxylic acid with molecular formula C₆H₁₂O₇. This indicates the presence of an aldehyde group since only the aldehyde group can be oxidised to an acid by gaining one oxygen atom without losing any hydrogen atoms.

Question 7.
Glucose and gluconic acid on oxidation with nitric acid give saccharic acid. What information do you get from this reaction about the structure of glucose ?
Answer:
On oxidation with nitric acid, glucose as well as gluconic acid both yield a dicarboxylic acid called saccharic acid with molecular formula C₆H₁₀O₈. This indicates the presence of a primary alcoholic group.

Question 8.
Glucose reacts with acetic anhydride to form penta acetate. What do you understand about the structure of glucose from his reaction ?
Answer:
Acetylation of glucose with acetic anhydride gives glucose penta acetate which confirms the presence of five -OH groups. Since it exists as a stable compound the five -OH groups should be attached to different carbon atoms.

Question 9.
Give any two reasons to understand that glucose molecule has no open chain structure.
Answer:

1. Although glucose has am aldehyde group, it does not give the Schiff’s test and does not form addition product with sodium bisulphite.
2. Glucose pentaacetate does not react with hydroxylamine.
   These observations suggest that glucose has no open structure.

Question 10.
D – glucose means dextro rotatory glucose. Is it true ? why ?
Answer:
The letter D’ before the name of glucose does not mean dextro rotatory. ‘D’ represents relative configuration with reference to glyceraldehyde. On the other hand, ’+’ is used to represent the dextro rotatory nature of glucose.

Question 11.
What are anomers ?
Answer:
Glucose forms a six – membered ring involving the -OH at C – 5 and the -CHO group at C – 1. Thus glucose exists in two cyclic hemiacetal forms which differ in their stereo chemistry at C₁, called the anomeric carbon (former aldehyde carbon). Such isomers i.e., α – form and β – form, are known as anomers.

Question 12.
Write the ring structure of D – Glucose. What are their names ?
Answer:

Question 13.
Write the ring structure and open chain structure of fructose.
Answer:
Ring structure and open chain structures of fructose:

Question 14.
What do you understand by invert sugars?
Answer:
Sucrose is dextrorotatory but after hydrolysis gives an equimolar mixture of glucose which is dextrorotatory and fructose which is laevorotatory. The hydrolysis of sucrose brings about a change in the sign of rotation, from dextro (+) to laevo (-) and the product is called invert sugar.

Question 15.
What are amino acids? Give two examples.
Answer:
Compounds containing both amino (-NH₂) and carboxyl (-COOH) functional groups are called amino acids.
Ex: Glycine, Alanine.

Question 16.
Write the structure of alanine and aspartic acid.
Answer:
NH₂ – CH₂ – COOH Alanine

Question 17.
What do you mean by essential amino acids ? Give two examples for non – essential amino acids. [TS 16; IPE 14]
Answer:
Amino acids, which cannot be synthesised in the body and must be obtained through diet are called essential amino acids. Examples for non – essential amino acids : glycine, alanine.

Question 18.
What is Zwitter ion ? Give an example.
Answer:
In an amino acid both acidic (carboxyl group) and basic (amino group) groups are present in the same molecule. In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton, giving rise to a dipolar ion known as Zwitter ion.
Ex: H₂N – CH₂ – COOH ⇌ H₃ \(\stackrel{+}{\mathrm{N}}\) -CH₂ – \(\mathrm{CO} \overline{\mathrm{O}}\)

Question 19.
What are proteins ? Give an example.
Answer:
Proteins are polymers of α – amino acids. They are the most abundant biomolecules of the living system.
Ex : Insulin

Question 20.
What are fibrous proteins ? Give examples.
Answer:
Proteins in which the polypeptide chains run parallel and are held together by hydrogen and disulphide bonds will have fibre – like structure. Such proteins are called fibrous proteins.
Ex : Keratin (present in hair, wool, silk)

Question 21.
What are globular proteins? Give examples.
Answer:
Proteins in which the polypeptide chains coil around to give a spherical shape are called globular proteins.
Ex : Insulin

Question 22.
How are proteins classified with respect to peptide bond ?
Answer:
With respect to peptide bond, proteins are classified into dipeptides, tripeptides, tetra- peptides, pentapeptides, hexapeptides etc depending upon the number of amino acids linked by peptide bonds. When the number of amino acids is more than ten, then the products are called polypeptides.

Question 23.
What are the components of a nucleic acid?
Answer:
There are three components of a nucleic acid. They are:

1. A pentose sugar
2. Phosphoric acid and
3. Nitrogen containing heterocyclic base.

Question 24.
Write the names of three types of RNA.
Answer:

1. Messenger RNA ( m – RNA)
2. Ribosomal RNA (r – RNA) and
3. Transfer RNA (t – RNA)

Question 25.
Write the biological functions of nucleic acids.
Answer:
Biological functions of nucleic acids:
1) Nucleic acids control heredity at molecular level. DNA is the chemical basis of heredity. It may be regarded as the reserve of genetic information. DNA is exclusively responsible for maintaining the identity of different species of organisms over millions of years. A DNA molecule is capable of self duplication during cell division and identical DNA strands are transferred to daughter cells.

2) Nucleic acids have an important role in protein synthesis. Actually the proteins are synthesised by various RNA molecules in the cell but the message for the synthesis of a particular protein is present in DNA.

Question 26.
Name the vitamin responsible for the coagulation of blood.
Answer:
Vitamin K

Question 27.
What are monosaccharides ?
Answer:
Carbohydrates which cannot be hydrolysed into smaller units are called monosaccharides.

Question 28.
What are reducing sugars ?
Answer:
Carbohydrates which reduce Fehling’s solution and Tollen’s reagent are called reducing sugars.

Question 29.
Write two main functions of carbohydrates in plants.
Answer:
Carbohydrates mainly act as the food storage or structural materials in plants.

1. Starch is the main storage polysaccharide of plants.
2. Cell wall of plants is made up of cellulose.

Question 30.
Classify the following into monosaccharides and disaccharides.
i) Ribose
ii) 2 – deoxy ribose
iii) Maltose
iv) Fructose
Answer:
i) Ribose …………… Monosaccharide
ii) 2 – deoxy ribose …………… Monosaccharide
iii) Maltose …………… Disaccharide
iv) Fructose …………… Monosaccharide

Question 31.
What do you understand by the term glycosidic linkage ?
Answer:
The two monosaccharide units in a disaccharide are joined together by an oxide linkage formed by the loss of a water molecule. Such a linkage is called glycosidic linkage.

Question 32.
What is glycogen ? How is it different from starch ?
Answer:
Carbohydrates are stored in animal body as glycogen. It is also known as animal starch. The structure of glycogen is similar to amylopectin, a component of starch, but it is more highly branched.

Question 33.
What are the hydrolysis products of
i) sucrose and
ii) lactose ?
Answer:
i) Sucrose on hydrolysis gives D – (+) – glucose and D – (-) – fructose.
ii) Lactose on hydrolysis gives galactose and glucose.

Question 34.
What is the basic structural difference between starch and cellulose ?
Answer:
Starch is a polymer of α – glucose. The α – D – glucose units are joined by glycosidic linkage. In amylose, which constitutes about 15 – 20% of starch, 200 -1000 α – D(+) – glucose units are held by C – 1 to C – 4 glycosidic linkage in a long unbranched chain. Amylopectin, the other component which constitutes about 80 – 85% of starch is a branched chain polymer of α – D – glucose units in which the chain is formed by C -1 to C – 4 glycosidic linkage whereas branching occurs by C -1 to C – 4.

Cellulose is a straight chain polysaccharide composed only of β – D – glucose units which are joined by glycosidic linkage between C -1 of one glucose unit and C – 4 of the next glucose unit.

Question 35.
What happens when D’- glucose is treated with the following reagents.
i) HI
ii) Bromine water
iii) HNO₃
Answer:
i) On prolonged heating with HI, glucose forms n – hexane.

ii) When glucose is treated with bromine water it gets oxidised to gluconic acid, a monocarboxylic acid with the molecular formula, C₆H₁₂O₇.

iii) On oxidation with nitric acid, glucose gives a dicarboxylic acid called saccharic acid, with molecular formula, C₆H₁₀O₈.

Question 36.
Enumerate the reactions of D – glucose which cannot be explained by its open chain structure.
Answer:
The following reactions of D – glucose cannot be explained by its open chain structure.

1. Glucose does not give Schiff’s test and it does not form the hydrogen sulphite addition product with NaHSO₃.
2. The pentaacetate of glucose doesnot react with hydroxylamine indicating the absence of free -CHO group.

Question 37.
What are essential and non – essential amino acids ? Give one example for each.
Answer:
Amino acids which cannot be synthesised in the body and must be obtained through diet are called essential amino acids.
Ex: Valine

Amino acids which are synthesised in the body are known as non – essential amino acids.
Ex: Glycine

Question 38.
Define the following as related to proteins.
i) Peptide linkage
ii) Primary structure
iii) Denaturation
Answer:
i) Peptide linkage :
Peptide linkage is an amide formed between -COOH group of one amino acid molecule and the NH₂ group of another molecule of the same or different amino acid.

ii) Primary structure :
The primary structure of proteins refers to the sequence of amino acids held together by peptide linkages.

iii) Denaturation :
The loss of biological acitvity of a protein due to the destruction of the highly organized tertiary structure is called denaturation.
Ex: coagulation of egg by boiling

Question 39.
What are the common types of secondary structure of proteins ?
Answer:
The secondary structure of proteins is of two common types:

1. α – helix and
2. β – pleated sheet structure

Question 40.
What type of bonding helps in stabilizing the a – helix structure of proteins ?
Answer:
The spiral or α – helix is stabilized by interchain hydrogen bonding. The -NH group of each amino acid residue is hydrogen bonded to the = O of an adjacent turn of the helix in the same chain.

Question 41.
Differentiate between globular and fibrous proteins.
Answer:
In globular proteins the polypeptide chains coil around to give a spherical shape. These are usually soluble in water.

In fibrous proteins the polypeptide chains run parallel and are held together by hydrogen and disulphide bonds. These proteins have a fibre like structure. These are usually insoluble in water.

Question 42.
How do you explain the amphoteric behaviour of amino acids ?
Answer:
Amino acids show both acidic and basic properties. This is called amphoteric behaviour. This behaviour is due to the presence of both acidic (carboxyl group) and basic (amino group) groups in the same molecule. In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton giving rise to a dipolar ion known as zwitter ion.

Question 43.
Why are vitamin A and vitamin C essential to us ? Give their important sources.
Answer:
Vitamin A is essential tolls because its deficiency causes xerophthalmia (hardening of cornea of the eye) and night blindness.
Important sources of vitamin A : Fish liver oil, milk etc.
Vitamin C is essential to us because its deficiency causes scurvy (bleeding gums).
Sources of vitamin C: Citrus fruits, green leafy vegetables.

Question 44.
What are nucleic acids ? Mention their two important functions.
Answer:
Nucleic acids are chains of five – membered ring sugars linked by phosphate groups. The anomeric carbon of each sugar is bonded to a nitrogen of a heterocyclic compound in a β – glycosidic linkage. Functions of nucleic acids:

1. DNA is the chemical basis of heredity and may be regarded as the reserve of genetic information.
2. Nucleic acids have an important role in protein synthesis. Actually the proteins are synthesised by various RNA molecules in the cell but the message for the synthesis of a particular protein is present in DNA.

Question 45.
What is the difference between a nucleo-side and a nucleotide ?
Answer:
A unit formed by the attachment of a base to 1′ position of sugar is known as nucleoside.
Ex: Cytidine

When nucleoside is linked to phosphoric acid at 5′ position of sugar moity, we get a nucleotide.
Ex: Adinodine Triphosphate

Short Answer Questions (4 Marks)

Question 46.
How are the carbohydrates classified on the basis of their
a) Taste
b) Hydrolysis
c) Functional groups ?
Answer:
a) On the basis of taste carbohydrates are classified into sugars and non – sugars. Carbohydrates which are sweet to taste are called sugars while others are called non – sugars.

b) Carbohydrates are classified into three groups on the basis of their hydrolysis,

1. Monosaccharides
2. Oligosaccharides and
3. Polysaccharides.

c) Carbohydrates are classified into

1. aldoses and
2. ketoses on the basis of the functional group present in them.

Question 47.
Write a brief note on the structure of glucose.
Answer:
Glucose is an aldohexose.

1. The molecular formula of glucose is C₆H₁₂O₆.
2. On prolonged heating with HI, glucose forms n – hexane suggesting that all the carbon atoms are linked in a straight chain in its molecule.
3. Glucose reacts with hydroxylamine to form an oxime and adds a molecule of HCN to give cyanohydrin. These reactions show the presence of a carbonyl group ( = O) in glucose molecule.
4. Glucose on oxidation with bromine water, gives gluconic acid containing the same number of carbon atoms as in its molecule. This indicates that the carboxyl group is an aldehyde group.
5. Acetylation of glucose with acetic anhydride gives glucose pentaacetate which indicates the presence of five -OH group. Since glucose is a stable compound, the five -OH groups should be present on different carbon atoms.
6. Glucose, as well as gluconic acid, is oxidised with nitric acid to give saccharic acid, a dicarboxylic acid which has the same number of carbon atoms as glucose. This indicates the presence of a primary alcoholic (-OH) group in glucose.

The exact spacial arrangement of different -OH groups in glucose molecule was given by Fisher. The open chain structure of glucose is given as follows.

The above structure of glucose explained most of its properties. But it could not explain the following reactions.
1) Despite having the aldehyde group, glucose does not give Schiff’s test and does not form the addition product with sodium bisulphite.

2) The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free -CHO group.
Glucose is found to exist in two different crystalline forms named as α and β. To expLain all these observations the following six membered ring structure was proposed to Glucose.

Question 48.
Write short notes on Sucrose.
Answer:
Sucrose (C₁₂H₂₂O₁₁) is the most common disaccharide widely present in plants. It is mainly obtained from sugarcane or beetroot.

It is a colourless crystalline substance sweet to taste. It is dextrorotatory.

Sucrose on hydrolysis gives equimolar mixture of D – (+) – glucose and D – (-) – fructose. These two monosaccharide units are held together by a glycosidic linkage between C – 1 of α – glucose and C – 2 of β – fructose. Since the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a non – reducing sugar.

Question 49.
Write the structures of maltose and lactose. What are the products of hydrolysis of maltose and lactose?
Answer:
Structure of Maltose :

Hydrolysis products : Maltose on hydro-lysis gives α – D – glucose. Lactose on hydrolysis gives β – D – galactose and β – D – glucose.

Question 50.
Write about polysaccharides with starch and cellulose as examples.
Answer:
Carbohydrates containing large number of monosaccharide units joined together through glycosidic linkages are called polysaccharides. They act as food storages or structural materials.

Starch:
Starch is the main storage poly-saccharide of plants. It is the most important dietary source for human beings. Starch is found in cereals, roots, tubers and some vegetables. It is a polymer of α – glucose and consists of two components amylose and amylopectin. Amylose is a water soluble component which constitutes 15 – 20% of starch.

Chemically amylose is a long unbranched chain with 200-1000 α – D – (+) – glucose units held by C -1 to C – 4 glycosidic linkage. Amylopectin is insoluble in water and constitutes about 80 – 85% of starch. It is a branched chain polymer of α – D – glucose units in which the chain is formed by C -1 to C – 4 glycosidic linkage whereas branching occurs by C -1 to C – 6 glycosidic linkage.

Starch is a white amorphous powder almost insoluble in cold water but relatively more soluble in boiling water. Starch solution gives blue colour with iodine solution but the blue colour disappears on heating.

Cellulose :
Cellulose is the main structural component of vegetable matter. It is a straight chain polysaccharide composed only of β – glucose units which are joined by glycosidic link between C -1 of one glucose unit and C – 4 of the next glucose unit.

Question 51.
Write the importance of carbohydrates.
Answer:
Carbohydrates are essential for the life of both plants and animals. Honey is an instant source of energy. Carbohydrates form a major portion of our food. They are used as storage molecules as starch in plants and glycogen in animals. Cell walls of bacteria and plants are made up of cellulose.

Wood and cotton fibre contain cellulose. They provide raw materials for many important industries like textiles, paper, lacquers and breweries. D – ribose and 2 – deoxy – D – ribose are present in nucleic acids. Carbohydrates are present in biosystems in combination with many proteins and lipids.

Question 52.
Exaplain the classification of proteins as primary, secondary, tertiary and quarternary proteins with respect to their structure.
Answer:
The structure and shape of proteins can be studied at four different levels, i.e., primary, secondary, tertiary and quarternary.

i) Primary structure of proteins:
Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence. It is this sequence of amino acids that is said to be the primary structure. The primary structure represents the constitution of the protein.

ii) Secondary structure of proteins:
The secondary structure of proteins refers to the shape in which a long polypeptide chain can exist. These chains are found to exist in two different types of structures (1) α – helix and (2) β – pleated sheet structure. These structures arise due to the regular folding of the backbone of the polypeptide chain due to hydrogen bonding between = O and -NH groups of the peptide bond.

iii) Tertiary structure of proteins :
The tertiary structure of proteins represents overall folding of the polypeptide chains i.e., further folding of the secondary structure. It gives rise to two major molecular shapes viz. fibrous (thread like) and globular.

iv) Quarternary structure of proteins :
Some of the proteins are composed of two or more polypeptide chains referred to as sub – units. The special arrangement of these subunits with respect to each other is known as quarternary structure.

Question 53.
Explain the denaturation of proteins. [TS 16]
Answer:
When a protein in its native form, is subjected to physical change like change in temperature or chemical change like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix gets uncoiled and protein loses its biological activity. This is called denaturation of protein. During denaturation, secondary and tertiary structures are destroyed but the primary structure remains intact. Coagulation of egg white on boiling is a common example of denaturation.

Question 54.
What are enzymes ? Give examples.
Answer:
Enzymes are biocatalysts which speed up reactions in biosystems. They are very specific and selective in their action and chemically all enzymes are proteins.
Ex:
1) The enzyme that catalyses the hydrolysis of maltose into glucose is named maltase.

2) An aqueous solution of sucrose is fermented by yeast to give ethyl alcohol and carbon dioxide. The enzyme invertage present in yeast converts sucrose into glucose and fructose. These sugars are then decomposed by the enzyme zymase (also present in yeast) to give ethyl alcohol and carbondioxide.

Question 55.
Explain the role of sucrose in its hydrolysis.
Answer:
Sucrose is dextrorotatory but after hydrolysis gives an equimolar mixture of glucose which is dextrorotatory and fructose which is laevorotatory. Since laevorotation of fructose (-92.4°) is more than dextrorotation of glucose (+52.5°), the mixture is laevorotatory. Thus hydrolysis of sucrose brings about a change in the sign of rotation, from dextro (+) to laevo (-) and the product is called invert sugar.

Question 56.
Write notes on vitamins.
Answer:
Vitamins are organic compounds which are required in small quantities in our diet for the maintenance of normal health. Their deficiency causes specific diseases. Plants can synthesise almost all vitamins. Vitamins are designated by alphabets A, B, C, D, E, K. Some of them are further named as subgroups e.g. B₁, B₂, B₆, B₁₂ etc. Excess of vitamins is also harmful.

