Srinivas

Students must practice these Maths 1A Important Questions TS Inter 1st Year Maths 1A Functions Important Questions Long Answer Type to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Functions Important Questions Long Answer Type

Question 1.
If f : A → B, g : B → C are two bijective functions, then prove that gof : A → C is also a bijective function. [Mar. 18, 16 (AP), 09 ; May 13, 12, 10, 08, 06, 04 00, 96, 92]
Answer:
Since f: A →B is a bijective function
o f: A → B is both one-one and onto functions.
Since f: A → B is a one-one function
⇔ a₁, a₂ ∈ A, f(a₁) = f(a₂) ⇒ a₁ = a₂
Since f: A → B is a onto function ⇔ ∃ one element a ∈ A such that f(a) = b, ∀ b ∈ B.
Since g: B → C is a bijective function
⇔ g: B → C is both one-one and onto functions.
Since g: B → C is a one-one function
⇔ b₁, b₂ ∈ B, g (b₁) = g (b₂) ⇒ b₁ = b₂
Since g: B → C is an onto function ⇔ ∃ one element b e B such that g(b) = c, ∀ c ∈ C.
If f: A → B, g : B → C ⇒ gof: A → C.

To prove that gof: A → C is a one-one function:
If gof: A → C is a one-one function
⇔ a₁, a₂ ∈ A, (gof) (a₁) = (gof) (a₂) ⇒ a₁ = a₂
Now (gof) (a₁) = (gof) (a₂)
g [f(a₁)] = g [ f(a₂)] [∵ g is one-one]
f(a₁) = f(a₂) [∵ f is one-one]
a₁ = a₂
Hence, gof: A → C is a one-one function.

To prove that gof: A → C is an onto function:
Let c ∈ C
If gof: A → C is an onto function ⇔ ∃ one element a ∈ A, such that
(gof) (a) = c, ∀ c ∈ C.
Now (gof) (a) = g [f(a)] = g(b) = c
Thus for any element c ∈ C, there is an element a ∈ A such that (gof) (a) = c.
∴ gof: A → C is an onto function.
Since gof: A → C is both one-one function and onto function then
gof: A → C is a bijective function.

Question 2.
If f : A → B, g : B → C are two bijective functions, then prove that (gof)^-1 = f^-1og^-1. [Mar. ’16 (TS), 14, 11, 10, 06, 04, 02, 00, 92; May 15 (AP) 14, 11, 09, 02, 98, 94 Mar. 19 (AP) ]
Answer:
Since f: A → B, g : B → C are bijections
⇒ gof: A → C is a bijection
⇒ (gof)^-1: C → A is also a bijection
Since f: A → B is a bijective function
then f^-1: B → A is also a bijective function
Since g: B → C is a bijective function
then g^-1: C → B is also a bijective function
⇒ f^-1og^-1: C → A is also a bijection
Since the two functions (gof)^-1, f^-1og^-1 are from C → A their domains are same.
Let c ∈ C
Since f : A → B is onto ⇔ ∃ one element
a ∈ A such that
f(a) = b, ∀ b ∈ B
f(a) = b ⇒ f^-1 (b) = a
Since g : B → C is onto ⇔ ∃ one element be B such that g(b) = c, ∀ c ∈ C
g(b) = c ⇒ g^-1(c) = b
Now (gof) (a) = g[f(a)] = g (b) = c ⇒ a = (gof)^-1(c)
⇒ (gof)^-1(c) = a ………………. (1)
Also (f^-1og^-1) (c) = f^-1 [g^-1(c)] = f^-1(b) = a ……………… (2)
∴ From (1) and (2)
(gof)^-1(c) = (f^-1og^-1) (c)
∴ (gof)^-1 = f^-1og^-1.

Question 3.
Let f : A → B, is a function and I_A, I_B are identity functions on A and B respectively. Then prove that foI_A = f = I_Bof. [Mar. 18 (TS); Mar. 13, 08, 05; May 92]
Answer:
If f: A → B is a function.
If I_A and I_B are identity functions on A and B respectively.
i.e., I_A : A → B, I_B : B → B

(i) I_A: A → A, f: A → B ⇒ f o I_A : A → B
Hence, functions f o I_A and f are defined on same domain A.
Let a ∈ A
(f o I_A) (a) = f[I_A (a)] = f(a)
∴ f o I_A = f

(ii) f: A → B, I_B: B → B ⇒ I_B o f: A → B
The functions (I_B o f) and f are defined on the same domain A.
Let a ∈ A
Now (I_B o f)(a) = I_B[f(a)] = f(a)
∴ I_B o f = f ………………. (2)
From (1) and (2) we get
f o I_A = I_B o f = f

Question 4.
If f: A → B is a bijection, then prove that fof^-1 = I_B and f^-1 o f = I_A. [Mar. 17, 15 (AP); Mar. 12, 07, 03, 02; May 07, 05, 01 Mar. 19 (TS)]
Answer:

(i) Since f: A → B is a bijection ⇒ f^-1: B → A is also a bijection
I_A; f: A → B, f^-1: B → A ⇒ f^-1 o f: A → A is also bijection
Clearly I_A: A → A such that I_A(a) = a, ∀ a ∈ A
Let a ∈ A
Since f^-1: B → A is onto function ⇔ ∃ one element b ∈ B,
such that
f^-1(b) = a, ∀ a ∈ A
f^-1 (b) = a ⇒ f(a) = b
Now (f^-1of) (a) = f^-1[f(a)] = f^-1(b) = a = I_A(a)
∴ f^-1of = I_A

(ii) Since f: A → B is a bijection ⇒ f^-1: B → A is also bijection
I_B: f^-1:B → A, f: A → B = fof^-1:B → B is also a bijection
Clearly I_B: B → B such that I_B(b) = b, ∀ b ∈ B
Let b ∈ B
Since f^-1: B → A is an onto function ⇔ ∃
one element b ∈ B such that f^-1(b) = a, ∀ a ∈ A
f^-1(b) = a ⇒ f(a) = b
Now (fof^-1) (b) = f[f^-1(b)] = f(a) = b = I_B(b)
∴ fof^-1 = I_B

Question 5.
If f:A → B, g:B → A are two functions such that gof = I_A and fog = I_B, then prove that f is a bijection and g = f^-1. [May 15 (TS); Mar. 08, 01; May 03]
Answer:

(i) To prove that f is one-one
Let a₁, a₂ ∈ A and since f : A → B, f(a₁), f(a₂) ∈ B
Now f(a₁) = f(a₂ ) ⇒ g[f(a₁)] = g[f(a₂)]
⇒ (gof) (a₁) = (gof) (a₂)
⇒ I_A(a₁) = I_A(a₂)
∴ a₁ = a₂
∴ f is one-one

(ii) To prove that f is onto
Let b be an element of B
I_B (b) = (fog) (b)
⇒ b = f[g(b)] ⇒ f(g(b)) = b
i.e., there exists a pre-image g(b) ∈ A for b, under the mapping f.
∴ f is onto
Thus ‘f’ is one-one and onto hence, f^-1: B → A exists and is also one-one onto.

(iii) To prove g = f^-1
Now g:B → A and f^-1:B → A
Let a ∈ A and b be the f – image of a where b ∈ B
∴ f(a) = b ⇒ a = f^-1 (b)
Now g(b) = g[f(a)] (gof) (a) = I_AA(a) = a
⇒ a = f^-1 (b)
∴ g = f^-1

Question 6.
If f:A → B, g: B → C and h: C → D are three functions then prove that ho(gof) = (hog) of. That is composition of functions is associative. [May ‘99, ‘95]
Answer:
f:A → B, g:B → C, h:C → D be three functions.
f:A → B, and g:B → C = gof: A → C
Now gof: A → C and h:C → D ⇒ ho(gof): A → D
g:B → C and h:C → D = (hog):B → D
Now f: A → B ⇒ hog: B→D
(hog)of: A → D
Thus ho(gof) and (hog)of both exist and have the same domain A and co-domain D.
Let a ∈ A,
Hence ho(gof) = (hog) of ∈ A
Now [ho(gof)] (a) = h [(gof) (a)] = h[g(f(a))]
= (hog) [f(a)] = [(hog) of] (a)
∴ [ho(gof)] (a) = [(hog) of] (a)

Question 7.
If f: A → B, g: B → C be surjections, then show that gof: A → C is a surjection. [May 98, 97, 96, 94, 93, 91]
Answer:
Let c ∈ C
Since f: A → B is a onto ⇔ ∃ one element
a ∈ A such that f(a) = b,∀ b ∈ B
Since g: B → C is a onto ⇔ ∃ one element
b ∈ B such that g(b) = c, ∀ c ∈ C.
If f: A → B, g:B → C = gof: A → C
To prove that gof : A → C is a onto

If gof : A → C is a onto ⇔ ∃ one element a ∈ A
such that
(gof) (a) = c, ∀ c ∈ C.
Now (gof) (a) = g[f(a)] = g(b) = c
Thus for any element c € C, there is an
element a ∈ A such that (gof) (a) = c.
∴ gof: A → C is an onto function.

Question 8.
If f = ((1, a), (2, c), (4, d), (3, b)} and g^-1 = {(2, a), (4, b), (1, c), (3, d)}, then show that (gof)^-1 = f^-1og^-1. {Mar. 15 (TS); May 07, 93}
Answer:
Given
f = {(1, a), (2, c), (4, d), (3, b))
f^-1 = ((a, 1), (c, 2), (d, 4), (b, 3))
g = ((a, 2), (b, 4), (c, 1), (d, 3))
g^-1 = {(2, a), (4, b), (1, c), (3, d)}

gof:

∴ gof = {(1, 2), (2, 1), (3, 4), (4, 3)}
(gof)^-1 = {(2, 1), (1, 2), (4, 3), (3, 4)}
f^-1og^-1

f^-1og^-1 = {(1, 2), (2, 1), (3, 4), (4, 3)}
∴ (gof)^-1 = f^-1og^-1

Question 9.
If the function f is defined by

then find the values of
(i) f(3)
(ii) f(0)
(iii) f(-1.5)
(iv) f(2) + f(- 2)
(v) f(- 5).
Answer:
Given

(i) For x > 1; f(x) = x + 2; f(3) = 3 + 2 = 5
(ii) For – 1 ≤ x ≤ 1; f(x) = 2, f(0) = 2
(iii) For – 3 < x < – 1; f(x) = x – 1 ∴ f(- 1.5) = – 1.5 – 1 = – 2.5 (iv) For x > 1, f(x) = x + 2
f(2) = 2 + 2 = 4
For – 3 < x < – 1, f(x) = x – 1
∴ f(- 2) = – 2 – 1 = – 3
f(2) + f(- 2) = 4 – 3 = 1
(v) f(- 5) is not defined.

Question 10.
If A = {- 2, – 1, 0, 1, 2) and f: A → B is a surjection defined by f(x) = x² + x + 1 find B.
Answer:
Given, A = {- 2, – 1, 0, 1, 2)
f(x) = x² + x + 1
Since f : A → B is a surjection then f(A) = B
f(-2) = (- 2)² – 2 + 1 = 4 – 2 + 1 = 3
f(-1) = (- 1)² – 1 + 1 = 1 – 1 + 1 = 1
f(0) = 0² + 0 + 1 = 1
f(1) = 1² + 1 + 1 = 1 + 1 + 1 = 3
f(2) = 2² + 2 + 1 = 4 + 2 + 1 = 7
∴ B = f(A) = {3. 1, 7}

Question 11.
If A = {1, 2, 3, 4} and f:A → R is a function defined by f(x) = \(\frac{x^2-x+1}{x+1}\), then find the range of f.
Answer:
Given A = {1, 2, 3, 4) and f(x) = \(\frac{x^2-x+1}{x+1}\)
Since f: A → R is a function, then

Question 12.
If f: Q → Q, is defined by f(x) = 5x + 4 for all x ∈ Q, find f^-1. [Mar. 17 (TS)]
Answer:
Let y = f(x) = 5x + 4
y = f(x) ⇒ x = f^-1(y) ……………… (1)
y = 5x + 4 ⇒ y – 4 = 5x
x = \(\frac{y-4}{5}\) ………………… (2)
From (1) and (2),
f^-1(y) = \(\frac{y-4}{5}\) ⇒ f(x) = \(\frac{x-4}{5}\), ∀ x ∈ Q

Question 13.
If f(x) = \(\frac{x+1}{x-1}\) (x ≠ ± 1), then find (fofofof) (x).
Answer:
Given f(x) = \(\frac{x+1}{x-1}\) (x ≠ ± 1)
Now (fofofof) (x) = f[f[f{f(x)}]]

Question 14.
Find the domain of the real valued function f(x) = \(\sqrt{16-x^2}\).
Answer:
Given f(x) = \(\sqrt{16-x^2}\) ∈ R
⇒ 16 – x² ≥ 0
⇒ x² – 16 ≤ 0
⇒ (x + 4) (x – 4) ≤ 4
⇒ x ∈ [- 4, 4]
∴ Domain of ‘f’ is [- 4, 4]

Question 15.
Find the domain of the real valued function f(x) = \(\sqrt{9-x^2}\).
Answer:
Given f(x) = \(\sqrt{9-x^2}\) ∈ R
⇒ 9 – x² ≥ 0
⇒ x² – 9 ≤ 0
⇒ (x + 3) (x – 3) ≤ 0
⇒ x ∈ [- 3 ,3]
∴ Domain of f’ is [- 3, 3]

Question 16.
Find the domain of the real valued function f(x) = \(\frac{1}{6 x-x^2-5}\).
Answer:
Given f(x) = \(\frac{1}{6 x-x^2-5}\) ∈ R
⇒ 6x – x² – 5 ≠ 0
⇒ x² – 6x + 5 ≠ 0
⇒ x² – 5x – x + 5 ≠ 0
⇒ x(x – 5) – 1 (x – 5) ≠ 0
⇒ x – 1 ≠ 0 or x – 5 ≠ 0
⇒ x ≠ 1 or x ≠ 5
∴ x ≠ 1, 5
∴ Domain of ‘f’ is R – {1, 5}

Question 17.
Find the domain of the real valued function f(x) = \(\frac{2 x^2-5 x+7}{(x-1)(x-2)(x-3)}\).
Answer:
Given f(x) = \(\frac{2 x^2-5 x+7}{(x-1)(x-2)(x-3)}\) ∈ R
⇒ (x – 1) (x – 2) (x – 3) ≠ 0
⇒ x – 1 ≠ 0, x – 2 ≠ 0, x – 3 ≠ 0
⇒ x ≠ 1, x ≠ 2, x ≠ 3
∴ x ≠ 1, 2, 3
∴ Domain of ‘f’ is R – {1, 2, 3}

Question 18.
Find the domain of the real valued function f(x) = \(\frac{\sqrt{2+x}+\sqrt{2-x}}{x}\).
Answer:
Given f(x) = \(\frac{\sqrt{2+x}+\sqrt{2-x}}{x}\) ∈ R

⇒ 2 + x ≥ 0, 2 – x ≥ 0 and x ≠ 0
x ≥ – 2, 2 ≥ x and x ≠ 0
x ≤ 2 and x ≠ 0
⇒ x ∈ [- 2, 0) ∪ (0, 2]
∴ Domain of ‘f’ is [- 2, 0) ∪ (0, 2]

Question 19.
If f = {(1, 2), (2, – 3), (3, – 1)}, then find
(i) 2f
(ii) 2 + f
(iii) f²
(iv) √f
Answer:
Given f = {(1, 2), (2, – 3), (3, – 2)}
Domain of ‘f’ is A = {1, 2, 3}
f(1) = 2f(2) = – 3, f(3) = – 1

(i) (2f) (x) = 2f(x)
(2f) (1) = 2f(1) = 2(2) = 4
(2f) (2) = 2f(2) = 2(- 3) = – 6
(2f) (3) = 2f(3) = 2(- 1) = – 2
∴ 2f = {(1, 4), (2, – 6),(3, – 2)}

(ii) (2 + f) (x) = 2 + f(x)
(2 + f) (1) = 2 + f(1) = 2 + 2 = 4
(2 + f) (2) = 2 + f(2) = 2 – 3 = – 1
(2 + f) (3) = 2 + f(3) = 2 – 1 = 1
∴ 2 + f = {(1, 4), (2, – 1), (3, 1)}

(iii) (f²) (x) = [f(x)]²
(f²) (1) = [f(1)]² = 2² = 4
(f²) (2) = [f(2)]² = (- 3)² = 9
(f²) (3) = (f(3)]² = (- 1)² = 1
∴ f² = {(1, 4), (2, 9), (3, 1)}

(iv) (√f)(x) = √f(x)
(√f) (1) = √f(1) = √2
(√f) (2) = √f(2) = √- 3 (not valid)
(√f) (3) = √f(3) = √- 1 (not valid)
∴ √f = {(1, √2)}

Some More Maths 1A Functions Important Questions

Question 1.
If f(x) = \(\frac{\cos ^2 x+\sin ^4 x}{\sin ^2 x+\cos ^4 x}\), ∀ x ∈ R then show that f(2012) = 1.
Answer:

Question 2.
If f: R → R is defined by f(x) = \(\frac{1-x^2}{1+x^2}\), then show that f(tan θ) = cos 2θ.
Answer:
Given f: R → R, f(x) = \(\frac{1-x^2}{1+x^2}\)
LHS = f(tan θ)
= \(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\) = cos 2θ + RHS
∴ f(tan θ) = cos 2θ

Question 3.
If f: R – {±1} → R is defined by f(x) = log \(\left|\frac{1+x}{1-x}\right|\), then show that f(\(\left(\frac{2 x}{1+x^2}\right)\)) = 2f(x)
Answer:
Given f: R – {±1} → R

Question 4.
If A = {x/ – 1 ≤ x ≤ 1}, f(x) = x², g(x) = x³ which of the following are surjections?
(i) f : A → A
(ii) g: A → A
Answer:
(i) Given A = {x/ – 1 ≤ x ≤ 1}
∴ A = {- 1, 0, 1}
f(x) = x²
f(- 1) = (- 1)² = 1
f(0) = (0)² = 0
f(1) = (1)² = 1
∴ f = (- 1, 1), (0 , 0), (1, 1))

f: A → A

Range of f(A) = {0, 1) ≠ A (co-domain)
∴ f : A → A is not a surjection.

(ii) Given A = {x/ – 1 ≤ x ≤ 1}
∴ A = {- 1, 0, 1}
g(x) = x³
g(- 1) = (- 1)³ = – 1
g(0) = (0)³ = 0
g(1) = (1)³ = 1
∴ g = {(- 1, -1), (0, 0), (1, 1)}
g: A → A

Range of g(A) = {- 1, 0, 1} = A (co-domain)
∴ g is a surjection.

Question 5.
If f(x) = cos (log x) then show that
\(f\left(\frac{1}{x}\right) \cdot f\left(\frac{1}{y}\right)-\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f(x y)\right]\) = 0
Answer:
Given f(x) = cos (log x)
f\(\left(\frac{1}{x}\right)\) = cos\(\left(\log \frac{1}{x}\right)\) = cos (log x^-1)
= cos (- log x) = cos (log x)
Similarly f\(\left(\frac{1}{y}\right)\) = cos (log y)
f\(\left(\frac{x}{y}\right)\) = cos \(\left(\log \left(\frac{x}{y}\right)\right)\)
= cos (log x – log y)
f(xy) = cos (log xy)
= cos (log x + log y)
L.H.S: f\(\left(\frac{1}{x}\right) \cdot f\left(\frac{1}{y}\right)-\frac{1}{2}\left(f\left(\frac{x}{y}\right)+f(x y)\right)\)
= cos (log x) . cos (log y) – \(\frac{1}{2}\) [cos (log x – log y) + cos (log x + log y)]
= cos (log x) . cos (log y)

= cos (log x) . cos (log y) – cos (log x) . cos (log y) = 0
= R.H.S

Question 6.
Find the inverse function of f(x) = log₂x.
Answer:
Given f: (0, ∝) → R, f(x) = log₂x
Let y = f(x) = log₂x
y = f(x) = x = f^-1(y) ……………. (1)
y = log₂x ⇒ x = 2^y (2)
From (1) & (2)
f^-1(y) = 2y
⇒ f^-1(x) = 2x

Question 7.
If f(x) = 1 + x + x² +…….. for |x| < 1, then show that f^-1(x) = \(\frac{\mathbf{x}-1}{\mathbf{x}}\).
Answer:
Given that f(x) = 1 + x + x² + ………….
f(x) = \(\frac{1}{1-\mathrm{x}}\)

Question 8.
Find the domain of the real valued function f(x) = \(\frac{1}{\sqrt{\mathbf{x}^2-a^2}}\) (a >0). [Mar.15 (AP)]
Answer:
Given f(x) = \(\frac{1}{\sqrt{x^2-a^2}}\) ∈ R
⇒ x² – a²
⇒ (x + a) (x – a) > 0
⇒ x < – a or x > a
⇒ x ∈ (- ∝, – a) ∪ (a, ∝)
∴ Domain of f’ is (- ∝, – a) ∪ (a, ∝)

Question 9.
Find the domain of the real valued function f(x) = \(\sqrt{(\mathbf{x}-\alpha)(\beta-\mathbf{x})}\) (0 < α < β).
Answer:
Given f(x) = \(\sqrt{(\mathbf{x}-\alpha)(\beta-\mathbf{x})}\) ∈ R
⇒ (x – α) (x – β) ≥ 0
⇒ (x – α) (x – β) ≤ 0
⇒ α ≤ x ≤ β
⇒ x ∈ [α, β]
∴ Domain of ‘f’ is [α, β]

Question 10.
Find the domain of the real valued function f(x) = \(\sqrt{2-x}+\sqrt{1+x}\).
Answer:
Given f(x) = \(\sqrt{2-x}+\sqrt{1+x}\) ∈ R
⇒ 2 – x ≥ 0 and 1 + x ≥ 0
⇒ 2 ≥ x and x ≥ – 1
⇒ x ≤ 2 and x ≥ – 1
⇒ x ∈ [- 1, 2]
∴ Domain of ‘f’ is [-1, 2]

Question 11.
Find the domain of the real valued function f(x) = \(\sqrt{|\mathbf{x}|-\mathbf{x}}\)
Answer:
Given f(x) = \(\sqrt{|\mathbf{x}|-\mathbf{x}}\) ∈ R
⇒ |x| – x ≥ 0
⇒ |x| ≥ x
⇒ x ∈ R
∴ Domain of ‘f’ is ‘R.

Question 12.
Find the domain and range of the real valued function f(x) = \(\frac{2+x}{2-x}\)
Answer:
Given f(x) = \(\frac{2+x}{2-x}\) ∈ R
⇒ 2 – x ≠ 0 ⇒ x ≠ 2
Domain of T is R – { 2 }.
Let y = f(x) = \(\frac{2+x}{2-x}\)
y = \(\frac{2+x}{2-x}\)
2yx – xy = 2 + x
2y – 2 = x + xy
2y – 2 = x(1 + y)
x ∈ R – {2}, y + 1 ≠ 0
y ≠ – 1
∴ Range of ‘f’ is R – {- 1}.

