Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 66

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Question 1
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The $$'\sigma '$$ of a material is 0.20. If a longitudinal strain of $$4.0\times { 10 }^{ -3 }$$ is caused, by what percentage will the volume change -

A
0.48 %

B
0.32 %

C
0.24 %

D
0.50 %

Solution

$$\begin{array}{l}\text { Longitudenal } \text { Strain }=4\times 10^{-3} \\\text {Lateral Strain }=\sigma \times 0.4\% \\=0.08 \%\end{array}$$

$$\begin{aligned}\text { Volumetric strain } &=\text { long}\text {itudinal Strain }-2 \times \text { lateral Strain } \\&=4 \times 10^{-3}-2 \times 8 \times 10^{-4}\\&=0.24\%\end{aligned}$$
Question 2
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When a wire is stretched, an amount of work is done. What is the amount of work done in stretching a wire through $$0.1mm$$, if its length is $$2m$$ and area of cross - section, $$10^-6m^2 ( Y=2 \times10^11 N/m^2)$$

A
$$5 \times 10^-1 J$$

B
$$5 \times 10^-2 $$

C
$$5 \times 10^-3 J $$

D
$$ 5 \times 10^-4 J $$

Solution

Young Modulus of Elasticity
$$Y=\frac{FL}{Al}$$
$$F=\frac{YAl}{L}$$
$$F=\frac{2\times 10^{11}\times 1\times 10^{-6}\times 10^{-3}\times 10^{-1}}{2}$$
$$F= 100\,N$$
Work Done = $$\frac{1}{2}\times 100\times 1\times 10^{-3}\times 10^{-1}$$
$$ = 0.005 \, J$$
Work done = $$5\times 10^{-3}\, joule$$
Question 3
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A square frame of ABCD consisting of five steel bars of cross section area $$400 \,mm^2$$ and joined by pivot is subjected to action of two forces $$P =40 \, kN$$ in the direction of the diagonal as shown.Find change in angle at $$A$$ if Young's modulus $$Y=2\times 10^5 \,N/min.$$

A
$$\dfrac{1}{2000}$$rad

B
$$\dfrac{1}{1000}$$rad

C
$$\dfrac{\sqrt{2}}{1000}$$rad

D
$$\dfrac{\sqrt{3}}{1000}$$rad
Question 4
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A wire is subjected to a tensile stress. If A represents area of cross-section, L represents original length, I represents extension and Y is Young's modulus of elasticity, then elastic potential energy of the stretched wire is

A
$$U=\frac{2L}{AY} I^2$$

B
$$U=\frac{AL}{2Y} I^2$$

C
$$U=\frac{AY}{2L} I^2$$

D
$$U=\frac{1}{4} \frac{AY}{L} I^2$$

Solution

Work done $$=\dfrac{1}{2}\times stress \times strain \times volume $$

$$=\dfrac{1}{2}\times Y\times \left( strain \right)^2\times volume$$

  $$=\dfrac{1}{2}\times Y\times \left( \dfrac{I}{L}\right)^2 \times AL$$

  $$=\dfrac{1}{2}\times Y\times \dfrac{I^2}{L^2}\times AL$$

  $$=\dfrac{YI^2A}{2L}$$

  $$=\dfrac{YI^2A}{2L}$$
Hence, the answer is $$\dfrac{YI^2A}{2L}.$$
Question 5
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A bar is subjected to an axial forces as shown in figure. Find the total elongation in the bar. (E is the modulus of elasticity of the bar and A is its cross-section)

A
$$\dfrac{FI}{AE}$$

B
$$\dfrac{2FI}{AE}$$

C
$$\dfrac{3FI}{AE}$$

D
$$\dfrac{4FI}{AE}$$

Solution

In given figure,
Considering the two halves separately.
Since, this heavy need to be in equilibrium.
An additional force facts towards light.
So, $$\Delta h = \frac{F}{A} \times \frac{l}{E} = \frac{{3Fl}}{{AE}} = \frac{{3Fl}}{{AE}}$$
The other haly becomes.
$$\Delta {h_2} = \frac{{Fl}}{{AE}}$$
Net elongation $$ = \frac{{4Fl}}{{AE}}$$
Question 6
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For maximum extension with given load for a wire with same value of Young's modulus of elasticity, which is true?