Vitamins are classified into two groups depending upon their solubility in water or fat.

1. Fat soluble vitamins: These are soluble in fat and oils but insoluble in water. Vitamins A, D, E and K are fat soluble vitamins.
2. Water soluble vitamins : B group vitamins and vitamin C are soluble in water.

Some vitamins, their sources and deficiency diseases:
Vitamin A:
Sources: Fish liver oil, butter and milk. Deficiency diseases: Xerophthalmia, Night blindness.
Vitamin B₁ (Thiamine):
Sources: Milk, green vegetables and cereals.
Deficiency diseases: Beri beri

Vitamin B₂ (Riboflavin):
Sources: Milk, egg white, liver.
Deficiency diseases: Cheilosis

Vitamin C:
Sources: Citrus fruits, amla and green leafy vegetables.
Deficiency diseases: Scurvy

Question 57.
What do you understand by The two strands of DNA are complementary to each other” ? Explain.
Answers:
Nucleic acids have a secondary structure also besides the primary structure. DNA was given a double strand helix structure. Two nucleic acid chains are wound about each other and held together by hydrogen bonds between pairs of bases. The two strands are complementary to each other because the hydrogen bonds are formed between specific pairs of bases. Adenine forms hydrogen bonds with thymine whereas cytosine forms hydrogen bonds with guanine.

Question 58.
What are hormones ? Give one example for each. [Mar. 19, 18 ; AP & TS]

1. Steroid hormones
2. Poly peptide hormones and
3. Amino acid derivatives.

Answer:
Hormones are molecules that act as intercellular messengers. They have several functions in the body.

1. Steroid hormones:
   a) Adrenal cortical hormones : Ex : Corti- coids
   b) Sex hormones
   a) Male sex hormones: Ex: Testosterone
   b) Female sex hormones : Ex : Estrone
2. Poly peptide hormones : Ex : Insulin
3. Amino acid derivatives: Ex: Thyroxine

Question 59.
Give the sources of the following vitamins and name the diseases caused by their deficiency (a) A (b) D (c) E and (d) K. [AP 16; 15; TS 15]
Answer:

Long Answer Questions (8 Marks)

Question 60.
Explain the classification of carbohydrates.
Answers:
1) Classification on the basis of hydrolysis:
Carbohydrates are classified into three types on the basis of their hydrolysis.

1. Monosaccharides
2. Oligosaccharides and
3. Polysaccharides.

Monosaccharides:
Carbohydrates which cannot be hydrolysed into smaller units are called monosaccharides.
Ex : Glucose, fructose, ribose etc.

Oligosaccharides:
Carbohydrates that yield two to ten monosaccharide units on hydrolysis are called oligosaccharides. They are further classified into disaccharides, trisaccharides, tetrasaccharides etc., depending upon the number of monosaccharides they provide on hydrolysis. Disaccharides, on hydrolysis, give two monosaccharide units which may be same or different.
Ex : Sucrose on hydrolysis gives one molecule of glucose and one molecule of fructose.

Maltose on hydrolysis gives two molecules of glucose only.

Polysaccharides:
Carbohydrates which yield a large number of monosaccharide units on hydrolysis are called polysaccharides. Starch, cellulose, glycogen etc., are examples for polysaccharides.

2) Classification on the basis of reducing properties:
Carbohydrates are also classified into reducing and non – reducing sugars. All those carbohydrates which reduce Fehling’s solution and Tollen’s reagent are called reducing sugars.
Ex: Glucose, fructose, maltose, lactose.

Sugars which do not reduce Fehling’s solution etc are called non – reducing sugars. For example sucrose is a non – reducing sugar.

3) Classification based on taste :
Carbo-hydrates are also classified into sugars and non – sugars on the basis of their taste. Carbohydrates such as glucose, fructose, sucrose etc., which are sweet to taste are called sugars. Polysaccharides are not sweet to taste, hence they are called non – sugars.

Question 61.
Discuss the structure of glucose on the basis of its chemical properties.
Answer:
Glucose is an aldohexose.

1. The molecular formula of glucose is C₆H₁₂O₆.
2. On prolonged heating with HI, glucose forms n – hexane suggesting that all the carbon atoms are linked in a straight chain in its molecule.
3. Glucose reacts with hydroxylamine to form an oxime and adds a molecule of HCN to give cyanohydrin. These reactions show the presence of a carbonyl group ( = O) in glucose molecule.
4. Glucose on oxidation with bromine water, gives gluconic acid containing the same number of carbon atoms as in its molecule. This indicates that the carboxyl group is an aldehyde group.
5. Acetylation of glucose with acetic anhydride gives glucose pentaacetate which indicates the presence of five -OH group. Since glucose is a stable compound, the five -OH groups should be present on different carbon atoms.
6. Glucose, as well as gluconic acid, is oxidised with nitric acid to give saccharic acid, a dicarboxylic acid which has the same number of carbon atoms as glucose. This indicates the presence of a primary alcoholic (-OH) group in glucose.

The exact spacial arrangement of different -OH groups in glucose molecule was given by Fisher. The open chain structure of glucose is given as follows.

The above structure of glucose explained most of its properties. But it could not explain the following reactions.
1) Despite having the aldehyde group, glucose does not give Schiff’s test and does not form the addition product with sodium bisulphite.

2) The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free -CHO group.
Glucose is found to exist in two different crystalline forms named as α and β. To expLain all these observations the following six membered ring structure was proposed to Glucose.

Question 62.
Write notes on (a) fructose (b) sucrose (c) maltose (d) lactose.
Answer:
a) Fructose:
Fructose is found in fruits and honey. It is the sweetest sugar. It is obtained along with glucose by the hydrolysis of sucrose. It is an important ketohexose.

The molecular formula of fructose is C₆H₁₂O₆. On the basis of its reactions it is found to contain a ketonic functional group at C₂ and all she carbons in straight chain as in glucose. It is laevorotatory and belongs to D – series. Its open chain structure is Fructose also exists in two cyclic forms which are obtained by the addition of -OH at C₅ to the ( = O) group.
To explain all the properties of fructose, a furanose ring structure is suggested.

b) Sucrose:
Sucrose is a common disaccharide. It is obtained mainly from sugarcane or beet root. It is a colourless crystalline sweet substance. Sucrose on hydrolysis gives equimolar mixture of glucose and fructose.

The two monosaccharide units are held together by a glycosidic linkage between C₁ of α – glucose and C₂ of β – fructose. Since the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a non – reducing sugar. Sucrose is dextrorotatory but after hydrolysis gives a mixture of dextrorotatory glucose and laevorotatory fructose. The mixture is lae- vorotatory. The product of hydrolysis is called invert sugar.

c) Maltose :
Maltose is a disaccharide. It is composed of two α – D – glucose units in which C₁ of one glucose unit is linked to C₄ of another glucose unit. Maltose is a reducing sugar.

d) Lactose:
Lactose occurs in milk, hence it is also known as milk sugar. On hydrolysis lactose gives β – D -galactose and β – D – glucose. The linkage is between C₁ of galactose and C₄ of glucose. It is a reducing sugar.

Question 63.
Write notes on (a) starch (b) cellulose (c) importance of carbohydrates.
Answer:
Carbohydrates containing large number of monosaccharide units joined together through glycosidic linkages are called polysaccharides. They act as food storages or structural materials.

Starch:
Starch is the main storage poly-saccharide of plants. It is the most important dietary source for human beings. Starch is found in cereals, roots, tubers and some vegetables. It is a polymer of α – glucose and consists of two components amylose and amylopectin. Amylose is a water soluble component which constitutes 15 – 20% of starch.

Chemically amylose is a long unbranched chain with 200-1000 α – D – (+) – glucose units held by C -1 to C – 4 glycosidic linkage. Amylopectin is insoluble in water and constitutes about 80 – 85% of starch. It is a branched chain polymer of α – D – glucose units in which the chain is formed by C -1 to C – 4 glycosidic linkage whereas branching occurs by C -1 to C – 6 glycosidic linkage.

Starch is a white amorphous powder almost insoluble in cold water but relatively more soluble in boiling water. Starch solution gives blue colour with iodine solution but the blue colour disappears on heating.

Cellulose :
Cellulose is the main structural component of vegetable matter. It is a straight chain polysaccharide composed only of β – glucose units which are joined by glycosidic link between C -1 of one glucose unit and C – 4 of the next glucose unit.

Carbohydrates are essential for the life of both plants and animals. Honey is an instant source of energy. Carbohydrates form a major portion of our food. They are used as storage molecules as starch in plants and glycogen in animals. Cell walls of bacteria and plants are made up of cellulose.

Wood and cotton fibre contain cellulose. They provide raw materials for many important industries like textiles, paper, lacquers and breweries. D – ribose and 2 – deoxy – D – ribose are present in nucleic acids. Carbohydrates are present in biosystems in combination with many proteins and lipids.

Question 64.
Write notes on amino acids.
Answer:
Compounds containing both amino (-NH₂) and carboxyl (-COOH) functional groups are called amino acids.
Ex: Glycine, Alanine.

Amino acids which cannot be synthesised in the body and must be obtained through diet are called essential amino acids.
Ex: Valine
Amino acids which are synthesised in the body are known as non – essential amino acids.
Ex: Glycine

Amino acids show both acidic and basic properties. This is called amphoteric behaviour. This behaviour is due to the presence of both acidic (carboxyl group) and basic (amino group) groups in the same molecule. In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton giving rise to a dipolar ion known as zwitter ion.

Question 65.
Write notes on proteins.
Answer:
Proteins are polymers of α – amino acids. They are the most abundant biomolecules of the living system.
Ex : Insulin

Proteins in which the polypeptide chains run parallel and are held together by hydrogen and disulphide bonds will have fibre – like structure. Such proteins are called fibrous proteins.
Ex : Keratin (present in hair, wool, silk)

Proteins in which the polypeptide chains coil around to give a spherical shape are called globular proteins.
Ex : Insulin

The structure and shape of proteins can be studied at four different levels, i.e., primary, secondary, tertiary and quarternary.

i) Primary structure of proteins:
Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence. It is this sequence of amino acids that is said to be the primary structure. The primary structure represents the constitution of the protein.

ii) Secondary structure of proteins:
The secondary structure of proteins refers to the shape in which a long polypeptide chain can exist. These chains are found to exist in two different types of structures (1) α – helix and (2) β – pleated sheet structure. These structures arise due to the regular folding of the backbone of the polypeptide chain due to hydrogen bonding between = O and -NH groups of the peptide bond.

iii) Tertiary structure of proteins :
The tertiary structure of proteins represents overall folding of the polypeptide chains i.e., further folding of the secondary structure. It gives rise to two major molecular shapes viz. fibrous (thread like) and globular.

iv) Quarternary structure of proteins :
Some of the proteins are composed of two or more polypeptide chains referred to as sub – units. The special arrangement of these subunits with respect to each other is known as quarternary structure.

When a protein in its native form, is subjected to physical change like change in temperature or chemical change like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix gets uncoiled and protein loses its biological activity. This is called denaturation of protein. During denaturation, secondary and tertiary structures are destroyed but the primary structure remains intact. Coagulation of egg white on boiling is a common example of denaturation.

Question 66.
Write notes on (a) enzymes and (b) vitamins.
Answer:
Enzymes are biocatalysts which speed up reactions in biosystems. They are very specific and selective in their action and chemically all enzymes are proteins.
Ex:
1) The enzyme that catalyses the hydrolysis of maltose into glucose is named maltase.

2) An aqueous solution of sucrose is fermented by yeast to give ethyl alcohol and carbon dioxide. The enzyme invertage present in yeast converts sucrose into glucose and fructose. These sugars are then decomposed by the enzyme zymase (also present in yeast) to give ethyl alcohol and carbondioxide.

Vitamins are organic compounds which are required in small quantities in our diet for the maintenance of normal health. Their deficiency causes specific diseases. Plants can synthesise almost all vitamins. Vitamins are designated by alphabets A, B, C, D, E, K. Some of them are further named as subgroups e.g. B₁, B₂, B₆, B₁₂ etc. Excess of vitamins is also harmful.

Vitamins are classified into two groups depending upon their solubility in water or fat.

1. Fat soluble vitamins: These are soluble in fat and oils but insoluble in water. Vitamins A, D, E and K are fat soluble vitamins.
2. Water soluble vitamins : B group vitamins and vitamin C are soluble in water.

Some vitamins, their sources and deficiency diseases:
Vitamin A:
Sources: Fish liver oil, butter and milk. Deficiency diseases: Xerophthalmia, Night blindness.
Vitamin B₁ (Thiamine):
Sources: Milk, green vegetables and cereals.
Deficiency diseases: Beri beri

Vitamin B₂ (Riboflavin):
Sources: Milk, egg white, liver.
Deficiency diseases: Cheilosis

Vitamin C:
Sources: Citrus fruits, amla and green leafy vegetables.
Deficiency diseases: Scurvy

Question 67.
Explain the structures of DNA and RNA.
Answer:
There are three components of a nucleic acid.
i) A pentose sugar ii) Phosphoric acid and iii) Nitrogen containing amino acid.

A simplified version of nucleic acid chain is as follows.

Information regarding the sequence of nucleotides in the chain of a nucleic acid is called its primary structure. Nucleic acids have a secondary structure also.

James Watson and Francis Crick proposed a double strand helix structure for DNA. Two nucleic acid chains are wound about each other and held together by hydrogen bonds between pairs of bases. The two strands are complementary to each other because the hydrogen bonds are formed between specific pairs of bases. Adenine forms hydrogen bonds with thymine whereas cytosine forms hydrogen bonds with guanine.

In secondary structure of RNA, helices are present but they are only single stranded. Sometimes they fold back on themselves like a hairpin thus acquiring double helix structure possesing double stranded characteristics. In double stranded arrangement guanine pairs with cytosine and adenine with uracil.

Question 68.
Write notes on the functions of different hormones in the body.
Answer:
Functions of hormones :
I. Steroid hormones:
a) Sex hormones:
i) Male sex hormones (androgens):
Test-osterone is the principal male sex hormone produced by testis. This is responsible for the development of secondary male sexual characteristics such as deep voice and facial hair.

ii) Female sex hormones (estrogens) :
Estradiol is the main female sex harmone. It is responsible for development of secondary female sex characteristics such as breast development, small voice, long hair. It also takes part in control of menstrual cycle.

iii) Pregnancy hormones progesterone) :
Progesterone is useful for preparing the uterus for the implantation of the fertilised egg. Cortico steroids (adrenal cortical hormones) are released by the adrenal cortex.

b) Cortico steroids (adrenal cortical hor-mones :
These are released by the adrenal cortex.
i) Mineralo corticoids :
These hormones control the excretion of water and salt by the kidney.

ii) Glucocorticoids :
These control the carbohydrate metabolism, modulate inflammatory reactions and are involved in reactions to stress.

If the adrenal cortex does not function properly then one of the results may be Addison’s disease. The disease is fatal unless it is treated by glucocorticoids and mineralocorticoids.

II. Non – steroidal hormones :
i) Peptide hormones:
The most important peptide hormone is insulin. It has great influence on carbohydrate metabolism. It is released in response to the rapid release in the glucose level in the blood. On the other hand the hormone glucagon tends to increase the glucose level in the blood. The two hormones together regulate the glucose level in the blood.

ii) Amino acid derivations :
These are thyroidal hormones. Thyroxine produced in the thyroid gland is an iodinated derivative of amino acid tyrosine. Abnormally low level of thyroxine leads to hypothyroidism which is characterised by Tetha- rgynea and obesity. Increased level of . thyroxine causes hyperthyroidism.

Intext Questions – Answers

Question 1.
Glucose and sucrose are soluble in water but cyclohexane and benzene (simple six membered ring compounds) are insoluble in water. Explain.
Answer:
Water is a polar solvent. Glucose and sucrose are polar compounds. Hence they are soluble in water. The partial charges of water interact with the partial charges on glucose and sucrose.

On the other hand, cyclohexane and benzene are non – polar compounds. Water molecules are not attracted to them because they have no net charge. Hence they are insoluble in water.

Question 2.
What are the expected products of hydrolysis of lactose ?
Answer:
The expected products of hydrolysis of lactose are β – D – galactose and β – D – glucose.

Question 3.
How do you explain the formation of furanose structure for fructose while glucose forms pyranose structure with the same molecular formula C₆H₁₂O₆ ?
Answer:
Fructose exists in two cyclic forms which are obtained by the addition of -OH at C₅ to the ( = O) group. The ring thus formed is a five membered ring and is named as furanose with analogy to the heterocyclic compound furan which has a five membered ring with one oxygen and four carbon atoms.

It was found that glucose forms a six – membered ring in which -OH at C₅ is involved in ring formation. The -OH at C₅ forms a cyclic hemiacetal structure with the -CHO group. The six membered cyclic structure of glucose is called pyranose structure in analogy with pyran.

Question 4.
The melting points and solubility of amino acids in water are generally higher than those of the corresponding halo acids. Explain.
Answer:
Amino acids are crystalline solids. They are water soluble and have high melting points. They behave like salts rather than simple amines or carboxylic acids. This behaviour is due to the presence of both acidic (carboxyl group) and basic (amino group) groups in the same molecule. In aqueous solution the carboxyl group can lose a proton and the amino group can accept a proton, giving rise to a dipolar ion known as Zwitter ion.

The corresponding haloacids do not show salt – like behaviour and hence their melting points and solubility in water are less.

Question 5.
Temperature and pH affect the native proteins. Explain.
Answer:
Protein found in biological system with a unique three – dimensional structure and biological activity is called as native protein. When a protein in its native form is subjected to change in temperature or pH the hydrogen bonds are disturbed. Due to this, globules unfold and helix gets uncoiled and protein loses its biological activity. This is called denaturation of protein.

Question 6.
Why can’t vitamin C be stored in our body?
Answer:
Vitamin C is a water soluble vitamin. It is readily excreted in urine and cannot be stored in our body.

Question 7.
What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
Answer:
Adenine and thymine would be formed when a nucleotide from DNA containing thymine is hydrolysis because adenine always pairs up with thymine.

Question 8.
When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA ?
Answer:
RNA molecule is a single strand complementary to only one of the two strands of a gene. Its guanine content does not necessarily be equal to its cytocine content, nor does its adenine contents to its uracil content. Hence, when RNA is hydrolysed there is no relationship among the quantities of different bases obtained.

Question 9.
What is DNA finger printing ? Mention its uses.
Answer:
Every individual has unique finger prints. These occur at the tips of fingers and have been used for identification. A sequence of bases on DNA is unique for a person and information regarding this is called DNA finger printing. It is same for every cell and can not be altered.

Importance : DNA finger printing is used
a) in forensic laboratories for identification of criminals.
b) to determine parternity of an individual.
c) to identify dead bodies in accident cases.
d) to identify racial groups to rewrite biological evolution.

Telangana TSBIE TS Inter 2nd Year Chemistry Study Material 10th Lesson Chemistry in Everyday Life Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material 10th Lesson Chemistry in Everyday Life

Very Short Answer Questions (2 Marks)

Question 1.
What are drugs ?
Answer:
Drugs are chemicals of low molecular masses (~ 100 to 500 u), that interact with macro-molecular targets and produce a biological response.