Question 13.
Find the domain and range of the real valued function f(x) = \(\sqrt{9-x^2}\) [Mar. 15 (TS)]
Answer:
Given f(x) = \(\sqrt{9-x^2}\) ∈ R
⇒ 9 – x² ≥ 0
⇒ x² – 9 ≤ 0
⇒ (x + 3) (x -3) ≤ 0
⇒ x ∈ [- 3, 3]
∴ Domain of ‘f’ is [- 3, 3]
Let y = f(x) = \(\sqrt{9-x^2}\)
y = \(\sqrt{9-x^2}\)
y² = 9 – x²
x² = 9 – y²
x = \(\sqrt{9-y^2}\) ∈ R
⇒ 9 – y² ≥ 0
⇒ y² – 9 ≤ 0
⇒ (y + 3) (y – 3) ≤ 0
⇒ y ∈ [- 3, 3]
But f(x) attains only non-negative values.
∴ Range of f = [0, 3].

Question 14.
Determine whether the function f(x) = x\(\left(\frac{e^x-1}{e^x+1}\right)\) is even or odd.
Answer:

Since f(- x) = f(x) then f is an even function.

Question 15.
Determine whether the function f(x) = log(x + \(\sqrt{x^2+1}\)) is even or odd.
Answer:

Since f(- x) = – f(x) then f(x) is an odd function.

Question 16.
Find the domain of the real valued function f(x) = log [x – (x)].
Answer:
Given f(x) = log [x – (x)] ∈ R
⇒ x – (x)> 0
⇒ x > (x)
Then x is a non – integer.
∴ Domain of ‘f’ is R – Z.

Question 17.
Find the domain of the real valued function f(x) = \(\frac{1}{\log (2-x)}\).
Answer:
Given f(x) = \(\frac{1}{\log (2-x)}\) ∈ R
⇒ log (2 – x) ≠ 0 and 2 – x > 0
⇒ log (2 – x) ≠ log 1 and 2 > x
⇒ 2 – x ≠ 1 and x < 2
⇒ x ≠ 1
∴ Domain of ‘f’ is (- ∝, 1) ∪ (1, 2)

Question 18.
Find the domain of the real valued function f(x) = \(\sqrt{\mathbf{x}-[\mathbf{x}]}\).
Answer:
Given f(x) = \(\sqrt{\mathbf{x}-[\mathbf{x}]}\) ∈ R
⇒ [x] – x ≥ 0 ⇒ x ≥ [x] ⇒ x ∈ R
∴ Domain of ’f is Z.

Question 19.
Find the domain of the real valued function f(x) = \(\sqrt{[\mathbf{x}]-\mathbf{x}}\).
Answer:
Given f(x) = \(\sqrt{[\mathbf{x}]-\mathbf{x}}\) ∈ R
⇒ [x] – x ≥ 0 ⇒ [x] ≥ x ⇒ x ∈ Z
∴ Domain of ‘f’ is Z.

Question 20.
If f and g are real valued functions defined by f(x) = 2x – 1 and g(x) = x² then find
(i) (3f – 2g)(x)
(ii) (fg) (x)
(iii) \(\left(\frac{\sqrt{f}}{g}\right)\)(x)
(iv) (f + g + 2) (x)
Answer:
Given f(x) = 2x – 1 and g(x) = x²
Domain of f = domain of g R
Hence the domain of all the functions is R.
(i) (3f – 2g) (x) = 3f(x) – 2g(x)
= 3(2x – 1) – 2(x²)
= 6x – 3 – 2x²
= – 2x² + 6x – 3

(ii) (fg)(x) f(x) . g(x)
= (2x – 1)(x²) = 2x³ – x².

(iii) \(\left(\frac{\sqrt{f}}{g}\right)\) (x) = \(\frac{\sqrt{f(x)}}{g(x)}\) = \(\frac{\sqrt{2 x-1}}{x^2}\)

(iv) (f + g + 2) (x) = f(x) .g(x) + 2
= 2x – 1 + x² + 2
= x² + 2x + 1 = (x + 1)²

Question 21.
Find the domain of the real valued function f(x) = \(\sqrt{x^2-3 x+2}\).
Answer:
Given f(x)= \(\sqrt{x^2-3 x+2}\) ∈ R
⇒ x² – 3x + 2 ≥ 0
⇒ x² – 2x – x + 2 ≥ 0
⇒ x(x – 2) – 1(x – 2) ≥ 0
⇒ (x – 1) (x – 2) ≥ 0
⇒ x ≤ 1 or x ≥ 2
⇒ x ∈ (- ∝, 1] ∪ [2, ∝)
∴ Domain of ‘f’ is (- ∝, 1] ∪ [2, ∝)

Question 22.
f:R → R defined by f(x) = \(\frac{2 x+1}{3}\), then this function Is injection or not ? Justify. (Mar. 15 (TS)
Answer:
Given that f(x) = \(\frac{2 x+1}{3}\)
Let x₁, x₂ ∈ R.
Take f(x₁) = f(x₂) ⇒ \(\frac{2 x_1+1}{3}=\frac{2 x_2+1}{3}\)
⇒ 2x₁ + 1 = 2x₂ + 1 ⇒ 2x₁ = 2x₂ = x₁ = x₂
∴ f(x₁) = f(x₂) ⇒ x₁ = x₂
⇒ f is one – one.

Question 23.
If f = {(4, 5), (5, 6),(6, – 4)} and g = ((4, – 4), (6, 5), (8, 5)) then find f + g and fg. [Mar. ‘17(TS)]
Answer:
Given f = {(4, 5), (5, 6), (6, – 4)} and
g = {(4, – 4), (6, 5), (8, 5’)) then domain of f = {4, 5, 6) and Range of f = {4, 6, 8}
Domain of f + g = A ∩ B = {4, 6}
= (domain of f) ∩ (domain of g)
(i) f.g={(4, 5, – 4), (6, – 4 + 5)}
= {(4, 1), (6, 1)}

(ii) Domain of fg = (domain of f) ∩ (domain of g)
= A ∩ B = (4, 6)
= {(4, 5 × – 4).(6, – 4 × 5)}.
= {(4, – 20), (6, – 20)}

Question 24.
If f(x) = 2x – 1, g(x) = \(\frac{x+1}{2}\) for all x ∈ R, are two functions, then find,
(i) (gof) (x)
(ii) (fog) (x) [Mar. 19(TS)]
Answer:
Given f(x) = 2x – 1, g(x) = \(\frac{x+1}{2}\)

(i) (gof) (x) = g[ f(x) ]
= g[2x – 1] = \(\frac{2 x-1+1}{2}\) = \(\frac{2 x}{2}\) = x

(ii) (fog) (x) = f [g(x)]
= \(f\left(\frac{x+1}{2}\right)\) = 2\(\left(\frac{\mathrm{x}+1}{2}\right)\) – 1 = x + 1 – 1 = x

Telangana TSBIE TS Inter 1st Year Physics Study Material 7th Lesson Systems of Particles and Rotational Motion Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 7th Lesson Systems of Particles and Rotational Motion

Very Short Answer Type Questions

Question 1.
Is it necessary that a mass should be present at the centre of mass of any system? [AP May. ’16; May ’14]
Answer:
No. It is not necessary to present some mass at centre of mass of the system.
Ex: At the centre of ring (or) bangle, there is no mass present at centre of mass.

Question 2.
What is the difference in the positions of a girl carrying a bag in one of her hands and another girl carrying a bag in each of her two hands?
Answer:
i) a) When she carries a bag in one hand her centre of mass will shift to the side of the hand that carries the bag.
b) When a bag is in one hand some unbalanced force will act on her and it is difficult to carry.

ii) If she carries two bags in two hands then her centre of mass remains unchanged. Force on two hands are equal i.e. balanced so it is easy to carry the bags.

Question 3.
Two rigid bodies have same moment of inertia about their axes of symmetry. Of the two, which body will have greater kinetic energy?
Answer:
Relation between angular momentum and kinetic energy is, KE = \(\frac{L^2}{2I}\)

Because moment of inertia is same the body with large angular momentum will have larger kinetic energy.

Question 4.
Why are spokes provided in a bicycle wheel? [AP May ’14]
Answer:
The spokes of cycle wheel increase its moment of inertia. The greater the moment of inertia, the greater is the opposition to any change in uniform rotational motion. As a result the cycle runs smoother and speeder. If the cycle wheel had no spokes, the cycle would be driven in jerks and hence unsafe.

Question 5.
We cannot open or close the door by applying force at the hinges. Why? [AP May ’16]
Answer:
To open or close a door, we apply a force normal to the door. If the force is applied at the hinges the perpendicular distance of force is zero. Hence, there will be no turning effect however large force is applied.

Question 6.
Why do we prefer a spanner of longer arm as compared to the spanner of shorter arm?
Answer:
The turning effect of force, τ = \(\overline{\mathrm{r}}\times\overline{\mathrm{F}}\). When arm of the spanner is long, r is larger. Therefore smaller force (F) will produce the same turning effect. Hence, the spanner of longer arm is preferred as compared to the spanner of shorter arm.

Question 7.
By spinning eggs on a table top, how will you distinguish an hard boiled egg from egg? [AP Mar. ’13]
Answer:
To distinguish between a hard boiled egg and a raw egg, we spin each on a table top. The egg which spins at a slower rate shall be a raw egg. This is because in a raw egg, liquid matter inside tries to get away from the axis of rotation. Therefore, its moment of inertia ‘I’ increases. As τ = Iα = constant, therefore, α decreases i.e., raw egg will spin with smaller angular acceleration.

Question 8.
Why should a helicopter necessarily have two propellers?
Answer:
If the helicopter had only one propeller, then due to conservation of angular momen¬tum, the helicopter itself would turn in the opposite direction. Hence, the helicopter should necessarily have two propellers.

Question 9.
If the polar ice caps of the earth were to melt, what would the effect of the length of the day be?
Answer:
Earth rotates about its polar axis. When ice of polar caps of earth melts, mass concen¬trated near the axis of rotation spreads out. Therefore, moment of inertia ‘I’ increases.

As no external torque acts, L = Iω = I(\(\frac{2 \pi}{T}\)) = constant

with increase of I, T will increase i.e., length of the day will increase.

Question 10.
Why is it easier to balance a bicycle in motion?
Answer:
A bicycle in motion is in rotational equilibrium. From principles of Dynamics of rotational bodies is that the forces that are perpendicular to the axis of rotation will try to turn the axis of rotation but necessary forces will arise it cancel these forces due to inertia of rotation and fixed position of axis is maintained. So it is easy to balance a rotating body.

Short Answer Questions

Question 1.
Distinguish between centre of mass and centre of gravity. [AP Mar. 18, 17, 16, 15, 14, 13, May 17; June 15 : TS Mar. 16. 15, May 18, 17]
Answer:

Centre of mass	Centre of gravity
1) A point inside a body at which the whole mass is supposed to be concentrated. A force applied at this point produces translatory motion.	1) A point inside a body through which the weight of the body acts.
2) It pertains (or) contain to mass of the body.	2) It refers to weight acting on all particles of the body.
3) In case of small bodies centre of mass and centre of gravity coincide. (Uniform gravitational field)	3) In case of a huge body centre of mass and centre of gravity may not coincide. (Non uniform gravitational field)
4) Algebraic sum of moments of masses about centre of mass is zero.	4) Algebraic sum of moments of weights about centre of gravity is zero.
5) Centre of mass is used to study translatory motion of a body when it is in complicated motion.	5) Centre of gravity is used to know the stability of the body where it is to be supported.

Question 2.
Show that a system of particles moving under the influence of an external force, moves as if the force is applied at its centre of mass. [AP May ’18]
Answer:
Consider a system of particles of masses m₁, m₂, ……….. mn moves with velocity

But Force (F) = ma, so total force on the body is
F = Ma_C.M = m₁a₁ + m₂a₂ + m₃a₃ + ……….. + m_n a_n
or Total Force F = Ma_C.M = F₁ + F₂ + F₃ + ……… + F_n

Hence, total force on the body is the sum of forces on individual particles and it is equals to force on centre of mass of the body.

Question 3.
Explain about the centre of mass of earth-moon system and its rotation around the sun.
Answer:
The interaction of earth and moon does not effect the motion of centre of mass of earth and moon system around the sun. The gravitational force between earth and moon is internal force. Internal forces cannot change the position of centre of mass.

The external force acting on the centre of mass of earth and moon system is force between the sun and C.M. of earth, moon system. Motion of centre of mass depends on external force. Hence, earth moon system continues to move in an elliptical path around the sun. It is irrespective of rotation of moon around earth.

Question 4.
Define vector product. Explain the properties of a vector product with two examples. [AP Mar. ’17, ’15 ; TS Mar. ’17, ’16, ’15; APMay ’18. ’17; TS May ’18. ’16]
Answer:
Vector product (or) cross product :
If the product of two vectors (say \(\overline{\mathrm{A}}\) and \(\overline{\mathrm{B}}\)) gives a vector then that multiplication of vectors is called cross product or vector product of vectors.

Properties of cross product:
1. Cross product is not commutative i.e.

2. Cross product obeys distributive law i.e.,

3. If any vector is represented by the combination of \(\overline{\mathrm{i}},\overline{\mathrm{j}}\) and \(\overline{\mathrm{k}}\) then cross product will obey right hand screw rule.
4. The product of two coplanar perpendicular unit vectors will generate a unit vector perpendicular to that plane

5. Cross product of parallel vectors is zero

Examples of cross product:
1) Torque (or) moment of force (\(\overline{\mathrm{\tau}}\)) :
It is defined as the product of force and perpendicular distance from the point of application.
∴ Torque τ = \(\overline{\mathrm{r}}\times\overline{\mathrm{F}}\)

2) Angular momentum and angular velocity :
For a rigid body in motion, Angular momentum (\(\overline{\mathrm{L}}\)) = radius (\(\overline{\mathrm{r}}\)) x momentum (\(\overline{\mathrm{P}}\))
∴ Angular momentum (\(\overline{\mathrm{L}}\)) = \(\overline{\mathrm{r}}\) × (m\(\overline{\mathrm{v}}\)) = m(\(\overline{\mathrm{r}}\times\overline{\mathrm{v}}\))

Question 5.
Define angular velocity. Derive v = r ω. [TS Mar. 19,’ 17, 16; AP Mar. 19, May. 16; May 14]
Answer:
Angular velocity (ω) :
Rate of change of angular displacement is called angular velocity.

Relation between linear velocity (v) and angular velocity (ω) :
Let a particle P is moving along circumference of a circle of radius r1 with uniform speed v. Let it is initially at the position A, during a small time ∆t it goes to a new position say C from B. Angle subtended during this small interval is say dθ.

By definition angular velocity,

Question 6.
State the principle of conservation of angular momentum. Give two examples.
Answer:
Law of conservation of angular momentum:
When no external torque is acting on a body then the angular momentum of that rota-ting body is constant.
i.e., I₁ω₁ = I₂ω₂ (when τ = 0)

Example -1:
A boy stands over the centre of a horizontal platform which is rotating freely with a speed ω₁ (n₁revolutions/sec.) about a vertical axis passing through the centre of the platform and straight up through the boy. He holds two bricks in each of his hands close to his body. The combined moment of inertia of the system is say I₁. Let the boy stretches his arms to hold the masses far away from his body. In this position the moment of inertia increases to I₂ and let ω₂ is his angular speed.

Here ω₂ < ω₁ because moment of inertia increases.

Example – 2 :
An athlete diving off a high spring board can bring his legs and hands close to the body and performs Somersault about a horizontal axis passing through his body in the air before reaching the water below it. During the fall his angular momentum remains constant.

Question 7.
Define angular acceleration and torque. Establish the relation between angular acceleration and torque. [TS Mar. ’18, ’17; TS May ’17, June ’15; AP Mar. ’19, June ’15]
Answer:
Angular acceleration (α) :
Rate of change of angular velocity is called angular acceleration.

Torque :
It is defined as the product of the force and the perpendicular distance of the point of application of the force from that point.

Relation between angular acceleration and Torque:
We know that, L = Iω

On differentiating the above expression with respect to time ‘t’

But \(\frac{dL}{dt}\) is the rate of change of angular momentum called ‘Torque (τ)”.

and \(\frac{d \omega}{dt}\) is the rate of change of angular velocity called “angular acceleration (α)”

∴ The relation between Torque and angular acceleration is, τ = lα

Question 8.
Write the equations of motion for a particle rotating about a fixed axis.
Answer:
Equations of rotational kinematics :
If ‘θ’ is the angular displacement, Wj is the initial angular velocity, ω_f is the final angular velocity after a time ‘t’ seconds and ‘α’ is the angular acceleration, then the equations of rotational kinematics can be written as,

Question 9.
Derive expressions for the final velocity and total energy of a body rolling without slipping.
Answer:
A rolling body has both translational kinetic energy and rotational kinetic energy. So the total K.E energy of a rolling body is,

Long Answer Questions

Question 1.
(a) State and prove parallel axis theorem.
(b) For a thin flat circular disk, the radius of gyration about a diameter as axis is k. If the disk is cut along a diameter AB as shown into two equal pieces, then find the radius of gyration of each piece about AB.

Answer:
a) Parallel axis theorem :
The moment of inertia of a rigid body about an axis passing through a point is the sum of moment of inertia about parallel axis passing through centre of mass (I_G) and mass of the body multiplied by Square of distance (MR²) between the axes i.e.,
I = I_G + MR²

Proof :
Consider a rigid body of mass M with ‘G’ as its centre of mass. Iq the moment of inertia about an axis passing through centre of mass. I = The moment of inertia about an axis passing through the point ’O’ in that plane.

Let perpendicular distance between the axes is OG = R (say)

Consider point P in the given plane. Join OP and GP. Extend the line OG and drop a normal from ’P’ on to it as shown in figure.

The moment of inertia about the axis passing through centre of mass G.
(I_G) = ∑mGP² ……….. (1)

M.O.I. of the body about an axis passing through ‘O’ (I) = ∑mOP² ………… (2)
From triangle OPD
OP² = OD² + DP²
⇒ OD = OG + GD
∴ OD² = (OG + GD)² = OG² + GD² + 2OG. GD ………….. (3)
From Equations (2) and (3)
I = ∑mOP² = Em [ (OG² + GD² + 2OG. GD) + DP²]
∴ I = ∑m {GD² + DP² + OG² + 20G. GN}
But GD² + DP² = GP²
∴ I = ∑m {GP² + OG² + 20G. GD}
∴ I = ∑m GP² + ∑mOG² + 20G ∑mGD ………. (4)

But the terms ∑mGP² = I_G
∑mOG² = MR² (∵ ∑m = M and OG = R)
The term 20G ∑mGD = 0. Because it represents sum of moment of masses about centre of mass. Hence its value is zero.
∴ I = I_G + MR²

Hence parallel axis theorem is proved.

b) Moment of inertia of a disc of mass ‘M’ and radius ‘R’ about its diameter is,
I = \(\frac{MR^2}{4}\)

If ‘k’ is radius of gyration of disc then, I = Mk²
∴ Mk² = \(\frac{MR^2}{4}\) ⇒ k = R/2

After cutting along the diameter, mass M of each piece = \(\frac{M}{2}\)
Moment of inertia of each piece,

Question 2.
(a) State and prove perpendicular axis theorem.
(b) If a thin circular ring and a thin flat circular disk of same mass have same moment of inertia about their respective diameters as axis. Then find the ratio of their radii.
Answer:
a) Perpendicular axis theorem :
The moment of inertia of a plane lamina about an axis perpendicular to its plane is equal to the sum of the moment of inertia about two perpendicular axis concurrent with perpendicular axis and lying in the plane of the body.
∴ I_z = I_x + I_y

Proof :
Consider a rectangular plane lamina. X and Y are two mutually perpendicular axis in the plane. Choose another perpendicular axis Z passing through the point ‘O’.

Consider a particle P in XOY plane.
Its co-ordinates are (x, y).
Moment of inertia of particle about
X-axis is I_X = ∑my².
M.O.I about Y-axis is I_Y = ∑mx²
M.O.I about Z axis is I_Z = ∑ m . OP²
From triangle OAP,
OP² = OA² + AP² = y² + x²
∴ Iz = ∑ mOP² = ∑ m (y² + x²)
∴ Iz = X my² + ∑ mx²
But ∑ my² = I_x and ∑ mx² = I_y

∴ Moment of Inertia about a perpendicular axis passing through ‘O’ is I_Z = I_X + I_Y
Hence perpendicular axis theorem is proved.

b) Moment of inertia of a thin circular ring about its diameter is, I₁ = m₁ R²₁

Moment of inertia of a flat circular disc about its diameter is, I₂ = \(\frac{m_2R^{2}_{2}}{2}\)
Given that two objects having same moment of inertia i.e., I₁ = I₂

Question 3.
State and prove the principle of conservation of angular momentum. Explain the principle of conservation of angular momentum with examples. [AP Mar. ’16]
Answer:
Law of conservation of angular momentum: When no external torque is acting on a body then the angular momentum of that rotating body is constant.
i.e., I₁ω₁ = I₂ω₂ (when τ = 0)

Explanation:
Here I₁ and I₂ are moment of inertia of rotating bodies and ω₁ and ω₂ are their initial and final angular velocities. If

Example -1 :
A boy stands over the centre of a horizontal platform which is rotating freely with a speed ω₁ (n₁ revolutions/sec.) about a vertical axis passing through the centre of the platform and straight up through the boy. He holds two bricks in each of his hands close to his body. The combined moment of inertia of the system is say I₁. Let the boy stretches his arms to hold the masses far away from his body. In this position the moment of inertia increases to I₂ and let ω₂ is his angular speed.

Here ω₂ < ω₁ because moment of inertia increases.

Example – 2 :
An athlete diving off a high spring board can bring his legs and hands close to the body and performs Somersault about a horizontal axis passing through his body in the air before reaching the water below it. During the fall his angular momentum remains constant.

Position of centre of mass of some symmetrical bodies :

Shape of the body	Position of centre of mass
1. Hollow sphere (or) solid sphere	At the centre of sphere
2. Circular ring	At the centre of the ring
3. Circular disc	At the centre of disc
4. Triangular plate	At the centroid
5. Square plate	At the point of intersection of diagonals
6. Rectangular plate	At the point of intersection of diagonals
7. Cone	At \(\frac{3h}{4}\) th of its height from its apex on its own axis
8. Cylinder	At the midpoint of its own axis.

Comparison of translatory and rotatory motions :

Problems

Question 1.
Show that a • (b × c) is equal in magnitude to the volume of the parallelopiped formed on the three vectors a, b and c. (IMP)
Solution:
Let a parallelopiped be formed on three

Now \(\hat{a}\) • (\(\hat{b}\times\hat{c}\) x c) = \(\hat{a}\) • be \(\hat{n}\) = (a) (be) cos 0° – abc

Which is equal in magnitude to the volume of parallelopiped.

