Mechanical Properties Of Solids(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

MECHANICAL PROPERTIES OF SOLIDS MCQs

Total Questions : 40 | Page 4 of 4 pages

Question 31. Two wires of the same material and length are stretched by the same force. Their masses are in the ratio 3:2. Their elongations are in the ratio

3 : 2
9 : 4
2 : 3
4 : 9

Discuss Question

Answer: Option C. -> 2 : 3
:
C

Y = \frac{F l}{A △ l} o r △ l \propto \frac{1}{A}

Again,

m = A l p, m \propto A

△ l \propto \frac{1}{m}

Therefore

\frac{△ l_{1}}{△ l_{2}} = \frac{m_{2}}{m_{1}} = \frac{2}{3}

Question 32. A wire is stretched 1 mm by a force of 1 KN. How far would a wire of the same material and length but of four times that diameter be stretched by the same force?

12mm
12mm
18mm
116mm

Discuss Question

Answer: Option D. -> 116mm
:
D

Y = \frac{F l}{A △ l}

Y,l and F are constants
Therefore

△ l \propto \frac{1}{D^{2}}

\frac{△ l_{2}}{△ l_{1}} = \frac{D_{1}^{2}}{D_{2}^{2}} = \frac{1}{16}

Therefore

△ l_{2} = \frac{1}{16}

Question 33. The ratio of diameters of two wires of same materials is n : 1.The length of each wire is 4 m. On applying the same load, the increase in length of thin wire will be

(n > 1)

n2 times
n times
2n times
(2n + 1) times

Discuss Question

Answer: Option A. -> n2 times
:
A

Y = \frac{\frac{F}{A}}{\frac{△ l}{l}} = \frac{F l}{A △ l}

Y = \frac{F l X 4}{π D^{2} X △ l} o r △ l \propto \frac{1}{D^{2}} o r \frac{△ L_{2}}{△ L_{1}} = \frac{D_{1}^{2}}{D_{1}^{2}} = \frac{n^{2}}{1}

Question 34. Two rods A and B of the same material and length have their radii

r_{1}

and

r_{2}

respectively. When they are rigidly fixed at one end and twisted by the same couple applied at the other end, the ratio of the angle of twist at the end of A and the angle of twist at the end of B is

r42r41
r41r42
r22r21
r21r22

Discuss Question

Answer: Option A. -> r42r41
:
A

τ = \frac{π n r^{4}}{21} θ

In the given problem,

r^{4} θ =

constant
Therefore

\frac{θ_{A}}{θ_{B}} = \frac{r_{2}^{4}}{r_{1}^{4}}

Question 35. Two wires of the same material have lengths in the ratio 1 : 2 and their radii are in the ratio 1 :

\sqrt{2}

. If they are stretched by applying equal forces, the increase in their lengths will be in the ratio

√2:2
2: √2
1 : 1
1 : 2

Discuss Question

Answer: Option C. -> 1 : 1
:
C

Y = \frac{F l}{π r^{2} △ l} o r △ l = \frac{F}{π r^{2} y}

△ l \propto \frac{1}{r^{2}}

△ l^{'} \propto \frac{21}{(\sqrt{2} r)^{2}} o r △ l^{'} \propto \frac{1}{r^{2}}

Therefore

\frac{△ l}{△ l^{'}}

Question 36. Two wires of the same material (Young’s modulus Y) and same length L but radii R and 2R respectively are joined end to end and a weight w is suspended from the combination as shown in the figure. The elastic potential energy in the system is

3w2L4π R2Y
3w2L8π R2Y
5w2L8π R2Y
w2Lπ R2Y

Discuss Question

Answer: Option C. -> 5w2L8π R2Y
:
C

K_{1} = \frac{y π (2 R)^{2}}{L}

K_{2} = \frac{y π (R)^{2}}{L}

Equivalent

\frac{1}{K_{1}} + \frac{1}{K_{2}} = \frac{1}{4 y π R^{2}} + \frac{L}{y π R^{2}}

Since,

K_{1} x_{1} = K_{2} x_{2} = w

Elastic potential energy of the system
U=

\frac{1}{2} k_{1} x_{1} + \frac{1}{2} k_{2} x_{2}^{2}

= \frac{1}{2} k_{1} (\frac{W}{k_{1}})^{2} + \frac{1}{2} k_{2} (\frac{W}{k_{2}})^{2}

= \frac{1}{2} W^{2} + (\frac{1}{k_{1}} + \frac{1}{k_{2}})

= \frac{1}{2} W^{2} + (\frac{5 L}{4 y π R^{2}})

(\frac{5 w^{2} L}{8 π y R^{2}})

Question 37. An aluminium rod, Young’s modulus

7 \times 10^{9} N m^{- 2},

has a breaking strain of 0.2%. The minimum cross sectional area of the rod in

m^{2}

in order to support a load of

10^{4}

N is

1× 10−2
1.4× 10−3
1× 10−3
7.1× 10−4

Discuss Question

Answer: Option D. -> 7.1× 10−4
:
D

Y = \frac{\frac{F}{A}}{B r e a k i n g s t r a i n}

a = \frac{F}{Y \times B r e a k i n g s t r a i n} = \frac{10^{4} \times 100}{7 \times 10 \times 0.2}

= 0.71 \times 10^{- 3} = 7.1 \times 10^{- 4}

Question 38. What amount of work is done in increasing the length of a wire through unity?

YL2A
YL22A
YA2L
YLA

Discuss Question

Answer: Option C. -> YA2L
:
C
Work done

= \frac{1}{2} F \times

extension

= \frac{1}{2} \times \frac{Y A}{L} \times 1 y = \frac{F \times L}{A \times 1}

= \frac{Y A}{2 L} h e n c e F = \frac{Y A}{2 L}

Question 39. Two wires of same material and radius have their lengths in ratio 1 : 2. If these wires are stretched by the same force, the strain produced in the two wires will be in the ratio

2 : 1
1 : 1
1 : 2
1 : 4

Discuss Question

Answer: Option C. -> 1 : 2
:
C
l=

\frac{F L}{A Y}

\propto \frac{1}{r^{2}}

\frac{l_{1}}{l_{2}} = \frac{L_{1}}{L_{2}} \times \frac{r_{2}^{2}}{r_{1}^{2}}

o r \frac{l_{1}}{l_{2}} = \frac{1}{2}

Question 40. A wire extends by 1 mm when a force is applied. Double the force is applied to another wire of same material and length but half the radius of cross section. The elongation of the wire in mm will be

Discuss Question

Answer: Option A. -> 8
:
A