Question 2.
When are the drugs called medicines ?
Answer:
The drugs are called medicines when their biological response is therapeutic and useful.

Question 3.
Define the term chemotherapy.
Answer:
Chemotherapy is defined as the use of chemicals for therapeutic effect.

Question 4.
Name the macromolecules that are chosen as drug targets.
Answer:
Carbohydrates, lipids, proteins and nucleic acids are the macromolecules that are chosen as drug targets.

Question 5.
What are enzymes and receptors ?
Answer:
Proteins which perform the role of biological catalysts in the body are called enzymes. Receptors are proteins that are crucial to body’s communication process.

Question 6.
Which forces are involved in holding the drug to the active site of enzymes ?
Answer:
Forces such as ionic bonding, hydrogen bonding, van der Waals interaction or dipole – dipole interaction are involved in holding the drug to the active site of enzymes.

Question 7.
What are enzyme inhibitors ?
Answer:
Drugs which can block the binding site of the enzyme and prevent the binding of substrate or can inhibit the catalytic activity of the enzyme are called enzyme inhibitors.

Question 8.
What is allosteric site ?
Answer:
Allosteric site is a site in the enzyme that is different from the active site. The binding of the inhibitor at the allosteric site changes the shape of the active site in such a way the substrate cannot recognise it.

Question 9.
What are antagonists and agonists ?
Answer:
Drugs that bind to the receptor site and inhibit its natural function are called antagonists. Drugs that mimic the natural messenger by switching on the receptor are called agonists.

Question 10.
Why do we need to classify the drugs in different ways ?
Answer:
The classification of drugs in different ways is useful to the doctors because it provides them the whole range of drugs available for the treatment of a particular disease.

Question 11.
What are antacids ? Give example. [IPE 14]
Answer:
Drugs which are used for the treatment of over production of acid in the stomach (acidity) are called antacids.
Ex : Ranitidine (Zantac), Omeprazole, Lansoprazole.

Question 12.
What are antihistamines ? Give example.
Answer:
Drugs which interfere with the natural action of histamine by competing with histamine for binding sites of receptor where histamine exerts its effects are called antihistamines.
Ex : Brompheniramine, Dincetane, terfenadine.

Question 13.
While antacids and antiallergic drugs interfere with the function of histamines, why do not these interfere with the function of each other ?
Answer:
Antacids and antiallergic drugs work on different receptors hence they do not interfere with each other’s function.

Question 14.
What are tranquilizers ? Give example.
Answer:
Tranquilizers are a class of chemical compounds used for the treatment of stress and mild or even severe mental diseases.
Ex: Iproniazid, Luminol, Second, Babituric acid.

Question 15.
What are barbiturates ?
Answer:
Derivatives of barbituric acid which con-stitute an important class of tranquilizers are called barbiturates.
Ex: veronal, amytal and seconal.

Question 16.
What are analgesics? How are they classified ?
Answer:
Drugs which reduce or abolish pain without causing disturbances of nervous system like impairment of consciousness, mental confusion, in coordination, paralysis etc., are called analgesics Analgesics are classified as

1. Narcotic analgesics and
2. Non – Narcotic analgesics.

Question 17.
What are narcotic analgesics ? Give example.
Answer:
Morphine and many of its homologues, when administered in medicinal doses, relieve pain and produce sleep. These are called narcotic analgesics because they have addictive properties.
Ex : Heroin, Morphine, Codeine.

Question 18.
What are non – narcotic analgesics ? Give example.
Answer:
Analgesics having no addictive properties are called non – narcotic analgesics.
Ex: Aspirin, Ibuprofen.

Question 19.
What are antimicrobials ?
Answer:
Drugs that tend to destroy / prevent development or inhibit the pathogenic action of microbes such as bacteria, fungi, virus or any other parasites are called antimicrobials.
Ex : Lysozyme, Lactic acids, etc.

Question 20.
What are antibiotics ? Give example. [Mar. 2018, 16-A.P.]
Answer:
Antibiotics are substances produced wholly or partly by chemical synthesis which in low concentrations inhibit the growth or destroy micro organisms by intervening in their metabolic processes.
Ex : Penicillin, Chloramphericol, sulphadiazine.

Question 21.
What are antiseptics ? Give example. [AP Mar. 19; (AP & TS 15)]
Answer:
Antiseptics are chemicals which either kill or prevent the growth of microorganisms.
Ex: Furacin, Dettol, Bithional.

Question 22.
What are disinfectants ? Give example.
Answer:
Disinfectants are chemicals which either kill or prevent the growth of microorganisms.
Ex: 1% solution of phenol, formalin (4% ag of soln of formaldehyde), 0.3 ppm chlorine.

Question 23.
Name a substance which can be used as an antiseptic as well as disinfectant.
Answer:
Phenol (0.2 % phenol antiseptic and 1% phenol as dis infectant).

Question 24.
What is the difference between antiseptics and disinfectants ?
Answer:
Antiseptics are applied to living tissues such as wounds, cuts, ulcers and diseased skin surfaces whereas disinfectants are applied to inanimate objects such as floors, drainage system, instruments etc.

Question 25.
What are the main constituents of dettol ?
Answer:
Chloroxylenol and terpineol are the main constituents of dettol.

Question 26.
What is tincture of iodine ? What is its use ?
Answer:
2-3% solution of iodine in alcohol – water mixture is known as tincture of iodine. It is used as an antiseptic.

Question 27.
What are antifertility drugs ? Give example.
Answer:
Drugs used to control population are called antifertility drugs.
Ex : Norethindrone, Mifepristone, Ethynyl estradiol.

Question 28.
Why chemicals are added to food ?
Answer:
Chemicals are added to food for

1. their preservation
2. enhancing their appeal and
3. adding nutritive value.

Question 29.
Name different categories of food additives.
Answer:
The main categories of food additives are:

1. Food colours
2. Flavours and sweeteners
3. Fat emulsifiers and stabilising agents
4. Flour improvers – antistaling agents and bleaches
5. Antioxidants
6. Preservatives
7. Nutritional supplements

Question 30.
What are artificial sweetening agents ? Give example.   [A.P. 16, 15; T.S. 15]
Answer:
Artificial sweetening agents are chemicals, used in place of sugar, which decrease the intake of calories and at the same time several times sweeter than sucrose.
Ex: Aspartame

Question 31.
Why do we require artificial sweetening agents ?
Answer:
Natural sweeteners e.g., sucrose add to calorie intake. Hence diabetic persons and ’ people who need to control intake of calories require artificial sweeteners.

Question 32.
Why is the use of aspartame limited to cold foods and drinks ?
Answer:
Use of aspartame is limited to cold foods and drinks because it is unstable at cooking temperature.

Question 33.
Name the sweetening agent used in the preparation of sweets for a diabetic patient
Answer:
Saccharin

Question 34.
What problem does arise in using alitame as artificial sweetener ?
Answer:
Alitame is a high potency sweetener and control of sweetness of food is difficult while using it.

Question 35.
What are food preservatives? Give example.
Answer:
Food preservatives are substances which prevent spoilage of food due to microbial growth.
Ex : Sodium benzoate, Sodium, metabisulphate.

Question 36.
Name two most familiar antioxidants used as food additives.
Answer:
Butylated hydroxy toluene (BHT) and butylated hydroxy anisole (BHA) are the two most familiar antioxidants used as food additives.

Question 37.
What is saponification ?
Answer:
Soaps containing sodium salts are formed by heating fat (i.e., glyceryl ester of fatty acid) with aqueous sodium hydroxide solution. This reaction is known as saponification.

Question 38.
What are soaps chemically ?
Answer:
Soaps are sodium or potassium salts of long chain fatty acids, e.g., oleic and palmitic acids.

Question 39.
Why do soaps not work in hard water ?
Answer:
Hard water contains calcium and magnesium ions. These ions form insoluble calcium and magnesium soaps respectively when sodium or potassium soaps are dissolved in hard water.

Question 40.
What are synthetic detergents ?
Answer:
Synthetic detergents are cleansing agents which have all the properties of soaps, but which actually do not contain any soaps.

Question 41.
What is the difference between a soap and a synthetic detergent ? [(Mar. 2018 AP)(IPE: 14) ]
Answer:
Soap does not work in hard water whereas a synthetic detergent can be used both in soft and hard water.

Question 42.
How are synthetic detergents better than soaps ?
Answer:
Synthetic detergents with straight chain of hydrocarbons are biodegradable and hence prevent pollution.

Question 43.
Name the different categories of synthetic detergents.
Answer:
Synthetic detergents are classified into three categories.

1. Anionic detergents
2. Cationic detergents and
3. Non – ionic detergents.

Question 44.
Can you use soaps and synthetic detergents to check the hardness of water ?
Answer:
No

Question 45.
If water contains dissolved calcium hydrogen carbonate, out of soaps and synthetic detergents which one will you use for cleaning clothes and why ?
Answer:
If water contains calcium ions, these ions form insoluble calcium soap when sodium or potassium soaps are dissolved in water. This insoluble soap separates as scum in water and is useless as cleansing agent. In such a case a synthetic detergent is preferred as it gives foam even in the presence of calcium ions.

Short Answer Questions (4 Marks)

Question 46.
Explain the term target molecules or drug targets as used in medicinal chemistry.
Answer:
Drugs usually interact with biomolecules such as carbohydrates, Lipids, proteins and nucleic acids. These are called target molecules or drug targets.

Question 47.
Explain the catalytic action of enzymes as drug targets.
Answer:
The enzymes perform two major functions in their catalytic activity.
The first function of an enzyme is to hold the substrate for a chemical reaction. Active sites of enzymes hold the substrate molecule in a suitable position, so that it can be attacked by the reagent effectively. Substrates bind to the active site of the enzyme through a variety of interactions such as ionic bonding, hydrogen bonding, van der Waals interaction or dipole – dipole interaction.

The second function of an enzyme is to provide functional groups that will attack the substrate and carryout chemical reaction.

Question 48.
Explain the drug – enzyme interaction.
Answer:
Drugs inhibit the activities of enzymes. They can block the binding site of the enzyme and prevent the binding of substrate or can inhibit the catalytic activity of the enzyme. Such drugs are called enzyme inhibitors.

Drugs inhibit the attachment of substrate on active site of enzymes in two different ways.

1. Drugs compete with the natural substrate for their attachment on the active sites of enzymes.
2. Some drugs bind to a different site of the enzyme called allosteric site. This binding of inhibitor at allosteric site changes the shape of the active site in such a way that substrate cannot recognise it.

Question 49.
Why are cimetidine and ranitidine better antacids than sodium hydrogen carbonate or magnesium hydroxide or aluminium hydroxide.
Answer:
Antacids such as sodium hydrogen carbo-nate or magnesium hydroxide or aluminium hydroxide control only the symptoms but not the cause. On the otherhand cimetidine and ranitidine prevent the interaction of histamine with the receptors present in the stomach wall. This results in release of lesser amount of acid. Hence these drugs are better antacids.

Question 50.
Low level of noradrenaline is the cause of depression. What type of drugs are needed to cure this problem ? Name two drugs.
Answer:
Antidepressant drugs, which inhibit the enzymes that catalyse the degradation of noradrenaline are needed to cure the pro-blem of depression.
Ex : Iproniazid and phenelzine are antidepressant drugs.

Question 51.
What are analgesics? How are they classified ? Give examples. [Mar. 2018 TS)(AP 15)]
Answer:
Drugs which reduce or abolish pain without causing disturbances of nervous system like impairment of consciousness, mental confusion, incoordination, paralysis etc., are called analgesics. Analgesics are classified as

1. Non – narcotic (non – addictive) analgesics and
2. Narcotic drugs.

Non – Narcotic analgesics: Aspirin and paracetamol belong to this class of analgesics. These drugs relieve pain and have other effects such as reducing fever and preventing platelet coagulation.
Ex : Morphine, Codine.

Narcotic analgesics: Morphin and many of its homologues belong to this category. When administered in medicinal doses they relieve pain and produce sleep.
Ex: Aspirin, Ibuprofen.

Question 52.
What are different types of microbial drugs ? Give one example for each.
Answer:
Antibiotics, antiseptics and disinfectants are different types of microbial drugs.
Antibiotics : Ex: Penicillin
Antiseptics : Ex: Furacin
Disinfectants : Ex: \% solution of phenol.

Question 53.
Write the characteristic properties of antibiotics.
Answer:

1. An antibiotics must be a product of metabolism.
2. An antibiotic should be effective in low concentrations.
3. An antibiotic should retard the growth or survival of microorganisms.
4. An antibiotic should be synthetic substance produced as a structural analogue of naturally occurring antibiotic.

Question 54.
What is meant by the term ‘broad spectrum antibiotics’ ? Explain.
Answer:
Antibiotics which kill or inhibit a wide range of gram – positive and gram – negative bacteria are said to be broad spectrum antibiotics.

Antibiotics which are effective mainly against gram positive or gram negative bacteria are narrow spectrum antibiotics. Those effective against a single organism or disease are referred to as limited spectrum antibiotics.

Question 55.
What are broad spectrum and narrow spectrum antibiotics ? Give one example for each.
Answer:
Antibiotics which kill or inhibit a wide range of gram – positive and gram – negative bacteria are called broad spectrum antibiotics.
Ex : Chloramphenicol, Ampicillin, Amoxycilin.

Antibiotics which are effective mainly against gram – positive or gram – negative bacteria are called narrow spectrum antibiotics.
Ex: Penicillin G

Question 56.
Write notes on antiseptics and disinfectants. [TS 15]
Answer:
Antiseptics and disinfectants are chemicals which either kill or prevent the growth of microorganisms.

Antiseptics are applied to the living tissues such as wounds, cuts, ulcers and diseased skin surfaces. Ex: Furacin, soframicine, dettol.

Disinfectants are chemicals which either kill or prevent the growth of microorganisms. These are applied to inanimate objects such as floors, drainage systems, instruments etc. Ex : 1 % solution of phenol.

Same substance can act as an antiseptic as well as disinfectant by varying concentration. For example 0.2% solution of phenol is an antiseptic while its 1% solution is disinfectant.

Question 57.
How do antiseptics differ from disinfectants ? Does the same substance be used as both ? Give one example for each.
Answer:
Antiseptics differ from disinfectants in the manner of application. Antiseptics are applied to the living tissues such as wounds, cuts, ulsers and diseased skin surfaces where as disinfectants are applied to inanimate objects such as floors, drainage system, instruments etc.

Same substance can be used as an anti-septic as well as disinfectant by varying concentration.
Ex : 0.2% solution of phenol is an anti-septic while its 1% solution is disinfectant.

Question 58.
What are the main categories of food additives ?
Answer:
The main categories of food additives are:

1. Food colours
2. Flavours and sweeteners
3. Fat emulsifiers and stabilising agents
4. Flour improvers – antistaling agents and bleaches
5. Antioxidants
6. Preservatives
7. Nutritional supplements such as minerals, vitamins and amino acids.

Question 59.
Write notes on antioxidants in food.
Answer:
Antioxidants are important food additives. They help in food preservation by retarding the action of oxygen in food. These are more reactive towards oxygen than the food material they are protecting.

Butylated hydroxytoluene (BHT) and butylated hydroxy anisole (BHA) are the two most familiar antioxidants. Sometimes BHT and BHA are mixed with citric acid or ascorbic acid to produce more effect. Sulphur dioxide and sulphite are used as antioxidants for wine, beer, sugar syrup, cut, peeled or dried fruits and vegetables.

Question 60.
Name different types of soaps.
Answer:
There are different types of soaps such as

1. Toilet soaps
2. Soaps that float in water
3. Transparent soaps
4. Medicated soaps
5. Shaving soaps
6. Laundry soaps
7. Soap granules
8. Soap powders
9. Scouring soaps

Question 61.
Explain the following terms with suitable examples.
i) Cationic detergents
ii) Anionic deter-gents
iii) Non – ionic detergents
Answer:
i) Cationic detergents: Cationic detergents are quartemary ammonium salts of amines with acetates, chlorides or bromides as anions. Cationic part possesses a long hydrocarbon chain and a positive charge on nitrogen atom.
Ex:
Cetyl trimethyl ammonium bromide is a popular cationic detergent. It is used in hair conditioners.

ii) Anionic detergents: Anionic detergents are sodium salts of sulphonated long chain alcohols or hydrocarbons. In these detergents, the anionic part of the molecule is involved in the cleansing action. Sodium salts of alkyl benzene sulphonates are an important class of anionic detergents.
Ex: Sodium dodecylbenzene sulphonate.

iii) Non – ionic detergents: Non – ionic detergents do not contain any ion in their constitution.
Ex : A non – ionic detergent is formed when stearic acid reacts with polyethylene glycol.
CH₃ (CH₂)₁₆ COOH + HO (CH₂ CH₂O)_n
CH₂ CH₂OH CH₃ (CH₂)₁₆ COO (CH₂ CH₂O)_n CH₂ CH₂OH
Liquid dish washing detergents are non-ionic type. Just like soaps, these detergents remove grease and oil by micelle formation.

Question 62.
What are biodegradable and non – biodegradable detergents ? Give one example for each.
Answer:
Detergents which are metabolised by natural bacteria are called biodegradable detergents. On the otherhand, detergents which are not degraded by bacteria easily leading to their accumulation are called non – biodegradable detergents. Branched chain detergents are non – biodegradable and unbranched hydrocarbon detergents are biodegradable.
Ex:

Question 63.
Explain the cleansing action of soaps.
Answer:
Soaps are sodium or potassium salts of fatty acids. They may be represented as \(2 \mathrm{RO} \overline{\mathrm{O}}\)\(\stackrel{+}{\mathrm{N}}\)a.
Ex: Sodium stearate, CH₃ (CH₂)₁₆ \(\mathrm{CO} \overline{\mathrm{O}}\stackrel{+}{\mathrm{N}}\)a. When dissolved in water the soap dissociates into \(\mathrm{RO} \overline{\mathrm{O}}\) and \(\stackrel{+}{\mathrm{N}}\)a ions. The \(\mathrm{RO} \overline{\mathrm{O}}\) ions consist of two parts – a long hydrocarbon chain R (also called non – polar tail) which is hydrophobic (water repelling), and a polar group \(\mathrm{CO} \overline{\mathrm{O}}\) (also called polar ionic head) which is hydrophilic (water loving).

Long chain carboxylate ions do not exist as individual ions in aqueous solution, instead they arrange themselves in spherical clusters called micelles. Each micelle contains 50-100 long – chain carboxylate ions. The polar carboxylate end of each ion is on the outside of the micelle because of its attraction for water.

The non – polar end is in the interior of the miscelle to minimise its contact with water. Because the surface of the miscelle is negatively charged, the individual miscelles repel each other instead of clustering to form larger agregates. The cleansing action of soap results from the fact that non – polar oil molecules, which carry dirt, dissolve in the non- polar interior of the miscelle and are carried away with the soap during rinsing.