Question 2.
A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope ? Assume that there is no slipping.
Soution:
Here M = 3 kg ; R = 40 cm = 0.4 m
Moment of inertia of the hollow cylinder about its axis, I = MR² = 3(0.4)² = 0.48 kg m²
Force applied, F = 30 N
∴ Torque, τ = F × R = 30 × 0.4 = 12 N – m

If α is angular acceleration produced, then from τ = Iα

Linear acceleration, a = Ra = 0.4 × 25
= 10 ms^-2.

Question 3.
A coin is kept a distance of 10 cm from the centre of a circular turn table. If the coefficient of static friction between the table and the coin is 0.8. Find the frequency or rotation of the disc at which the coin will just begin to slip.
Solution:
Distance of coin = r = 10 cm = 0.1 m.
Coefficient of friction µ = 0.8.
Frequency of rotation = number of rotations per second.

Question 4.
Find the torque of a force \(\mathbf{7} \overrightarrow{\mathbf{i}}+\mathbf{3} \overrightarrow{\mathbf{j}}-5 \overrightarrow{\mathbf{k}}\) about the origin. The force acts on a particle whose position vector is \(\overrightarrow{\mathbf{i}}-\overrightarrow{\mathbf{j}}+\overrightarrow{\mathbf{k}}\). [AP Mar. ’14, ’13; May ’13]
Answer:

Question 5.
Particles of masses 1g, 2g, 3g….., 100g are kept at the marks 1 cm, 2 cm, 3 cm…, 100 cm respectively on a meter scale. Find the moment of inertia of the system of particles about a perpendicular bisector of the meter scale.
Solution:
Given Masses of 1g, 2g, 3g 100 g are 1, 2, 3 ……….. 100 cm on a scale.

i) Sum of masses 2m = \(\sum_{i=1}^n\)ni
Sum of n natural numbers S

∴ Total mass M = 5051 gr = 5.051 kg → (1)

ii) Centre of mass of all these masses is given by

iii) M.O.I. = I

Sum of cubes of 1st n natural numbers is

M.O.I. about C.M. = I_G = I – MR²
= 2.550 – 5.05 × 0.67 × 0.67 = 2.550 – 2.267 = 0.283 kg.m2

iv) Perpendicular bisector is at 50 CM.
So shift M.O.I from centre of mass to
x₁ = 50cm point from x = 67 CM
∴ Distance between the axis
R = 67 – 50 = 17cm = 0.17M
M.O.I. about this axis I = I_G + MR²
= 0.283 + 5.05 × 0.17 × 0.17
= 0.283 + 0.146 = 0.429 kgm²
∴ M.O.I. about perpendicular bisector of scale = 0.429 kg – m²

Question 6.
Calculate the moment of inertia of a fly wheel, if its angular velocity is increased from 60 r.p.m. to 180 r.p.m. when 100 J of work is done on it. [TS May ’16]
Solution:
W = 100 J, ω₁ = 60 RPM = 1 R.P.S = 2π Rad.
ω₂ = 180 R.P.M. = 3 R.P.S = 6π Rad.

Question 7.
Three particles each of mass 100 g are placed at the vertices of an equilateral triangle of side length 10 cm. Find the moment of inertia of the system about an axis passing through the centroid of the triangle and perpendicular to its plane.
Solution:
Mass of each particle m = 100 g; side of equilateral triangle = 10 cm.
In equilateral triangle height of angular bisector CD = \(\frac{\sqrt{3}}{2}\)l

Centroid will divide the angular bisector in a ratio 2 : 1

So X distance of each mass from vertex to centroid is 2.\(\frac{\sqrt{3}}{2}\)l = \(\frac{\sqrt{3}}{2}\)l
Moment of Inertia of the system

Question 8.
Four particles each of mass 100g are placed at the corners of a square of side 10 cm. Find the moment of inertia of the system about an axis passing through the centre of the square and perpendicular to its plane. Find also the radius of gyration of the system.
Solution:
Mass of each particle, m = 100 g = 0.1 kg.
Length of side’of square, a = 10 cm = 0.1 m
In square distance of corner from centre of square = \(\frac{1}{2}\) diagonal = \(\frac{\sqrt{2}a}{2}=\frac{a}{\sqrt{2}}\)

∴ Total moment of Inertia

Question 9.
Two uniform circular discs, each of mass 1 kg and radius 20 cm, are kept in contact about the tangent passing through the point of contact. Find the moment of inertia of the system about the tangent passing through the point of contact.
Solution:
Mass of disc = M = 1 kg.
Radius of disc = 20 cm = 0.2 m

They are in contact as shown.
M.O.I of a circular disc about a tangent parallel to its plane = \(\frac{5}{4}\) MR²
Total M.O.I. of the system

Question 10.
Four spheres each diameter 2a and mass ‘m’ are placed with their centres on the four corners of a square of the side b. Calculate the moment of inertia of the system about any side of the square.
Solution:
Diameter of sphere = 2a ⇒ radius = a.
Side of square = b.
For spheres 1 and 2 axis of rotation is same and passing through diameters. M.O.I. of solid sphere about any diameter = \(\frac{2}{5}\)MR² (put M = m and R = a)

Transfer this M.O.I. on to the axis using
Parallel axis theorem.

Total M.O.I. of the system

Question 11.
To maintain a rotor at a uniform angular speed or 200 rad s^-1, an engine needs to transmit a torque of 180 Nm. What is the power required by the engine? (Note : uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.
Solution:
Given uniform angular speed (ω) = 200 rad s^-1
Torque, τ = 180 N – m ; But power p = τω
∴ P =180 × 200 = 36000 watt = 36 kW

Question 12.
A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?
Solution:
Let m be the mass of the stick concentrated at C, the 50 cm mark

For equilibrium
about C’, i.e. at the 45 cm mark,
10g (45 – 12) = mg (50 – 45)
10g × 33 = mg × 5
m = \(\frac{10\times33}{5}\) = 66 grams

Question 13.
Determine the kinetic energy of a circular disc rotating with a speed of 60 rpm about an axis passing through a point on its circumference and perpendicular to its plane. The circular disc has a mass of 5 kg and radius 1 m.
Solution:
Mass of disc, M = 5 kg; Radius R = 1 m.
Angular velocity, co = 60 RPM = \(\frac{60\times 2\pi}{60}\) = 2πRad/sec

M.O.I. of disc about a point passing through circumference and perpendicular to the plane.

Question 14.
Two particles, each of mass m and speed u, travel in opposite directions along para¬llel lines separated by a distance d. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momemtum is taken.
Solution:
Angular momentum, L = mvr
Choose any axis say ‘A’
Let at any given time distance between m₁ & m₂ = L = L₁ + L₂
About the axis ‘A’ both will rotate in same direction See fig.

∴ Total angular momentum
L = L₁ + L₂ = muL₁ + muL₂ = mu (L₁ + L₂) = muL

about any new axis say B distance of m₁ and m₂
are say L’₁ and L’₂
Total angular momentum,
L = mu L’₁ + mu L’₂
or L = mu(L’₁ + L’₂) = muL (∵ L’₁ + L’₂ = L)
Hence, total angular momentum of the system is always constant.

Question 15.
The moment of inertia of a fly wheel making 300 revolutions per minute is 0.3 kgm². Find the torque required to bring it to rest in 20s.
Solution:
M.O.I, I = 0.3 kg. ; time, t = 20 sec.,
ω₁ = 300 R.P.M. = \(\frac{300}{60}\) = 5. R.P.S.; ω₂ = 0

Torque, τ = Iα = 0.3 × \(\frac{5\times 2\pi}{20}\) = 0.471 N – m.

Question 16.
When 100J of work is done on a fly wheel, its angular velocity is increased from 60 rpm to 180 rpm. What is the moment of inertia of the wheel?
Solution:
W=100J, ω₁ = 60 RPM = 1 R.PS = 2π Rad.
ω₂ =180 R.P.M. = 3 R.P.S = 6π Rad.

Question 17.
Find the centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are lOOg, 150g and 200g respectively. Each side of the equilateral triangle is 0.5 m long, lOOg mass is at origin and 150g mass is on the X-axis. [TS Mar. 18, June 15; AP Mar. ’18]
Solution:
Mass at A = 100g ; Coordinates = 0, 0
Mass at B = 150 g; Coordinates = (0.5, 0)
Mass at C = 200g; Coordinates (0.25,0.25 √3 )
Coordinates x_cm

Students must practice these Maths 1A Important Questions TS Inter 1st Year Maths 1A Functions Important Questions Very Short Answer Type to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Functions Important Questions Very Short Answer Type

Question 1.
If the function f is defined by

then find the 2x +1, x < -3 values if exist of f(4), f(2.5), f(- 2), f(- 4), f(0), f(- 7). [Mar 14]
Answer:
(i) f(4) For x > 3, f(x) = 3x – 2
f(4) = 3 (4) – 2 = 12 – 2 = 10

(ii) f(2.5) is not defined.

(iii) f(-2)
For – 2 ≤ x ≤ 2, f(x) = x² – 2
f(- 2) = (- 2)² – 2 = 4 – 2 = 2

(iv) f(-4)
For x < – 3, f(x) = 2x + 1
f(- 4) = 2(- 4) + 1 = – 8 + 1 = – 7

(v) f(0)
For – 2 ≤ x ≤ 2, f(x) = x² – 2
f(0) = 0² – 2 = – 2

(vi) f(- 7)
For x < – 3, f(x) = 2x + 1
f(- 7) = 2 (- 7) + 1 = – 14 + 1 = – 13

If the function f is defined by

then find the values of
(i) f(3)
(ii) f(0)
(iii) f(-1.5)
(iv) f(2) + f(-2)
(v) f(- 5).
Answer:
(i) 5
(ii) 2
(iii) – 2.5
(iv) 1
(v) not defined

Question 2.
If A = \(\left\{0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}\right\}\) and f: A → B is a surjection defined by f(x) = cos x then find B. [Mar. (TS) 17, 16 (AP), 11 May 15 (AP), 15 (TS), 11]
Answer:
Given A = \(\left\{0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}\right\}\)
f(x) = cos x
Since f: A → B is a surjection then f(A) = B
f(0) = cos 0 = 1

∴ B = Range . f(A) = {1, \(\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}\), o}

If A = {- 2, – 1, 0, 1, 2} and f: A → B is a surjection defined by f(x) = x² + x + 1, then find B. [Mar. (AP) 19, 17] [Mar 16 (TS): May 14, 10]
Answer:
{3, 1, 7}

If A = {1, 2, 3, 4} and f: A → R is a function defined by f(x) = \(\frac{x^2-x+1}{x+1}\), then find the range of f.
Answer:
\(\left\{1, \frac{1}{2}, \frac{7}{4}, \frac{13}{5}\right\}\)

Question 3.
Determine whether the function f: R → R defined by

is an injection or a surjection or a bijection.
Answer:
Given f: R → R

If x = 3 > 2 then f(3) = 3
If x = 1 <2 then f(1) = 5(1) – 2 = 5 – 2 = 3
∴ 1 and 3 have same f image,
∴ f is not an injection.
If y ∈ R (co-domain) then y = x
x = y
then f (x) = x
f(x) = y
If y ∈ R (co-domain) then y = 5x – 2
y + 2 = 5x
x = \(\frac{y+2}{5}\)
then f (x) = 5x – 2

= y + 2 – 2 = y
∴ f is a surjection since f is not an injection then it is not a bijection.

Question 4.
If f: R → R, g : R → R are defined by f(x) = 4x – 1 and g(x) = x² + 2 then find [May 09. Mar. 05, 04]
(i)(gof) (x)
(ii) (gof) \(\left(\frac{a+1}{4}\right)\)
(iii) (fof) (x)
(iv) [go (fof)] (0)
Answer:
Given f: R → R, g: R → R
f(x) = 4x – 1 and g(x) = x² + 2

(i) (gof) (x) = g [f(x)] = g[4x – 1]
= (4x – 1)² + 2
= 16x² + 1 – 8x + 2
= 16x² – 8x + 3

(ii) (gof) \(\left(\frac{a+1}{4}\right)\) = g \(\left[\mathrm{f}\left(\frac{\mathrm{a}+1}{4}\right)\right]\)
= g\(\left[4\left(\frac{a+1}{4}\right)-1\right]\)
= g[a + 1 – 1 ]
= g(a) = a² + 2

(iii) (fof) (x) = f [f(x)] = f[4x – 1]
= 4 (4x – 1) – 1
= 16x – 4 – 1
= 16x – 5

(iv) [go (fof)] (0)
Now (fof) (0) = f[f(0)] = f[4(0) – 1] = f(- 1)
= 4( – 1) – 1 = – 4 – 1 = – 5
[go (fof)] (0) = go [(fof)(0)]
= g [- 5] = (- 5)² + 2
= 25 + 2 = 27

Question 5.
If f: Q → Q is defined by f(x) = 5x + 4 for all x ∈ Q, show that ‘f is a bijection and find f^-1. [Mar. 17 (TS). 16 (AP)]
Answer:
Given f: Q → Q, f(x) = 5x + 4, ∀ x ∈ Q
Let a₁, a₂ ∈ Q
f(a₁) = f(a₂)
5a₁ + 4 = 5a₂ + 4
5a₁ = 5a₂
a₁ = a₂
∴ f: Q → Q is an one – one function.
Let y ∈ Q (co-domain) then y = 5x + 4
y – 4 = 5x

f(x) = y
∴ f: Q → Q is an onto function.
∴ f: Q → Q is a bijection.
∴ f^-1: Q → Q is a bijection.

Question 6.
If f : R → R, g : R → R are defined by f(x) = 3x – 1, g(x) = x² + 1, then find (fog) (2). [Mar. 18 (AP) May 13; Mar 13]
Answer:
Given f: R → R and g : R → R defined by
f(x) = 3x – 1 ; g(x) = x² + 1
(fog) (2) = f[g(2)] = f(2² + 1)
= f(5) = 3(5) – 1 = 14

Question 7.
If f(x) = \(\frac{1}{x}\), g (x) = √x for all x ∈ (0, ∞), x then find (gof) (x).
Answer:
Given
f(x) = \(\frac{1}{x}\), g(x) = √x ∀ x ∈ (0, ∞)
Now (gof) (x) = g[f(x)] = g \(\left[\frac{1}{x}\right]\) = \(\sqrt{\frac{1}{x}}=\frac{1}{\sqrt{x}}\)

Question 8.
If f(x) = 2x – 1, g(x) = \(\frac{x+1}{2}\) for all x ∈ R, then find (gof) (x). [Mar 19 (AP); Sep 92]
Answer:
Given f(x) = 2x – 1, g(x) \(\frac{x+1}{2}\) ∀ x ∈ R
Now (gof) (x) = g[f(x)] = g(2x – 1)
= \(\frac{2 \mathrm{x}-1+1}{2}\) = \(\frac{2 x}{2x}\) = x

Question 9.
If f(x) = 2, g(x) = x², h(x) = 2x for all x ∈ R, then find [fo(goh)] (x). [Mar. 17 (TS); July 01]
Answer:
Given f(x) = 2, g(x) = x², h(x) = 2x ∀ x ∈ R.
Now (goh) (x) = g[h(x)] = g[2x] = (2x)² = 4x²
[fo(goh)](x) = f[(goh) (x)] = f[4x²] = 2

Question 10.
Find the inverse function of f(x) = ax + b, (a ≠ 0); a, b ∈ R [Mar. 18 (TS); Mar. 13]
Answer:
Given, a, b ∈ R, f: R → R and
f(x) = ax + b
Let y = f(x) = ax + b
y = f(x) ⇒ x = f ^-1 (y) …………….. (1)
y = ax + b ⇒ ax = y – b ⇒ x = \(\frac{\mathrm{y}-\mathrm{b}}{\mathrm{a}}\) ……………… (2)
From (1) and (2)
f^-1(y) = \(\frac{y-b}{a}\) ⇒ f^-1(x) = \(\frac{x-b}{a}\)

If f: Q → Q is defined by f(x) = 5x + 4, for all x ∈ Q, find f^-1. [Mar. 12, 10]
Answer:
\(\frac{x-4}{5}\)

Question 11.
Find the inverse function of f(x) = 5^x [Mar. 15 (AP); Mar. ’11’, 06]
Answer:
Given f(x) = 5^x
Let y = f(x) = 5^x
y = f(x) ⇒ x = f^-1(y) …………………. (1)
y = 5x ⇒ x = log₅y ……………………. (2)
From (1) and (2),
f^-1(y) = log₅y ⇒ f^-1 (x) = log₅x

Question 12.
If f: R → R, g: R → R defined by f(x) = 3x – 2, g(x) = x² + 1, then find (i) (gof^-1) (2) (ii) (gof) (x – 1). [Mar. 08; May 06]
Answer:
Given f: R → R, g: R → R, f(x) = 3x – 2, g(x) = x² – 1
Let y = f(x) = 3x – 2
y = f(x) ⇒ x = f^-1 ……………. (1)
y = 3x – 2 ⇒ y + 2 = 3x
x = \(\frac{\mathrm{y}+2}{3}\) ………………….. (2)
From (1) & (2)
f^-1(y) = \(\frac{\mathrm{y}+2}{3}\)
⇒ f^-1 (x) = \(\frac{\mathrm{x}+2}{3}\)

(i) (gof^-1) (2)
= g[f^-1 (2)]

(ii) (gof) (x – 1)
= g[f(x -1)]
= g[3(x – 1) – 2]
= g[3x – 3 – 2]
= g(3x – 5)
= (3x – 5)² + 1
= 9x² + 25 – 30x + 1
= 9x² – 30x + 26

Question 13.
If f(x) = \(\frac{x+1}{x-1}\) (x ≠ ± 1), then find (fofof) (x) [Mar. 05]
Answer:
Given f(x) = \(\frac{x+1}{x-1}\) (x ≠ ± 1)

If f(x) = \(\) (x ≠ ± 1), then find (fofofof) (x)
Answer:
x

Question 14.
Find the domain of the real valued function f(x) = \(\sqrt{a^2-x^2}\) [June 04]
Answer:
Given f(x) = \(\sqrt{a^2-x^2}\) ∈ R
⇒ a² – x² ≥ 0
⇒ x² – a² ≤ 0
⇒ (x + a) (x – a) ≤ 0
– a ≤ x ≤ a

⇒ x ∈ [-a, a]
∴ Domain of ‘f is [-a, a]

Find the domain of the real valued function f(x) = \(\sqrt{16-x^2}\).
Answer:
[- 4, 4]

Find the domain of the real valued function f(x) = \(\sqrt{9-x^2}\).
Answer:
[-3, 3]

Question 15.
Find the domain of the real valued function
f(x) = \(\frac{1}{\left(x^2-1\right)(x+3)}\). [May 14, 93; Mar. 14]
Answer:
Given f(x) = \(\frac{1}{\left(x^2-1\right)(x+3)}\)
f(x) = \(\frac{1}{\left(x^2-1\right)(x+3)}\) ∈ R
⇒ (x² – 1) (x + 3) ≠ 0
x² – 1 ≠ 0, x + 3 ≠ 0
x² ≠ 1, x ≠ – 3
x ≠ ± 1
∴ x ≠ -3, -1, 1
∴ Domain of ‘f is R – {-3, -1, 1}

Find the domain of the real valued function f(x) = \(\frac{1}{6 x-x^2-5}\).
Answer:
R – {1, 5}

Find the domain of the real valued function f(x) = \(\frac{2 x^2-5 x+7}{(x-1)(x-2)(x-3)}\)
Answer:
R- {1, 2, 3}

Question 16.
Find the domain of the real valued function f(x) = \(\sqrt{4 x-x^2}\). [May 12, 10] [Mar; 18 (TS)]
Answer:
Given, f(x) = \(\sqrt{4 x-x^2}\) ∈ R
⇒ x(4 – x) ≥ 0

⇒ x(4 – x) ≥ 0
⇒ x(x-4) ≤ 0
⇒ (x – 0) (x – 4) ≤ 0
⇒ 0 ≤ x ≤ 4
∴ Domain of ‘f’ is [0, 4].

Question 17.
Find the domain of the real valued function f(x) = \(\frac{1}{\sqrt{1-x^2}}\)
Answer:
Given f(x) = \(\frac{1}{\sqrt{1-x^2}}\) ∈ R

⇔ 1 – x² > 0 “
⇔ (1 + x) (1 – x) > 0
⇔ x ∈ (-1, 1)
∴ Domain of T = {x/x ∈ (-1, 1)}

Question 18.
Find the domain of the real valued function f(x) = \(\sqrt{\mathbf{x}^2-25}\) [May 15 (AP); Mar. 12]
Answer:
Given f(x) = \(\sqrt{x^2-25}\) ∈ R
⇒ x² – 25 ≥ 0
⇒ (x + 5) (x – 5) ≥ 0
⇒ x < -5 or x > 5
⇒ x ∈ (- α, -5] ∪ [5, α)
∴ Domain of T is (-α, -5] ∪ [5, α).

Question 19.
Find the domain of the real valued function f(x) = log (x² – 4x + 3) [Mar. 16 (TS), 10, ‘08 ; May 11, 07]
Answer:
Given f(x) = log (x² – 4x + 3) ∈ R
⇒ x² – 4x + 3 > 0
⇒ x² – 3x – x + 3 > 0
⇒ x (x – 3) – 1 (x – 3) > 0
⇒ (x – 1) (x – 3) > 0
⇒ x < 1 or x > 3
⇒ x ∈ (- α, 1) ∪ (3, α)

Question 20.
Find the domain of the real valued function f(x) = \(\frac{\sqrt{3+x}+\sqrt{3-x}}{x}\)
Answer:
Given f(x) = \(\frac{\sqrt{3+x}+\sqrt{3-x}}{x}\) ∈ R
⇒ 3 + x ≥ 0 and 3 – x ≥ 0, x ≠ 0
x ≥ – 3 and 3 ≥ x, x ≠ 0, x ≤ 3
x ∈ [- 3, ∝) ∩ (-∝, 3) – {0}
⇒ x ∈ [- 3, 3] – {0}
(or)
⇒ x ∈ [- 3, 0) ∪ (0, 3]
∴ Domain of ‘f is [- 3, 3] – {0}

Find the domain of the real valued function f(x) = \(\frac{\sqrt{2+x}+\sqrt{2-x}}{x}\).
Answer:
[- 2, 0) ∪ (0, 2]

Question 21.
Find the range of the reed valued function, log |4 – x²|.
Answer:
Let y = f(x) = log |4 – x²|
f(x) ∈ R ⇒ 4 – x² ≠ 0
x² ≠ 4 ⇒ x ≠ ± 2
∴ Domain of ‘f is R – {- 2, 2}
∴ y = log_e |4 – x²|
|4 – x²| = e^y
⇒ e^y > 0, ∀ y ∈ R
∴ Range of T is R.

Question 22.
Find the range of the real valued function. \(\frac{x^2-4}{x-2}\) [May 03, 97]
Answer:
Let y = f(x) = \(\frac{x^2-4}{x-2}\)
f(x) ∈ R ⇒ x – 2 ≠ 0 ⇒ x ≠ 2
∴ Domain of ‘f is R – {2}
Let y = \(\frac{x^2-4}{x-2}\), if x ≠ 2 then y = x + 2
If x = 2, then y = 2 + 2 = 4
y is not defined at x = 2, then y cannot be equal to 4.
∴ Range of T is R – {4}.

