Mechanical Properties Of Solids(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

MECHANICAL PROPERTIES OF SOLIDS MCQs

Total Questions : 40 | Page 1 of 4 pages

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 2.

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 3.

Two wires are made of the same material and have the same volume. However, wire 1 has cross sectional area 3A. If length of wire 1 increased by x on applying force F, how much force is needed to stretch wire 2 by the same amount?

Discuss Question

Answer: Option C. -> 9F
:
C

$Y = \frac{F l}{A △ l}$
so, $F = \frac{Y A^{2} △ l}{A l} = \frac{Y A^{2} △ l}{V}$
where, $A l = V =$ Volume of wire
hence $F \propto A^{2}$
$\frac{F^{'}}{F} = \frac{(3 A)^{2}}{A^{2}} = 9$
or F'=9F

Question 4.

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 5.

A wire is stretched under a force. If the wire suddenly snaps, the temperature of the wire

remains the same
decreases
increases
first decreases then increases

Discuss Question

Answer: Option C. -> increases
:
C

The work done on the wire to produce a strain in it will be stored as energy which is converted into heat, when wire snaps suddenly

Due to this the temperature increases

Question 6.

A student performs an experiment to determine the Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the students measures the extension in the length of the wire to be 0.8 mm with an uncertainty of +/- 0.05 mm at a load of exactly 1kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of +/-0.01 mm.The Young’s modulus obtained from the reading is $(g = 9.8 m s^{- 2})$

$(2.0 + / - 0.3) \times 10^{11} N m^{- 2}$
$(2.0 + / - 0.2) \times 10^{11} N m^{- 2}$
$(2.0 + / - 0.1) \times 10^{11} N m^{- 2}$
$(2.0 + / - 0.05) \times 10^{11} N m^{- 2}$

Discuss Question

Answer: Option B. ->

(2.0 + / - 0.2) \times 10^{11} N m^{- 2}

:
B

$Y = \frac{4 F l}{π D^{2} △ l}$
$= \frac{4 \times (1 X 9.8) X 2}{(\frac{22}{7}) (0.4 \times 10^{-} 3)^{2} X (0.8 X 10^{- 3})}$
$= 2 \times 10^{11} N m^{- 2}$
Fractional error in y is
$\frac{△ y}{y} = + / - (\frac{2 △ D}{D} + \frac{△ l}{l})$
$o r △ y = + / - (\frac{2 X 0.01}{0.4} + \frac{0.05}{0.8}) \times 2 \times 10^{11}$
$= + / - 0.2 \times 10^{11} N m^{- 2}$
Therefore $y = (2 \times 10^{11} + / - 0.2 \times 10^{11}) N m^{- 2}$

Question 7.

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 8.

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
$α_{1}$ : $α_{2}$ = 2:3, the thermal stress developed in the two rods are equal provided $Y_{1} : Y_{2}$ equal to

2 : 3
4 : 9
1 : 2
3 : 2

Discuss Question

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

From the formula for thermal stress
$(F / A)_{1} = α_{1} y_{1} △ T$
$(F / A)_{2} = α_{2} y_{2} △ T$
Since,temperature $△ T$ is same
$\frac{(\frac{F}{A})_{1}}{(\frac{F}{A})_{2}} = \frac{α_{1} y_{1}}{α_{2} y_{2}}$
For thermal stress to be equal
$α_{1} y_{1} = α_{2} y_{2}$
$o r \frac{y_{1}}{y_{2}} = \frac{α_{2}}{α_{1}} = \frac{3}{2}$ since $\frac{α_{1}}{α_{2}} = 2 : 3)$

Question 9.

The pressure of a medium is changed from $1.01 \times 10^{5}$ Pa to $1.165 \times 10^{5}$ Pa and change in volume is 10% keeping temperature constant. The bulk modulus of the medium is

$204.8 \times 10^{5}$ Pa
$102.4 \times 10^{5}$ Pa
$51.2 \times 10^{5}$ Pa
$1.55 \times 10^{5}$ Pa

Discuss Question

Answer: Option D. ->

1.55 \times 10^{5}

Pa
:
D

Here
$△ P = (1.165 \times 10^{5} - 1.01 \times 10^{5})$
$= 0.155 \times 10^{5}$ Pa
$\times V / V = 10 / 100 = 1 / 10$
Bulk modulus $k = \frac{△ p}{△ v / v}$
$= \frac{0.155 \times 10^{5}}{\frac{1}{10}}$
$= 1.55 \times 10^{5} P a$

Question 10.

If S is stress and Y is Young’s modulus of material of a wire, the energy stored in the wire per unit volume is

$\frac{S}{2 y}$
$\frac{2 y}{S^{2}}$
$\frac{S^{2}}{2 y}$
$2 S^{2} Y$

Discuss Question

Answer: Option C. ->

\frac{S^{2}}{2 y}

:
C

Energy stored per unit volume
$U = \frac{1}{2} s t r e s s \times s t r a i n$
$= \frac{1}{2} s t r e s s \times \frac{s t r a i n}{y}$
$= \frac{1}{2} S \times \frac{S}{y}$
$= \frac{S^{2}}{2 y}$