Question 64.
Label the hydrophillic and hydrophobic parts in the following compounds.
i) CH₃ (CH₃)₁₀ CH₂ OSO₃ ^–Na⁺
ii) CH₃ (CH₂)₁₅ N⁺ (CH₂)₃ Br^–
iii) CH₃ (CH₂)₁₆ COO(CH₂CH₂O)_nCH₂CH₂ OH
Answer:

Question 65.
Draw the structures of the following.
i) Serotonin
ii) Bithionol
iii) Chloramphenicol
iv) Saccharin
Answer:

Long Answer Questions (8 Marks)

Question 66.
Describe the classification of drugs into different classes.
Answer:
Drugs are classified on the basis of various criteria.
i) On the basis of pharmacological effect:
Drugs can be classified on the basis of their pharmacological effect. For example, analgesics are pain killers and antiseptics kill or stop the growth of microorganisms.

ii) On the basis of drug action:
Based on the action of drugs on particular biochemical processes they are classified as antihistamines, sedatives, cardiovascular drugs etc. All antihistamines inhibit the action of the compound, histamine which causes inflammation in the body.

iii) On the basis of chemical structure:
Drugs classified on the basis of structure share common structural features and have similar pharmacological activity. For example, sulphonamides have common structural feature given below.

iv) On the basis of molecular targets:
Drugs usually interact with biomolecules like carbohydrates, lipids, proteins, nucleic acids etc. These biomolecules are called drug targets or target molecules. Drugs with common structural features may have the same mechanism of action on targets.

Question 67.
Describe briefly the therapeutic action of different classes of drugs.
Answer:
Therapeutic action of different classes of drugs:

Antacids :
Over production of acid in the stomach causes irritation and pain. Antacids are used for the treatment of acidity. Antacids such as sodium bicarbonate or a mixture of aluminium and magnesium hydroxides control only the symptoms and not the cause. Drugs such as cimetidine (Tegamet) and ranitidine (Zantac) prevent the interaction of histamine with the receptors present in the stomach wall resulting in the release of less amount of acid.

Antihistamines :
Drugs such as brom- phenaramine (Dimetapp) and terfenadine (seldane) act as antihistamines. They interfere with the natural action of histamine by competing with histamine for binding sites of receptor where histamine exerts its effect.

Neurologically active drugs:
Tranquilizers and analgesics are neurologically active drugs. These affect message transfer mechanism from nerve to receptor. Tranquilizer are a class of compounds used for the stress and mild or even severe mental diseases. These relieve anxiety, stress, irri-tability or excitement by inducing a sense of well being.

If the level of noradrenaline is low the person suffers from depression. In such situations antidepressant drugs such as iproniazid and phenelzine are used. These drugs inhibit the enzymes which catalyse the degradation of noradrenaline and counteract the effect of depression.

[Mar. 2018-TS]
Analgesics:
Analgesics reduce or abolish pain without causing disturbances of nervous system. They are two classes of analgesis i) Non -narcotic (non addictive) analgesics and ii) Narcotic analgesics. Aspirin and paracetamol are non – narcotic analgesics. Aspirin inhibits the synthesis of chemicals known as prostaglandins which stimulate inflammation in the tissue and cause pain.

Morphine and many, of its homologues are narcotic analgesics. When administered in medicinal doses they relieve pain and produce sleep.

Antimicrobials:
An antimicrobial tends to destroy / prevent development or inhibit pathogenic action of microbes such as bac-teria, fungi, virus or other parasites. Anti-biotics, antiseptics and disinfectants are antimicrobial drugs.

Antibiotics in low concentrations inhibit the growth or destroy microorganisms by intervening in their metabolic process.

Broad spectrum antibiotics like chloramphenicol kill or inhibit a wide range of gram – positive and gram – negative bacteria. Narrow spectrum antibiotics like penicillin G are effective mainly against gram – positive or gram – negative bacteria.

Antiseptics and disinfectants :
These are chemicals which either kill of prevent the growth of microorganisms. Antiseptics are applied to the living tissues such as wounds, cuts and diseased skin surfaces.
Ex: Furacin.

Disinfectants are applied to inanimate objects such as floors, drainage system, instruments etc. Some substances can act as antiseptic as well as disinfectant by varying concentration.

Antifertility drugs :
These drugs are used to control population. Birth control pills essentially contain a mixture of estrogen and progesterone derivatives. Progesterone suppresses ovulation.

Question 68.
Write an essay on antimicrobials.  [TS Mar. 19]
Answer:
Diseases in human beings and animals may be due to the action of a variety of microorganisms like bacteria, virus, fungi and other pathogens.

An antimicrobial tends to destroy / prevent development or inhibit the pathogenic action of microbes selectively. Antibiotics, antiseptics and disinfectants are antimicrobial drugs.

Antibiotics :
Antibiotics are chemical substances produced by microorganisms (bacteria, fungi and mould) which inhibit the growth or destroy microorganisms. To cover synthetic antibiotics the definition for antibiotics is given as “An antibiotic is a substance that is produced wholly or partly by chemical synthesis and in low concentrations inhibits the growth or destroys microorganisms by interfering in their metabolic process.”

Paul Ehrlich developed salvarsan, an arsenic based drug for the treatment of siphilis caused by bacteria, spirochete. He also developed the first effective antibacterial agent, prontosil which was converted into sulphanilamide in the body. Later a large range of sulpha drugs were synthesised.

Despite the success of sulfonamides, the real success in antibacterial therapy began with the discovery of penicillin by Alexander Fleming in 1929.

Antibiotics have either a cidal (killing) effect or a static (inhibitory) effect on microbes. For example, penicillin, amino glycosides and ofloxacin are bactericidal whereas erythromycin, tetracycline and chloramphenicol are bacteriostatic.

The range of bacteria or other micro-organisms that are affected by a certain antibiotic is expressed as its spectrum of action. Broad spectrum antibiotics kill or inhibit a wide range of gram positive and gram negative bacteria. For example, Ampicillin and Amoxycillin are broad spectrum antibiotics. Antibiotics that are effective mainly against gram – positive or gram – negative bacteria are called narrow spectrum antibiotics. Penicillin G has a narrow spectrum.

Antiseptics and disinfectants :
Anti-septics and disinfectants are chemicals which either kill or prevent the growth of microorganisms.

Antiseptics are applied to the living tissues such as wounds, cuts, ulcers and diseased skin surfaces. Ex : Furacin, Soframi- cine etc.

Iodine is a powerful antiseptic. 2-3% solution of iodine in aqueous alcohol is known as tincture of iodine. It is applied on wounds.

Disinfectants are applied to inanimate objects such as floors, drainage system, instruments etc.

1% solution of phenol is used as disinfectant.

Question 69.
Write notes on the following
i) Artificial sweetening agents
ii) Food preservatives
iii) Antioxidants in food.
Answer:
i) Artificial sweetening agents :
Natural sweeteners like sucrose add to calorie intake . Hence artificial sweeteners are used by many people. Orthosulphobenzimide, also called saccharin, is a popular artificial sweetening agent. It is 550 times as sweet as sucrose. It is excreted through urine. It is inert and harmless when taken. It has great value in controlling calories and it is useful to diabetic persons.

Aspartame is the most successful and widely used artificial sweetener. It is about 100 times as sweet as sucrose. Use of aspartame is limited to cold foods and soft drinks because it is unstable at cooking temperature. Alitame is a high potency sweetener about 2000 times sweeter than sucrose. Control of sweetness of food is difficult while using it. Sucralose is similar to sugar in appearance and taste. It is stable at cooking temperature. It is 600 times as sweet as cane sugar.

ii) Food preservatives:
Food preservatives are added to the food to prevent spoilage due to microbial growth. The most commonly used preservatives include table salt, sugar, vegetable oils and sodium benzoate. Salts of sorbic acid and propanoic acid are also used as food preservatives, sodium benzoate is the most important food preservative. It is metabolised by conversion into hippuric acid which in finally excreted in the urine.

iii) Antioxidants in food:
Antioxidants are important food additives. They help in food preservation by retarding the action of oxygen in food. These are more reactive towards oxygen than the food material they are protecting.

Butylated hydroxy toluene (BHT) and butylated hydroxy anisole (BHA) are the two most familiar antioxidants. Sometimes BHT and BHA are mixed with citric ; acid or ascorbic acid to produce a better effect. Sulphur dioxide and sulphite are used as antioxidants for wine, beer, sugar syrup, cut, peeled or dried fruits and vegetables.

Question 70.
Write notes on the following :
i) Soaps
ii) Synthetic detergents.
Answer:
i) Soaps: Soaps are sodium or potassium salts of long chain fatty acids e.g. stearic, oleic and palmitic acids. Soaps containing sodium salts are obtained by heating fats and oils (glyceryl esters of fatty acids) with sodium hydroxide solution. This reaction is known as saponification.

In this reaction esters of fatty acids are hydrolysed and the soap obtained remains in colloidal form. It is precipitated from the solution by adding sodium chloride. Generally potassium soaps are soft to the skin than sodium soaps. These can be prepared by using potassium hydroxide instead of sodium hydroxide.

Different types of soaps are made by using different raw materials. Toilet soaps, transparent soaps, medicated soaps, laundry soaps, shaving soaps, soaps that float on water etc., are different types of soaps.

Soaps do not work in hard water. Hard water contains calcium and magnesium ions. When sodium soaps are dissolved in it, these ions form calcium and magnesium soaps which are insoluble. These insoluble soaps separate as scum. This results in the waste of sodium soap.

Soaps remove grease and oil by micelle formation.

ii) Synthetic detergents:
Synthetic deter-gents are cleansing agents which have all the properties of soaps, but actually do not contain any soap. These can be used both in soft and hard water.

Synthetic detergents are classified into three categories. i) Anionic detergents ii) Cationic detergents and iii) Non – ionic detergents.

Anionic detergents are sodium salts of long chain alcohols or hydrocarbons. These are formed by treating long chain alcohols with concentrated sulphuric acid and neutralising the alkyl hyrogen sulphates formed with sodium hydroxide. Alkyl benzene sulphonates are obtained by neutralising alkyl benzene sulphonic acids with sodium hydroxide. In anionic detergents the anionic part of the molecule is involved in the cleansing action. These are mostly used in household work.
Ex: Sodium lauryl sulphate, sodium dodecyl benzene sulphonate.

Cationic detergents:
Cationic detergents are quarternary ammonium salts of amines with acetates, chlorides or bromides as anions. Cationic part possesses a long hydrocarbon chain and a positive charge on nitrogen atom.
Ex : Cetyltrimethyl ammonium bromide

Non – ionic detergents :
Non – ionic detergents do not contain any ion in their constitution. Liquid dishwashing detergents are non – ionic type. Branched chain detergents are non – biodegradable and unbranched hydrocarbon detergents me biodegradable and hence pollution is prevented.

Intext Questions – Answers

Question 1.
Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it is not advisable to take in doses without consultation with the doctor. Why ?
Answer:
If sleeping pills are taken in doses higher than those recommended, they may cause harmful effects and act as poisons. Hence, it is not advisable to use them without consulting the doctor.

Question 2.
With reference to which classification has the statement, ‘ranitidine is an antacid’ been given?
Answer:
This statement is given based on the drug action. Ranitidine counteracts the effects of excess acid in the stomach. Hence it is an antacid.

Question 3.
Why do we require artificial sweetening agents?
Answer:
Natural sweeteners like sucrose add to calorie intake. Hence diabetic persons and people who need to control intake of calories require artificial sweetening agents.

Question 4.
Write the chemical equation for preparing sodium soap from glyceryloleate and glyceryl palmitate. Structural formulas of these compounds are given below.
i) (C₁₅H₃₁COO)₃C₃H₅ – glyceryl palmitate
ii) (C₁₇H₃₂COO)₃ C₃H₅ – glyceryl oleate.
Answer:

Question 5.
Following type of non – ionic detergents are present in liquid detergents, emulsifying agents and wetting agents. Label the hydrophilic and hydrophobic parts in the molecule. Identify the functional group(s) in the molecule.

Answer:

b) Functional group: Ethen and alcohol.

Telangana TSBIE TS Inter 2nd Year Chemistry Study Material 11th Lesson Haloalkanes and Haloarenes Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material 11th Lesson Haloalkanes and Haloarenes

Very Short Answer Questions (2 Marks)

Question 1.
Write the structures of the following compounds. [A.P. 15]
i) 2-Chloro-3-methyl pentane
ii) 1-Bromo-4-sec-butyl-2-methyl benzene
Answer:

Question 2.
Which one of the following has highest dipole moment ?
i) CH₂Cl₂
ii) CHCl₃
iii) CCl₄
Answer:
CH₂Cl₂ (μ = 1.62 D)

Question 3.
What are ambident nucleophiles ?
Answer:
Groups like cyanides and nitrites which possess two nucleophilic centres are called ambident nucleophiles. Ex : Cyanide ion, Nitrite ion.

Question 4.
Write the isomers of the compound having molecular formula C₄H₉Br.
Answer:
The isomers of the compound having the molecular formula C₄H₉Br are
i) CH₃ CH₂ CH₂ CH₂ Br
n-butyl bromide

ii)

Question 5.
Which compound in each of the following pairs will react faster in S_N² reaction with [AP Mar. 19; (IPE 14)]
i) CH₃Br or CH₃I
ii) (CH₃)₃ CCl or CH₃Cl
Answer:
i) CH₃I
ii) CH₃Cl

Question 6.
Explain why the alkyl halides though polar are immiscible with water.
Answer: In order for a haloalkane to dissolve in water, energy is required to overcome the attractions, between the haloalkane molecules and to break the hydrogen bonds between water molecules. Less energy would be released when new attractions are set up between the haloalkane and the water molecules, as these are not as strong as the original hydrogen bonds in water. Consequently the solubility of haloalkanes in water is low.

Question 7.
Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅, which is more easily hydrolysed by aqueous KOH?
Answer:
Under S_N² conditions, C₆H₅CH₂Cl is more easily hydrolysed than C₆H₅CHCl C₆H₅. Whereas under S_N¹ conditions C₆H₅CHCI C₆H₆ is more easily hydrolysed than C₆H₅CH₆CI.

Question 8.
Treatment of alkylhaiides with aq. KOH leads to the formation of alcohols, while in presence of ale. KOH what products are formed ?
Answer:
When an alkyl halide is heated with ale. KOH an alkene is formed due to elimination of hydrogen halide from adjacent carbon atoms.

Question 9.
What is the stereochemical result of S_N¹ and S_N² reactions?
Answer:
An optically active alkyl halide undergoes inversion of configuration in S_N² reaction whereas S_N¹ reaction of an optically active alkyl halide is accompanied by recemisation.

Question 10.
What type of isomerism is exhibited by o, m and p-chlorobenzenes ?
Answer:
Position isomerism as the isomers differ in the positions occupied by the two chlorine atoms.

Question 11.
What are Enantiomers ? [IPE 14] [TS Mar. 19]
Answer:
Stereoisomers related to each other as non-superimposable mirror images are called enantiomers.

Short Answer Questions (4 Marks)

Question 12.
Give the IUPAC names of the following compounds:
i) CH₃CH(Cl) CH(I) CH₃
ii) ClCH₂CH = CH CH₂Br
iii) (CCl₃)₃CCl
iv) CH₃C(p-Cl-C₆H₄)₂ CH(Br) CH₃
Answer:

iii) (CCl₃)₃CCl
2-(tricholor methyl) 1, 1, 1, 2, 3, 3, 3 hepta chloropropane
iv)

2-bromo-3, 3 – di – p – chlorophenyl butane

Question 13.
Write the structures of the following organic halides. [AP 15]
i) 1-Bromo-4-sec-butyl-2-methylbenzene
ii) 2-Choro-1-phenylbutane
iii) p-bromochlorobenzene
iv) 4-t-butyl-3-iodoheptane
Answer:

Question 14.
A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single mono- chloro compound C₅H₉Cl in bright sunlight. Identify the hydrocarbon.
Answer:
The hydrocarbon is cyclopentane.

In cyclopentane, C₅H₁₀, all hydrogens are chemically identical. Hence it gives a single monochloro compound, C₅H₉Cl.

Question 15.
Which compound in each of the following parts will react faster in S_N² reaction with ^–OH?
i) CH₃Br or CH₃I
ii) (CH₃)₃ CCl or CH₃Cl
Answer:
i) CH₃I will react faster than CH₃Br in S_N² reaction with – OH because Iodine is a better leaving group because of its larger size.

ii) Of the two compounds CH₃Cl and (CH₃)₃CCl, the former i.e., CH₃Cl will react faster in S_N² reaction with – OH because there are only three small hydrogen atoms on carbon. (CH₃)₃CCl is a tertiary alkyl halide. It is least reactive under S_N² conditions because the bulky groups hinder the approaching nucleophiles.

Question 16.
Predict the alkenes that would be formed in the following reactions and identify the major alkene.

Answer:

Question 17.
How will you carry out the following conversions ?
i) Ethane to bromoethane
ii) Toluene to benzyl alcohol
Answer:
i) Free radical bromination of ethene gives bromoethane.

ii) Toluene is converted to benzyl chloride by reaction with chlorine in sunlight Benzyl chloride on reaction with NaOH solution gives benzyl alcohol.

Question 18.
Explain why the dipolemoment of chloro-benzene is lower than that of cyclohexyl-chloride.
Answer:
In cyclohexyl chloride, chlorine is bonded to an sp³ carbon. In chlorobenzene chlorine is bonded to an sp² carbon, sp² carbon is more electronegative than sp³ carbon. Hence the C – Cl bond in cyclohexyl chloride is more polar than the C – Cl bond in chloro-benzene. Thus, the dipole moment of chloro-benzene is lower than that of cyclohexyl chloride.

Question 19.
Write the mechanism of the following reaction.

Answer:

n-Butyl bromide is a primary alkyl halide. It undergoes reaction with CN by S_N² mechanism.

Long Answer Questions (8 Marks)

Question 20.
Name the following halides according to IUPAC system and classify them as primary, secondary, tertiary, vinyl or aryl halides.
i) CH₃CH(CH₃)CH(Br)CH₃
ii) CH₃C(Cl) (C₂H₅) CH₂CH₃
iii) m – ClCH₂C₆H₄CH₂C(CH₃)₃
iv) O – Br – C₆H₄CH (CH₃) CH₂CH₃
Answer:
i)

IUPAC name:
2 – Bromo – 3 – Methyl butane
It is a secondary alkyl halide.

ii)

IUPAC name:
3 – Chloro – 3 – Methyl pentane
It is a tertiary alkyl halide.

iii)

IUPAC name:
3 – neopentyl – 1 – chloromethyl benzene
It is an aromatic primary alkyl halide.

iv)

2 – sec. butyl bromo benzene
It is an aryl halide.

Question 21.
Write the structures of the following organic halogen compounds.
i) 2-Bromo-3-methylhexane
ii) 2-(2-chlorophenyI)-1-iodooctane
iii) 4-tertiary-butyl-3-iodo-1-chloro benzene
iv) 1-Bromo-4-sec-butyl-2-methylbenzene
Answer:

Question 22.
Discuss the physical properties of haloalkanes.
Answer:
General physical properties of haloalkanes.
1) State:
Lower members like methyl chloride, methyl bromide and ethyl chloride are colourless gases at room temperature. Some of the higher members are colourless sweet smelling liquids but still higher homologues are colourless solids.

2) Melting and boiling points:
For the same alkyl group, the boiling points of alkyl halides decrease in the order : RI> RBr > RCl > RF. This Is because with the increase in size and mass of halogen atom, the magnitude of van der Waal forces increases.

The boiling points of isomeric habalkanes decrease with increase in branching.