Question 23.
Find the domain and range of the function f(x) = \(\frac{x}{2-3 x}\)
Answer:
Given f(x) = \(\frac{x}{2-3 x}\) ∈ R
⇒ 2 – 3x ≠ 0
⇒ 2 ≠ 3x
⇒ x ≠ \(\frac{2}{3}\)
∴ Domain of ‘f’ is R – \(\left\{\frac{2}{3}\right\}\)
Let y = f(x) = \(\frac{x}{2-3 x}\)
∴ y = \(\frac{x}{2-3 x}\)
2y – 3xy = x ⇒ 2y = x + 3x ⇒ 2y = x(1 + 3y)

Question 24.
If f = {(4, 5), (5, 6), (6, – 4)} and g = {(4, -4), (6, 5), (8, 5)}, then find
(i) f + 4
(ii) fg
(iii) √f
(iv) f².
Answer:
(i) f + 4
Domain of f + 4 = A = {4, 5, 6}
(f + 4) (x) = f(x) + 4
(f + 4) (4) = f(4) + 4 = 5 + 4 = 9
(f + 4) (5) = f(5) + 4 = 6 + 4= 10
(f + 4) (6) = f(6) + 4 = -4 + 4 = 0
∴ f + 4 = {(4, 9), (5, 10), (6, 0)}

(ii) fg
Domain of fg = A ∩ B = {4, 6}
(fg) (x) = f(x) . g(x)
(fg) (4) = f(4) . g(4) = 5(-4) = -20
(fg) (6) = f(6) . g(6) = (-4) (5) = -20
∴ fg = {(4, -20), (6, – 20)}

(iii) √f
Domain of √f = {4, 5, 6} = A
√f (x) = √f(x)
√f(4) = √f(4) = √5
√f (5) = √f(5) = √6
√f (6) = √f(6) = √-4 (does not exist)
∴ √f = {(4, √5), (5, √6)}

(iv) f²
Domain of f² = A = {4, 5, 6}
f²(x) = [f(x)]²
f²(4) = [f(4)]² = (5)² = 25
f²(5) = [f(5)]² = (6)² = 36
f²(6) = [f(6)]² = (- 4)² = 16
∴ f² = {(4, 25), (5, 36), (6, 16)}

If f = {(1, 2), (2, -3), (3, -1)}, then find
(i) 2f [Mar. 12; 94, 90; May 94]
(ii) 2 + f [Mar. 12; May 08]
(iii) f² [Mar. 08, May. 95, 90]
(iv) √f
Answer:
(i) {(1, 4), (2, -6), (3, -2)}
(ii) {(1, 4), (2, -1), (3, 1)}
(iii) {(1, 4), (2, 9), (3, 1)}
(iv) {(1, √2)}

Telangana TSBIE TS Inter 1st Year Physics Study Material 6th Lesson Work, Energy and Power Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 6th Lesson Work, Energy and Power

Very Short Answer Type Questions

Question 1.
If a bomb at rest explodes into two pieces, the pieces must travel in opposite directions. Explain.
Answer:
Explosion is due to internal forces. In law of conservation of linear momentum internal forces cannot change the momentum of the system. So after explosion m₁v₁ + m₂v₂ = 0 or m₁v₁ = – m₂v₂ ⇒ they will fly in opposite directions.

Question 2.
State the conditions under which a force does no work.
Answer:

1. When force (F) and displacement (S) are mutually perpendicular then work done is zero.
   ∵ W = \(\overline{\mathrm{F}}.\overline{\mathrm{S}}\) = |F| |S| cos θ when θ = 90° work W = 0
2. Even though force is applied if displacement is zero then work done W = 0.

Question 3.
Define Work, Power and Energy. State their S.I. units.
Answer:
Work :
The product of force and displacement along the direction of force is called work.
Work done W = \(\overline{\mathrm{F}}.\overline{\mathrm{S}}\)
= \(|\overline{\mathrm{F}}||\overline{\mathrm{S}}|\) cos θ
S.I. unit of work is Joule.
Dimensional formula : ML²T^-2.

Power :
The rate of doing work is called power.

S.I. unit: Watt;
D.F. : ML²T^-3

Energy :
It is the capacity or ability of the body to do work. By spending energy we can do work or by doing work energy contentment of the body will increase.
S.I. unit: Joule ; D.F. : MML²T^-2

Question 4.
State the relation between the kinetic energy and momentum of a body.
Answer:
Kinetic energy K.E = \(\frac{1}{2}\)mv² ;
momentum \(\overline{\mathrm{p}}\) = mv
Relation between P and KE is
K.E = p²/2m. ⇒ P = \(\sqrt{K.E.2m}\)

Question 5.
State the sign of work done by a force in the following.
a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
b) Work done by gravitational force in the above case.
Answer:
a) When a bucket is lifted out of well work is done against gravity so work done is negative.

b) Work done by gravitational force is positive.

Question 6.
State the sign of work done by a force in the following.
a) work done by friction on a body sliding down an inclined plane.
b) work done gravitational force in the above case.
Answer:
a) Work done by friction while sliding down is negative. Because it opposes downward motion of the body.

b) Work done by gravitational force when a body is sliding down is positive.

Question 7.
State the sign of work done by a force in the following.
a) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.
b) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Answer:
a) Work done against the direction of motion of a body moving on a horizontal plane is negative.

b) In pendulum a∝ – y. So work done by air resistance to bring it to rest is considered as positive.

Question 8.
State if each of the following statements is true or false. Give reasons for your answer.
a) Total energy of a system is always conserved, no matter what internal and external forces on the body are present
b) The work done by earth’s gravitational force in keeping the moon in its orbit for its one revolution is zero.
Answer:
a) Law of conservation of energy states that energy can be neither created nor destroyed. This rule is applicable to internal forces and also for external forces when they are conservative forces.

b) Gravitational forces are conservative forces. Work done by conservative force around a closed path is zero.

Question 9.
Which physical quantity remains constant (i) in an elastic collision (ii) in an inelastic collision?
Answer:
In elastic collision :
P and K.E. are conserved, (remains constant)

In inelastic collision :
only momentum is conserved, (remains constant)

Question 10.
A body freely falling from a certain height ‘h’, after striking a smooth floor rebounds and h rises to a height h/2. What is the coefficient of restitution between the floor and the body?
Answer:
Given that, h₁ = h and h₂ = \(\frac{h}{2}\)
We know that coefficient of restitution.

Question 11.
What is the total displacement of freely falling body, after successive rebounds from the same place of ground, before it comes to stop? Assume that V is the coefficient of restitution between the body and the ground.
Answer:
Total displacement of a freely falling body after successive rebounds from the same place of ground, before it comes to stop is equal to height (h) from which the body is dropped.

Short Answer Questions

Question 1.
What is potential energy? Derive an expression for the gravitational potential energy.
Answer:
Potential energy :
It is the energy possessed by a body by the virtue of its position.
Ex: Energy stored in water a over head tank, wound spring.

Equation for potential energy :
Let a body of mass m is lifted through a height ‘h’ above the ground. Where ground is taken as refe-rence. In this process we are doing some work.

Work done against gravity W = m.g.h.
i. e., Force × displacement along the direction of force applied. This work done is stored in the body in the form of potential energy. Because work and energy can be interchanged.
∴ Potential Energy P.E. = mgh.

Question 2.
A lorry and a car moving with the same momentum are brought to rest by the application of brakes, which provide equal retarding forces. Which of them will come to rest in shorter time? Which will come to rest in less distance?
Answer:
Momentum (\(\overline{\mathrm{P}}\) = mv) is same for both lorry and car.
Work done to stop a body = Kinetic energy stored
∴ W = F. S = \(\frac{1}{2}\) mv² = K.E. But force applied by brakes is same for lorry and car.
Relation between \(\overline{\mathrm{P}}\) on K.E. is

So lighter body (car) will travel longer distance when P, F are same.
Then mS = constant.
∴ So car travels longer distance than lorry before it is stopped.

Question 3.
Distinguish between conservative and non-conservative forces with one example each.
Answer:
i) Conservative forces :
If work done by the force around a closed path is zero and it is independent of the path then such forces are called conservative forces.

Example:

1. Work done in lifting a body in gravitational field. When the body returns to its original position work done on it is zero.
   So gravitational forces are conservative forces.
2. Let a charge ‘q’ is moved in an electric field on a closed path then change in its electric potential i.e., static forces are conservative forces.

ii) Non-conservative forces :
For non-conservative forces work done by a force around a closed path is not equal to zero and it is dependent on the path.
Ex: Work done to move a body against friction. While taking a body between two points say A & B. We have to do work to move the body from A to B and also work is done to move the body from B to A. As result, the work done in moving the body in a closed path is not equals to zero. So frictional forces are non-conservative forces.

Question 4.
Show that in the case of one dimensional elastic collision, the relative velocity of approach of two colliding bodies before collision is equal to the relative velocity of separation after collision.
Answer:
To show relative velocity of approach of two colliding bodies before collision is equal to relative velocity of separation after collision.

Let two bodies of masses m₁, m₂ are moving with velocities u₁, u₂ along the straight line in same direction collided elastically.

Let their velocities after collision be v₁ and v₂.

According to the law of conservation of linear momentum
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ or m₁(u₁ – v₁) = m₂(v₂ – u₂) ………… (1)

According to law of conservation of kinetic energy
\(\frac{1}{2}\)m₁u²₁ + \(\frac{1}{2}\)m₂u²2 = \(\frac{1}{2}\)m₁v²₂ + \(\frac{1}{2}\)m₂v²₂
m₁(u²₁ – v²₁) = m₂(v²₂ – u²₂) ………. (2)
Dividing eqn. (2) by (1)

u₁ + v₁ = v₂ + u₂ ⇒ u₁ – u₂ = v₂ – v₁ ……. (3)
i.e., relative velocity of approach of the two bodies before collision = relative velocity of separation of the two bodies after collision. So coefficient of restitution is equal to ‘1’.

Question 5.
Show that two equal masses undergo oblique elastic collision will move at right angles after collision, if the second body initially at rest.
Answer:
Consider two bodies possess equal mass (m) and they undergo oblique elastic collision.

Let the first body moving with initial velocity ‘u’ collides with the second body at rest.

In elastic collision, momentum is conserved. So, conservation of momentum along X-axis yields.
mu = mv₁ cos θ₁ + mv₂ cos θ₂.
(i.e.) u = v₁ cos θ₁ + v₂ cos θ₂ ……. (1)
along Y-axis
0 = v₁ sin θ₁ – v₂ sin θ₂ ……… (2)
squaring and adding eq. (1) and (2) we get
u² = v²₁ + v²₂ + 2v₁v₂ cos (θ₁ + θ₂) …. (3)
As the collision is elastic,
Kinetic Energy (K.E.) is also conserved.

From eq. (3) and (4) 2v₁v₂ cos(θ₁ + θ₂) = 0
As it is given that v₁ ≠ 0 and v₂ ≠ 0
∴ cos(θ₁ + θ₂) = 0 or θ₁ + θ₂ = 90°.
The two equal masses undergoing oblique elastic collision will move at right angles after collision, if the second body initially at rest.

Question 6.
Derive an expression for the height attained by a freely falling body after ‘n’ number of rebounds from the floor.
Answer:
Let a small ball be dropped from a height ‘h’ on a horizontal smooth plate. Let it rebounds to a height ‘h₁‘.

Velocity with which it strikes the plate u₁ = \(\sqrt{2gh}\)
Velocity with which it leaves the plate v₁ = \(\sqrt{2gh_1}\)

The velocity of plate before and after collision is zero i.e., u₂ = 0, v₂ = 0
Coefficient of restitution,

For 2nd rebound it goes to a height
h₂ = e²h₁ = e²e²h = e⁴h

For 3rd rebound it goes to a height
h₃ = e²h₂ = e²e⁴h = e⁶h

For nth rebound height attained
h_n = e²ⁿh.

Question 7.
Explain the law of conservation of energy.
Answer:
Law of conservation of energy:
Wien forces doing work on a system are conservative then total energy of the system is constant i.e., energy can neither be created nor destroyed.
i.e., Total energy = (K + u) = constant form.

Explanation :
Consider a body undergoes small displacement ∆x under the action of conservative force F. According to work energy theorem.
Change in K.E = work done
∆K = F(x)∆x ………….. (1)
but Potential energy Au = -F(x)∆x ………….. (2)
from (1) and (2) = ∆K = – ∆u
⇒ ∆(K + u) = 0
Hence (K + u) = constant
i.e., sum of the kinetic energy and potential energy of the body is a constant

Since the universe may be considered as an isolated system, the total energy of the universe is constant.

Long Answer Questions

Question 1.
Develop the notions of work and kinetic energy and show that it leads to work- energy theorem. State the conditions under which a force does no work. [AP Mar. I 7, 15, May 1 7; TS Mar. 15]
Answer:
Work :
The product of component of force in the direction of displacement and the magnitude of displacement is called work.
W = \(\overline{\mathrm{F}}.\overline{\mathrm{S}}\)
When \(\overline{\mathrm{F}}\) and \(\overline{\mathrm{S}}\) are parallel W = \(|\overline{\mathrm{F}}|\times|\overline{\mathrm{S}}|\)
When \(\overline{\mathrm{F}}\) and \(\overline{\mathrm{S}}\) has some angle 6 between them
W = \(\overline{\mathrm{F}}.\overline{\mathrm{S}}\) cos θ

Kinetic energy :
Energy possessed by a moving body is called kinetic energy (k)

The kinetic energy of an object is a measure of the work that an object can do by the virtue of its motion.

Kinetic energy can be measured with equation K = \(\frac{1}{2}\)mv²
Ex : All moving bodies contain kinetic energy.

Work energy theorem (For variable force):
Work done by a variable force is always equal to the change in kinetic energy of the body.
Work done W = \(\frac{1}{2}\)mV² – \(\frac{1}{2}\)mV²₀? = K_f – K_i

Proof :
Kinetic energy of a body K = \(\frac{1}{2}\)mv²
Time rate of change of kinetic energy is

When force is conservative force F = F(x)
∴ On integration over initial position (x₁) and final position x₂

i.e., work done by a conservative force is equal to change in kinetic energy of the body.
Condition for Force not to do any work.

When Force (\(\overline{\mathrm{F}}\)) and displacement (\(\overline{\mathrm{S}}\)) are perpendicular work done is zero, i.e., when
θ = 90° then W = \(\overline{\mathrm{F}}.\overline{\mathrm{S}}\) = 0

Question 2.
What are collisions? Explain the possible types of collisions? Develop the theory of one dimensional elastic collision. [TS Mar.’ 18; AF Mar. 19. May 14]
Answer:
A process in which the motion of a system of particles changes but keeping the total momentum conserved is called collision.

Collisions are two types :

1. elastic
2. inelastic.

To show relative velocity of approach before collision is equal to relative velocity of separation after collision.

Let two bodies of masses m₁, m₂ are moving with velocities u₁, u₂ along the same line in same direction collided elastically.

Let their velocities after collision are v₁ and v₂.

According to the law of conservation of linear momentum
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
or m₁ ( u₁ – v₁ ) = m₂ ( v₂ – u₂ ) ……… (1)

According to law of conservation of kinetic energy

i.e., In elastic collisions relative velocity of approach of the two bodies before collision = relative velocity of separation of the two bodies after collision.

Velocities of two bodies after elastic collision:

Question 3.
State and prove law of conservation of energy in case of a freely falling body. [TS Mar. ’19, ’17, ’16, May ’18, ’17, ’16, June ’15; AP Mar. ’18, ’16, ’15, May ’18, ’16, June ’15, May ’13]
Answer:
Law of conservation of energy :
Energy can neither be created nor destroyed. But it can be converted from one form into the another form so that the total energy will remains constant in a closed system.

Proof : In case of a freely fidling body :
Let a body of mass is dropped from a height H’ at point A.

Forces due to gravitational field are conservative forces, so total mechanical energy (E = P.E + K.E.) is constant i.e., neither destroyed nor created.

The conversion of potential energy to kinetic energy for a ball of mass ra dropped from a height H

1. At point H : Velocity of body v = 0
⇒ K = 0
Potential energy (u) = mgH
where H=height above the ground
T.E = u + K = mgH (1)

2. At point 0 :
i.e., just before touching the ground :
A constant force is a special case of specially dependent force F(x) so mechanical energy is conserved.
So energy at H = Energy at 0 = mgH

Proof:
At point ‘0’ height h = 0 ⇒
⇒ v = \(\sqrt{2gH}\) ; u = 0
K₀ = \(\frac{1}{2}\) mv² = \(\frac{1}{2}\) m2gH = mgH
Total energy E = mgH + 0 = mgH ………….. (2)

3. At any point h:
Let height above ground = h
u = mgh, K_h = \(\frac{1}{2}\)mV²
where v = \(\sqrt{2g(h – x)}\)
∴ Velocity of the body when it falls through a height (h – x) is \(\sqrt{2g(h – x)}\)
∴ Total energy =mgh + \(\frac{1}{2}\)m2g(H – h)
⇒ E = mgh + mgH – mgh = mgH ………… (3)
From eq. 1, 2 & 3 total energy at any point is constant.
Hence, law of conservation of energy is proved.

Conditions to apply law of conservation of energy:

1. Work done by internal forces is conservative.
2. No work is done by external force.

When the above two conditions are satisfied then total mechanical energy of a system will remain constant.

Problems

Question 1.
A test tube of mass 10 grams closed with a cork of mass 1 gram contains some ether. When the test tube is heated the cork flies out under the presssure of the ether gas. The test tube is suspended horizontally by a weight less rigid bar of length 5 cm. What is the minimum velocity with which the cork should fly out of the tube, so that test tube describing a full vertical circle about the point O. Neglect the mass of ether.
Solution:
Length of bar, L = 5 cm, = \(\frac{5}{100}\), g = 10m/s²
For the cork not to come out minimum velocity at lowest point is, v = \(\sqrt{5gL}\). At this condition centrifugal and centripetal forces are balanced.

Question 2.
A machine gun fires 360 bullets per minute and each bullet travels with a velocity of 600 ms^-1. If the mass of each bullet is 5 gm, find the power of the machine gun? [AP May ’16, ’13, June ’15, Mar. ’14; AP Mar. ’18. ’16; TS May ’18]
Solution:
Number of bullets, n = 360
Time, t = 1 minute = 60s
Velocty of the bullet, v = 600 ms^-1 ; Mass of each bullet, m = 5gm = 5 × 10^-3 kg

⇒ P = 5400W = 5.4KW

Question 3.
Find the useful power used in pumping 3425 m³ of water per hour from a well 8 m deep to the surface, supposing 40% of the horse power during pumping is wasted. What is the horse power of the engine?
Solution:
Mass of water pumped, m = 3425 m³
= 3425 × 10³ kg.
Mass of lm³ water = 1000 kg
Depth of well d = 8 m., Power wasted = 40%
∴ efficiency, η = 60%
time, t = 1 hour = 3600 sec.

Question 4.
A pump is required to lift 600 kg of water per minute from a well 25m deep and to eject it with a speed of 50 ms^-1. Calculate the power required to perform the above task? (g = 10 m sec^-2) [TS Mar. ’19, ’16; AP May 18, Mar. 15, June 15]
Solution:
Mass of water m = 600 kg; depth = h = 25 m
Speed of water v = 25 m/s; g = 10 m/s², time t = 1 min = 60 sec.
Power of motor P = Power to lift water (P₁) + Kinetic energy of water (K.E) per second.
Power to lift water

∴ Power of motor P = 2500 + 3125 = 5625 watt 5.625 K.W.

Question 5.
A block of mass 5 kg initially at rest at the origin is acted on by a force along the X-positive direction represented by F=(20 + 5x)N. Calculate the work done by the force during the displacement of the block from x = 0 to x = 4m.
Solution:
Mass of block, m = 5 kg
Force acting on the block, F = (20 + 5x) N
If ‘w’ is the total amount of work done to displace the block from x = 0 to x = 4m then,

Question 6.
A block of mass 5 kg is sliding down a smooth inclined plane as shown. The spring arranged near the bottom of the inclined plane has a force constant 600 N/m. Find the compression in the spring at the moment the velocity of the block is maximum?

Solution:
Mass of the block, m = 5kg
Force constant, K = 600 N m^-1
From figure, sin θ = \(\frac{3}{5}\)
Force produced by the motion in the block,
F = mg sin0 ⇒ F = 5 × 9.8 × \(\frac{3}{5}\) = 29.4 N
But force constant K = \(\frac{F}{x}\) x
∴ x = \(\frac{F}{K}=\frac{29.5}{600}\) = 0.05m = 5cm

Question 7.
A force F = – \(\frac{K}{x^2}\) (x ≠ 0) acts on a particle along the X-axis. Find the work done by the force in displacing the particle from x = + a to x = + 2a. Take K as a positive constant.
Solution:
Force acting on the particle, F = –\(\frac{K}{x^2}\)
Total amount of work done to displace the particle from x = + a to x = + 2a is,

Question 8.
A force F acting on a particle varies with the position x as shown in the graph. Find the work done by the force in displacing the particle from x = – a to x = + 2a?

Solution:
Average force acting on the particle, F = \(\frac{F}{K}\)

Amount of work done by the force to displace the particle from x = -a to x = +2a is,

Question 9.
From a height of 20 m above a horizontal floor, a ball is thrown down with initial velocity 20 m/s. After striking the floor, the ball bounces to the same height from which it was thrown. Find the coefficient of restitution for the collision between the ball and the floor? (g = 10 m/s²)
Solution:
Initial velocity = u¹ = 20 m/s, h – 20 m,
g = 10 m/s²

Velocity of approach,
u² = u^1² = u^1² +2as = 400+ 2 × 10 × 20
⇒ u² = 400 + 400 = 800 ⇒ u = 20√2
Height of rebounce = h = 20 m.
∴ Velocity of separation

Question 10.
A ball falls from a height of 10 m on to a hard horizontal floor and repeatedly bounces. If the coefficient of restitution is \(\frac{1}{\sqrt{2}\), then what is the total distance travelled by the ball before it ceases to rebound?
Solution:
Height from which the ball is allowed to fall, h = 10 m
Coefficient of restitution between the hard horizontal floor and the ball, e = \(\frac{1}{\sqrt{2}\)
∴ Total distance travelled by the ball before it ceases to rebound,

Question 11.
In a ballistics demonstration, a police officer fires a bullet of mass 50g with speed 200 msr1 on soft plywood of thickness 2 cm. The bullet emerges with only 10% of its initial kinetic energy. What is the emergent speed of the bullet?
Answer:
Mass of bullet m = 50g = 0.05 kg
Initial velocity V₀ = 200 m/s

Question 12.
Find the total energy of a body of 5 kg mass, which is at a height of 10 in from the earth and foiling downwards straightly with a velocity of 20 m/s. (Take the acceleration due to gravity as 10 m/s²) [TS May ’16]
Answer:
Mass m = 5 kg; Height h = 10 m ; g = 10 m/s²
Velocity v = 20 m/s.