A
$$l=300 cm, d= 10 cm$$

B
$$l=200 cm, d=5 cm$$

C
$$l=100 cm, d=2 cm$$

D
$$l=50 cm, d=0.5 cm$$

Solution

$$ \begin{array}{l} \text { Given } \\ \text { - load is same for all } \\ \text { - young modulus of elasticity is also same } \end{array} $$
We know.
$$ Y=\frac{\text { stress }}{\text { strain }} $$
$$Y=\frac{F_{\text {load }}}{A\left(\frac{\Delta l}{l}\right)}$$
$$\left[\begin{array}{l}\text { Stress }=\frac{\text { Fioad }}{\text { Area of cross section }} \\ \text { Strain }=\frac{\text { change in length }}{\text { total length }}\end{array}\right.$$
$$ \Delta l=\frac{F_{\text {load }} l}{A Y}=\frac{4 F_{\text {load }} l}{\pi d^{2} Y} $$
$$ \begin{array}{l} \text { - For } \Delta l \text { to be maximum }\left(\frac{l}{d^{2}}\right) \text { should be maximum } \\ \text { By checking option (d) will be correct option } \end{array} $$
Question 7
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A wire of density $$ 9 \times 10^3 Kg/m^3 $$ is stretched between two clamps 1 m apart and is stretched to an extension of $$ 4.9 \times 10^{-4} $$ meter, Young's modulus of material is $$ 9 \times 10^{10} N/m^2 $$. Then

A
The lowest frequency of standing waveis 35 Hz

B
The frequency of 1 st overtone is 70 Hz

C
The frequency of 1 st overtone iis 105 Hz

D
A and B both

Solution

$$\begin{array}{l}\text { This is a fundamental standing wave } \\\qquad \begin{aligned}\Rightarrow L &=\frac{\lambda}{2}\\& \lambda=2 L \\\text { we know that, }&V=n\lambda\\\text { Also } & Y=\sqrt{\frac{T}{\mu}}\end{aligned}\end{array}$$

$$\begin{aligned}n_{0} &=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}\\\text { Given, } & \rho=a \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3} \\& l=1 \mathrm{~m} \\& \Delta l=4.9\times10^{-4} \mathrm{~m} \\&Y=9\times10^{10}\mathrm{~N} / \mathrm{m}^{2}\end{aligned}$$

$$\begin{aligned}n_{0} &=\frac{1}{2 \times 1} \sqrt{\frac{V{A }\Delta l}{l A \rho}} \\&=\frac{1}{2}\sqrt{\frac{9\times10^{10} \times 4.9 \times 10^{-4}}{1\times 9 \times 10^{3}}}{10}\\n_{0}&=35\mathrm{~Hz}\end{aligned}$$
Question 8
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The bulk modulus of a spherical object is B. If it is subjected to uniform pressure p, the fractional decrease in radius is

A
$$\frac{p}{B}$$

B
$$\frac{B}{3p}$$

C
$$\frac{3p}{B}$$

D
$$\frac{p}{3B}$$

Solution
Question 9
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When a rubber ball is taken to bottom of lake, its volume reduces by 0.1% then depth of lake (Bulk modulus rubber is $$6\times { 10 }^{ 8 }N/{ m }^{ 2 }$$)

A
60 m

B
120 m

C
150 m

D
30 m

Solution

Given decrease in volume $$\left(\frac{\Delta V}{v} \times 100\right)=-0.1 \%$$
Bulk modulus of rubber $$(\beta)=6 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}$$
depth of lake Let $$d$$
we Know Putting value $$ 6 \times 10^{8}=\frac{+10^{3} \times 10 \times d}{+0.1 \times 10^{-2}} \quad\left\{\begin{array}{l} P_{\text {water }}=10^{3} \mathrm{Kg} / \mathrm{m}^{3} \\ g=10 \mathrm{~m} / \mathrm{s}^{2} \end{array}\right. $$ $$ d=60 \mathrm{~m} $$
Question 10
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When a rubber cord is stretched, the change in volume with respect to change in its linear dimensions is negligible. The Poisson's ratio for rubber is

A
1

B
0.25

C
0.5

D
0.75

Solution

$$V = \pi {r^2}I$$
$$\frac{{\Delta V}}{V} = \frac{{\Delta \left( {\pi {r^2}I} \right)}}{{\pi {r^2}I}}$$
$$\frac{{\Delta V}}{V} = \frac{{{r^2}\Delta I + 2rI\Delta r}}{{{r^2}I}}$$
$$\frac{{\Delta V}}{V} = \frac{{\Delta I}}{I} + \frac{{2\Delta r}}{r}$$
But $$\frac{{\Delta V}}{V} = 0$$
therefore$$,$$ $$\frac{{\Delta I}}{I} = \frac{{ - 2\Delta r}}{r}$$
Now$$,$$ Poisson's ratio$$,$$ $$\sigma = \frac{{\frac{{ - \Delta r}}{r}}}{{\frac{{\Delta I}}{I}}}$$-------------------$$(1)$$
from equation $$(1),$$
$$\sigma = - \left( {\frac{{\frac{{\Delta r}}{r}}}{{\frac{{ - 2\Delta r}}{r}}}} \right) = \frac{1}{2} = 0.5$$
Hence,
option $$(C)$$ is correct answer.