Boiling points of isomeric dihalobenzenes are very nearly the same. However, the para Isomers are high melting as com pared to their ortho – and meta-Isomers. It Is due to symmetry of para-isomers that fits in crystal lattice better as compared to ortho-and meta-isomers.

3) Odour: Many volatile haloalkanes have sweet smell.

4) Solubility:
Haloalkanes are only very slightly soluble in water. However, they are soluble in organic solvents like alcohol, ether etc.

5) Density :
Bromo, iodo and polychloro derivatives are heavier than water. The density increases with increase in number of carbon atoms, halogen atoms and atomic mass of the halogen atoms.

6) Toxicity:
Haloalkanes are, in general, toxic compounds and cause general anaethesia when inhaled in larger amounts.

Question 23.
Explain the mechanism of Nucleophilic bimolecular substitution (S_N²) reaction with one example. [Mar. 2018-AP & TS]
Answer:
Nucleophilic bimolecular substitution (S_N²) reactions : The rate of S_N² reaction depends upon the concentration of both the reactants. For example, conversion of alkylhalide to alcohol by the action of alkali shows,
Rate ∝ [R – X] [OH^–]
The reaction between methyl chloride (CH₃Cl) and hydroxide ion (OH^–) to yield methanol and chloride ion follows second order kinetics.
Rate ∝ [CH₃Cl] [OH^–]
The incoming nucleophile (- OH^–) interacts with the alkyl halide. The bond between the nucleophile and the carbon atom starts forming, while the bond between carbon atom and leaving group weakens forming a transition state. As this happens, the configuration of carbon atom under attack inverts in much the same way as an umbrella is turned inside out when caught in a strong wind. This process is called inversion of configuration. In the transition state the carbon atom is simultaneously bonded to incoming nucleophile and the outgoing leaving group. The carbon atom in the transition state is simultaneously bonded to five atoms or groups and therefore is unstable and cannot be isolated.

The order of reactivity for S_N² reaction followed is.
Primary halide > Secondary halide > Tertiary halide.

Question 24.
Explain why allylic and benzylic halides are more reactive towards S_N¹ substitution while 1- halo and 2-halobutanes preferentially undergo S_N² substitution.
Answer:
S_N¹ reaction involves the ionisation of the C – X (X = halogen) bond, irt the first step, and an intermediate carbonium ion is produced. Allylic and benzylic halides show high reactivity towards S_N¹ substitution because the carbonium ion intermediates formed from them are stabilised by resonance.

1 – halobutane and 2-halobutane being primary and secondary alkyl halides respectively preferentially undergo S_N² substitution as the attacking nucleophile can easily approach the carbon that bears the halogen atom.

Question 25.
Describe the stereo chemical effect on the hydrolysis of 2-bromobutane.
Answer:
2-bromobutane is an optically active alkyl-halide. Hydrolysis of 2-bromobutane by S_N² mechanism results in the formation of 2-butanol which has inverted configuration compared to the reactant. This happens because the – OH group occupies the posi-tion opposite to what the bromide had occupied.

Question 26.
What is the criteria for optical activity ? Give two examples of chiral molecules.
Answer:
An organic compound will show optical activity if there is asymmetry in its molecule.

The spacial arrangement of four groups around a central carbon is tetrahedral. If all the substitutents attached to that carbon are different such a carbon is called asymmetric carbon or stereocentre. The resulting molecule would lack symmetry and is referred to as asymmetric molecule. The asymmetry is responsible for the optical activity in such organic compounds.

Hence if a substance has at least one asymmetric carbon atom it will be optically active. However, the presence of asymmetric carbon atoms may not make a compound necessarily optically active; what is essential is the asymmetry of the molecule as a whole.

Ex : 1. 2-butinol has four different groups attached to the central carbon. The molecule is asymmetric or chiral. Hence it is optically active.

Ex: 2. Lactic acid molecule is chiral.

Question 27.
Define the following:
i) Racemic mixture
ii) Retention of configuration
iii) Enantiomers
Answer:
i) Racemic mixture: A mixture containing two enantiomers in equal proportion will have zero optical rotation, as the rotation due to one isomer will be exactly cancelled by the rotation due to the other isomer. Such a mixture is known as racemic mixture.

ii) Retention of configuration : Retention of configuration is the preservation of the spacial arrangement of bonds to an asymmetric centre during a chemical reaction or transformation.

iii) Enantiomers : Enantiomers are stereo-mers related to each other as non-super-imposable mirror images.

Question 28.
Write the mechanism of dehydrohalo-genation of 2-bromobutane.
Answer:
When 2-bromobutane is heated with alcoholic solution of potassium hydroxide, there is elimination of hydrogen atom from β-carbon and the bromine atom on the α-carbon atom. As a result, 2-butene is formed as the product.

Here there is the possibility of the formation of 1-butene also due to the availability of hydrogen on antoher (3-carbon atom. However, according to Saytzeff rules (also called Zaitsev rule), “in dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms”. Thus 2- bromobutane gives 2-butene as the major product. 1-butene is formed only as a minor product.

Question 29.
Explain the Grignard reagents preparation and application with suitable example. [AP 15] [TS Mar. 19]
Answer:
Alkyl magnesium halides R MgX are called Grignard reagents.

Preparation : Grignard reagents are prepa-red by the reaction of alkylhalides with magnesium metal in dry ether.

For example, ethylbromide reacts with magnesium metal in dry ether to give ethyl magnesium bromide.

Applications of Grignard reagents : Grignard reagents are highly reactive and enter into reaction with wide variety of substrates yielding many types of organic compounds.

Ex: Reaction with active hydrogen corn-pounds. Grignard reagents react with alcohols, water, ammonia, amines etc, to form a hydrocarbon corresponding to the alkyl group of the Grignard reagent.

Question 30.
A primary alkyl halide C₄H₉Br (A) reacted with alcoholic KOH to give compound B. B on reaction with HBr yields C which is an isomer of A. When A is reacted with sodium metal forms D, C₈H₁₈ which is different from the compound formed when n-butylbromide is reacted with sodium. Give the structural formulae of A-D and write equations for all the reactions.
Answer:
The structural formulas of compounds A – D are :
A CH₃CH₂CH₂CH₂Br
n-Butyl bromide

B CH₃CH₂CH = CH₂
1- Butene

A and C are position isomers.
n-butyl bromide reacts with sodium to give n-octane (Wurtz reaction). This is different from the compound obtained from C.

Question 31.
Account for the following statements:
i) Arylhalides are extremely less reactive towards Nucleophilic substitution reactions.
ii) p-Nitrochlorobenzene and o, p-dinitrochlorobenzene undergo Nucleophilic substitution readily compared to chlorobenzene.
Answer:
i) Aryl halides are extremely less reactive towards’ nucleophilic substitution reactions for two reasons.

Resonance effect: The electron pairs of halogen atom are in conjugation with π-electrons of the ring and the following reso-nance structure are possible.

The C – X (X = halogen) bond acquires a partial double bond character due to reasonance. As a result, the bond cleavage in haloarene is more difficult than in haloalkane and therefore, they are less reactive towards nucleophilic substitution reaction.

Difference in hybridisation of carbon atom in C – X bond :
In haloalkane the carbon atom attached to halogen is sp³ hy-bridised whereas in haloarene the carbon atom attached to halogen is sp² hybridised. The sp² hybridised carbon with a greater s-character is more electronegative and can hold the electron pair of the C – X bond more tightly than sp³ hybridised carbon in haloalkane with less s-character. Hence C – X bond length in haloarene is less than that in haloalkane. Since it is more difficult to break a shorter bond than a longer bond, haloarenes are less reactive than halo-alkane toward nucleophilic substitution reaction.

ii) The reactivity of haloarenes such as chlorobenzene will increase if an electron withdrawing group (-NO₂) is present at ortho-and para-positions. Hence p-nitro-chlorobenzene and o, p – dinitrochlorobenzene undergo nucleophilic substitution more readily than chlorobenzene.

Question 32.
Explain how the following conversions are carried out:
i) Propene to Propanol
ii) Ethanol to but-1-yne
iii) 1-Bromopropane to 2-Bromopropane.
iv) Aniline to Chlorobenzene
Answer:
i) Propene reacts with HBr to give 2-Bromo-propane as major product. 2-Bromopro-pane reacts with aq.KOH to yield 2-propanol.

ii) Ethanol reacts with HCl to give ethyl chloride. Ethyl chloride reacts with sodium acetalide to give but-1-yne.

iii) 1-Bromopropane when heated with alcoholic KOH undergoes dehydrobromination to give propene which when reacted with HBr gives 2-Bromopropane as the major product.

iv) Aniline on diazotisation gives benzene diazonium chloride which reacts with cuprous chloride in HCl to give chlorobenzene.

Question 33.
What happens when –
i) n-butylchloride is treated with ale. KOH.
ii) Bromobenzene is treated with Mg in presence of dry ether.
iii) Methylbromide is treated with sodium in presence of dry ether.
Answer:
i) When n-butyl chloride is treated with ale. KOH 1-butene is formed.

ii) When bromobenzene is treated with Mg in presence of dry ether, phenyl magnesium bromide (Grignard reagent) is formed.

iii) When methyl bromide is treated with sodium in presence of dry ether, ethane is obtained.

Question 34.
Write the reactions showing the major and minor products when chlorobenzene is reacted with CH₃Cl and CH₃COCl in presence of AlCl₃.
Answer:

Intext Questions – Answers

Question 1.
Write the structures of the following compounds.
i) 2-Chloro-3-methylpentane
ii) 1-Chloro-4-ethylcydohexane
iii) 4-tert. Butyl-3-iodoheptane
iv) 1,4-dibromobut-2-ene
v) 1-Bromo-4-sec. butyl-2-methylbenzene
Answer:

Question 2.
Why is H₂SO₄ not used duting the reaction of alcohols with KI?
Answer:
H₂SO₄ converts KI to the corresponding acid. HI which is then oxidised by it to I₂. Hence H₂SO₄ cannot be used along with KI in the conversion of an alcohol to alkyl iodide.

Question 3.
Write the structures of different dihalogen derivatives of propane.
Answer:

1. CH₃CH₂CHX₂
2. X CH₂ CH₂ CH₂ X
3. CH₃ CX₂ CH₃
4. X CH₂ CHX CH₃
   (X = halogen).

Question 4.
Among the isomeric alkanes of molecular formula C₅H₁₂, identify the one that on photochemical chlorination yields
i) A single monochloride
ii) Three ishmeric monochloridea
iii) Four isomeric monochlorides.
Answer:
i)

All the hydrogen atoms are equivalent in neopentane (C₅H₁₂). It gives a single mono chloride by replacement of any hydrogen.

ii)

There are three, groups of equivalent hydrogens marked a, b and c. The replacement of equivalent protons will yield three isomeric monochlorides.

iii)

There are four groups of equivalent hydrogens, marked a, b, c and d. Replacement of equivalent hydrogens by chlorine yields, four isomeric monochlorides.

Question 5.
Draw the structures of major monohalo products in each of the following reactions:


Answer:

Question 6.
Arrange each set of compounds in order of increasing boiling points.
i) Bromomethane, Bromoform, Chloro- methane, Dibromomethane.
ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.
Answer:
i) Boiling points of haloalkanes increase with increase in molecular mass as inter- molecular forces of attraction increase. Hence the boiling points of the compounds increase in the order.

Chloromethane < Bromomethane < Dibromomethane < Bromoform.

ii) The boiling points of isomeric haloalkanes decrease with increase in branching. Hence the boiling points increase in the order.

Isopropyl chloride < 1-chloropropane < 1-chlorobutane

Question 7.
Which alkyl halide from the following pairs would you expect to react more rapidly by an S_N² mechanism ? Explain your answer.

Answer:
i) CH₃CH₂CH₂CH₂Br-is a primary alkyl halide. As there is no steric hindrance for the attacking nucleophile, it undergoes S_N² reaction more rapidly than which is a secondary alkyl halide.

ii) being a secondary alkyl halide reacts more rapidly by an S_N² mechanism than which is a tertiary alkyl halide.

iii) reacts more rapidly by S_N² mechanism because in the other compound the methyl group is closer to the halogen atom and increases steric hindrance causing a decrease in the reaction rate.

Question 8.
In the following pairs of halogen compounds, which compound undergoes fester S_N¹ reaction?

Answer:
i) undergoes faster S_N¹ reaction because of the greater stability of the carbocation formed from it.

ii)
is a secondary alkyl halide whereas CH₃ (CH₂)₄ CH₂Cl is a primary alkyl halide. undergoes faster S_N¹ reaction because of the greater stability of secondary carbocation than primary carbocation.

Question 9.
Identify A, B, C, D, E, R and R’ in the following.

Answer:

Telangana TSBIE TS Inter 2nd Year Chemistry Study Material Lesson 12(b) Aldehydes, Ketones, and Carboxylic Acids Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material Lesson 12(b) Aldehydes, Ketones, and Carboxylic Acids

Very Short Answer Questions (2 Marks)

Question 36.
Arrange the following compounds in increasing order of their property indicated –
i) Acetaldehyde, Acetone, Methyl t-butyl ketone reactivity towards HCN.
ii) Floroacetic acid, monochloroacetic acid, Acetic acid and Dichloroacetic acid (acid strength).
Answer:

Question 37.
Write the reaction showing a-haloge-nation of carboxylic acid and give its name.
Answer:

The reaction is known as Hell-Volhard-Zelinsky reaction.

Question 38.
Although phenoxide ion has more number of resonating structures than carboxylate ion carboxylic acid is a stronger acid than phenol. Why ?
Answer:
The answer lies mostly in the relative stability of the negative charge of the phenoxide and carboxylate ions. In carboxylate ion, the negative charge is divided between two electronegative oxygen atoms. Whereas it less effectively delocalised over one oxygen atom and less electronegative carbon atoms in phenoxide ion. Thus the carboxylate ion is more stable than phenoxide ion. Hence carboxylic acids are more acidic than phenols.

Question 39.
How do you distinguish acetophenone and benzophenone ?
Answer:
Acetophenone being a methyl ketone responds to iodoform test while benzophenone does not.

Question 40.
Explain the position of electrophilic substitution in benzoic acid.
Answer:
The carboxyl group in benzoic acid acts as a deactivating and meta-directing group. Hence electrophilic substitution occurs in the meta position.

Question 41.
Write equations showing the conversion of –
i) Acetic acid to Acetyl chloride
ii) Benzoic acid to Benzamide
Answer:
i) CH₃COOH + PCl₅ → CH₃COCl + POCl₃ + HCl
ii)

Question 42.
An organic acid with molecular formula C₈H₈O₂ on decarboxylation forms Toluene. Identify the organic acid.
Answer:

Question 43.
List the reagents needed to reduce carboxylic acid to alcohol.
Answer:
The following reagents can be used to reduce carboxylic acid to primary alcohol.

1. Lithium aluminium hydride or
2. Diborane

Question 44.
Write the mechanism of esterification.
Answer:

Question 45.
Compare the acidic strength of Acetic acid, Chloroacetic acid, Benzoic acid and Phenol. [IPE 14]
Answer:
Acidic strength of the acids decreases in the order
Chloroacetic acid > Benzoic acid > Acetic acid > Phenol

Short Answer Questions (4 Marks)

Question 46.
Write the equations for the reaction of any aldehyde with Fehling’s reagent.
Answer:
Fehling’s solution contains complex cupric ions. It is prepared by adding Fehiing A solution which contains copper sulphate, to Fehiing B solution which contains sodium hydroxide and s Rochelle salt (sodium potassium tartarate). During the oxidation of aldehydes to acids, the cupric ions are reused to cuprous ions which are precipitated as red cuprous oxide.
RCHO + 2Cu²⁺ + 3OH^– → RCOO^– + 2Cu⁺ + 2H₂O
2Cu⁺ + 2OH^– → Cu₂O ↓ + H₂O Cuprous oxide (red)

Question 47.
What is Tollens’reagent ? Explain its reaction with Aldehydes.
Answer:
Ammonical silver nitrate solution is known as Tollens’ reagent.

On warming an aldehyde with freshly prepared ammonical silver nitrate solution, (Tollens reagent), a bright silver mirror is produced due to the formation of silver metal. The aldehyde is oxidised to the corresponding carboxylate anion. The reaction occurs in alkaline medium.
RCHO + 2 [Ag(NH₃)₂]+ + 3OH^– → RCOO^– + 2Ag ↓ + 2H₂O + 4NH₃

Question 48.
Write the oxidation product of: Acetaldehyde, Acetone and Acetophenone.
Answer:
Acetaldehyde on oxidation gives acetic acid.
CH₃CHO CH₃COOH
Acetone on oxidation with a strong oxi-dising agent like KMnO₄ gives acetic acid.
CH₃COCH₃ CH₃COOH + CO₂ + H₂O
Acetophenone on oxidation with potassium permanganate gives benzoic acid.

Question 49.
Explain why Aldehydes and Ketones undergoes nucleophilic addition while alkenes undergoes electrophilic addition though both are unsaturated compounds.
Answer:
The real cause of the reactivity of the carbonyl group in aldehydes and ketones towards nucleophiles is the tendency of the oxygen atom to acquire electrons, i.e., its ability to carry a negative charge and impart positive charge to the carbonyl carbon. Hence carbonyl compounds – undergo nucleophilic addition.

In alkenes, the carbon-carbon double bond consists of a strong c-bond and a weaker π- bond. The π-electrons are loosely held between the carbon nuclei and are, therefore readily polarisable. The loosely held π-electrons are particularly available to an electrophilic reagent. Hence alkenes undergo electrophilic addition reactions.

Question 50.
Write the IUPAC names of following:
i) CH₃CH₂CH(Br) CH₂COOH
ii) Ph.CH₂COCH₂COOH
iii) CH₃ . CH(CH₃)CH₂COOC₂H₅
Answer:

Question 51.
Arrange the following in the increasing order of their acidic strength :
Benzoic acid, 4- Methoxybenzoic acid, 4-Nitroben-zoic acid and 4-Methylbenzoic acid.
Answer:
4-Methoxybenzoic acid < 4-Methylbenzoic acid < Benzoic acid < 4-Nitrobenzoic acid

Question 52.
Describe the following :
i) Cross aldol condensation
ii) Decarboxylation
Answer:
i) Cross Aldol condensation : When aldol condensation is carried out between two different aldehydes and / or ketones, it is called cross aldol condensation. If both of them contain a-hydrogen atoms, it gives a mixture of four products.

For example a mixture of acetalde-hyde and propionaldehyde gives four products.


Decarboxylation : Carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with sodalime (NaOH + CaO). This reaction is known as decarboxylation.

Question 53.
Explain the role of electron withdrawing and electron releasing groups on the acidity of carboxylic acids.
Answer:
Electron withdrawing groups increase the acidity of carboxylic acids by stabilising the conjugate base through delocalisation of the negative charge by inductive and/or resonance effects. On the other hand, electron donating groups decrease the acidity by destabilising the conjugate base.

Question 54.
Draw the structures of the following derivatives :
i) Acetaldehyde dimethylacetal
ii) The ethylene ketal of hexan-3-one
iii) The methyl hemiacetal of formaldehyde.
Answer:

Question 55.
An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollens’ reagent but forms sodium hydrogensulphite adduct and gives +ve iodoform test. On vigorous oxidation forms ethanoic and propanoic acids. Write the possible structure of the compound.
Answer:
The compound contains 69.77% carbon, 11.63% hydrogen and 100 – (69.77 + 11.63) = 18.60%. oxygen. From the percentage composition the emperical formula is calculated.