Telangana TSBIE TS Inter 1st Year Physics Study Material 5th Lesson Laws of Motion Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 5th Lesson Laws of Motion

Very Short Answer Type Questions

Question 1.
Why are spokes provided in a bicycle wheel? [AP May ’14. ’13]
Answer:
The spokes of cycle wheel increase its moment of inertia. The greater the moment of inertia, the greater is the opposition to any change in uniform rotational motion. As a result the cycle runs smoother and steadier. If the cycle wheel had no spokes, the cycle would be driven with jerks and hence unsafe.

Question 2.
What is inertia? What gives the measure of inertia? [TS ‘Mar. 17; AP Mar. 19, 14]
Answer:
The inability of a body to change its state by itself is known as inertia.

Mass of a body is a measure for its Inertia.
Types of inertia

1. Inertia of rest
2. Inertia of motion
3. Inertia of direction.

Question 3.
According to Newton’s third law, every force is accompanied by an equal and opposite force. How can a movement ever take place? [AP May 17, June 15]
Answer:
From Newton’s third law action = – reaction. But action and reaction are not working on the same system. So they will not cancel each other. Hence, motion is possible.

Question 4.
When a bullet is fired from a gun, the gun gives a kick in the backward direction. Explain. [AP Mar. ’15]
Answer:
Firing of a gun is due to internal forces. Internal forces do not change the momentum of the system. Before firing m₁u₁ + m₂u₂ = 0. Since system is at rest after firing m₁v₁ + m₂v₂ = 0 (or) m₁v₁ = – m₂v₂. So gun and bullet will move in opposite directions to satisfy law of conservation of linear momentum.

Question 5.
Why does a heavy rifle not recoil as strongly as a light rifle using the same cartridges?
Answer:
Velocity (or) recoil v = \(\frac{mv}{M}\) i.e., ratio of momentum of bullet to mass of gun. If mass of gun is high then velocity of recoil is less with same cartridge.

Question 6.
If a bomb at rest explodes into two pieces, the pieces must travel in opposite directions. Explain. [TS Mar. 16, 15, June 15]
Answer:
Explosion is due to internal forces. From law of conservation of linear momentum, internal forces cannot change the momentum of the system. So after explosion m₁v₁ + m₂v₂ = 0 (or) m₁v₁ = – m₂v₂. According to law of conservation of linear momentum they will fly in opposite directions.

Question 7.
Define force. What are the basic forces in nature?
Answer:
Force is that which changes (or) tries to change the state of a body.

The basic forces in nature are :

1. Gravitational forces,
2. Electromagnetic forces,
3. Nuclear forces.

Question 8.
Can the coefficient of friction be greater than one?
Answer:
Yes. Generally coefficient of friction between the surfaces is always less than one. But under some special conditions like on extreme rough surfaces coefficient of friction may be greater than one.

Question 9.
Why does the car with a flattened tyre stop sooner than the one with inflated tyres?
Answer:
Due to flattening of tyres, frictional force increases. Because rolling frictional force between the surfaces is proportional to area of contact. Area of contact increases for flattened tyres. So rolling frictional force increases and the car will be stopped quickly.

Question 10.
A horse has to pull harder during the start of the motion than later. Explain. [AP Mar. 18, May 16, Mar. 13]
Answer:
To start motion in a body we must apply force to overcome static friction (F_s = µ_smg). When once motion is started between the bodies then kinetic frictional force comes into act. Kinetic friction (F_k = µ_kmg) is always less than static friction. So it is tough to start a body from rest than to keep it in motion.

Question 11.
WHat happens to the coefficient of friction if the weight of the body is doubled? [TS Mar. 19; AP Mar. 16, May 14]
Answer:
When weight of the body is doubled still then there is no change in coefficient of friction. Because frictional force cc normal reaction. So when weight of a body is doubled then frictional force and normal reaction will also becomes doubled and coefficient of friction remains constant.

Short Answer Questions

Question 1.
A stone of mass 0.1 kg is thrown vertically upwards. Give the magnitude and direction of the net force on the stone (a) during its upward motion, (b) during its downward motion, (c) at the highest point, where it momentarily comes to rest.
Answer:
Mass of stone, m = 0.1 kg.
a) During upward motion force acts downwards due to acceleration due to gravity.
Magnitude of force F = mg = 0.1 × 9.8
= 0.98 N (↓)
b) During downward motion force acts downward. Magnitude of force F = mg
= 0.1 × 9.8 = 0.98N (↓)

c) At highest point velocity v = 0. But still g will act on it only in downward motion so resultant force F = 0.98 N. downward.
Note : In the entire journey of the body force due to gravitational pull acts only in downward direction.

d) If the body is thrown with an angle of 30° with horizontal then vertical component of gravitational force does not change, hence in this case downward force F = mg = 0.1 × 9.8
= 0. 98 newton.

Question 2.
Define the terms momentum and impulse. State and explain the law of conservation of linear momentum. Give examples. [TS May 18, June 15]
Answer:
Momentum (\(\overline{\mathrm{P}}\)) : It is the product of mass and velocity of a body.

Momentum (\(\overline{\mathrm{P}}\)) = mass (m) × velocity (v)
∴ (\(\overline{\mathrm{P}}\)) = m\(\overline{\mathrm{v}}\)

Impulse (J) :
When a large force (F) acts on a body for small time (t) then the product of force and time is called Impulse.
Impulse (J) = Force (F) × time (t)
∴ Impulse (J) = F × t

Law of conservation of linear momentum:
There is no external force act on the system. The total linear momentum of the isolated system remains constant.

Proof :
Let two bodies of masses say A and B are moving with initial momenta PA and PB collided with each other. During collision they are in contact for a small time say ∆t. During this time of contact they will exchange their momenta. Let final momenta of the bodies are P¹_A and P¹_B. Let force applied by A on B is F_AB and force applied by B on A is F_BA.

From Newton’s 3rd Law F_AB = F_BA or F_AB ∆t = F_BA ∆t

From 2nd Law F_AB∆t = P¹_A – P_A change in momentum of A.
F_BA ∆t = P¹_B – P_B change in momentum of B.
∴ P¹_B – P_A = P¹_B – P_B or P_A + P_B = P¹_B + P¹_B
i. e., sum of momentum before collision is equals to sum of momentum after collision.

Question 3.
Why are shock absorbers used in motor cycles and cars? [AP June ’15]
Answer:
When vehicles are passing over the vertocies and depressions of a rough road they will collides with them for a very short period. This causes impulse effect. Due to large mass and high speed of the vehicles the magnitude of impulse is also high. Impulse may cause damage to the car or even to the passengers in it.

The bad effects of impulse is less if time of contact is more. Impulse J = F.t. For the same magnitude of impulse (change in momentum) if time of contact is high force acting on the vehicle is less. Shock absorbers will absorb the impulse and releases the same force slowly. This is due to large time constant of the springs.

So shock absorbers are used in vehicles to reduce impulse effects.

Question 4.
Explain the terms limiting friction, dynamic friction and rolling friction.
Answer:
Limiting friction :
Frictional forces always oppose relative motion between the bodies. These forces are self adjusting forces. Their magnitude will increase upto some extent with the value of applied force.

The maximum frictional force between the bodies at rest is called “limiting friction”.

Dynamic (or) kinetic friction :
When applied force is equal to or greater than limiting friction then the body will move. When once motion is started then frictional , force will abruptly falls to a minimum value.

Frictional force between moving bodies is called dynamic (or) kinetic friction. Kinetic friction is always less than limiting friction.

Rolling friction :
The resistance encountered by a rolling body on a surface is called rolling friction.

Question 5.
Explain the advantages and disadvantages of friction. [TS Mar. ’17, ’15; AF Mar. ’15]
Answer:
Advantages of friction :

1. We are able to walk because of friction.
2. It is impossible for a car to move on a slippery road.
3. Breaking system of vehicles works with the help of friction.
4. Friction between roads and tyres provides the necessary external force to accelerate the car.
5. Transmission of power to various parts of a machine through belts is possible by friction.

Disadvantages of friction:

1. In many cases we will try to reduce friction because it dissipates energy into heat.
2. It causes wear and tear to machine parts which causes frequent replacement of machine parts.

Question 6.
Explain Friction. Mention the methods used to decrease friction. [TS May, ’17, ’16; Mar. ’19, ’16; AP Mar. ’18. ’14; May ’18, ’14)
Answer:
Friction :
It is a contact force parallel to the surfaces in contact Friction will always oppose relative motion between the bodies.

Friction is a necessary evil. Friction is a must at some places and it must be reduced at some places.

Methods to reduce friction :
1) Polishing :
Friction causes due to surface irregularities. So by polishing friction can be reduced to some extent.

2) Lubricants :
By using lubricants friction can be reduced. Lubricants will spread as an ultra thin layer between the surfaces in contact and in friction decreases.

3) A thin cushion of air maintained between solid surfaces reduces friction.
Ex : Air pressure in tyres.

4) Ball bearings :
Ball bearings are used to reduce friction between machine parts.

Ball bearings will convert sliding motion into rolling motion. As a result friction is reduced.

Question 7.
State the laws of rolling friction.
Answer:
When a body is rolling over the other, then friction between the bodies is known as rolling friction.

Rolling friction coefficient,

Laws of rolling friction :

1. Rolling friction will develop a point contact between the surface and the rolling sphere. For objects like wheels line of contact will develop.
2. Rolling friction(f_r) has least value for given normal reaction when compared with static friction (f_s) or kinetic friction (f_k)
3. Rolling friction is directly proportional
   to normal reaction, f_r ∝ N.
4. In rolling friction the surfaces in contact will get momentarily deformed a little.
5. Rolling friction depends on area of contact. Due to this reason friction increases when air pressure is less in tyres (Flattened tyres).
6. Rolling friction is inversely proportional to radius of rolling body µ_r ∝ \(\frac{1}{r}\)

Question 8.
Why is pulling the lawn roller preferred to pushing it?
Answer:
Let a lawn roller is pulled by means of a force F with some angle θ to the horizontal. By resolving the force into two components.

1. Horizontal component F cos θ is useful to pull the body.
2. The vertical component F sin θ opposes the weight

So N.R. = mg – F sin θ
But frictional force = µ. N.R.
∴ Frictional force [µ(mg – F sin θ)] decreases.

So it is easier to pull the body.
When the lawn roller is pushed by a force, the vertical component F sin θ causes the apparent increase of weight of the object. So the normal reaction N.R. = mg + F sin θ.
∴ Frictional force [µ(mg + F sin θ)] increases and it will be difficult to pull the body.

Long Answer Questions

Question  1.
a) State Newton’s second law of motion. Hence, derive the equation of motion F = ma from it. [AP Mar. ’19, ’17, ’16; AP May ’17. ’16; May ’13]
b) A body is moving along a circular path such that its speed always remains constant. Should there be a force acting on the body?
Answer:
a) Newton’s 2nd law :
The rate of change of momentum of a body is proportional to external force and acts along the direction of force applied.
i.e., \(\frac{dp}{dt}\) ∝ F

Derivation of equation F = ma:
According Newton’s 2nd law.
We know


Here, k = constant.
The proportional constant is made equal to one, by properly selecting the unit of force.
∴ F = ma

b) Force on a body moving in a circular path :
Let a body of mass’m’ is moving in a circular path of radius V with constant speed. The velocity of the body is given by the tangent drawn at that point. Since velocity is changing continuously the body will have acceleration.

So the body will experience some acceleration. This is called normal acceleration (or) centripetal acceleration.

Question 2.
Define Angle of friction and Angle of repose. Show that angle of friction is equal to angle of repose for a rough inclined plane.
A block of mass 4 kg is resting on a rough horizontal plane and is about to move when a horizontal force of 30 N is applied on it. If g = 10 m/s². Find the total contact force exerted by the plane on the block.
Answer:
Angle of friction :
The angle made by the resultant of the Normal reaction and the limiting friction with Normal reaction is called angle of friction (Φ).

Angle of reppse :
Let a body of mass m is placed on a rough inclined plane. Let the angle with the horizontal ‘θ’ is gradually increased then fora particular angle of inclination (say α) the body will just slide down without acceleration. This angle θ = α is called angle of repose. At this stage the forces acting on the body are in equilibrium.

Equation for angle of repose :
Force acting on the body in vertically downward direction = W = mg.
By resolving this force into two components.

1. Force acting along the inclined plane in downward direction = mg sin θ.
   This component is responsible for downward motion.
2. The component mg cos θ . which is balanced by the normal reaction.

If the body slides down without acceleration resultant force on the body is zero, then
mg sin θ = Frictional force (f_k)
mg cos θ = Normal reaction (N.R.)
But coefficient friction

Hence θ = α is called angle of repose.
∴ µ_k = tan α

Hence tangent of angle of repose (tan θ) is equal to coefficient of friction (f_k) between the bodies.

b) When the block rests on the horizontal surface, it is in equilibrium under the action of four forces. They are
i) Normal reaction (N)
ii) Weight of the block (mg)
iii) Horizontal force (30 N)
iv) Limiting frictional force (f_L)

If the applied horizontal force is equal to the limiting frictional force, then only the block will be ready to move on the rough horizontal surface, i.e., f_L = horizontal force applied.
∴ Total contact force = 30 N.

Problems

Question 1.
The linear momentum of a particle as a function of time ‘t’ is given by, p = a + bt, where a and b are positive constants. What is the force acting on the particle?
Solution:
Linear momentum of a particle, p = a + bt
We know that force acting on a particle is equal to rate of change of linear momentum.

Question 2.
Calculate the time needed for a net force of 5 N to change the velocity of a 10 kg mass by 2 m/s. [TS May ’16]
Solution:
Force, F = 5N
Change in velocity, v – u = 2ms ^-1
Mass, m= 10 kg
From Newton’s second law of motion,

Question 3.
A ball of mass ‘m’ is thrown vertically upward from the ground and reaches a height ‘h’ before momentarily coming to rest, If ‘g’ is acceleration due to gravity. What is the impulse received by the ball due to gravity force during its flight?
Solution:
Impulse, J = force × time

Question 4.
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s^-1 to 3.5 m s^-1 in 25 s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force?
Solution:
Mass of the body, m = 3.0 kg ;
Initial velocity of the body, u = 2.0 ms^-1
Final velocity of the body, v = 3.5 ms^-1
Time, t = 25 s
From Newton s second law of motion,

∴ Magnitude of force acting on the body, F = 0.18 N. The direction of force acting on the body is along the direction of motion of the body because force is positive.

Question 5.
A man in a lift feels an apparent weight ‘W’ when the lift is moving up with a uniform acceleration of 1/3^rd of the acceleration due to gravity. If the same man was in the same lift now moving down with a uniform acceleration that is 1/2 of the acceleration due to gravity, then what is his apparent weight?
Solution:
Case (i) :
When lift is moving upwards :
Apparent weight of the man = W
Acceleration, a = g/3
Apparent weight of the man when the lift is moving upwards is,

Case (ii) : When lift is moving downwards:
Let W’ be the apparent weight of the man Acceleration, a = g/2
Apparent weight of the man when the lift is moving downwards is,
W’ = m(g – a) = m (g – g/2)

Question 6.
A container of mass 200 kg rests on the back of an open truck. If the truck accelerates at 1.5 m/s², what is the minimum coefficient of static friction between the container and the bed of the truck required to prevent the container from sliding off the back of the truck?
Solution:
Mass of the container, m = 200 kg
Acceleration of truck, a = 1.5 ms^-2
Coefficient of static friction, µ_s = \(\frac{a}{g}\)
\(\frac{1.5}{9.8}\) = 0.153

Question 7.
A bomb initially at rest at a height of 40 m above the ground suddenly explodes into two identical fragments. One of them starts moving vertically downwards with an initial speed of 10m/s. If acceleration due to gravity is 10m/s², What is the separation between the fragments 2s after the explosion?
Solution:
Case (i): (for downward moving fragment)
Initial velocity, u = 10 ms^-1
Acceleration, a = +g = 10 ms^-2
Time, t = 2s
From the equation of motion, s = ut + \(\frac{1}{2}\) at²
the distance moved in downward direction is,
s₁ = 10 × 2 + \(\frac{1}{2}\) × 10 × (2)² = 40 m

Case (ii) (for upward moving fragment)
Given that two fragments are identical hence, after explosion the fragments move in opposite directions. Here the first fragment moves in downward direction, hence, second fragment moves upward direction.
Again from s = ut = \(\frac{1}{2}\)at² we can write,
s₂ = – 10 × 2 + \(\frac{1}{2}\) × 10 × 4 = -20 + 20 = 0m
∴ Separation between the fragments 2s after the explosion = S₁ ~ S₂ = 40 – 0 = 40m

Question 8.
A fixed pulley with a smooth grove has a light string passing over it with a 4 kg attached on one side and a 3 kg on the other side. Another 3 kg is hung from the other 3 kg as shown with another light string. If the system is released from rest, find the common acceleration? (g = 10 m/s²)

Solution:
Here, m₁ = 3 + 3 = 6 kg; m₂ = 4 kg ;
g = 10 ms^-2

Acceleration of the system,

Question 9.
A block of mass of 2 kg slides on an inclined plane that makes an angle of 30° with the horizontal. The coefficient of friction between the block and the surface is √3/2.
a) What force should be applied to the block so that it moves down without any acceleration?
b) What force should be applied to the block so that it moves up without any acceleration?
Solution:
Mass of the block, m = 2kg
Angle of inclination, θ = 30°
Coefficient of friction between the block and the surface, µ = \(\frac{\sqrt{3}}{2}\)

a) The required force to move the block down without acceleration is,

b) The required force to move the block up without any acceleration is,

Question 10.
A block is placed on a ramp of parabolic shape given by the equation y = x²/20, sec Figure.

If µ_s = 0.5, what is the maximum height above the ground at which the block can be placed without slipping?
(tan θ = µ_s = \(\frac{dy}{dx}\))
Solution:
For the body not to drop
mg cos θ = µ mg sin θ
⇒ tan θ = µ given µ = 0.5 dy
But tan θ = \(\frac{dy}{dx}\) slope of parabolic region
⇒ \(\frac{dy}{dx}\) = µ = 0.5 …………… (1)

Question 11.
A block of metal of mass 2 kg on a horizontal table is attached to a mass of 0.45 kg by a light string passing over a frictionless pulley at the edge of the table. The block is subjected to a horizontal force by allowing the 0.45 kg mass to fall. The coefficient of sliding friction between the block and table is 0.2.
Calculate (a) the initial acceleration, (b) the tension in the string, (c) the distance the block would continue to move if, after 2 s of motion, the string should break.

Solution:
Mass of first block, m₁ = 0.45kg
Mass of second block, m₂ = 2 kg
coefficient of slidding friction between the block and table, µ = 0.2

b) Tension in the string

c) Velocity of string after 2 sec = u in this case; u’ = 0
∴ u = u’ + at = 0 + 0.2 × 2 = 0.4 m/s

Question 12.
On a smooth horizontal surface, a block A of mass 10 kg is kept. On this block, a second block B of mass 5 kg is kept. The coefficient of friction between the two blocks is 0.4. A horizontal force of 30 N is applied on the lower block as shown. The force of friction between the blocks is (take g = 10 m/s²)

Solution:
Mass of block ‘A’ is, m_A = 10 kg
Mass of block ‘B’ is, m_B = 5 kg
Applied horizontal force, F = 30 N
Coefficient of friction between two blocks, µ = 0.4

Frictional force of block ‘B’ is f = µmg
⇒ f = 0.4 × 5 × 10 = 20N
∴ The frictional force acting between the two blocks, = F – f = 30 – 20 = 10 N

Question 13.
A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 ms^-1. If the mass of the ball is 0.15 kg., determine the impulse imparted to the ball. (Assume linear motion of the ball). [AP Mar. ’17]
Solution:
Impulse = change in momentum
= (0.15 × 12) – (- 0.15 × 12) = 3.6 NS
in the direction from the batsman to the bowler.

Question 14.
A force \(2\overline{\mathrm{i}}+\overline{\mathrm{j}}-\overline{\mathrm{k}}\) Newton acts on a body which is initially at rest. At the end of 20 seconds the velocity of the body is \(4\overline{\mathrm{i}}+2\overline{\mathrm{j}}-2\overline{\mathrm{k}}\) m/s. What is the mass of the body? [AP May ’16]
Answer:
Force F = \(2\overline{\mathrm{i}}+\overline{\mathrm{j}}-\overline{\mathrm{k}}\), time t = 20 sec.
Initial velocity u₀ = 0.
Final velocity U = \(4\overline{\mathrm{i}}+2\overline{\mathrm{j}}-2\overline{\mathrm{k}}\)

Additional Problems

Question 1.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms-1. How long does the body take to stop?
Solution:
Here, F = – 50N, m = 20 kg
u = 15 ms^-1, v = 0, t = ?

Question 2.
A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.
Solution:

This is the direction of resultant force and hence the direction of acceleration of the body as shown in figure.

Question 3.
The driver of a three-wheeler moving with a speed of 36 km / h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle ? The mass of the three wheeler is 400 kg and the mass of the driver is 65 kg.
Solution:
Here, u = 36 km/h = 10 m/s, v = 0, t = 4s
m = 400 + 65 = 465 kg

Question 4.
A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 ms^-2. Calculate the initial thrust (force) of the blast.
Solution:
Here, m = 20000 kg = 2 × 10⁴ kg
Initial acceleration, a = 5 ms^-2;
Thrust, F = ?

Clearly, the thrust should be such that it overcomes the force of gravity besides giving it an upward acceleration of 5 ms^-2.

Thus the force should produce a net acceleration of 9.8 + 5.0 = 14.8 ms^-2.
As thrust = force = mass × acceleration
∴ F = 2 × 10⁴ × 14.8 = 2.96 × 10⁵N

Question 5.
A man of mass 70 kg stands on a weighing scale in a lift which is moving
a) upwards with a uniform speed of 10 ms^-1,
b) downwards with a uniform acceleration of 5 ms^-2,
c) upwards with a uniform acceleration of 5 ms^-2,
What would be the readings on the scale in each case?
d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
Solution:
Here, m = 70 kg, g = 10 m/s²
The weighing machine in each case measures the reaction R i.e., the apparent weight.
a) When the lift moves upwards with a uniform speed, its acceleration is zero.
R = mg = 70 × 10 = 700 N

b) When the lift moves downwards with a = 5 ms^-2
R = m(g – a) = 70 (10 – 5) = 350 N

c) When the lift moves upwards with a = 5 ms^-2
R = m (g + a) = 70 (10 + 5) = 1050 N

d) If the lift were to come down freely under gravity, downward acceleration. a = g
∴ R = m (g – a) = m (g – g) = Zero.

Question 6.
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string, a horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?
Solution:
Here, F = 600 N m₁= 10 kg, m2 = 20 kg

Let T be the tension in the string and a be the acceleration of the system, in the direction of force applied.
∴ a = \(\frac{F}{m_1+m_2}=frac{600}{10+20}\) = 20 m/s²

i) When force is applied on lighter block A, Fig (i).
T = m₂ a = 20 × 20 N’= 400 N

ii) When force is applied on heavier block B, Fig (ii).
T = m₁a = 10 × 20 NT = 200 N
Which is different from value of T in case (i). Hence our answer depends on which mass end, the force is applied.

