Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

The stress required to double the length of wire (or) to produce $$100\%$$ longitudinal strain is:

A
$$Y$$

B
$$\cfrac{Y}{2}$$

C
$$2Y$$

D
$$3Y$$

Solution

$$\cfrac{Stress}{Strain}=Young's$$ $$modulus$$
$$Longitudinal$$ $$strain=\cfrac{Change\quad in\quad length}{Initial\quad length}$$
Here, $$\cfrac{2L-L}{L}=\cfrac{L}{L}=1$$
$$\therefore Y=\cfrac{Stress}{1}$$
Question 2
1 / -0

A uniform rod of length $$60 cm$$ and mass $$6kg$$ is acted upon by two forces as shown in the diagram. The force exerted by $$45 cm$$ part of the rod on $$15 cm$$ part of the rod is

A
$$9 N $$

B
$$ 18 N $$

C
$$27 N $$

D
$$30 N $$

Solution

The force exerted by $$45\;{\rm{cm}}$$ part of the rod on $$15\;{\rm{cm}}$$art of the rod is given as,

$$F = {F_1} \times \frac{{15}}{{60}} + {F_2} \times \frac{{45}}{{60}}$$

$$F = 21 \times \frac{{15}}{{60}} + 45 \times \frac{{45}}{{60}}$$

$$F = 45 \times \frac{{15}}{{60}} + 21 \times \frac{{45}}{{60}}$$

$$F = 27\;{\rm{N}}$$
Question 3
1 / -0

Two wires A and B are identical in shape and size are stretched by same magnitude of force. Then the extensions are found to be 0.2% and 0.3% respectively. Find the rate of their Young's modulii

A
2 : 3

B
3 : 2

C
4 : 9

D
9 : 4

Solution

$$\gamma{A}=\dfrac{stress}{strain}$$
$$stress_{A}=stress_{B}$$
$$\dfrac { { \gamma }_{ A } }{ { \gamma }_{ B } } =\dfrac { { strain }_{ B } }{ { stress }_{ A } } =\dfrac { 0.3\times { 10 }^{ -2 } }{ 0.2\times { 10 }^{ -2 } } $$
$$=3:2$$
Question 4
1 / -0

In the Searle's method to determine the Young's modulus of a wire, a steel wire of length $$156\ cm$$ and diameter $$0.054\ cm$$ is taken as experimental wire. The average increase in length for $$1.5\ kg\ wt$$ is found to be $$0.050\ cm$$. Then the Young's modulus of the wire is

A
$$3.002\times 10^{11}N/m^{2}$$

B
$$1.002\times 10^{11}N/m^{2}$$

C
$$2.002\times 10^{11}N/m^{2}$$

D
$$2.5\times 10^{11}N/m^{2}$$

Solution

Young’s modulus is given by

$$ Y=\dfrac{FL}{A\Delta l} $$

$$ Y=\dfrac{1.5\times 10\times 156}{3.14\times 2.7\times {{10}^{-4}}\times 5\times {{10}^{-4}}} $$

$$ Y=1.002\times {{10}^{-11}}\,N/{{m}^{2}} $$
Question 5
1 / -0

For a material $$Y={ 6.6\times 10 }^{ 10 }\ { N/m }^{ 2 }$$ and bulk modulus $$K{ 11\times 10 }^{ 10 }\ { N/m }^{ 2 }$$, then its Poisson's ratio is:

A
$$0.8$$

B
$$0.35$$

C
$$0.7$$

D
$$0.4$$

Solution

Given that,

Young’s modulus $$Y=6.6\times {{10}^{10}}\,N/{{m}^{2}}$$

Bulk modulus $$B=11\times {{10}^{10}}\,N/{{m}^{2}}$$

We know that,

$$ Y=3K\left( 1-2\mu \right) $$

$$ 6.6\times {{10}^{10}}=3\times 11\times {{10}^{10}}-66\times {{10}^{10}}\mu $$

$$ -\mu =\dfrac{\left( 6.6-33 \right)\times {{10}^{10}}}{66\times {{10}^{10}}} $$

$$ \mu =0.4 $$

Hence, the poisson’s ratio is $$0.4$$
Question 6
1 / -0

A copper wire and an aluminium wire have lengths in the ratio 3:2, diameters in the ration 2:3 and forces applied in the ration 4:5. Find the ratio of the increase in the length of the two wires.
$$\left( { Y }_{ cu }=1.1\times { 10 }^{ 11 }N{ m }^{ -2 },\quad { Y }_{ Al }=0.70\times { 10 }^{ 11 }N{ m }^{ -2 }\quad \right) $$

A
110:189

B
180:110

C
189:110

D
80:11

Solution

Youngs Modulus $$=\dfrac{FL}{A\Delta L}$$
$$\Rightarrow \Delta L=\dfrac{FL}{AY}$$
Given :- $$L_{cu}:L_m=3:2$$
$$d_{cu}:d_m=2:3$$

$$A_{cu}:A_{Al}=\dfrac{\pi}{4}(d_{cu})^2/\dfrac{\pi}{4}(d_{Al})^2$$

$$=\left(\dfrac{2}{3}\right)^2$$

$$F_{cu}:F_{Al}=4:5$$

Ratio of increase in length of wires $$=\left(\dfrac{F_{cu}}{F_{Al}}\right)\left(\dfrac{L_{cu}}{L_{Al}}\right)\left(\dfrac{A_{Al}} {A_{Cu}}\right)\left(\dfrac{Y_{Al}}{Y_{Cu}}\right)$$

$$=\left(\dfrac{4}{5}\right)\left(\dfrac{3}{2}\right)\left(\dfrac{9} {4}\right)\left(\dfrac{7}{11}\right)$$

$$=\dfrac{189}{110}$$
Question 7
1 / -0

The breaking stress of aluminium is $$7.5\times 10^{7}Nm^{-2}$$. The greatest length of aluminium wire that can hang vertically without breaking is(Density of auminium is $$2.7\times 10^{3}\ kg\ m^{-3}$$)