The emperical formula = C₅H₁₀O
Emperical formula weight = 5 × 12 + 10 × 1 + 16 = 86. Since the emperical formula weight is the same as the molecular mass, the molecular formula of the compound is C₅H₁₀O.

Since the compound does not reduce tollens’ reagent, it is not an aldehyde. It gives product with sodium bisulphite. It contains a carbonyl group. It may be a ketone. Since it gives iodoform test, it may be a methyl ketone. On vigorous oxidation it gives ethanoic acid and propanoic acid. The compound must be 2-pentanone.

Long Answer Questions (8 Marks)

Question 56.
Explain the following terms. Give an example of the reaction in each case. ’
i) Cyanohydrin
ii) Acetal
iii) Semicarbazone
iv) Aldol
v) Hemiacetal
vi) Oxime
Answer:
i) Cyanohydrin : Aldehydes and ketones react with hydrogen cyanide (HCN) to yield cyanohydrins.

The addition of HCN is catalysed by cyanide ion, but HCN is too weak an acid to provide enough : C ≡ N: for the reaction to proceed at a reasonable rate. Cyanohydrins are therefore normally prepared by adding an acid to a solution containing the carbonyl compound and sodium or potassium cyanide.

ii) Acetal: In the presence of dry hydrogen chloride, aldehydes react with one equivalent of monohydric alcohol to yield a product called hemiacetal. It further reacts with one more molecule of alcohol to form a gem-dialkoxy compound called acetal.

When the carbonyl compound is a ketone instead of an aldehyde, the addition products are called a hemiketal and a ketal respectively.

Acetal formation requires an acid catalyst. The acid protonates the carbonyl oxygen, making it easier for the alcohol to attack the carbonyl carbon.

iii) Semicarbazone : Semicarbazide undergoes acid catalysed nucleophilic addition reaction with the = O group of aldehydes and ketones but the reaction is followed by elimination of H₂O molecule from the addition product. The product is a well defined crystalline compound known as semicarbazone.

iv) Aldol: Aldehydes and ketones having at least one a-hydrogen undergo a reaction in the presence of dilute alkali as catalyst to form β-hydroxy aldehydes (aldol) or (β-hydroxy ketones (ketols). This is known as Aldol reaction.

The name aldol is derived from the names of the two functional groups, aldehyde and alcohol.

v) Hemiacetal: Aldehydes react with one mole of a monohydric alcohol in the presence of dry HC/ to yield alkoxyalcohol intermediate, known as hemiacetal.

Hemi is the Greek prefix for “half”. When one equivalent of alcohol has added to an aldehyde or ketone the compound is halfway to the final acetal or ketal, which contains two groups from two equivalents of alcohol.

vi) Oxime : Aldehydes and ketones, when treated with hydroxylamine, form oximes which are known as aldoximes and ketoximes respectively. These oximes are usually well-defined crystalline solids.

Question 57.
Name the following compounds according to IUPAC system of nomenclature :
i) CH₃CH(CH₃)CH₂CH₂CHO
ii) CH₃CH₂COCH(C₂H₅)CH₃CH₃Cl
iii)CH₃CH = CHCHO
iv) CH₃COCH₂COCH₃
Answer:

Question 58.
Draw the structures of the following compounds.
i) 3-Methylbutanal
ii) p-Nitropropiophenone
iii) P-Methylbenzaldehyde
iv) 3-Bromo-4-phenylpentanoic acid
Answer:

Question 59.
Write the IUPAC names of following ketones and aldehydes. Wherever possible, give also common names,
i) CH₃CO(CH₂)₄CH₃
ii) CH₃CH₂CHBrCH₂CH(CH₃)CHO
iii) CH₃(CH₂)₅CHO
iv) PhCH = CHCHO
v)

vi) PhCOPh
Answer:

iii) CH₃(CH₂)₅CHO
Heptanal
(Enanthaldehyde)

iv) PhCH = CHCHO
3 – phenyl prop-2-enal
(Cinnamaldehyde)

v)

Cyclopentanecarbaldehyde

vi) PhCoph
Benzophenone
(Diphenyl ketone)

Question 60.
Draw the structure of the following derivatives.
i) The 2,4-dinitrophenylhydrazone of benzaldehyde
ii) Cyclopropanone oxime
iii) Acetaldehyde hemiacetal
iv) The Semicarbazone of cyclobutanone
Answer:

Question 61.
Predict the products formed when Cyclohexanecarbaldehyde reacts with the following reagents.
i) PhMgBr and then H₃O⁺
ii) Tollens’ reagent
iii) Semicarbazide and weak acid
iv) Zinc amalgam and dilute HCl
Answer:
i) Phenylcyclohexylmethanol, C₆H₁₁CH(OH)Ph
ii) Cyclohexane carboxylic acid (C₆H₁₁COOH) and silver
iii) Semicarbazone
iv) Cyclohexylmethanol

Question 62.
Which of the following compounds would undergo aldol condensation ? Write the structures of the products expected.
i) 2-Methylpentanal
ii) 1-Phenylpropanone
iii) Phenyl acetaldehyde
iv) 2,2-Dimethylbutanal
Answer:
Compounds (i), (ii) and (iii) contain at least one a-hydrogen. Hence they undergo Aldol reaction. 2, 2-Dimethyl butanal does not undergo Aldol condensation as it has no α-hydrogen.

iv) 2, 2-dimethyl butanal: Not involved in aldol condensation due to the absence of ‘α’ hydrogen.

Question 63.
An organic compound A(C₉H₁₀O) forms 2,4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizaro reaction. On vigorous oxidation it gives 1,2-benzenedicarboxylic acid. Identify the compound.
Answer:
Molecular formula of the compound A is C₉H₁₀O. Since it gives 2, 4-DNP derivative it is a carbonyl compound, aldehyde or ketone. As it reduces Tollens’ reagent, it must be an aldehyde. It undergoes Cannizzaro reaction. It has no α-hydrogen. Since it gives 1,2-benzene dicarboxylic acid, the compound is o-ethyl benzaldehyde.

Question 64.
How do you distinguish the following pairs of compounds ?
i) Propanal and propanone
ii) Acetophenone and benzophenone
iii) Phenol and benzoic acid
iv) Pentan-2-one and Pentan-3-one
Answer:
i) Propanal (CH₃CH₂CHO) gives silver mirror with Tollens’ reagent while propanone, CH₃ – CO – CH₃ does not. This test is used to distinguish between the two compounds.

ii) Acetophenone (C₆H₅COCH₃) being a methyl ketone gives yellow crystals of iodoform (CHI₃) on heating with iodine in the presence of sodium hydroxide. Benzophenone (C₆H₆COC₆H₅) does not respond to iodoform test. This test is used to distinguish between acetophenone and benzophenone.

iii) Phenol is less acidic than benzoic acid. Benzoic acid gives brisk effervescence with sodium bicarbonate solution due to evolution of carbondioxide. Phenol does not react with sodium bicarbonate. This test is used to distinguish between phenol and benzoic acid.

iv) Pentan-2-one is a methylketone. It gives positive iodoform test.
Pentan-3-one does not give iodoform.

Question 65.
How are the following conversions carried in not more than two steps ?
i) Ethanol to 3-hydroxybutanal
ii) Bromobenzene to 1-Phenylethanol
iii) Benzaldehyde to ±-Hydroxyphenylacetic acid
iv) Benzaldehyde to benzophenone
Answer:
i) Ethanol is first oxidised to acetaldehyde which undergoes Aldol condensation in the presence of dilute sodium hydroxide to give 3-hydroxybutanol.

ii) Bromobenzene is reacted with magnesium in dry ether to get phenyl magnesium bromide (Grignard reagent). This on reaction with acetaldehyde followed by hydrolysis gives 1-phenyl ethanol.

iii) Benzaldehyde reacts with hydrogen cyanide to give cyanohydrin which on hydrolysis gives α-hydroxyphenyl acetic acid.

iv) Benzaldehyde on reaction with chlorine in the absence of halogen carrier forms benzoyl chloride which reacts with benzene in the presence of anhydrous aluminium chloride to give benzophenone (Friedel Craft’s reaction)

Question 66.
Describe the following. [A.P. Mar. 19]
i) Acetylation
ii) Cannizaro reaction
iii) Cross aldol condensation
iv) Decarboxylation
Answer:
i) Acetylation : When benzene or substituted benzene is treated with acetyl chloride in the presence of anhydrous aluminium chloride, the corresponding ketone is obtained. This is called acetylation reaction. Substitution of hydrogen atom by acetyl group (CH₃CO -) is acetylation.

Phenols, primary and secondary amines also undergo acetylation with acetyl chloride or acetic anhydride.

ii) Cannizaro reaction: Aldehydes which donot have α-hydrogen atom, undergo self oxidation and reduction (disproportionation) reaction on heating with concentrated alkali. In this reaction one molecule of the aldehyde is reduced to alcohol while another is oxidised to carboxylic acid salt.

iii) Cross aldol condensation : Aldol condensation between two different aldehydes and / or ketones is called cross aldol condensation. If both of them contain α-hydrogen, a mixture of four products will be obtained.

For example, a mixture of acetaldehyde and propionaldehyde undergoes cross aldol condensation in the presence of dilute sodium hydroxide solution to give a mixture of four products.

iv) Decarboxylation: Carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda lime (NaOH + CaO). This reaction is known as decar-boxylation.

Question 67.
Complete each synthesis by giving the missing starting material, reagent or product.

Answer:

Question 68.
Explain how methyl ketones are distinguished from other ketones. Write the equations showing it
Answer:
Methyl ketones are distinguished from other ketones by haloform reaction. Methyl ketones are oxidised by sodium hypohalite (halogen in the presence of NaOH) to salts of the corresponding carboxylic acid having one carbon atom less than that of the carbonyl compound. The methyl group is converted to haloform.

In this reaction the three hydrogen atoms of the methyl group are replaced by halogen atoms. The trihalo ketone so obtained further reacts with aqueous alkali. The ^–OH attacks the carbon atom of the trihaloketone and causes a cleavage of the carbon-carbon bond between the carbonyl group and the trihalomethyl group. The ultimate products are the carboxylate ion and the haloform (CHCl₃, CHBr₃ or CHI₃).

Question 69.
Write the equations showing the conversion of the following along with reagents :
i) 1-phenylpropane to Benzoic acid
ii) Benzamide to Benzoic acid
iii) Ethyl butanoate to Butanoic acid.
Answer:

Question 70.
Write the products and reagents needed for the below given conversions :
i) 3-Nitrobromobenzene to 3-Nitrobenzoic acid
ii) 4-Methylacetophenone to Benzene-1,4-dicarboxylic acid
Answer:
i) 3-Nitrobromobenzene reacts with magnesium in dry ether to give the corresponding Grignard reagent which reacts with dry ice to give a product which on acidic hydrolysis gives 3-nitrobenzoic acid.

ii) 4-methyl acetophenone is converted to dipotassium benzene-1,4-dicarboxylate on oxidation with alkaline potassium permanganate which on acidification gives Benzene-1,4-dicarboxylic acid (Terephthalic acid).

Intext Questions – Answers

Question 1.
Write the structures of the following compounds.
i) α-methoxypropionaldehyde
ii) 3-Hydroxybutanaol
iii) 2-Hydroxycyclopentane carbaldehyde
iv) 4-oxopentanal
v) Di-sec. butyl ketone
vi) 4-Fluroacetophenone
Answer:

Question 2.
Write the structure of products of the following questions.

Answer:

Question 3.
Arrange the following compounds in increasing order of their boiling points.
CH₃CHO, CH₃CH₂OH, CH₃OCH₃, CH₃CH₂CH₃.
Answer:
CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CHO < CH₃CH₂OH

Question 4.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.
i) Ethanal, propanal, propanone, butanone.
ii) Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone.
Answer:
i) Butanone < Propanone < Propanal < Ethanal
ii) Acetophenone < p-Tolualdehyde, Benzaldehyde < p-Nitrobenzaldehyde

Question 5.
Predict the products of the following reactions.

Answer:

Question 6.
Give the IUPAC names of the following compounds :
i) PhCH₂CH₂COOH
ii) (CH₃)₂C = CHCOOH

Answer:
i) 3-phenylpropanoic acid
ii) 3-methylbut-2-enoic acid
iii) 2-methylcyclopentane carboxylic acid
iv) 2, 4, 6-Trinitrobenzoic acid

Question 7.
Show how the following compounds can be converted to benzoic acid.
i) Ethylbenzene
ii) Acetophenone
iii) Bromobenzene
iv) Phenylethene (Styrene)
Answer:

Question 8.
Which acid of each pair shown here would you expect to be stronger ?
i) CH₃COOH or CH₂FCOOH
ii) CH₂FCOOH or CH₂ClCOOH
iii) CH₂FCH₂CH₂COOH or CH₃CHFCH₂COOH
iv) F₃C COOH or H₃C COOH .
Answer:
i) CH₃COOH
ii) CH₂FCOOH
iii) CH₃CHFCH₂COOH
iv) F₃C COOH

Telangana TSBIE TS Inter 2nd Year Accountancy Study Material 1st Lesson Depreciation Textbook Questions and Answers.

TS Inter 2nd Year Accountancy Study Material 1st Lesson Depreciation

Short Answer Questions

Question 1.
Write the need or significance of depreciation.
Answer:
Need or significance or providing depreciation:
1. To ascertain true results of business operations: As depreciation is treated as expenditure, it must be charged to profit and loss account against income to assess the real profit/loss of the business.

2. To show the fixed asset at their original worth in the balance sheet: If depreciation is not provided, the value of the asset shown in the balance sheet is not correct. Hence, the real value of the asset must be shown in the balance sheet after deducting depre¬ciation from its book value.

Question 2.
Briefly explain different methods of providing depreciation.
Answer:
The various methods pf providing depreciation are given below:
1) Fixed Instalment method: Fixed instalment method is also called as equal instalment method or original cost method or straight-line method under this method depreciation is charged on the original cost of the asset every year. In this method annual depreciation is fixed or uniform.

2) Diminishing Balance method: Diminishing balance method is also called as written down value method (WDU) or reducing balance method under this method deprecia¬tion is calculated at a fixed percentage on the diminishing value of the asset.

3) Annuity method: This method takes into account the interest lost on the acqisition of an asset. Interest is calculated on the book value of the asset in the beginning of every year, and is debited to asset account. The amount of depreciation to be charged is uniform and is calculated on the basis of annuity tables.

4) Depreciaton fund method: This method provides funds for replacement of assets. The amount writtenoff as depreciation is kept aside and that amount is invested in certain securities. When the life of the asset expires, the securities are sold and a new asset is purchased with the help of sale proceeds. Depreciation amount is calculated with the help of Sinking Fund Tables.

5) Depletion method: Under this method total quantity of output likely to be available is estimated. Depreciation rate is calculated by dividing the cost of the asset by the esti-mated quantity of product likely to be available.
Suitability: It is used for mines and quarries.

6) Machine Hour Bate method: In this method the life of asset is estimated in hours. In order to calculate depreciation the actual number of hours in a particular year is multiplied by the hourly rate.

Suitability: This method is Used for Machines.

Question 3.
Write the merits and demerits of straight-line method.
Answer:
Merits:

1. This method is very easy to understand.
2. It is very simple to calculating depreciation & rate of depreciation.
3. Under this method, assets can be written off completely. That is upto zero, so the value of asset is equally spread out over the useful life of the asset.
4. This method is suitable for small businesses.
5. This method is useful to those assets having lower value and fixed life.

Demerits:

1. The method is illogical, because, depreciation is charged on original cost every year, but to the balance of asset is declining every year.
2. This method is not recognised by the income tax authorities.
3. This method is not useful incase any additions and expansions made to assets.
4. In put pressure on the asset in the slack years, in the end of the asset’s life, it bears more repairs and maintence charges but the depreciation charge is equal for all years.

Question 4.
What are the advantages and limitations of the diminishing balance method?
Answer:
Advantages:

1. This method is easy to calculate depreciation when there are no additions to the asset.
2. It equalises the burden on profit and loss account in respect of repairs and deprecia¬tion put together.
3. This method is accepted by income tax authorities under the income tax Act 1961.
4. This method is logical in the sence that as the asset value decreases the amount of depreciation also goes on decreasing.
5. This method can be used where obsolescence rare is high.

Limitations:

1. It is very difficult to determine the appropriate rate of depreciation.
2. In this method the value of the asset cannot be reduced to zero. Some balance will be left at the end of its useful life.
3. This method is not suitable for the assets which have shorter life.
4. It does not provide for the replacement of the asset.

Question 5.
Distinguish between stright line method and diminishing balance method.
Answer:

Very Short Answer Questions

Question 1.
Define the term Depreciation.
Answer:

1. Depreciation is a permanent, continuous and gradual shrinkage in the value of fixed asset.
2. Depreciation means a fell in the value or quality of an asset.
3. According to companies Act 2013, “Depreciation is the systematic allocation of the depreciable amount of an asset over its useful life”.

Question 2.
What is obsolescene?
Answer:

1. Obsolescene means diminution in the value of fixed asset due to new inventions, new improvements, change in fashions, change in customer’s tastes and preferences.
2. For Example: Invention of computer made typewriters becomes obsiete or out dated.

Question 3.
What are the causes of depreciation?
Answer:
Causes of depreciation are:

1. Wear and Tear,
2. Depletion,
3. Accidents,
4. Obsolescence,
5. Effluxion of time/passage of time, 6) Fluctuations.

Question 4.
What is Depletion?
Answer:

1. Depletion means loss of natural resources mineral wealth due to extraction of raw material from mines, quarries, oil wells etc.
2. It refers to the physical deterioration by the exhaustion of natural resources like oil wells, mines quarries etc.

Question 5.
What is the straight-line method?
Answer:

1. The method under which, the same amount of depreciation is charged on the original cost of the asset every year throughout the life of the asset, is called straight line method.
2. Straight line method is also known as “Fixed Instalment Method” or Equal Instalment method” or original cost method”.

Question 6.
What is diminishing Balance Method?
Answer:

1. A method under which depreciation is calculated at a fixed percentage on the original cost of the asset in the first year and on written down value in the subsequent year is called dininishing balance method.
2. It is also called reducing balance method (or) written down value method.