Question 7.
Two masses 8 kg and 12 kg are connected at the two ends of a light in extensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

Solution:
Here, m₂ = 8kg, ; m₁ = 12 kg

Question 8.
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
Solution:
Let m₁, m₂ be the masses of products and \(\overrightarrow{\mathrm{v_1}},\overrightarrow{\mathrm{v_2}}\) be their respective velocities. Therfore, total linear momentum after disintegration = \(m_1\overrightarrow{\mathrm{v_1}}+m_2\overrightarrow{\mathrm{v_2}}\). Before disintegra-tion, the nucleus is at rest. Therefore, its linear momentum before disintegration is zero.

According to the principle of conservation of linear momentum,

Question 9.
Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 ms^-1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
Solution:
Here, initial momentum of the ball
A = 0.05 (6) = 0.3 kg ms^-1

As the speed is reversed on collision, final momentum of the ball A = 0.05 (-6)
= – 0.3 kg ms^-1

Impulse imparted to ball A = change in momentum of ball A = final momentum – initial momentum = – 0.3 – 0.3 = – 0.6 kg ms^-1.

Question 10.
A shell of mass 0.02 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms^-1, what is the recoil speed of the gun?
Solution:
Here, mass of shell, m = 0.02 kg
mass of gun, M = 100 kg
muzzle speed of shell, V = 80 ms^-1
recoil speed of gun, v = ?
According to the principle of conservation of linear momentum, mV + Mυ = 0

Question 11.
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
Solution:
Here, m = 0.25 kg, r = 1.5 m ;

Question 12.
Explain why
a) a horse cannot pull a cart and run in empty space,
b) passengers are thrown forward from their seats when a speeding bus stops suddenly,
c) it is easier to pull a lawn mover than to push it,
d) a cricketer moves his hands backwards while holding a catch.
Solution:
a) While trying to pull a cart, a horse pushes the ground backwards with a certain force at an angle. The ground offers an equal reaction in the opposite direction, on the feet of the horse. The forward component of this reaction is responsible for motion of the cart. In empty space, there is no reaction and hence, a horse cannot pull the cart and run.

b) This is due to “inertia of motion”.
When the speeding bus stops suddenly, lower part of the bodies in contact with the seats stop. The upper part of the bodies of the passengers tend to maintain the uniform motion. Hence, the passengers are thrown forward.

c) While pulling a lawn mover, force is applied upwards along the handle. The vertical component of this force is upwards and reduces the effective weight of the mover, Fig (a). While pushing a lawn mover, force is applied downwards along the handle. The vertical component of this force is downwards and increases the effective weight of the mover, Fig (b). As the effective weight is lesser in case of pulling than in case of pushing, therefore, “pulling is easier than pushing”.

d) While holding a catch, the impulse received by the hands, F × t = change in linear momentum of the ball is constant. By moving his hands backwards, the cricketer increases the time of impact (t) to complete the catch. As t increases, F decreases and as a reaction, his hands are not hurt severely.

Question 13.
A stream of water flowing horizontally with a speed of 15 ms^-1 pushes out of a tube of cross-sectional area 10^-2 m², and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?
Solution:
Here, v = 15 ms^-1
Area of cross section, a = 10^-2 m²
Volume of water pushing out/sec = a × v
= 10^-2 × 15 m³ s^-1

As density of water is 10³ kg/m³, therefore, mass of water striking the wall per sec.
m = (15 × 10^-2) × 10³ = 150 kg/s.

Question 14.
Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m.
Give the magnitude and direction of
a) the force on the 7^th coin (counted from the bottom) due to all the coins on its top,
b) the force on the 7^th coin by the eighth coin,
c) the reaction of the 6^th coin on the 7^th coin.
Solution:
a) The force on 7^th coin is due to weight of the three coins lying above it. Therefore,
F = (3 m) kgf = (3 mg) N
where g is acceleration due to gravity. This force acts vertically downwards.

b) The eighth coin is already under the weight of two coins above it and it has its own weight too. Hence force on 7^th coin due to 8^th coin is sum of the two forces i.e.
F = 2m + m = (3m) kgf = (3 mg) N
The force acts vertically downwards.

c) The sixth coin is under the weight of four coins above it.
Reaction, r = -F = -4m (kgf) = – (4 mg) N

Minus sign indicates that the reaction acts vertically upwards, opposite to the weight.

Question 15.
An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?
Solution:
Here θ = 15°

Question 16.
A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 10⁶ kg. What provides the centripetal force required for this purpose – The engine or the rails? What is the angle of banking required to prevent wearing out of the rail?
Solution:
The centripetal force is provided by the lateral thrust exerted by the rails on the wheels. By Newton’s 3rd law, the train exerts an equal and opposite thrust on the rails causing its wear and tear.

Obviously, the outer rail will wear out faster due to the larger force exerted by the train on it.

Question 17.
A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding?

Solution:
Here, mass of block, m = 25 kg
Mass of man, M = 50 kg
Force applied to lift the block
F = mg = 25 × 9.8 = 245 N
Weight of man W = Mg = 50 × 9.8 = 490 N.

a) When block is raised by man as shown in Fig. (a), force is applied by the man in the upward direction. This increases the apparent weight of the man. Hence action on the floor.
W’ = W + F = 490 + 245 = 735 N

b) When block is raised by man as shown in Fig. (b), force is applied by the man in the downward direction. This decreases the apparent weight of the man. Hence, action on the floor in this case would be W’ = W – F = 490 – 245 = 245 N.

As the floor yields to a normal force 700 N, the mode (b) has to be adopted by the man to lift the block.

Question 18.
A monkey of mass 40 kg climbs on a rope (Fig) which can stand a maximum tension of 600 N. In which of the following cases will the rope break : the monkey {LAWS OF MOTION )
a) climbs up with an acceleration of 6 ms^-2
b) climbs down with an acceleration of 4 ms^-2
c) climbs up with a uniform speed of 5 ms^-1
d) falls down the rope nearly freely under gravity?
(Ignore the mass of the rope).
Solution:
Here, mass of monkey, m = 40 kg
Maximum tension the rope can stand, T = 600 N.
In each case, actual tension in the rope will be equal to apparent weight of monkey (R), The rope will break when R exceeds T.
a) When monkey climbs up with a = 6 ms^-2,
R = m (g + a) = 40 (10 + 6) = 640 N (which is greater than T).
Hence the rope will break.

b) When monkey climbs down with a = 4 ms^-2
R = m (g – a) = 40 (10 – 4) = 240 N, which is less than T
∴ The rope will not break.

c) When monkey climbs up with a uniform speed v = 5 ms^-1,
its acceleration a = 0 ∴ R = mg = 40 × 10 = 400 N, which is less than T
∴ The rope will not break.

d) When monkey falls down the rope nearly freely under gravity, a = g
∴ R = m (g – a) = m (g – g) = 0 (Zero.)
Hence the rope will not break.

Question 19.
A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 in rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?
Solution:
Here, m = 70 kg, r = 3 m
n = 200, rpm = \(\frac{200}{60}\) rps, p = 0.15, ω = ?

The horizontal force N by the wall on the man provides the necessary centripetal force = m r ω². The frictional force (f) in this case is vertically upwards opposing the weight (mg) of the man.

After the floor is removed, the man will remain stuck to the wall, when mg = f < µ N, i.e. mg < µ m r ω² or g < µ r ω²
∴ Minimum angular speed of rotation of

Question 20.
A thin circular loop of radius R rotates about its vertical diameter with an angular frequency ω. Show that a small bead on the wire loop remains at its lowermost point for to ω ≤ √g/R. What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for to ω = √2g/R? Neglect friction.
Solution:
In Figure we have shown that radius vector joining the bead to the centre of the wire makes an angle 0 with the verticle downward direction. If N is normal reaction, then as is clear from the figure,
mg = N cos θ —- (i)
m r ω² = N sin θ —- (ii)
or m (R sin θ) ω² = N sin θ or m R ω² = N
from (i), mg = m R ω² cos θ or

Telangana TSBIE TS Inter 1st Year Physics Study Material 4th Lesson Motion in a Plane Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 4th Lesson Motion in a Plane

Very Short Answer Type Questions

Question 1.
Write the equation for the horizontal range covered by a projectile and specify when it will be maximum. [TS May ’16]
Answer:
Range of a projectile (R) = \(\frac{u^2 \sin 2 \theta}{g}\)
When θ = 45° Range is maximum.
Maximum Range (R_max) = \(\frac{u^2}{g}\)

Question 2.
The vertical component of a vector is equal to its horizontal component. What is the angle made by the vector with x-axis? [AP Mar. ’19; TS May ’18]
Answer:
Let R be a vector.
Vertical component = R sin θ;
Horizontal component = R cos θ
∴ R sin θ = R cos θ.
So sin θ = cos θ ⇒ θ = 45°

Question 3.
A vector V makes an angle θ with the horizontal. The vector is rotated through an angle α. Does this rotation change the vector V?
Answer:
Magnitude of vector = V ;
Let initial angle with horizontal = θ
Angle rotated = α
So new angle with horizontal = θ + α
Now horizontal component,
V_α = V cos (θ + α)
Vertical component, V_y = V sin (θ + α)
Magnitude of vector, V = \(\sqrt{V^{2}_{x}+V^{2}_{y}}\) = V
So rotating the vector does not change its magnitude.

Question 4.
Two forces of magnitudes 3 units and 5 units act at 60° with each other. What is the magnitude of their resultant? [AP Mar. 17. 15; May 17, 16]
Answer:
Given \(\overline{\mathrm{P}}\) = 3 units, \(\overline{\mathrm{Q}}\) = 5 units and θ = 60°

Question 5.
A = \(\overrightarrow{\mathrm{i}}+\overrightarrow{\mathrm{j}}\) What is the angle between the vector and x-axis? [TS Mar. ’17; AP Mar. ’14; May ’13]
Answer:
Given that, \(\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{i}}+\overrightarrow{\mathrm{j}}\)

If ‘θ’ is the angle made by the vector with x-axis then,

Question 6.
When two right angled vectors of magnitude 7 units and 24 units combine, what is the magnitude of their resultant? [AP Mar. 18, 16; May 18. 14]
Answer:
Given \(\overline{\mathrm{P}}\) = 7 units; \(\overline{\mathrm{Q}}\) = 24 units; 0 = 90°

Question 7.
If \(\overline{\mathrm{P}}\) = 2i + 4j + 14k and \(\overline{\mathrm{Q}}\) = 4i + 4j + 10k, find the magnitude of \(\overline{\mathrm{P}}+\overline{\mathrm{Q}}\). [TS Mar. ’16, ’15]
Answer:

Question 8.
Can a vector of magnitude zero have non-zero components?
Answer:
A vector with zero magnitude cannot have non-zero components. Because magnitude of given vector \(\overline{\mathrm{V}}\) = \(\sqrt{V^{2}_{x}+V^{2}_{y}}\) must be zero. This is possible only when V²_x and V²_y are zero.

Question 9.
What is the acceleration of a projectile at the top of its trajectory? [TS Mar. ’19]
Answer:
At highest point acceleration, a = g. In projectile, motion acceleration will always acts towards centre of earth. It is irrespective of its position, whether it is at highest point or somewhere.

Question 10.
Can two vectors of unequal magnitude add up to give the zero vector? Can three unequal vectors add up to give the zero vector?
Answer:
No. Two unequal vectors can never give zero vector by addition. But three unequal vectors when added may give zero vector.

Short Answer Questions

Question 1.
State parallelogram law of vectors. Derive an expression for the magnitude and direction of the resultant vector. [TS Mar. ’17, ’16, May ’17; AP Mar. ’14, ’13]
Answer:
Parallelogram Law :
If two vectors are represented by the two adjacent sides of a parallelogram then the diagonal passing through the intersection of given vectors represents their resultant both in direction and magnitude.

Proof :
Let \(\overline{\mathrm{P}}\) and \(\overline{\mathrm{Q}}\) be two adjacent vectors ‘θ’ be angle between them. Construct a parallelogram OACB as shown in figure. Extend the line OA and draw a normal D from C. The diagonal OC = the resultant \(\overline{\mathrm{R}}\) both in direction and magnitude.

In figure OCD = right angle triangle
⇒ OC = OD² + DC²

Angle of resultant with adjacent side ‘α’

Question 2.
What is relative motion? Explain it.
Answer:
Relative velocity is the velocity of a body with respect to another moving body.

Relative velocity in two dimensional motion :
Let two bodies A and B are moving with velocities \(\overrightarrow{\mathrm{V}}_A\) and \(\overrightarrow{\mathrm{V}}_B\) then relative velocity of Aw.r.t B is \(\overrightarrow{\mathrm{V}}_{AB}=\overrightarrow{\mathrm{V}}_{A}+\overrightarrow{\mathrm{V}}_{B}\)
Relative velocity of B w.r.t. A is
\(\overrightarrow{\mathrm{V}}_{BA}=\overrightarrow{\mathrm{V}}_{B}-\overrightarrow{\mathrm{V}}_{A}\)

Procedure to find resultant :
To find rela-tive velocity in two dimensional motion use vectorial subtraction of V_A or V_B. Generally to find relative velocity one vector \(\overrightarrow{\mathrm{V}}_A\) or \(\overrightarrow{\mathrm{V}}_B\) is reversed (as the case may be) and parallelogram is constructed. Now resultant of that parallelogram is equal to \(\overrightarrow{\mathrm{V}}_A-\overrightarrow{\mathrm{V}}_B\) or \(\overrightarrow{\mathrm{V}}_B-\overrightarrow{\mathrm{V}}_A\) (8° one vector is reversed V_A is taken as –\(\overrightarrow{\mathrm{V}}_A\) or \(\overrightarrow{\mathrm{V}}_B\) is taken as –\(\overrightarrow{\mathrm{V}}_B\))
In figure relative velocity of B w.r.t A is V_BA =V_R = V_B – V_A.

Question 3.
Show that a boat must move at an angle with respect to river water in order to cross the river in minimum time.
Answer:
Motion of a boat in a river :
Let a boat can travel with a speed of V_bE in still water w.r. to earth. It is used to cross a river which flows with a speed of V_WE with respect to earth. Let width of river is W.

We can cross the river in two different ways.
1) in shortest path 2) in shortest time.

To cross the river in shortest time :

To cross the river in shortest time boat must be rowed along the width of river i.e., boat must be rowed perpendicular to the bank or 90° with the flow of water. Because width of river is the shortest distance. So velocity must be taken in that direction to obtain shortest time. In this case V_bE and V_WE are perpendicular and boat will travel along AC. The distance BC is called drift. So to cross the river in shortest time angle with flow of water = 90°.

Question 4.
Define unit vector, null vector and position vector. [AP June ’15]
Answer:
Unit vector :
A vector whose magnitude is one unit is called unit vector.

Let a is \(\overline{\mathrm{a}}\) given vector then unit vector

Null vector :
A vector whose magnitude is zero is called null vector. But it has direction.

For a null vector the origin and terminal point are same.
Ex : Let \(\overline{\mathrm{A}}\times\overline{\mathrm{B}}=\overline{\mathrm{0}}\) . Here magnitude of \(\overline{\mathrm{A}}\times\overline{\mathrm{B}}=\overline{\mathrm{0}}\) . But still it has direction perpendicular to the plane of \(\overline{\mathrm{A}}\) and \(\overline{\mathrm{B}}\).

Position vector :
Any vector in space can be represented by the linear combination

Question 5.
If |\(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\)| = |\(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}\)|, prove that the angle between \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{b}}\) is 90°. [TS Mar., May ’18]
Answer:
Let \(\overrightarrow{\mathrm{a}}\), \(\overrightarrow{\mathrm{b}}\) are the two vectors.
Sum of vectors

by squaring on both sides,
a² + b² + 2ab cos θ = a² + b² – 2ab cos θ
∴ 4 ab cos θ = 0 or θ = 90°

Question 6.
Show that the trajectory of an object thrown at certain angle with the horizontal is a parabola. [AP Mar. ’18, ’17. ’16. ’15, May ’18, ’17, ’14. ’13; June ’15; TS Mar. ’18, ’15, May ’16, June ’15]
Answer:
Projectile :
A body thrown into the air same angle with the horizontal, (other tan 90°) its motion under the influence of gravity is called projectile. The path followed by it is called trajectory.

Let a body is projected from point O, with velocity ‘u’ at an angle θ with horizontal. The velocity u’ can be resolved into two rectangular components u_x and u_y along x-axis and y-axis.
u_x = u cos θ and u_y = u sin θ

After time t, Horizontal distance travelled x = u cos θ . t ……….. (1)

After a time t’ sec; vertical displacement

The above equation represents “parabola”. Hence the path of a projectile is a parabola.

Question 7.
Explain the terms the average velocity and instantaneous velocity. When are they equal ?
Answer:
Average velocity :
It is the ratio of total displacement to total time taken.

Average velocity is independent of path followed by the particle. It just deals with initial and final positions of the body.

Instantaneous velocity :
Velocity of a body at any particular instant of time is defined as instantaneous velocity.

as instantaneous velocity.
For a body moving with uniform velocity its average velocity = Instantaneous velocity.

Question 8.
Show that the maximum height and range

respectively where the terms have their regular meanings.
Answer:
Let a body is projected with an initial velocity ‘u’ and with an angle θ to the horizontal. Initial velocity along x direction, u_x = u cos θ Initial velocity along y direction, u_y = u sin θ

Horizontal Range :
It is the distance covered by projectile along the horizontal between the point of projection to the point on ground, where the projectile returns again.

It is denoted by R. The horizontal distance covered by the projectile in the to time of flight is called horizontal range. Therefore, R = u cos θ × t.

Angle of projection for maximum range:
For a given velocity of projection, the horizontal range will be maximum, when sin 2θ = 1.
∴ Angle of projection for maximum range is 2θ = 90° or θ = 45°
∴ R_max = \(\frac{u^2}{g}\)

Maximum height :
The vertical distance covered by the projectile until its vertical component becomes zero.

Question 9.
If the trajectory of a body is parabolic in one reference frame, can it be parabolic in another reference frame that moves at constant velocity with respect to the first reference frame? If the trajectory can be other than parabolic, what else can it be?
Answer:
Yes. According to Newton’s first law, a body at rest or a body moving with uniform velocity are treated as same. Both of them belong to inertial frame of reference.

If a frame (say 1) is moving with uniform velocity with respect to other, then that second frame must be at rest or it maintains a constant velocity w.r.t the first. So both frames are inertial frames. So if trajectory of a body in one frame is a parabola, then trajectory of that body in another frame is also a parabola.

Question 10.
A force 2i + j – k newton acts on a body which is initially at rest. At the end of 20 seconds the velocity of the body is 4i + 2j – 2k ms-1. What is the mass of the body? [AP May ’16]
Answer:
Force, F = 2i + j – k
time, t = 20
Initial velocity, u = 0
Final velocity, v = 4i + 2j – 2k = 2(2i + j – k)

Problems

Question 1.
Ship A is 10 km due west of ship B. Ship A is heading directly north at a speed of 30 km/h, while ship B is heading in a direction 60° west of north at a speed of 20 km/h.
(i) Determine the magnitude of the
(ii) What will be their distance of closest approach?
Answer:
Velocity of A = 30 kmph due North
∴ V_A = 30\(\hat{\mathbf{j}}\)
Velocity of B = 20 kmph 60° west of North
∴ V_B = -20sin60° + 20 cos60° = 10√3\(\hat{\mathbf{i}}\) + 10\(\hat{\mathbf{j}}\)

Shortest distance :
In ∆le ANB shortest distance, AN = AB sin θ
But distance, AB = 10 km
∴ AN = 10 × \(\frac{20}{10\sqrt{7}}=\frac{20}{\sqrt{7}}\) = 7.56 km

Question 2.
If θ is the angle of projection, R the range, h the maximum height, T the time of flight, then show that (a) tan θ = 4h/R and (b)h = gT²/8
Answer:
(a) Given angle of projection = θ,

Question 3.
A projectile is fired at an angle of 60° to the horizontal with an initial velocity of 800 m/s:
(i) Find the time of flight of the projectile before it hits the ground.
(ii) Find the distance it travels before it hits the ground (range).
(iii) Find the time of flight for the projectile to reach its maximum height.
Answer:
Angle of projection, θ = 60°.
Initial velocity, u = 800 m/s

iii) Time of flight to reach maximum height = \(\frac{T}{2}\)

Question 4.
For a particle projected slantwise from the ground, the magnitude of its position vector with respect to the point of projection, when it is at the highest point of the path is found to be √2 times the maximum height reached by it. Show that the angle of projection is tan^-1 (2).
Answer:
Position vector of h (max point) from 0, is

Question 5.
An object is launched from a cliff 20. m above the ground at an angle of 30° above the horizontal with an initial speed of 30 m/s. How far horizontally does the object travel before landing on the ground? (g = 10 m/s²)
Answer:
Height of cliff = 20m
Angle of projection, θ = 30°
Velocity of projection, u = 30 m/s
Total horizontal distance travelled = OC = OB’ + B’C

b) Distance B’C = Range of a horizontal projectile.
∴ Range = u cos θ t
u. cos θ = 30.\(\frac{\sqrt{3}}{2}\) = 15√3 .
Time taken to reach the ground, t = ?
Given S_y = 20, u_y = u sin θ = 30 sin 30° = 15 m/s
∴ S_y = 20 = 15t + \(\frac{10}{2}\)t² ⇒ 5t² + 15t – 20 = 0
or t² + 3t – 4 = 0 or (t + 4) (t – 1) = 0
∴ t = – 4 or t = 1 ∴ t is Not – ve use t = 1
∴ Range = 4 . cos θ . t = 15√3 → (2)
Total distance travelled before reaching the ground = 45√3 +15√3 = 60√3 m.

Question 6.
‘O’ is a point on the ground chosen as origin. A body first suffers a displacement of 10√2 mm North-East, next 10 m North and finally 10√2 North-West. How far it is from origin? [TS Mar. ’19]
Answer:
a) 10√2 m North-East
b) 10m North
c) 10√2 m North-West

From figure total displacement from origin ‘O’ is OC
ButOC = OA’ + A’B’ + B’C =10 + 10 + 10 = 30 m.

Question 7.
From a point on the ground a particle is projected with initial velocity u, such that its horizontal range is maximum. Find the magnitude of average velocity during its ascent.
Answer:
Velocity of projection = u.
Range is maximum ⇒ θ = 45°
During time of ascent ⇒ when h = h_max
⇒ u_x = V_x = u . cos θ

Average velocity, V_A = \(\sqrt{V^{2}_{x}+V^{2}_{y}}\)
V_x = Average velocity along x-axis

Average velocity during time of ascent

Question 8.
A particle is projected from the ground with some initial velocity making an angle of 45° with the horizontal. It reaches a height of 7.5 m above the ground while it travels a horizontal distance of 10 m from the point of projection. Find the initial speed of projection (g = 10 m/s2).
Anwser:
Angle of projection = 45°
Vertical height, h_y = 7.5 m
Horizontal distance, h_x = 10 m

Question 9.
Wind is blowing from the south at 5 ms^-1. To a cyclist it appears to be blowing from the east at 5 ms^-1. Show that the velocity of the cyclist is ms^-1 towards north-east.
Answer:
Direction of wind South to North 5 m/s.