A
$$283\times 10^{3}\ m$$

B
$$28.3\times 10^{3}\ m$$

C
$$2.83\times 10^{3}\ m$$

D
$$0.283\times 10^{3}\ m$$

Solution

$$ \begin{array}{l} \text { Given, breaking stress of aluminium }=7.5 \times 10^{7}\mathrm{~N}/\mathrm{m}^{2}\\\text { Density of aluminium }==2.7 \times 10^{3}\mathrm{~kg}/\mathrm{m}^{3}\\\text { Stress }=\frac{F}{A}=\frac{m g}{A} \\ d=\frac{m}{A \times L}\Rightarrow\frac{m}{A}=d x L \\ \therefore \quad 7.5\times 10^{73}=2 \cdot 7 \times 10^{3} \times L \times 100\\L=2.83\times 10^{3} \mathrm{~m} \text { . } \end{array} $$
Question 8
1 / -0

A steel wire $$2\ m$$ in length is stretched by $$1\ mm$$. The cross-sectional area of the wire is $$2\ mm^{2}$$. If Young's modulus of steel is $$2\times 10^{11}\ N/m^{2}$$, then the elastic potential energy stored in the wire is

A
$$0.1\ J$$

B
$$0.2\ J$$

C
$$0.3\ J$$

D
$$0.4\ J$$

Solution

given, Steel wire length
$$=2 \mathrm{~m}$$
$$ Area=2 \mathrm{~mm}^{2} $$
Young modulus of steel $$=2 \times 10^{11}$$
Using the spring equivalent concept: $$ \begin{aligned} U &=\frac{1}{2} K(\Delta L)^{2} \quad \text { where } \quad K=\frac{Y A}{L} \\ &=\frac{1}{2}\left(\frac{Y A}{L}\right)(\Delta L)^{2} \\ &=\frac{1}{2} \times\left(\frac{2 \times 10^{11} \times 10^{-6} \times 2}{2}\left(10^{-6}\right)\right.\\ U &=0.1\mathrm{~J}\end{aligned} $$ Elastic potential energy stored is $$\underline{0.1 \mathrm{I}}$$
Question 9
1 / -0

Three bars having length $$l,2l$$ and $$3l$$ and area of cross-section $$A,2A$$ and $$3A$$ are joined rigidly and to end. Compound rod is subjected to a stretching force $$F$$. The increase in length of rod is (Young's modulles of material is $$Y$$ and bars are massless)

A
$$\dfrac {13Fl}{2\ AY}$$

B
$$\dfrac {3Fl}{AY}$$

C
$$\dfrac {9Fl}{AY}$$

D
$$\dfrac {13\ Fl}{AY}$$

Solution

Changed in length of compound bar $$=\Delta {l}_{1}+\Delta {l}_{2}+\Delta {l}_{3}$$
$$\Delta l=\cfrac{PL}{AE}=\cfrac{FL}{AE}=\cfrac{FL}{AY}$$ ($$E=$$Young's modulus)
$$=\cfrac{Fl}{AY}+\cfrac{F(2l)}{2AY}+\cfrac{F(3l)}{3AY}$$
$$=\cfrac{3Fl}{AY}$$
Question 10
1 / -0

A hydraullic press contains $$250 lit$$ of oil. Find the decrease in volume of the oil when its pressure increases to $$ 10^7 Pa$$ . The bulk modulus of the oil is $$ K = 5 \times 10^5 Pa$$

A
$$- 0.8 lit$$

B
$$- 0.5 lit$$

C
$$- 0.6 lit$$

D
$$- 0.9 lit$$

Solution

Bulk modulus $$=-\dfrac{volume \times \Delta Pressure}{\Delta volume}$$

$$\Delta volume$$ $$=-\dfrac{volume \times \Delta Pressure}{Bulk \quad modulus}$$

$$\Delta volume=-\dfrac{250\times 10^7}{5\times 10^9}=-0.5 $$ liters.

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Mechanical Properties of Solids Test - 57

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Mechanical Properties of Solids Test - 57

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Question 1

$$Y$$

$$\cfrac{Y}{2}$$

$$2Y$$

$$3Y$$

Solution

Question 2

$$9 N $$

$$ 18 N $$

$$27 N $$

$$30 N $$

Solution

Question 3

2 : 3

3 : 2

4 : 9

9 : 4

Solution

Question 4

$$3.002\times 10^{11}N/m^{2}$$

$$1.002\times 10^{11}N/m^{2}$$

$$2.002\times 10^{11}N/m^{2}$$

$$2.5\times 10^{11}N/m^{2}$$

Solution

Question 5

$$0.8$$

$$0.35$$

$$0.7$$

$$0.4$$

Solution

Question 6

110:189

180:110

189:110

80:11

Solution

Question 7

$$283\times 10^{3}\ m$$

$$28.3\times 10^{3}\ m$$

$$2.83\times 10^{3}\ m$$

$$0.283\times 10^{3}\ m$$

Solution

Question 8

$$0.1\ J$$

$$0.2\ J$$

$$0.3\ J$$

$$0.4\ J$$

Solution

Question 9

$$\dfrac {13Fl}{2\ AY}$$

$$\dfrac {3Fl}{AY}$$

$$\dfrac {9Fl}{AY}$$

$$\dfrac {13\ Fl}{AY}$$

Solution

Question 10

$$- 0.8 lit$$

$$- 0.5 lit$$

$$- 0.6 lit$$

$$- 0.9 lit$$

Solution

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