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 1 of 4 pages

Question 1. The bulk modulus for an incompressible liquid is

zero
unity
infinity
between 0 and 1

Discuss Question

Answer: Option C. -> infinity
:
C
The bulk modulus is

K = - \frac{p v}{△ v}

If liquid is incompressible so

△ V = 0

Hence, K=

- \frac{p v}{0}

=

infinity

Question 2. The Young's modulus of a rubber string 8 cm long and density 1.5

\frac{k g}{m^{3}}

is 5

\times 10^{8} \frac{N}{m^{2}}

. It is suspended from the ceiling in a room. The increase in length due to its own weight will be

9.6× 10−5 m
9.6× 10−11m
9.6× 10−3m
9.6 m

Discuss Question

Answer: Option B. -> 9.6× 10−11m
:
B
b):l =

\frac{L^{2} d g}{2 Y}

\frac{(8 \times 10^{- 2})^{2} \times 1.5 \times 9.8}{2 \times 5 \times 10^{8}}

= 9.6

\times 10^{- 11} m

Question 3. The coefficient of linear expansion of brass and steel are

α_{1}

and

α_{2}

. If we take a brass rod of length

l_{1}

and steel rod of length

l_{2} a t 0^{\circ} C,

their difference in length

(l_{2} - l_{1})

will remain the same at a temperature if

α1l2=α2l1
α1l22=α2l21
α21l1=α22l2
α1l1=α2l2

Discuss Question

Answer: Option D. -> α1l1=α2l2
:
D

d) L_{2} = l_{2} (l + α_{2} △ θ)

and

L_{1} = l_{1} (l + α_{1} △ θ)

\Rightarrow (L_{2} - L_{1}) = (l_{2} - l_{1}) + △ θ (l_{2} α_{2} - l_{1} α_{1})

Now

(L_{2} - L 1) = (l_{2} - l_{1})

so,

l_{2} α_{2} - l_{1} α_{1} = 0

Question 4. When a rod is heated but prevented from expanding, the stress developed is independent of

material of the rod
rise in temperature
length of rod
none of these

Discuss Question

Answer: Option C. -> length of rod
:
C

L_{t} = L_{0} (1 + α △ θ)

△ L = L_{t} - L_{0} = L_{0} α △ θ

If the same rod of lenghth

L_{0}

is subjected to stress along its length, then extension in length can be calculated by Hooke's law

Y = \frac{s t r e s s}{s t r a i n} = \frac{s t r e s s}{\frac{△ L}{L_{n}}}

= \frac{L_{0} \times s t r e s s}{△ L}

Therefore

△ L = \frac{L_{0} \times s t r e s s}{Y}

If the rod is prevented from expanding, we have

L_{o} α △ θ = \frac{L_{o} \times s t r e s s}{Y}

Therefore stress

= Y α △ θ

= (independent of

L_{0})

Question 5. The Poisson’s ratio of the material is 0.5. If a force is applied to a wire of this material, there is a decrease in the cross sectional area by 4%. The percentage increase in its length is

1%
2%
2.5%
4%

Discuss Question

Answer: Option D. -> 4%
:
D
Given Poisson's ration= 0.5
It shows that the density of material is constant
Therefore the change in volume of the wire is zero
Thus

V = A \times l

=constant
log V=log A+log |
therefore

\frac{△ V}{V} = 0 = \frac{△ A}{A} + \frac{△ l}{l}

\frac{△ l}{l} = - \frac{△ A}{A}

or % increase in length=

\frac{△ l}{l} \times 100

=-(-4)
=4%

Question 6. Two identical wires of rubber and iron are stretched by the same weight, then the number of atoms in the iron wire will be

equal to that of rubber
less than that of the rubber
more than that of rubber
none of the above

Discuss Question

Answer: Option C. -> more than that of rubber
:
C
Iron is more elastic than rubber, hence the number of atoms per unit volume will be more in case of iron.

Question 7. The diameter of a brass wire is 0.6 mm and Y is

9 \times 10^{6} N m^{- 2} .

The force which will increase its length by 0.2% is about

100 N
51 N
25 N
none of these

Discuss Question

Answer: Option B. -> 51 N
:
B

F = \frac{Y A △ l}{l}

= 9 \times 10^{10} \times \frac{22}{7} \times \frac{(0.6 \times 10^{- 3})^{2}}{4} \times \frac{0.2}{100} N

=51 N approx

Question 8. A cube is compressed at

0^{\circ} C

equally from all sides by an external pressure p. By what amount should the temperature be raised to bring it back to the size it had before the external pressure was applied? (Given K is bulk modulus of elasticity of the material of the cube and

α

is the coefficient of linear expansion)

pKα
p3Kα
3παp
K3p

Discuss Question

Answer: Option B. -> p3Kα
:
B

k = \frac{p V}{△ v} = \frac{p V}{y △ T} = \frac{p}{3 α T}

T = \frac{p}{3 k α}

Question 9. A wire 3 m in length and 1 mm in diameter at 303K is kept at a low temperature of 103K and is stretched by hanging a weight of 10 kg at one end. The change in length of the wire is

(Y = 2 \times 10^{11} N m^{- 2}, g = 10 m s^{- 2}

and =

1.2 \times 10^{- 5} K^{- 1})

5.2 mm
2.5 mm
52 mm
25 mm

Discuss Question

Answer: Option A. -> 5.2 mm
:
A
The contraction in the length of the wire due to change in the temperature

= α L Δ T

= 1.2 \times 10^{- 5} \times 3 \times (103 - 303)

= - 7.2 \times 10^{- 3} m

The expansion in the length of wire due to stretching force

= \frac{F L}{A Y} = \frac{(10 X 10) X 3}{(0.75 \times 10^{- 6}) (2 \times 10^{11})}

= 2 \times 10^{- 3} m

Resultant change in length

= - 7.2 \times 10^{- 3} + 2 \times 10^{- 3}

= - 5.2 \times 10^{- 3}

=-5.2 mm
Negative sign shows contraction

Question 10. Two rods of different materials having coefficients of linear expansion

α_{1}

and

α_{2}

and Young’s modulus

Y_{1}

and

Y_{2}

respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of rods. If