Apparent direction is from East to West 5 m/s.

This is relative velocity.
To find velocity of cyclist reverse the direction of resultant vector OB and find resultant
∴ Velocity of cyclist = \(\sqrt{5^2+^2+0}\) = 5√2 m/s

Question 10.
A person walking at 4 m/s finds rain drops falling slantwise into his face with a speed of 4 m/s at ah angle of 30° with the vertical. Show that the actual speed of the rain drops is 4 m/s.
Answer:
Velocity of man = 4 m/sec
Apparent velocity of rain drop = 4 m/sec with θ = 30° with vertical. This is relative velocity V_B.

Additional Problems

Question 1.
The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 ms-1 can go without hitting the ceiling of the hall?
Solution:
Here, u = 40 ms^-1; H = 25m, R = ?
Let θ be the angle of projection with the horizontal direction to have the maximum range, with maximum height = 25 m.

Question 2.
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?
Solution:
Here, r = 80 cm = 0.8 m; o = 14/25 s^-1.

The centripetal acceleration,
a = ω²r = (\(\frac{88}{25}\))² × 0.80 = 9.90m/s²
The direction of centripetal acceleration is along the string directed towards the centre of circular path.

Question 3.
An aircraft executes a horizontal loop of radius 1 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.
Solution:
Here, r = 1 km = 1000 m;
v = 900 km h^-1 = 900 × (1000m) × (60 × 60s)^-1
= 250 ms^-1

Question 4.
An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the air-craft positions 10.0 s apart is 30°, what is the speed of the aircraft?
Solution:
In Fig, O is the observation point at the ground. A and B are the positions of aircraft for which ∠AOB = 30°. Draw a perpendicular OC on AB. Here OC = 3400 m and ∠AOC = ∠COB = 15°. Time taken by aircraft from A to B is 10 s.

Question 5.
A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to be fixed, and neglect air resistance.
Solution:

Since the muzzle velocity is fixed, therefore, Max. horizontal range,
R_max = \(\frac{u^2}{g}\) = 2√3 = 3.464 m.
So, the bullet cannot hit the target.

Telangana TSBIE TS Inter 1st Year Physics Study Material 3rd Lesson Motion in a Straight Line Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 3rd Lesson Motion in a Straight Line

Very Short Answer Type Questions

Question 1.
The states of motion and rest are relative. Explain.
Answer:
REST :
If the position of a body does not change with respect to surroundings, it is said to be at “rest”.

MOTION :
If the position of a body changes with respect to surroundings, it is said to be in “motion”.

By definitions rest and motion are relative with respect to surroundings.

Question 2.
How is average velocity different from instantaneous velocity? [AP Mar. 19, 13, May 17]
Answer:
Average velocity :
It is the ratio of total displacement to total time taken. It is independent of path of the body.

∴ Average velocity = \(\frac{\mathrm{s}_2-\mathrm{s}_1}{\mathrm{t}_2-\mathrm{t}_1}\)

Velocity of a particle at a particular instant of time is known as instantaneous velocity. Here time interval is very small.

Only in uniform motion, instantaneous velocity = average velocity. For all other cases instantaneous velocity may differ from average velocity.

Question 3.
Give an example where the velocity of an object is zero but its acceleration is not zero. [AP May ’17, Mar. ’13]
Answer:
In case of VPB at maximum height its velocity v = 0. But acceleration due to gravity ‘g’ is not zero.

So even though velocity v = 0 ⇒ acceleration is not zero.

Question 4.
A vehicle travels half the distance L with speed v₁ and the other half with speed v₂. What is the average speed?
Answer:
The average speed of a vehicle for the two equal parts.

Question 5.
A lift coming down is just about to reach the ground floor. Taking the ground floor as origin and positive direction upwards for all quantities, which one of the following is correct?
a) x < 0, v < 0, a > 0
b) x > 0, v < 0, a < 0
c) x > 0, v < 0, a > 0
d) x > 0, v > 0, a > 0
Answer:
As the lift is coming down, the value of x become less hence negative, i.e., x < 0.

Velocity is downwards (i.e., negative). So v < 0. Just before reaching ground floor, lift is retarded, i.e., acceleration is upwards. Hence a > 0.

We can conclude that x < 0, v < 0 and a > 0.
∴ (a) is correct.

Question 6.
A uniformly moving cricket ball is hit with a bat for a very short time and is turned back. Show the variation of its acceleration with time taking the acceleration in the backward direction as positive.
Answer:
For a ball moving with uniform velocity acceleration is zero. But during time of contact between ball and bat acceleration is applied in opposite direction. The shape of acceleration – time graph is as shown.

Question 7.
Give an example of one-dimensional motion where a particle moving along the positive x-direction comes to rest periodically and moves forward.
Answer:
When length of pendulum is high and amplitude is less then its motion is along a straight line. The pendulum will come to a stop at extreme position and moves back in forward direction (‘x’ + ve) periodically.

Question 8.
An object falling through a fluid is observed to have an acceleration given by a = g – bv, where g is the gravitational acceleration and b, is a constant. After a long time it is observed to fall with a constant velocity. What would be the value of this constant velocity?
Answer:
Acceleration, a = g – bv when moving with constant velocity, a = 0 ⇒ 0 = g – bv
∴ Constant velocity, v = \(\frac{g}{b}\) m/sec.

Question 9.
If the trajectory of a body is parabolic in one frame, can it be parabolic in another frame that moves with a constant velocity with respect to the first frame? If not, what can it be?
Answer:
If the trajectory of a body is parabolic with reference frames one and two then those two frames are of rest or moving with uniform velocity.

If they are not parabolic then for that reference frame it may be in straight line path.
Ex : When a body is dropped from a moving plane its path is parabolic for a person outside the plane. But for the pilot in the plane it is falling vertically downwards.

Question 10.
A spring with one end attached to a mass and the other to a rigid support is stretched and released. When is the magnitude of acceleration a maxium?
Answer:
Maximum restoring force setup in the spring, when stretched by a distance ’r’, is F = – kr

Potential energy of stretched spring = \(\frac{1}{2}\) kx²

As F ∝ r and this force is directed towards equilibrium position, hence if mass is left free, it will execute damped SHM due to gravity pull.

Magnitude of acceleration in the mass attached to one end of spring when just released is

a = \(\frac{F}{m}=\frac{-k}{m}\) r = (Maximum)

The magnitude of acceleration of the spring will be maximum when just released.

Question 11.
Define average velocity and average speed. When does the magnitude of average velocity become equal to the average speed?
Answer:
Average velocity :
It is defined as the ratio of total displacement to total time taken.

Average velocity

Average velocity is independent of path followed by the particle. It just deals with initial and final positions of the body. Average Speed: The ratio of total path length travelled to the total time taken is known as “average speed”.

Speed and average speed are scalar quantities so no direction for these quantities.

When the body is along with the straight line its average velocity and average speed are equal.

Short Answer Questions

Question 1.
Can the equations of kinematics be used when the acceleration varies with time? If not, what form would these equations take?
Answer:
a) The equations of motion are
1) v =u + at
2) s = ut + \(\frac{1}{2}\) at² and 3) v² – u² = 2as. All these three equations applicable body moves with uniform acceleration ‘a’.

No, the equations of are not applicable when the acceleration varies with time.

Question 2.
A particle moves in a straight line with uniform acceleration. Its velocity at time t = 0 is v₁ and at time t₂ = t is v₂. The average velocity of the particle in this time interval is (v₁ + v₁)/ 2. Is this correct? Substantiate your answer.
Answer:
t₁ = 0 ⇒ u = v₁
t₂ = t ⇒ v = v₂

Question 3.
Can the velocity of an object be in a direction other than the direction of acceleration of the object? If so, give an example.
Answer:
Yes. Velocity of a body and its acceleration may be in different directions.

Explanation:

1. Incase of vertically projected body ⇒ velocity of body is in the upward direction and acceleration is in a downward direction.
2. When brakes are applied the velocity of body before coming to rest is opposite to retarding acceleration.

Question 4.
A parachutist flying in an aeroplane jumps when it is ata height of 3 km above ground. He opens his parachute when he is about 1 km above ground. Describe his motion.
Answer:
a) Height of fall before opening, h = 2 km
= 2000 m
∴ Velocity at a height of 1 km

b) After parachute is opened it touches the ground with almost zero velocity.
∴ u = 200 m/sec, v = 0, S = h = 1000 m From v² – u² = 2as

The Motion is as shown in figure.

Question 5.
A bird holds a fruit in its beak and flies parallel to the ground. It lets go of the fruit at some height. Describe the trajectory of the fruit as it falls to the ground as seen by (a) the bird (b) a person on the ground.
Answer:
a) As the bird is flying parallel to the ground, it possesses velocity in horizontal direction. Hence the fruit also possess velocity in horizontal direction and acceleration in downward direction. Hence the path of the fruit is a straight line with respect to the bird.

b) With respect to a person on the ground, the fruit seems to be in a parabolic path.

Question 6.
A man runs across the roof of a tall building and jumps horizontally on to the (lower) roof of an adjacent building. If his speed is 9 ms^-1 and the horizontal distance between the buildings is 10 m and the height difference between the roofs is 9 m, will he be able to land on the next building? (take g = 10 ms^-2) [TS Mar. ’18]
Answer:
Given that,
initial speed, u = 9 ms^-1 ; g = 10m/s² height difference between the roofs, h = 9 m
horizontal distance between two buildings, d = 10 m

Range of the man = R = u × T = 9 × 1.341
= 12.069 m

Since R > d, the man will be able to land on the next building.

Question 7.
A ball is dropped from the roof of a tall building and simultaneously another ball is thrown horizontally with some velocity from the same roof. Which ball lands first? Explain your answer. [TS June ’15]
Answer:
Let ‘h’ be the height of the tall building.

For dropped ball:
Let ‘t₁‘ be the time taken by the dropped ball to reach the ground.

Initial velocity, u = 0 ; Acceleration, a = + g
Distance travelled, s = h; Time of travel, t = t₁

From the equation of motion, s = ut + \(\frac{1}{2}\) at²
we can write,

For horizontally projected ball:
If the ball is thrown horizontally then its initial velocity along vertical direction is zero and in this case let ‘t₂‘ be the time taken by the ball to reach the ground.
Again from the equation of motion,

From equations (1) and (2) t₁ = t₂
i.e., both the balls reach the ground in the same time.

Question 8.
A ball is dropped from a building and simultaneously another ball is projected upward with some velocity. Describe the change in relative velocities of the balls as a function of time.
Answer:
a) For a body dropped from building its velocity, v₁ = gt → (1) (∵ u₁ = 0)

b) For a body thrown up with a velocity ‘u’ its velocity, v₂ = u – gt → (2)
∵ The two balls are moving in opposite direction the relative velocity,
V_R = v₁ + v₂
∴ v_R =gt + u – gt = u

Here the relative velocity remains constant, but velocity of one body increases at a rate of g’ m/sec and velocity of another body decreases at a rate of ‘g’m/sec.

Question 9.
A typical raindrop is about 4 mm in diameter. If a raindrop falls from a cloud which is at 1 km above the ground, estimate its momentum when it hits the ground.
Answer:
Diameter, D = 4 m
⇒ radius, r = 2mm = 2 × 10^-3 m
mass of rain drop = volume × density = \(\frac{4}{3}\)πr³ × 1000 m
(∵ mass of one m³ of water = 1000 kg)

Question 10.
Show that the maximum height reached by a projectile launched at an angle of 45° is one quarter of its range. [AP May ’16, Mar. ’14]
Answer:

∴ When θ = 45° maximum height reached is one quarter of maximum range.

Question 11.
Derive the. equation of motion x = v₀t + \(\frac{1}{2}\) at² using appropriate graph. [TS Mar. ’19, May ’16]
Answer:
The velocity-time graph of a body moving with initial velocity u’ and with uniform acceleration a’ as shown. Let ‘v’ be the velocity of the body after a time t.

In v – t graph area of velocity-time graph = total displacement travelled by it. Area under velocity – time graph = area of OABCD
∴ Area of Rectangular part OACD = Area of OACD + Area of ABC.
A₁ = OA × OD = v₀.t. ……….. (1)

2) Area of triangle ABC = A₂
A₂ = \(\frac{1}{2}\)Base × height
= \(\frac{1}{2}\)AC × BC
= \(\frac{1}{2}\) t(v – v₀).
But v – v₀ = at
A₂ = \(\frac{1}{2}\)t.at = \(\frac{1}{2}\)at².
∴ Total area under graph = s = A₁ + A₂
s(n) = v₀t + \(\frac{1}{2}\)at².
∴ s = ut + \(\frac{1}{2}\)at² is graphically proved.

Problems

Question 1.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^-1. What is the (a) magnitude of average velocity and (b) average speed of the man over the time interval 0 to 50 minutes? [AP Mar. ’19. May ’18; TS Mar. ’18]
Solution:
Time taken by man to go from his home to

Time take by man to go from market to his home, t₂ = \(\frac{2.5}{7.5}=\frac{1}{3}\)h
∴ Total time taken = t, + to = \(\frac{1}{2}+\frac{1}{3}=\frac{5}{6}h\)
= 50 min.
In time interval 0 to 50 min,
Total distance travelled = 2.5 + 2.5 = 5 km.
Total displacement = zero.

Question 2.
A stone is dropped from a height 300 in and at the same time another stone is projected vertically upwards with a velocity of 100 m/sec. Find when and where the two stones meet. [AP Mar. ’16]
Solution:
Height h = 300 m ;
Initial velocity U₀ = 100 m/s
Let the two stones will meet at a height ‘x’ above the ground.
For 1st stone h – x = \(\frac{1}{2}\) gt² …………. (1)
For 2nd stone x = u₀t – \(\frac{1}{2}\) gt²
⇒ \(\frac{1}{2}\) gt² = u₀t – x ……….. (2)

Since t is same for the two stones
From equations 1 & 2.
h – x = u₀t – x
⇒ u₀t = h or time t = \(\frac{h}{u_0}=\frac{300}{100}\) = 3 sec.
∴ The two stones will meet 3 seconds after the 1st stone is dropped or 2nd stone is thrown up.

Question 3.
A car travels the first third of a distance with a speed of 10 kmph, the second third at 20 kmph and the last third at 60 kmph. What is its mean speed over the entire distance? [TS Mar. ’16; AP May ’14, AP Mar. ’18]
Solution:
Total distance = s;
distance travelled, s₁ = \(\frac{s}{3}\) ;
velocity, v₁ = 10 kmph
distance, s₂ = \(\frac{s}{3}\)
velocity, v₂ = 20 kmph
distance, s₃ = \(\frac{s}{3}\)
velocity, v₃ = 60 kmph

Question 4.
A bullet moving with a speed of 150 m s^-1 strikes a tree and penetrates 3.5 cm before stopping. What is the magnitude of its retardation in the tree and the time taken for it to stop after striking the tree?
Solution:
Velocity of bullet, u = 150 m/s;
Final velocity, v = 0
Distance travelled, s = 3.5 cm = 3.5 × 10^-2 m,

Question 5.
A motorist drives north for 30 min at 85 km/h and then stops for 15 min. He continues travelling north and covers 130 km in 2 hours. What is his total displacement and average velocity?
Solution:
In first part:
Velocity, v₁ = 85 kmph
Time, t₁ = 30 min
Distance travelled, s₁ = v₁ t₁
= 85 × \(\frac{30}{60}\) = 42.5 km

In second part:
Distance travelled, s₂ = 0 ;
Time, t₂ = 15.0 min.

In third part:
Distance travelled, s₃ = 130 km ;
Time, t₃ = 120 min = 2 hours
a) Total distance of the motorist,
s = s₁ + s₂ + s₃ = 42.5 + 0 + 130 = 172.5 km

b) Total time travelled,
t = t₁ + t₂ + t₃ = 30 + 15 + 120
= 165 minutes
= 2 hrs 45 minutes
= 2\(\frac{3}{4}\)hrs. = \(\frac{11}{4}\) hrs.
∴ Average velocity,

Question 6.
A ball A is dropped from the top of a building and at the same time an identical ball B is thrown vertically upward from the ground. When the balls collide the speed of A is twice that of B. At what frac¬tion of the height of the building did the collison occur?
Solution:
Given at time of collision velocity of A = V_A
= 2 × V_B (velocity of B)
Let the body be dropped from a height h’.
Let the two stones collide at x from ground.
For the body dropped,
s = h – x = \(\frac{1}{2}\)gt² → (1)
For the body thrown up,
x = ut – \(\frac{1}{2}\)gt² → (2)
For the body dropped,
v = u + at ⇒ V_A = gt → (3)
For the body thrown up,
v = u – gt ⇒ V_B = u – gt → (4)
Given V_A = 2V_B
⇒ gt = 2 (u – gt) or u = \(\frac{3gt}{2}\) → (5)
Divide equation (1) with equation (2)

∴ Fraction of height of collision = \(\frac{2}{3}\)

Question 7.
Drops of water fall at regular intervals from the roof of a building of height 16 m. The first drop strikes the ground at the same moment as the fifth drop leaves the roof. Find the distances between successive drops.
Solution:
Height of building, h = 16 m

Number of drops, n = 5
∴ Number of intervals = n – 1 = 5 – 1 = 4

Time interval between drops = \(\frac{1.8}{4}\)
= 0.45 sec
Time of travel of 1st drop, t₁ = 4 × 0.45 = 1.8
∴ Distance travelled by
1st drop, S₁ = \(\frac{1}{2}\)gt²₁ = \(\frac{1}{2}\) × 9.8 × 1.8 × 1.8= 16 m

For 2nd drop, t₂ = 3 × 0.45 = 1.35 sec.,
∴ S₂ = \(\frac{1}{2}\) × 9.8 × 1.35²
= 4.9 × 1.822 ≅ 1.822 ≅ 9m

For 3rd drop, t₃ = 2 × 0.45 = 0.9 sec.
Distance, S₃ = \(\frac{1}{2}\)gt²₃ = \(\frac{1}{2}\) × 9.8 × 0.9² = 3.97≅4 m

For 4th drop, t₄ = 1 × 0.45 = 0.45 sec
Distance travelled, S₄ = \(\frac{1}{2}\)gt²₄ = \(\frac{1}{2}\) × 9.8 × (0.45)² = 1

For 5 th drop, t₅ = 0 ⇒ S₅ = 0
Distance between 1^st and 2^nd drop
S_{1, 2} = S₁ – S₂ = 16 – 9 = 7 m

Distance between 2^nd and 3^rd drop
S_{2, 3} = S₂ – S<₃ = 9 – 4 = 5m

Distance between 3^rd and 4^th drop
S_{3, 4} = S₃ – S4 = 4 – 1= 3 m

Distance between 4^th and 5^th drop
S_{4, 5} = S₄ – S₅ = 1 – 0 = 1 m

∴ Distances between successive drops are 7m, 5m, 3m and lm.

Question 8.
Rain is falling vertically with a speed of 35 ms^-1. A woman rides a bicycle with a speed of 12 ms^-1 in east to west direction. What is the direction in which she should hold her umbrella? [TS June ’15]
Solution:
Velocity of rain V_R = 35 m/s (vertically)
Velocity of women V_w = 12 m/s (towards east)
Resultant angle θ = tan^-1 \(\frac{V_W}{V_R}=\frac{12}{35}\)
∴ θ = tan^-1\(\frac{12}{35}\) = 0.343. or q = 19° (Nearly)
She should hold umbrella at an angle of 19c with east.

Question 9.
A hunter aims a gun at a monkey hanging from a tree some distance away. The monkey drops from the branch at the moment he fires the gun hoping to avoid the bullet. Explain why the monkey made a wrong move.
Solution:
Let the bullet is fired with an angle α and distance from hunter’s rifle to monkey = x
Vertical component of velocity v_xy = v sin α
when exactly aimed at monkey s_y = v sin α
t = h

But due to acceleration due to gravity
h₁ = u sin α t – \(\frac{1}{2}\)gt² = h – \(\frac{1}{2}\)gt² → (1)
So bullet passes through a height of \(\frac{1}{2}\)gt² below the monkey.
But when the monkey is falling freely height of fall during time t = \(\frac{1}{2}\)gt²
So new height is \(\frac{1}{2}\)gt² → (2)

From equations (1) & (2) h₁ is same i.e., if the monkey is dropped from the branch bullet will hit it exactly.

Question 10.
A food packet is dropped from an aero-plane, moving with a speed of 360 kmph in a horizontal direction, from a height of 500m. Find (i) its time of descent (ii) the horizontal distance between the point at which the food packet reaches the ground and the point above which it was dropped.
Solution:
Velocity of plane, V = 360 kmph
= 360 × \(\frac{5}{18}\) = 100 m/s

Height above ground, h = 500 m;
g = 10 m/s²

i) Time of descent,

ii) Horizontal distance between point o{ dropping and point where it reaches the ground = Range R

Question 11.
A ball is tossed from the window of a building with an initial velocity of 8 ms^-1 at an angle of 20° below the horizontal. It strikes the ground 3 s later. From what height was the ball thrown? How far from the base of the building does the ball strike the ground?
Solution:
Initial velocity, u = 8 m/s;
Angle of projection, θ = 20°
Time taken to reach the ground, t = 3 sec
Horizontal component of initial velocity,
u_x = u. cos θ = 8 cos 20°
= 8 × 0.94 = 7.52 m/s

Vertical component of initial velocity,
v_y = u sin θ = 8 sin 20°
= 8 × 0.342 = 2.736 m/s

a) From equation of motion, s = ut + \(\frac{1}{2}\)at²
we can write
h = (u sin θ)t + \(\frac{1}{2}\)gt²
⇒ h = (2.736)3 + \(\frac{1}{2}\)9.8 × (3)²
⇒ h = 8.208+ 4.9 × 9
⇒ h = 8.208 + 44.1 or h = 52.308 m

b) Horizontal distance travelled, s_x = v_x x t = 7.52 × 3 = 22.56 m

Question 12.
Two balls are projected from the same point in directions 30° and 60° with respect to the horizontal. What is the ratio of their initial velocities if they (a) attain the same height? (b) have the same range?
Solution:
Angle of projection of first ball, θ₁ = 30°
Angle of projection of second ball, θ₂ = 60°
Let u₁ and u₂ be the velocities of projections of the two balls.
i) Maximum height of first ball,

ii) If the balls have same range, then R₁ = R₂

Question 13.
A ball is thrown vertically upwards with a velocity of 20 ins’1 from the top of a multistorey building. The height of the point from where the ball is thrown is 25.0 m from the ground. [TS May ’17; AP & TS Mar. ’15]
(a) How high will the ball rise?
(b) How long will it be before the ball hits the ground?
Take g = 10 ms^-2 [Actual value of ‘g’ is 9.8 ms^-2]
(OR)
When a ball is thrown vertically upwards with a velocity of 20 ms^-1 from the top of a multistorey building, the height of the point from where the ball is thrown is 25.0 m from the ground. [TS Mar. ’15]
a) How high will the ball rise? and
b) How long will it be before the ball hits the ground?
Solution:
Initial velocity V₀ = 20 m/s;
height above ground h₀ = 2.50 m ;
g = 10 m/s²

a) For a body thrown up vertically height of rise

b) Time spent in air (t) is y₁ – y₀ = V₀t + \(\frac{1}{2}\)gt²
Where y₁ = Total displacement of the body from ground = 0
∴ 0 = y₀ + V₀t+ \(\frac{1}{2}\)gt² = 25 + 20t – \(\frac{1}{2}\). 10 . t²
[∵ g = – 10 m/s² while going up]
∴ 0 = – 5t² + 20t + 25 (or) t² – 4t – 5 = 0
i.e., (t – 5) (t + 1) = 0 ⇒ t = 5 (or) t = – 1
But time is not – ve.
∴ Time spent in air t = 5 sec

Question 14.
A parachutist flying in an aeroplane jumps when it is at a height of 3 km above the ground. He opens his parachute when he is about 1 km above ground. Describe his motion.
Answer:
Initially the path is a parabola as seen by an observer on the ground. It is a vertical straight line as seen by the pilot. He opens his parachute, it is moving vertically downwards with decreasing velocity and finally it reaches the ground.