Question 7.
Distinguish depletion from depreciation.
Answer:

Textual Problems

Question 1.
Straight Line Method:
Sri sai & Co purchased a machine for Rs. 2,50,000 on 1st April 2009. Estimated life of the machine is 10 years. The scrap value at the end of its life is Rs. 50,000. Calculate the annual depreciation and rate of depreciation assuming that accounts are closed on 31st March every year. Show the machinery a/c for the first three years under fixed instalment method.
Answer:
Calculation of depreciation and rate of depreciation:

Question 2.
Ganesh bought a machine or Rs. 17,000 and paid for its installation Rs. 3,000 on 30th Sept. 2017. Depreciation is provided at 20% under the straight-line method. Prepare machine a/c up to 31st March, 2020.
Answer:

Question 3.
Kiran & Sons purchased a machine for 142,000 and paid Rs. 3,000 for its erection on 1st April 2017. Additions are made to the machine on 31st March 2018 for Rs. 20,000. Accounts are closed at the end of financial year. Depreciation is allowed at 10% under fixed instalment method. Prepare machine a/c for three years.
Answer:

Question 4.
Rajesh & Sons purchased a machine on 1st October, 2017 for Rs. 20,000. Depreciation is provided at 15% on original cost method. Another machine was bought on 1st April, 2019 for Rs. 30,000. Prepare machine a/c upto 31st March, 2020.
Answer:

Question 5.
Ramana & Brothers purchased furniture for Rs. 22,000 on 1st July 2017. Errection charges paid for Rs. 3,000 and paid for carriage Rs. 5,000. Depreciation is to be charged at the rate of 10% on original cost method. Additions are made to the asset for Rs. 10,000 on 1st April 2018. Show the furniture a/c for three years assuming that accounts are closed on 31st March every year.
Answer:

Question 6.
Karthik & Company bought a machine on 1st January, 2017 for Rs. 1,00,000. Depreciation is provided at 15% under straight line method. Another machine was purchased on 1st October 2019 for Rs. 50,000. Company closes its accounts every year at the end of the financial year. Prepare machine a/c for four years.
Answer:

Question 7.
On 1st April, 2017 a firm purchased a machine for Rs. 80,000. Additional machinery was purchased on 30th September, 2018 for Rs. 40,000 and on April 1st 2019 for Rs. 20,000. Depict iation is charged at the rate of 10% under straight line method. Show the machine a/c for 2018, 2019 and 2020 assuming that accounts are closed on 31st March,
Answer:

Question 8.
Raghava bought a plant and machine on 1st April, 2017 for Rs. 23,000 and paid Rs. 2,000 for its installation. Depreciation is to be allowed at 10% under straight line method. On 31st March 2020 the plant was sold for Rs. 8,000. Assuming that the accounts are closed at the end of the financial year. Prepare plants machine a/c.
Answer:

Working Note:
Cost of the plant & machine = 25,000
Loss calculation on the date of sale of machine:
Accumulated depreciaton = 7,500
(2,500 + 2,500 + 2,500)
Bank value = 17,500
Sale Price of the Machine = 8,000
Loss = 9,500

Question 9.
Vardhan purchased a machine on 31st March 2017 for Rs. 70,000. Depreciation is charged at 10% under original cost method. After three years he found that the machine was not suitable and sold for Rs. 55,000. Show the machine a/c.
Answer:

Working Note:
Cost of the Machine = 70,000
Less: Accumulated Depreciation = 21,000
[7,000 + 7,000 + 7,000]
Book value = 49,000
Profit on sale of Machine = Sales price – Book value
= 55,000 – 49,000
= 6,000

Question 10.
Neelima traders purchased furniture Rs. 20,000 on 1st April, 2016. Additions are made to the furniture on 30th September, 2017 for Rs. 10,000. On 31st December, 2019, the furniture purchased on 1st April 2016 was sold for Rs. 7,000. Depreciation is charged at 10% on fixed instalment method. The firm closes its books at the end of financial year. Prepare furniture a/c for four years.
Answer:


Working Note:
Calculation of profit / loss on sale of Furniture:
Cost of the furniture = 20,000

Diminishing Balance Method:

Question 11.
Madhu and Co. purchased a machine on 1st April 2017 for Rs. 20,000. Depreciation is charged at 10% under dimishing balance method. The company prepares accounts at the end of the financial year. Prepare machine a/c for three years.
Answer:

Question 12.
A firm purchased a machine on 1st April 2017 for Rs. 30,000 and paid errection charges Rs. 5,000. Depreciation is provided at 20% under diminishing balance method. Prepare machine a/c upto 31st March, 2020.
Answer:

Question 13.
On 1st October 2017, Raju traders bought a machine for Rs. 15,000. Additions made to the machine on 1st April, 2019 were Rs. 10,000. Depreciation is charged at 10% under reducing balance method. Prepare machine a/c for three years assuming that accounts are closed on 31st March, every year.
Answer:

Question 14.
Dinesh & Company purchased a machine on 1st July, 2016 for Rs. 1,00,000. Depreciation is provided at 20% under diminishing balance method. On 1st October 2016 another machine was bought for Rs. 20,000. Prepare machine a/c for three years by closing books on 81st March, every year.
Answer:

Question 15.
Siva traders bought a machine for Rs. 75,000 on 1st April, 2017. Depreciation was provided at 10% under diminishing balance method. On 31st March, 2020, the machine became obsolute and sold for t 30,000. Prepare machine a/c.
Answer:

Working Note:
Calculate profit/loss on sale of machine:

16. On January 1st 2017 Krishna & Company purchased a second hand machine for Rs. 65,000 and paid Rs. 15,000 for errection. Depreciation is provided at 10% under diminishing balance method on 31st March every year. Another machine was bought on 1st April 2018 for Rs. 40,000. On 31st December 2019, the machine bought on January 1st 2017 was sold for Rs. 60,000. Prepare machine a/c.
Answer:
Working Note:
Calculation of profit/loss on sale of I machine:

Textual Examples

Question 1.
Sri Raghavendra Traders purchased machinery costing Rs. 25,000 on 01-04-2020. Depreciation is provided annually under (equal) fixed instalment method. Estimated life of the machine is 10 years. The residual value at the end of its life is Rs. 5,000. Calculate annual depreciation and rate of depreciation assuming that accounts are closed on 31st March every year.
Answer:

Cost of the machine = Rs. 25,000
Residual scrap value = Rs. 5,000
Life of the Asset =10 years.

Question 2.
Srinivas bought a machine for Rs. 18,000 on 1st October 2015 and paid Rs. 2,000 for its installation. Depreciation is charged at 10% under original cost method. Every year accounts and closed at the end of financial year. Record necessary journal entries and also prepare Machine a/c for the first three years.
Answer:
Cost of the Asset = Rs. 18,000 + Rs. 2,000
= Rs. 20,000

Note: 1. Generally, accounts are closed only at the end of the March, that is the end of the financial year.
2. a) If the asset is purchased at the beginning of the year, depreciation is to be calculated for the full year.
b) If the asset is purchased in the middle of the year, depreciation is to be calculated from the date of purchase to the date of closing of books.

Question 3.
Sharada & Sons acquired a machine for Rs. 5,00,000 on 1st April, 2017 and spent 150,000 for its installation. The scrap value of the plant after is useful life of 10 years is Rs. 10,000. Prepare Machine a/c, Depreciation a/c for the first four years assuming that depreciation is provided under straight line method. Accounts are closed every year on 31st March.
Answer:
Cost = Rs. 5,00,000 + 50,000 = Rs. 5,50,000

Question 4.
Rama and Company purchased machine for Rs. 60,000 on 1st April, 2017 and spent Rs.5,000 for its errection. Depreciation is charged at the rate of 10% under straight line method. Additional machinery purchased on 1st Oct. 2018 for Rs. 20,000. Prepare Machine a/c for first three years assuming that accounts are closed at the end of March 31st every year.
Answer:

Working Notes:
Cost of the asset = Rs. 60,000 + Rs. 5,000 = Rs. 65,000
Rate of Depreciation = 10%

Note: For the calculation of depreciation, one must be careful about the following dates
(i) Date of purchase of asset,
(ii) Date of commencement of the year and
(iii) Date of closing of books.

Example: A machine is purchased on 1st April 2015 for Rs. 20,000. Depreciation is to be , provided th<; rate of 10% under straight line method. The machine was sold on 31st March 2019 for Rs. 9,000. Find out the profit or loss on sale of asset.
Cost of the Machine = 20,000
Depreciation rate = 10%

(a) Depreciation for the year 2015-2016 = 20,000 x 10/100 = Rs. 2,000
(b) Depreciation for the year 2016-2017 = 20,000 x 10/100 = Rs. 2,000
(c) Depreciation for the year 2017-2018 = 20,000 x 10/100 = t 2,000
(d) Depreciation for the year 2018-2019 = 20,000 x 10/100 = Rs. 2,000

Total Depreciation = Rs. 8,000
Book value of the asset on the date of sale:
= Cost – Total Depreciation (Accumulated Depreciation)
= 20,000 – (2,000 + 2,000 + 2,000 + 2,000) = 12,000

Loss on Sale of Asset:
= Book value – Selling price of the machine = 12,000 – 9,000 = 3,000.

Journal Entry:

5. Raghu & Company purchased furniture on 1st April, 2016 for Rs. 40,000. Depreciation is provided at the rate of 10% under straight fine method. On 31st March, 2020 the scrap of the furniture was sold for Rs. 18,000. Prepare Furniture a/c.
Answer:


Working Notes:
Cost of the asset = Rs. 40,000
Accumulated Depreciation = 4,000 + 4,000 + 4,000 + 4,000 = 16,000
Book value of the asset = Cost – Total Depreciation
= Rs. 40,000 – Rs. 16,000 = Rs. 24,000
Loss on Sale of Asset = Book value – Selling Price of the Furniture
= Rs. 24,000 – Rs. 18,000 = Rs. 6,000.

Question 6.
Dinesh & Co. Purchased a Machine on 31-12-2017 for Rs. 20,000. Depreciation provided at the rate of 15% under straight line method. On 1st July 2018 another machine was purchased for Rs. 10,000. On 31st March 2020, the machine purchased on 31st December 2017 was sold for Rs. 15,000. Prepare Machine a/c upto 31st March 2020.
Answer:


Working Notes:
Cost of the asset = Rs. 20,000
Rate of Depreciation = 15%
Depreciation (31-12-2017 to 31-03-2018) on 1st Machine = 20,000 x 15/100 x 3/12 = Rs. 750
Depreciation on 1st Machine (01-04-2018 to 31-03-2019) = 20,000 x 15/100 = Rs. 3,000
Depreciation on 2nd Machine 1st July, 2018 to 31st March 2019

Depreciation on 2nd Machine from 1st April, 2019 to 31st March 2020
= 10,000 x 15/100 = Rs. 1,500

Profit on Sale of Machine:
Cost of the Machine sold = Rs. 20,000
Accumulated Depreciation = (31st Dec. 2017 to 31st March 2020)
= 750 + 3,000 + 3,000 = 6,750
Book value of Machine = Rs. 20,000 – Rs. 6,750 = Rs. 13,250
Profit value = Selling price – Book value
= Rs. 15,000 – Rs. 13,250 = Rs. 1,750
It is to be credited to P & L a/c

Journal Entry:

Question 7.
Kiran Traders bought machinery on 1st July 2016 for Rs. 30,000. Depreciation is to be provided at the rate of 10% under straight line method. On 1st April. 2018, additional machinery was purchased for Rs. 20,000. On 1st Oct 2019 the machinery bought on 1st July, 2016 became obsolute and sold for Rs. 12,000. Prepare machinery a/c for four years assuming that books are closed every year on 31st March.
Answer:


Working Notes:
Cost of the asset = Rs. 30,000 Rate of Depreciation = 10%
1. Depreciation on 31-03-2017 = (July 1st, 2016 to March 31st, 2017 = 9 months)

2. Depreciation on 31 1 March 2018 = 30,000 x 10/100 = Rs. 3,000,
(1st April 2017 to March 31st, 2018 = 1 year)

3. Depreciation on 31st March 2019.
On 1st machine from 1st April 2018 to 31st March 2019 = 1 year
= 30,000 x 10/100 = Rs. 3,000
On 2nd machine from 1st April 2018 to 31st March 2019 = 1 year
= 20,000 x 10/100 = Rs. 2,000
Total Depreciation = Rs. 3,000 + Rs. 2,000 = Rs. 5,000.

4. Sale of Asset: On 1st Oct. the (1st Machine) machine bought on 1st July 2016 was sold for Rs. 12,000
On this machine depreciation from 1st April 2019 to 1st Oct 2019 (6 months)

Calculation of loss on sale of Machine:
Cost of the Machine = Rs. 30,000
Accumulated Depreciation = Rs. 2,250 + Rs. 3,000 + Rs. 3,000 + Rs. 1,500 = Rs. 9,750
Book value of the Asset = Rs. 30,000 – Rs. 9,750 = Rs. 20,250
Loss = Book value – Selling Price
= 20,250 – 12,000 = Rs. 8,250.

5. Depreciation 31st March 2020 Since, the first machine is sold depreciation is to be calculated only on the second machine.
i.e., 20,000 x 10/100 = Rs. 2,000

Question 8.
The Original cost of the asset is 9,00,000 and its scrap value is 50,000, expected working life is 16 years. Then the rate of depreciation is calculated as
Answer:

Illustrations on the Diminishing Balance Method:

Question 9.
On 1st April, 2017 Santhosh Brothers purchased furniture for Rs. 40,000. Installation charges paid by them Rs. 10,000. Depreciation is provided at 20% under diminishing balance method. Pass journal entries and prepare Furniture a/c for the first three years assuming that accounts are closed on 31st March every year.
Answer:
Journal entries in the Books of Santhosh Brothers.
Journal Entries

Working Notes:
Cost of the Furniture = Rs. 40,000 + Rs. 10,000 = Rs. 50,000

1. Depreciation on 31st March, 2011 = 50,000 x 20/100 = 10,000
2. Depreciation on 31st March, 2012 = 40,000 x 20/100 = Rs. 8,000
3. Depreciation on 31st March, 2013 = 32,000 x 20/100 = 6,400

Question 10.
Jagadesh & Co. purchased a machine on 1st July 2017 for Rs. 28,000. They spent Rs. 1,000 for carriage and Rs. 1,000 for installation. The machine is depreciated by 15% every year on written down value basis. Prepare machine a/c upto March 31st, 2020.
Answer:

Working Note:
Cost of the asset = Purchase price 4- Carriage + Installation charges
= Rs. 28,000 + Rs. 1,000 + Rs. 1,000 = Rs. 30,000

1. Depreciation on 31st March, 2018 = 30,000 x 15/100 x 9/2 = Rs. 3,375 (July 1st 2017 to 31st March 2018 = 9 months)
Write down value on 31st March 2018 = Rs. 30,000 – Rs. 3,375 = Rs. 26,625
2. Depreciation on 31st March 2019 = 26,625 x 15/100 = Rs. 3,994
Written down value on 31st March 2019 = Rs. 26,625 – Rs. 3,394 = Rs. 22,631
3. Depreciation on 31st March 2020 = Rs. 22.631 x 15/100 = Rs. 3,395

Question 11.
Ramesh & Company purchased a machine for Rs. 60,000 on 1st October 2017. Additions are made to the machine on 1st April 2009, for Rs. 20,000 and on 30th June 2019 for Rs. 10,000. Depreciation is provided at 10% under reducing balance method. Prepare machine a/c from 1st Oct. 2017 to 31st March 2020.
Answer:

Working Notes:

Year – 1: Depreciation on 31st March, 2018 = 60,000 x 10/100 x 6/12 = Rs. 3,000
Reduced Balance = Rs. 60,000 – Rs. 3,000 = Rs. 57,000

Year – 2: Depreciation on 31st March 2019:
Depreciation on 1 Machine = 57,000 x 10/100 = Rs. 5,700
Depreciation on 2nd Machine = 20,000 x 10/100 = Rs. 2,000
Total Depreciation = Rs. 7,700
Reduced Balance = Rs. 77,000 – Rs. 7,700 = Rs. 69,300

Year – 3: Depreciation on 31st March 2020:
On the reduced Balance of 1st and 2nd Machines
Depreciation = 69,300 x 10/100 = Rs. 6,930
Depreciation on 3rd machine from 2019 June 30th to 2020 March 31st. Since, it is bought on 30th June 2019 only for 9 months depreciation is to be calculated.

Question 12.
Sripathi traders bought a Machine on 1st April 2016 for Rs. 90,000. They paid for installation Rs. 10,000. Depreciation is to be provided at 20% on the diminishing balance method. On 31st March 2020, scrap of the Machine was sold for Rs. 50,000. Show the Machine a/c assuming that books are closed at the end of the financial year.
Answer:

Cost of the Machine = Rs. 90,000 + Rs. 10,000 = Rs. 1,00,000
a) Depreciation on 31st March 2017 = 1,00,000 x 20/100 = Rs. 20,000
Written down value on 31st March 2017 = Rs. 1,00,000 – Rs. 20,000 = Rs. 80,000

b) Depreciation on 31 * March, 2018 = 80,000 x 20/100 = Rs. 16,000
Written down value on 31st March 2018 = Rs. 80,000 – Rs. 16,000 = Rs. 64,000

c) Depreciation on 31st March, 2019 = 64,000 x 20/100 = Rs. 12,800
Written down value on 31st March 2019 = Rs. 64,000 – Rs. 12,800 = Rs. 51,200
d) Depreciation on 31st March, 2020 = 51,200 x 20/100 = Rs. 10,240

Calculation of Profit on Sold Machine:

Question 13.
Ashrith a business man purchased a machine for Rs. 40,000 on 1st, July, 2016. Additional machinery was purchased on 1st April 2017 for Rs. 20,000. Depreciation is provided at 10% under diminishing balance method. On 31st December 2019 the machine bought on 1st July 2016 was sold for Rs. 17,000. Prepare machine a/c assuming that books are closed on 31st March every year.
Answer:


Working Notes:
a) Depreciation on 31st March, 2017
Cost = Rs. 40,000,
Rate = 10%, Machine was bought on July 1st, 2016
Depreciation period from July 1st 2016 to March 31st 2017 is 9 months

b) Depreciation on 31st March, 2018
Written down value of the 1st Machine = Rs. 40,000 – Rs. 3,000 = Rs. 37,000
Depreciation on 1st Machine = 37,000 x 10/100 = Rs. 3,700
Depreciation on 2 d Machine = 20,000 x 10/100 = Rs. 2,000

c) Depreciation on 31st March, 2019
Written down value of the both machines = Rs. 57,000 – Rs. 5,700 = Rs. 51,300
Depreciation = 51,300 x 10/100 = Rs. 5,130

d) Calculation of loss on sale of Machine:

e) Depreciation on the second machine:

Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type

Question 1.
Evaluate \(\int \frac{d x}{4+5 \sin x}\). [Mar. ’05]
Solution:

Question 2.
Evaluate \(\int \frac{d x}{5+4 \cos x}\). [Mar. ’12, ’11, ’10]
Solution:
Write tan \(\frac{x}{2}\) = t
Differentiating on both sides with respect to ‘x’.