Telangana TSBIE TS Inter 1st Year Physics Study Material 2nd Lesson Units and Measurements Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 2nd Lesson Units and Measurements

Very Short Answer Type Questions

Question 1.
Distinguish between accuracy and precision. [AP Mar. ’15, ’16, May 16, 13, June 15; TS May ’18, Mar. ’15]
Answer:
Accuracy :
It indicates the closeness of a measurement to the true value of given quantity.
→ If the measurement is nearer to true value then accuracy is more.

Precision :
Precision of a measuring instrument depends on the limit (or) resolution of the quantity measured with that instrument.
→ If the least measurable value is less, then precision is more for that instrument.

Question 2.
What are the different types of errors that can occur in a measurement?
Answer:
Types of errors 1) Systematic errors and 2) Random errors.

Systematic errors are again divided into 1) Imperfectional errors 2) Environmental errors and 3) Personal errors.

Question 3.
How can systematic errors be minimised or eliminated? [AP May ’17; TS Mar. ’17; AP May ’17; TS Mar. ’17]
Answer:
Systematic errors can be minimised

1. by improving experimental techniques,
2. by selecting better instruments,
3. by taking mean value of number of readings and
4. by removing personal errors as far as possible.

Question 4.
Illustrate how the result of a measurement is to be reported indicating the error involved.
Answer:
Suppose length of an object is measured with a metre rod with least count equal to 0.1 cm. If the measured length is 62.5 cm, it has to be recorded as (62.5 ± 0.1) cm, stating the limits of error. Similarly, suppose time period of a pendulum is measured to be 2.0 sec, using a stopwatch of least count 0.1 sec, it has to be recorded as (2.0 ± 0.1) sec. It indicates that time period is in the range of 1.9 sec and 2.1 sec.

Question 5.
What are significant figures and what do they represent when reporting the result of a measurement? [TS Mar. ’18]
Answer:
Significant figures represents all practically measured digits plus one uncertain digit at the end.

When a result is reported in this way we can know up to what extent the value is reliable and also the amount of uncertainty in that reported value.

Question 6.
Distinguish between fundamental units and derived units. [TS Mar. ’16 ; AP May ’14]
Answer:
1) Fundamental units are used to measure fundamental quantities. Derived units are used to measure derived quantities.

2) Fundamental units are independent. Derived units are obtained by the combination of Fundamental units.
Ex: Metre is fundamental unit of length L’. metre/sec is derived unit of velocity which is a combination of fundamental unit metre and second.

Question 7.
Why do we have different units for the same physical quantity? [TS May ’16. June ’15]
Answer:
To measure the same physical quantity we have different units by keeping magnitude of the quantity to be measured.
Example:

1. The measure astronomical distances we will use light year.
   1 light year = 9.468 × 10¹⁵m.
2. To measure atomic distances we will use Angstrom A (or) Fermi.

Question 8.
What is dimensional analysis?
Answer:
Dimensional analysis is a tool to check the relations among physical quantities by using their dimensions.
Dimensional analysis is generally used to check the correctness of derived equations.

Question 9.
How many orders of magnitude greater is the radius of the atom as compared to that of the nucleus?
Answer:
Size of atom = 10^-10 m,
Size of atomic nucleus = 10^-14m.
Size of atom ÷ size of nucleus is \(\frac{10^{-10}}{10^{-14}}\) = 10⁴
∴ Size of atom is 10⁴ times greater than size of nucleus.

Question 10.
Express unified atomic mass unit in kg. [TS Mar. ’19]
Answer:
By definition,
1 a.m.u. = \(\frac{1}{12}\) × mass of an atom of ¹²₆C

Short Answer Questions

Question 1.
The vernier scale of an instrument has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm, then using this instrument what would be the minimum inaccuracy in the measurement of distance?
Answer:
Least count of Vernier Callipers = 1 MSD – 1 VSD
∴ L.C. = 1 MSD – \(\frac{49}{50}\)MSD = \(\frac{1}{50}\)MSD
= \(\frac{1}{50}\) × 0.5 = 0.01 m.m

Question 2.
In a system of units, the unit of force is 100N, unit of length is 10m and the unit of time is 100s. What is the unit of mass in this system?
Answer:
Here, F = MLT^-2 = 100 N → (1) ;
L = 10 m ; T = 100s
∴ From equation (1)
M × (10) × (100)^-2 = 100 ⇒ M × 10^-3 = 100
⇒ M = \(\frac{100}{10^{-3}}\) ⇒ M = 10⁵ kg.

Question 3.
The distance of a galaxy from Earth is of the order of 10²⁵m. Calculate the order of magnitude of the time taken by light to reach us from the galaxy.
Answer:
Size of galaxy = 10²⁵m,
Velocity of light, c = 3 × 10⁸ ms^-1
Time taken by light to reach earth,

While calculating the order of magnitude we will consider the powers of Ten only.
So order of magnitude of time taken by light to reach earth from galaxy is 1017 seconds.

Question 4.
The Earth-Moon distance is about 60 Earth radius. What will be the approximate diameter of the Earth as seen from the Moon?
Answer:
Earth moon distance, D = 60r.
Diameter of earth, b = 2r

Question 5.
Three measurements of the time for 20 oscillations of a pendulum give t₁ = 39.6 s, t₂ = 39.9 s and t₃ = 39.5 s. What is the precision in the measurements? What is the accuracy of the measurements?
Answer:
Precision is the least measurable value with that instrument in our case precision is ±0.1 sec.

Calculation of accuracy :
Average value of measurements

Error in each measurement =

Precision ± 1 sec. In these measurements only two significant figures are believable. 3rd one is uncertain.

Adjustment of ∆a_mean upto given significant figure = 0.156 adjusted to 0.2.

So our value is accurate upto ±0.2

So our result is 39.67 ± 0.2, when significant figures taken into account it is 39.7 ± 0.2 sec.

Question 6.
1 calorie = 4.2 J where 1J = 1 kg m²s^-2. Suppose we employ a system of units in which the unit of mass is α kg, the unit of length is β m and the unit of time γ s, show that a calorie has a magnitude 4.2 α^-1 β^-2 γ² in the new system.
Answer:
Here, 1 calorie = 4.2 J = 4.2 kg m² / s² → (1)
As new unit of mass = α kg
∴ 1 kg = \(\frac{1}{\alpha}\) new unit of mass
⇒ α^-1 new unit of mass

Similarly lm = β^-1 new unit of length and 1s = γ^-1 new unit of time

Putting these values in (1) we get
1 calorie = 4.2 (α^-1 new unit of mass) (β^-1 new unit of length)² (γ^-1 new unit of time)^-2
= 4.2 α^-1 β^-2 γ² new unit of energy, which was proved.

Question 7.
A new unit of length is chosen so that the speed of light in vacuum is 1 ms^-2. If light takes 8 min and 20s to cover this distance, what is the distance between the Sun and Earth in terms of the new unit?
Answer:
Given that velocity of light in vacuum,
c = 1 new unit of length s^-1
Time taken by light of Sun to reach Earth, t = 8 min 20s = 8 × 60 + 20 = 500 s
∴ Distance between the Sun and Earth, x = c × t
= 1 new unit of length s^-1 × 500s
= 500 new units of length.

Question 8.
A student measures the thickness of a human hair using a microscope of magnification 100. He makes 20 observations and finds that the average thickness (as viewed in the microscope) is 3.5 mm. What is the estimate of the thickness of hair?
Answer:

∴ Thickness of hair is 0.035 mm

Question 9.
A physical quantity X is related to four mea-surable quantities a, b, c and d as follows:
X = a²b³C^5/2d^-2
The percentage error in the measurement of a, b, c and d are 1%, 2%, 3% and 4% respectively. What is the percentage error in X?
Answer:
Here, X = a²b³C^5/2d^-2

The percentage error in X is ± 23.5 %

Question 10.
The velocity of a body is given by v = At² + Bt + C. If v and t are expressed in SI, what are the units of A, B and C?
Answer:
From principle of Homogeneity the terms At², Bt and C must have same dimensional formula of velocity ‘v’.
v = Velocity = LT^-1 ⇒ CT^-1 = A [T²]
∴ A = \(\frac{LT^{-1}}{T^2}\) = LT^-3 . So unit of A is m/sec³
LT^-1 = BT ⇒ B = LT^-2 So unit of B is m/sec²
LT^-1 = C So unit of C is m/sec.

Dimensional formulae of physical quantities

Problems

Question 1.
In the expression P = El² m^-5 G^-2 the quantities E, l, m and G denote energy, angular momentum, mass and gravitational constant respectively. Show that P is a dimensionless, quantity.
Solution:
Here, P = El² m^-5 G^-2
Here,
I = energy,
l = angular momentum
m = mass
G = gravitational constant
= [M L²T^-2][ML²T^-1]² [M]^-5 [M^-1L³T^-2]^-2
= M^1+2+5+2 L^2+4-6 T^-2-2+4
P = [M° L° T°]
Hence, P is a dimensionless quantity.

Question 2.
If the velocity of light c, Planck’s constant, h and the gravitational constant G are taken as fundamental quantities; then express mass, length and time in terms of dimensions of these quantities.
Solution:
Here, c = [L T^-1] ; h = [ML²T^-1]
G = [M^-1L³T^-2]

Applying the principle of homogeneity of dimensions, we get

y – z = 1 → (2) ; x + 2y + 3z = 0 → (3) ; – x – y – 2z = 0 → (4)
Adding eq. (2), eq. (3) and eq. (4),
2y – 1 ⇒ y = \(\frac{1}{2}\)
∴ From eq. (2) z = y – 1 = \(\frac{1}{2}\) – 1 = \(\frac{-1}{2}\)
From eq. (4) x = -y – 2z = \(\frac{-1}{2}\) + 1 = \(\frac{1}{2}\)
Substituting the values of x, y & z in eq. (1), we get

Question 3.
An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. Using dimensional analysis show that the period of the satellite.
T = \(\frac{k}{R} \sqrt{\frac{r^3}{g}}\)
where k is a dimensionless constant and g is acceleration due to gravity.
Solution:
Given that
T² ∝ r³ or T ∝ r^3/2 Also T is a function of g and R
Let T ∝ r^3/2 g^a R^b where a, b are the dimen¬sions of g and R.
(or) T = k r^3/2 g^a R^b → (1)
where k is dimensionless constant of proportionality
From equation (1)

Applying the principle of homogeneity of dimensions, we get

a + b + \(\frac{3}{2}\) = 0 → (2) ∴ -2a = 1 ⇒ a = \(\frac{-1}{2}\)
From eq (1), \(\frac{-1}{2}\) + b + \(\frac{3}{2}\) = 0 ⇒ b = -1
Substituting the values of a’ and b’ in eq. (1), we get

This is the required relation.

Question 4.
State the number of significant figures in the following
a) 6729 b) 0.024 c) 0,08240 d) 6.032 e) 4.57 x 108
Solution:
a) In 6729 all are significant figures.
∴ Number of significant figures Four.

b) In 0.024 the zeroes to the left of 1st non-zero digit of a number less than one are not significant.
∴ Number of significant figures Two.

c) 0.08240 – Significant figures Four.

d) In 6.032 the zero between two non-zero digits is significant.
So, number of significant figures in 6.023 are 4.

e) 4.57 × 10⁸ – Significant figures Three. [In the representation of powers of Ten our rule is only significant figures must be given].

Question 5.
A stick has a length of 12.132 cm and another has a length of 12.4 cm. If the two sticks are placed end and to what is the total length? If the two sticks are placed side by side, what is the difference in their lengths?
Solution:
a) When placed end to end total length is l = l₁ + l₂
l₁ = 12.132 cm and l₂ = 12.4 cm.
∴ l₁ + l₂ = 12.132 + 12.4 = 24.532 cm.

In addition final answer must have least number of significant numbers in that addition, i.e., one after decimal point.
So our answer is 24.5 cm. b) For difference use l₁ – l₂
i.e., 12.4- 12.132 = 0.268 cm.
In subtraction final answer must be adjusted to least number of significant figures in that operation.

Here least number is one digit after decimal. By applying round off procedure our answer is 0.3 cm.

Question 6.
Each side of a cube is measured to be 7.203 m. What is (i) the total surface area and (ii) the volume of the cube, to appropriate significant figures?
Solution:
Side of cube, a = 7.203 m.
So number of significant figures are Four.
i) Surface area of cube = 6a²
= 6x 7.203 × 7.203 = 311.299

But our final answer must be rounded to least number of significant figures is four digits.
So surface area of cube = 311.3 m²
ii) Volume of cube, V = a³ = (7.203)³
= 373.147
But the answer must be limited to Four significant figures.
∴ Volume of sphere, V = 373.1 m³.

Question 7.
The measured mass and volume of a body are 2.42 g and 4.7 cm³ respectively with possible errors 0.01 g and 0.1 cm³. Find the maximum error in density.
Solution:
Mass, m = 2.42 g ; Error, ∆m = 0.01 g.
Volume, V = 4.7 cm³, Error, ∆V = 0.1 cc.

Maximum % error in density = % error in mass + % error in volume Maximum percentage error in density
= \(\frac{1}{2.42}+\frac{10}{4.7}\) = 0.413 + 2.127 = 2.54%

Question 8.
The error in measurement of radius of a sphere is 1%. What is the error in the measurement of volume? [AP Mar. ’19]
Solution:

Question 9.
The percentage error in the mass and speed are 2% and 3% respectively. What is the maximum error in kinetic energy calculated using these quantities?
Solution:
Percentage change in mass = \(\frac{\Delta \mathrm{m}}{\mathrm{m}}\) × 100 = 2%

Question 10.
One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). If the size of the hydrogen molecule is about 1Å, what is the ratio of molar volume to the atomic volume of a mole of hydrogen?
Solution:
Size of Hydrogen atom ≈ 1Å = 10^-10 m = 10^-8cm
V₁ = Atomic volume = number of atoms × volume of atom.
One mole gas contains n’ molecules.
Avogadro Number, n = 6.022 × 10²³

V₂ = Molar volume of 1 mol. gas = 22.4 lit
= 2.24 × 10⁴C.C
∵ 1 lit = 1000 c.c.
∴ Ratio of molar volume to atomic volume = V₂ : V₁
= 2.24 × 10⁴ : 2.523 ≅ 10⁴ m.

Telangana TSBIE TS Inter 1st Year Physics Study Material 1st Lesson Physical World Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 1st Lesson Physical World

Very Short Answer Type Questions

Question 1.
What is Physics? [TS Mar. ’16]
Answer:
Physics is a branch of science which deals with the study of nature and natural phenomena.

Question 2.
What is the discovery of C.V. Raman? [AP Mar. ’18, 14; May 18, 16, 14; TS Mar. ’19, ’18, ’17]
Answer:
C.V. Raman’s contribution to physics is Raman effect. It deals with scattering of light by molecules of a medium when they are excited to vibrational energy levels.

Question 3.
What are the fundamental forces in nature? [TS May ’18]
Answer:
There are four fundamental forces in nature that govern the diverse phenomena of the macroscopic and the microscopic wu.m. These are the ‘gravitational force’, the ‘electromagnetic force’, the ‘strong nuclear force’, and the ‘weak nuclear force’.

Question 4.
Which of the following has symr etry?
a) Acceleration due to gravity.
b) Law of gravitation.
Answer:
Acceleration due to gravity varies from place to place. So it has no symmetry.
Law of gravitation has symmetry, because it does not depend on any physical quantity.

Question 5.
What is the contribution of S. Chandra Sekhar to Physics?
Answer:
S. Chandra Sekhar discovered the structure and evolution of stars. He defined “Chandra Sekhar limit” which is used in the study of black holes.

Question 6.
What is beta (β) decay? Which force is a function of it?
Answer:
In β-decay the nucleus emits an electron and an uncharged particle called neutrino.

β – decay is due to weak nuclear forces.

❖ Some physicists and their major contributions

Name	Major contribution/ Discovery
1. Archimedes	Principle of buoyancy, Principle of the lever
2. Galileo Galilei	Law of inertia
3. Isaac Newton	Universal law of gravitation; Laws of motion, Corpuscular theory of light; Reflecting telescope.
4. C.V.Raman	Inelastic scattering of light by molecules.
5. Edwin Hubble	Expanding universe
6. Hideki Yukawa	Theory of nuclear forces
7. S. Chandrasekhar	Chandrasekhar limit, structure and evolution of stars
8. Michael Faraday	Electromagnetic induction laws
9. James Clark Maxwell	Electromagnetic theory – light – electromagnetic waves
10. J.J.Thomson	Electron
11. Albert Einstein	Explanation of photoelectric effect and theory of relativity
12. R.A.Millikan	Measurement of charge of electron.
13. Ernest Rutherford	Nuclear model of atom
14. John Bardeen	Transistors; Theory of super conductivity.

❖ Fundamental forces of nature

Name	Relative strength (N)
1. Gravitational force	10-39
2. Weak nuclear forces	10-13
3. Electromagnetic forces	10-2
4. Strong nuclear forces	1

 

❖ Fundamental constants of Physics

Physical constant	Symbol	Value
1. Speed of light in vacuum	C	3 × 108 meter/sec
2. Planck’s constant	h	6.63 × 10-34 joule.sec
3. Molar gas constant	R	8.31 joule/mole.K
4. Avogadro’s number	NA	6.02 × 1023/ mol
5. Boltzmann’s constant	K	1.38 × 10-23/mol
6. Gravitational constant	G	6.67  10-11 Newton.m2/kg2
7. Mechanical equivalent of heat	J	4.185 joule/cal.
8. Triple point of water	Ttr	273.16 K
9. Density of water at 20° C	dω	103kg/m3
10. Density of mercury	dm	13.6 × 103 kg/m3
11. Density of dry air at N.T.P.	da	1.293 kg /m3
12. Specific heat of water	sω	1 cal./gm/°C
13. Latent heat of ice	Lf	80 cal./gm
14. Latent heat of steam	Lυ	540 cal/gm (or 539)
15. √5 = 2.236, √3 = 1-732, √10 = 3.162, loge 10 = 2.3026
16. π = 3.14, π2 = 9.87, √π = 1.7772, √2 = 1.414

❖ Conversion factors:
1 metre – 100 cm
1 millimeter – 10^-3m
1 inch – 2.54 × 10^-2m
1 micron (µ) – 10^-4cm
1 Angstrom (A°) – 10^-8cm
1 fermi (f) – 10^-13 cm
1 kilometer – 10³ m
1 light year – 9.46 × 1015 m
1 litre – 10³cm³
1 kilogram – 1000 gm
1 metricton – 1000 kg
1 pound – 453.6 gm
1 atomic mass
unit (a.m.u) 1.66 × 10^-27 kg
1 a.m.u – 931 MeV
1 day – 8.640 × 10⁴ seconds
1 km/hour – \(\frac{5}{18}\) m/sec (or)
0.2778 meter/sec.
1 Newton – 10⁵ dynes
1 gm wt – 980.7 dynes
1 kg.wt – 9.807 Newton
1 Newton/meter² – 1 pascal

1 Pascal – 10 dyne/cm²
1 Joule – 10⁷ erg
1 kilo watt hour – 3.6 × 10⁶ joule
1 electro volt (ev) – 1.602 × 10^-19 joule
1 watt – 1 joule / sec
1 horse power (HP) – 746 watt
1 degree (° ) – 60 minute (‘)
1 Radian – 57.3 degree ( ° )
1 Poise – 1 dyne . sec / cm²
1 Poiseuille – 10 poise
(Newton, sec/m² (or) Pascal sec.)

❖ Important Prefixes:

❖ The Greek Alphabet:

❖ Formulae of geometry :
1. Area of triangle = \(\frac{1}{2}\) × base × height
2. Area of parallelogram = base × height
3. Area of square = (length of one side)²
4. Area of rectangle = length × breadth
5. Area of circle = πr² (r = radius of circle)
6. Surface area of sphere = 4πr² (r = radius of sphere)
7. Volume of cube = (length of one side of cube)³
8. Volume of parallelopiped = length × breadth × height
9. Volume of cylinder = πr²l
10. Volume of sphere = \(\frac{4}{3}\)πr³
Circumference of square = 4l
11. Volume of cone = \(\frac{1}{3}\) πr²h
12. Circumference of circle = 2πr

❖ Formulae of algebra:
(a + b)² = a² + b² + 2ab
(a – b)² = a² + b² – 2ab
(a² – b²) = (a + b) (a – b)
(a + b)³ = a³ + b³ + 3ab (a + b)
(a – b)³ = a³ – b³ – 3ab (a – b)
(a + b)² – (a – b)² = 4ab
(a + b)² + (a – b)² = 2(a² + b²)

❖ Formulae of differentiation:
1. \(\frac{d}{dx}\) (constant) = 0
differentiation with respect to x = \(\frac{d}{dx}\)
2. \(\frac{d}{dx}\) (xⁿ) = n x^{n – 1}
3. \(\frac{d}{dx}\) (sin x) = cos x
4. \(\frac{d}{dx}\) (cos x) = – sin x dx

❖ Formulae of Integration:
Integration with respect to x = ∫dx
1. ∫dx = x
2. ∫xⁿ dx = r n + 1
3. ∫sin x dx = cos x + c
4. ∫cos x dx = sin x + c

❖ Formulae of logarithm :
1. log mn = (log m + log n)
2. log\(\frac{m}{n}\) = (log m – log n)
3. log mⁿ = n log m

❖ Value of trigonometric functions :

❖ Signs of trigonometrical ratios :
sin (90° – θ) = cos θ ; sin (180° – θ) = sin θ
cos (90° – θ) = sin θ ; cos (180° – θ) = – cos θ
tan (90° – θ) = cot θ ; tan (180° – θ) = – tan θ

❖ According to Binomial theorem :
(1 + x)ⁿ ≈ (1 + nx) if x < < 1

❖ Quadratic equation:
ax² + bx + c = 0