Question 3.
Evaluate \(\int \frac{1}{1+\sin x+\cos x} d x\). [(AP) Mar. ’20; (TS) ’15]
Solution:

Question 4.
Evaluate \(\int \frac{d x}{4 \cos x+3 \sin x}\). [(TS) Mar. ’18]
Solution:

Question 5.
Find \(\int \frac{d x}{3 \cos x+4 \sin x+6}\). [May ’15 (AP); Mar. ’13]
Solution:
Let tan \(\frac{x}{2}\) = t
Differentiating on both sides with respect to ‘x’

Question 6.
Evaluate \(\int \frac{1}{2-3 \cos 2 x} d x\). [May ’10, ’05]
Solution:

Question 7.
Evaluate \(\int \frac{d x}{5+4 \cos 2 x}\). [May ’13, Mar. ’11]
Solution:

Question 8.
Evaluate \(\int \frac{\cos x+3 \sin x+7}{\cos x+\sin x+1} d x\). [(AP) Mar. ’19; May ’06]
Solution:
Let cos x + 3 sin x + 7 = \(\frac{d}{dx}\) (cos x + sin x + 1) + µ (cos x + sin x + 1) + γ
= λ(-sin x + cos x) + µ (cos x + sin x + 1) + γ …….(1)
= -λ sin x + λ cos x + µ cos x + µ sin x + µ + γ
Comparing the coefficients of cos x on both sides, we get
λ + µ = 1 ……..(2)
Comparing the coefficients of sinx on both sides, we get
-λ + µ = 3 ………(3)
Solving (2) and (3)

Comparing constant terms on both sides, we get
µ + γ = 7
⇒ 2 + γ = 7
⇒ γ = 5



where k is an integration constant.

Question 9.
Evaluate \(\int \frac{2 \sin x+3 \cos x+4}{3 \sin x+4 \cos x+5} d x\). [Mar. ’16 (AP & TS); Mar. ’14, ’11]
Solution:
2 sin x + 3 cos x + 4 = λ(3 sin x + 4 cos x + 5) + µ(3 sin x + 4 cos x + 5) + γ
= λ(3 cos x – 4 sin x) + µ(3 sin x + 4 cos x + 5) + γ …….(1)
Comparing,
Coeff. of cos x, 3 = 3λ + 4µ
⇒ 3λ + 4µ – 3 = 0 ……….(2)
Coeff. of sin x, 2 = -4λ + 3µ
⇒ 4λ – 3µ + 2 = 0 ……..(3)
Solving (2) and (3),

Question 10.
Evaluate \(\int \frac{9 \cos x-\sin x}{4 \sin x+5 \cos x} d x\). [Mar. ’17 (TS), Mar. ’08]
Solution:
9 cos x – sin x = λ(4 sin x + 5 cos x) + µ(4 sin x + 5 cos x) + γ
= λ(4 cos x – 5 sin x) + µ(4 sin x + 5 cos x) + γ ………(1)
Comparing,
Coeff. of cos x, 9 = 4λ + 5µ
⇒ 4λ + 5µ – 9 = 0 ……….(2)
Coeff. of sin x, -1 = -5λ + 4µ
⇒ -5λ + 4µ + 1 = 0 ………(3)
Coeff. of constant, 0 = γ
Solving (2) & (3),

Question 11.
Evaluate \(\int \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x} d x\). [(TS) May ’19, ’16; (AP) Mar. ’18, ’15]
Solution:
Let 2 cos x + 3 sin x = λ \(\frac{d}{dx}\) (4 cos x + 5 sin x) + µ(4 cos x + 5 sin x) + γ
= λ(-4 sin x + 5 cos x) + µ(4 cos x + 5 sin x) + γ ………(1)
= -4λ sin x + 5λ cos x + 4µ cos x + 5µ sin x + γ
Comparing the coefficients of cos x on both sides, we get
5λ + 4µ = 2 ……..(2)
Comparing the coefficients of sinx on both sides, we get
-4λ + 5µ = 3 ……….(3)
Solving (2) and (3)

Comparing constant terms on both sides, we get X = 0
From (1),
2 cos x + 3 sin x = \(\frac{-2}{41}\) (-4 sin x + 5 cos x) + \(\frac{23}{41}\) (4 cos x + 5 sin x) + 0
= \(\frac{-2}{41}\) (-4 sin x + 5 cos x) + \(\frac{23}{41}\) (4 cos x + 5 sin x)
Now,

Put 4 cos x + 5 sin x = t
(-4 sin x + 5 cos x) dx = dt

Question 12.
Evaluate \(\int \frac{2 x+5}{\sqrt{x^2-2 x+10}} d x\). [(AP) May ’17; Mar. ’15 (TS)]
Solution:
Write 2x + 5 = A . \(\frac{d}{dx}\) (x2 – 2x + 10) + B
= A(2x – 2) + B ………(1)
Coeff. of x, 2 = 2A then A = 1
Constant, 5 = -2A + B then B = 5 + 2 = 7
From (1), 2x + 5 = (2x – 2) + 7

Question 13.
Evaluate \(\int \frac{x+1}{x^2+3 x+12} d x\). [(AP) Mar. ’17; May ’16]
Solution:
Write x + 1 = A \(\frac{d}{dx}\) [x2 + 3x + 12) + B
= A(2x + 3) + B ……..(1)
Coefficient of x, 1 = A(2) then A = \(\frac{1}{2}\)
Constant, 1 = 3A + B then B = 1 – \(\frac{3}{2}\) = \(\frac{-1}{2}\)
From (1), x + 1 = \(\frac{1}{2}\)(2x + 3) – \(\frac{1}{2}\)

Question 14.
Evaluate \(\int \sqrt{\frac{5-x}{x-2}} d x\). [(AP) May ’19, (TS) ’17]
Solution:

Question 15.
Evaluate \(\int(6 x+5) \sqrt{6-2 x^2+x} d x\). [(AP) & (TS) May ’18]
Solution:

Question 16.
Evaluate \(\int x \sqrt{1+x-x^2} d x\). [May ’12]
Solution:
Take x = A \(\frac{d}{dx}\) (1 + x – x²) + B
x = A(1 – 2x) + B …….(1)
= A – 2Ax + B
Comparing the coefficient of x on both sides, we get
-2A = 1
⇒ A = \(-\frac{1}{2}\)
Comparing constant terms on both sides, we get
A + B = 0
⇒ \(-\frac{1}{2}\) + B = 0
⇒ B = \(\frac{1}{2}\)
From (1), x = \(-\frac{1}{2}\)(1 – 2x) + \(\frac{1}{2}\)

Question 17.
Evaluate \(\int(3 x-2) \sqrt{2 x^2-x+1} d x\). [(TS) May ’15; May ’03]
Solution:
Take 3x – 2 = A \(\frac{d}{dx}\)(2x² – x + 1) + B
3x – 2 = A(4x – 1) + B …….(1)
3x – 2 = 4Ax – A + B
Comparing the coefficients of x on both sides, we get
4A = 3
⇒ A = \(\frac{3}{4}\)
Comparing the constant terms on both sides
-A + B = -2
⇒ \(-\frac{3}{4}\) + B = -2
⇒ B = \(\frac{-5}{4}\)
From (1), 3x – 2 = \(\frac{3}{4}\) (4x – 1) – \(\frac{5}{4}\)

Question 18.
Evaluate \(\int \frac{d x}{(1+x) \sqrt{3+2 x-x^2}}\). [(TS) Mar. ’20; May ’14, ’05]
Solution:

Question 19.
Evaluate \(\int \frac{d x}{(x+1) \sqrt{2 x^2+3 x+1}}\). [Mar. ’18 (TS)]
Solution:

Question 20.
Evaluate the Reduction formula for I_n = ∫sinⁿx dx and hence find ∫sin⁴x dx, ∫sin⁵x dx. [(TS) Mar. ’20, May 18; (AP) May ’19, 15; Mar. ’17; Mar. ’14, ’13]
Solution:

Question 21.
Obtain the Reduction formula for ∫cosⁿx dx for n ≥ 2 and deduce the value of ∫cos⁵x dx. [(AP) Mar. ’20, May ’18; (TS) May ’19, ’16; Mar. ’17]
Solution:

Question 22.
Find the Reduction formula of ∫tanⁿx dx for an integer n ≥ 2. And deduce the value of ∫tan⁶x dx. [(AP) Mar. ’18, ’15; May ’16; (TS) ’17; Mar. ’12; May ’13]
Solution:

Question 23.
Find the Reduction formula of ∫cotⁿx dx for an integer n ≥ 2. And deduce the value of ∫cot⁴x dx. [Mar. ’19 (TS); Mar. ’16 (AP); May ’11]
Solution:

Question 24.
Find the Reduction formula for ∫cosecⁿx dx for an integer n ≥ 2 and deduce the value of ∫cosec⁵x dx. [Mar. ’19 (AP); Mar. ’16 (TS); May ’14]
Solution:

Question 25.
Find the Reduction formula for ∫secⁿx dx for an integer n ≥ 2 and deduce the value of ∫sec⁵x dx. [(AP) May ’17; (TS) May ’15; Mar. ’04]
Solution:
\(I_n=\int \sec ^n x d x=\int \sec ^{n-2} x \cdot \sec ^2 x d x\)

Question 26.
Evaluate \(\int \frac{\sin 2 x}{a \cos ^2 x+b \sin ^2 x} d x\)
Solution:
Put a cos²x + b sin²x = t
then [a . 2 cos x (-sin x) + b . 2 sin x . cos x] dx = dt
⇒ [-a sin 2x + b sin 2x] dx = dt
⇒ (b – a) sin 2x dx = dt
⇒ sin 2x dx = \(\frac{1}{b-a}\) dt

Question 27.
Evaluate ∫x cos^-1x dx. [Mar. ’09]
Solution:

Question 28.
Evaluate ∫x sin^-1x dx. [Mar. ’04]
Solution:

Question 29.
Evaluate \(\int \frac{d x}{x^2+x+1}\)
Solution:

Question 30.
Evaluate \(\int \frac{d x}{\sqrt{1+x-x^2}}\)
Solution:

Question 31.
Evaluate \(\int \frac{\cos x}{\sin ^2 x+4 \sin x+5} d x\). [Mar. ’07, ’03]
Solution:
Put sin x = t then cos dx = dt

Question 32.
Evaluate \(\int \frac{\sin x \cos x}{\cos ^2 x+3 \cos x+2} d x\)
Solution:
Put cos x = t
⇒ -sin x dx = dt
⇒ sin x dx = -dt

Question 33.
Evaluate \(\int \frac{d x}{\left(x^2+a^2\right)\left(x^2+b^2\right)}\)
Solution:

Question 34.
Evaluate \(\int \frac{d x}{\left(x^2+a^2\right)^2}\)
Solution:

Question 35.
Evaluate ∫e^ax sin(bx + c) dx. [Mar. ’19 (TS)]
Solution:

Question 36.
Evaluate \(\int \sqrt{\mathbf{a}^2-x^2} d x\). [Mar. ’02]
Solution:

Question 37.
Evaluate \(\int \sqrt{1+3 x-x^2} d x\). [Mar. ’11]
Solution:

Question 38.
Evaluate \(\int \frac{d x}{\sin x+\sqrt{3} \cos x}\)
Solution:

Question 39.
Evaluate \(\int \frac{\tan ^{-1} x}{x^2} d x\). [May ’01]
Solution:
\(\int \frac{\tan ^{-1} x}{x^2} d x=\int \tan ^{-1} x \cdot x^{-2} d x\)

Question 40.
Evaluate \(\int \frac{1}{a \sin x+b \cos x} d x\). [May ’03]
Solution:
Let a = r cos θ, b = r sin θ then r = \(\sqrt{a^2+b^2}\)
Now a sin x + b cos x = r cos θ sin x + r sin θ cos x = r[sin (x + θ)]

Question 41.
Evaluate ∫e^x log(e^2x + 5e^x + 6) dx.
Solution:

Question 42.
Evaluate \(\int \frac{1}{(1+\sqrt{x}) \sqrt{x-x^2}} d x\)
Solution:
Put x = t², then dx = 2t dt

Question 43.
Evaluate \(\int \frac{1}{(x-a)(x-b)(x-c)} d x\)
Solution:

Question 44.
Evaluate \(\int \frac{d x}{x(x+1)(x+2)}\)
Solution:

Question 45.
Evaluate \(\int \frac{7 x-4}{(x-1)^2(x+2)} d x\)
Solution:
Let \(\frac{7 x-4}{(x-1)^2(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}\)
⇒ \(\frac{7 x-4}{(x-1)^2(x+2)}=\frac{A(x-1)(x+2)+B(x+2)+C(x-1)^2}{(x-1)^2(x+2)}\)
⇒ 7x – 4 = A(x – 1) (x + 2) + B(x + 2) + C(x – 1)²
If x = 1,
1 = 7(1) – 4 = B(1 + 2)
⇒ 3 = 3B
⇒ B = 1
If x = -2,
7(-2) – 4 = C(-2 – 1)²
⇒ -14 – 4 = C(-3)²
⇒ -18 = 9C
⇒ C = -2
(1) ⇒ 7x – 4 = Ax² + Ax – 2A + Bx + 2B + Cx² – 2Cx + C
Comparing the coefficients of x² on both sides, we get
A + C = 0
⇒ A – 2 = 0
⇒ A = 2

Question 46.
Evaluate \(\int \frac{x^2}{(x+1)(x+2)^2} d x\)
Solution:
Let \(\frac{x^2}{(x+1)(x+2)^2}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}\)
⇒ \(\frac{x^2}{(x+1)(x+2)^2}=\frac{A(x+2)^2+B(x+1)(x+2)+C(x+1)}{(x+1)(x+2)^2}\)
⇒ x² = A(x + 2)² + B(x + 1)(x + 2) + C(x + 1) …..(1)
If x = -1, then (-1)² = A(-1 + 2)²
⇒ 1 = A(1)²
⇒ A = 1
If x = -2, then (-2) = C(-2 + 1)
⇒ 4 = C(-1)
⇒ 4 = -C
⇒ C = -4
(1) ⇒ x² = Ax² + 4Ax + 4A + Bx² + 3Bx + 2B + Cx + C
Comparing the coefficients of x² on both sides, we get
1 = A + B
⇒ 1 + B = 1
⇒ B = 0

Question 47.
Evaluate \(\int \frac{2 x+3}{x^3+x^2-2 x} d x\)
Solution:
Given \(\int \frac{2 x+3}{x^3+x^2-2 x} d x=\int \frac{2 x+3}{x\left(x^2+x-2\right)} d x=\int \frac{2 x+3}{x(x+2)(x-1)} d x\)
Let \(\frac{2 x+3}{x(x+2)(x-1)}=\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x-1}\)
⇒ \(\frac{2 x+3}{x(x+2)(x-1)}=\frac{A(x+2)(x-1)+B x(x-1)+C x(x+2)}{x(x+2)(x-1)}\)
⇒ 2x + 3 = A(x + 2)(x – 1) + B(x – 1) x + Cx(x + 2)
If x = 0, then 2(0) + 3 = A(0 + 2) (0 – 1)
⇒ 3 = A(2)(-1)
⇒ 3 = -2A
⇒ A = \(\frac{-3}{2}\)
If x = -2, then 2(-2) + 3 = B(-2 – 1) (-2)
⇒ -4 + 3 = B(-2)(-3)
⇒ -1 = 6B
⇒ B = \(\frac{-1}{6}\)
If x = 1, then 2(1) + 3 = C(1)(1 + 2)
⇒ 5 = C(1)(3)
⇒ 3C = 5
⇒ C = \(\frac{5}{3}\)

Question 48.
Evaluate \(\int \frac{x+3}{(x-1)\left(x^2+1\right)} d x\). [May ’07]
Solution:
Let \(\frac{x+3}{(x-1)\left(x^2+1\right)}=\frac{A}{x-1}+\frac{B x+C}{x^2+1}\)
\(\frac{x+3}{(x-1)\left(x^2+1\right)}=\frac{A\left(x^2+1\right)+(B x+C)(x-1)}{(x-1)\left(x^2+1\right)}\)
x + 3 = A(x² + 1) + (Bx + C) (x – 1) ………(1)
If x = 1 then 1 + 3 = A(1² + 1)
⇒ 4 = A(2)
⇒ A = 2
from (1), x + 3 = Ax² + A + Bx² – Bx + Cx – C
Comparing x² coefficients on both sides, we get
A + B = 0
⇒ 2 + B = 0
⇒ B = -2
Comparing coefficients of x on both sides, we get
-B + C = 1
⇒ -(-2) + C = 1
⇒ 2 + C = 1
⇒ C = 1 – 2
⇒ C = -1

Question 49.
Evaluate \(\int \frac{2 x+3}{(x+3)\left(x^2+4\right)} d x\). [May ’02]
Solution:
Let \(\frac{2 x+3}{(x+3)\left(x^3+4\right)}=\frac{A}{x+3}+\frac{B x+C}{x^2+4}\)
\(\frac{2 x+3}{(x+2)\left(x^2+4\right)}=\frac{A\left(x^2+4\right)+(B x+C)(x+3)}{(x+3)\left(x^2+4\right)}\)
2x + 3 = A(x² + 4) + (Bx + C)(x + 3) …….(1)
If x = -3 then
2(-3) + 3 = A[(-3)² + 4]
⇒ -6 + 3 = A(9 + 4)
⇒ 13A = -3
⇒ A = \(\frac{-3}{13}\)
From (1),
2x + 3 = Ax² + 4A + Bx² + 3Bx + Cx + 3C
Comparing x² coefficients on both sides, we get
A + B = 0
⇒ \(\frac{-3}{13}\) + B = 0
⇒ B = \(\frac{3}{13}\)
Comparing x coefficients on both sides, we get
3B + C = 2
⇒ 3(\(\frac{3}{13}\)) + C = 2
⇒ C = 2 – \(\frac{3}{13}\)
⇒ C = \(\frac{17}{13}\)

Question 50.
Evaluate \(\int \frac{1}{(1-x)\left(4+x^2\right)} d x\)
Solution:
\(\frac{1}{(1-x)\left(4+x^2\right)}=\frac{A}{1-x}+\frac{B x+C}{x^2+4}\)
1 = A(x² + 4) + (Bx + C)(1 – x)
Put x = 1 then A = \(\frac{1}{5}\)
Coeff. of x², 0 = A – B ⇒ B = \(\frac{1}{5}\)
Constant, 1 = C + 4A ⇒ C = \(\frac{1}{5}\)

Question 51.
Evaluate \(\int \frac{d x}{x^3+1}\). [May ’03]
Solution:
\(\frac{1}{x^3+1}=\frac{1}{(x+1)\left(x^2-x+1\right)}=\frac{A}{x+1}+\frac{B x+C}{x^2-x+1}\) ……(1)
1 = A(x² – x + 1) + (Bx + C) (x + 1)
Put x = -1, 1 = A(3) ⇒ A = \(\frac{1}{3}\)
Coeff. of x², 0 = A + B ⇒ B = \(\frac{-1}{3}\)
Constant, 1 = A + C ⇒ C = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)

Question 52.
Evaluate \(\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x\)
Solution:

Question 53.
Find the Reduction formula for ∫sin^mx cosⁿx dx for a +ve integer and n ≥ 2.
Solution:

Question 54.
If I_n = ∫(log x)ⁿ dx, then show that I_n = x(log x)ⁿ – nI_n – 1 and find ∫(log x)⁴ dx.
Solution:

Now I₄ = ∫(log x)⁴ dx
= x(log x)⁴ – 4I₃
= x(log x)⁴ – 4[x(log x)³ – 4I₂]
= x(log x)⁴ – 4x(log x)³ + 16I₂
= x(log x)⁴ – 4x(log x)³ + 16[x(log x)² – 2I₁]
= x(log x)⁴ – 4x(log x)³ + 16x(log x)² – 32I₁
= x(log x)⁴ – 4x(log x)³ + 16x(log x)² – 32[x(log x) – x] + c
= x(log x)⁴ – 4x(log x)³ + 16x(log x)² – 32x log x + 32x + c