Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

The bulk modulus of water is $$2.1 \times 10^9 N/m^2$$. The pressure required to increase the density of water by $$0.1$$% is:-

A
$$2.1 \times 10^5 N/m^2$$

B
$$2.1 \times 10^3 N/m^2$$

C
$$2.1 \times 10^6 N/m^2$$

D
$$2.1 \times 10^7 N/m^2$$

Solution

$$\begin{array}{l} \frac { { \Delta V } }{ V } =\frac { { -\Delta P } }{ P } =-{ 10^{ -3 } } \\ \Rightarrow \Delta P=-B\frac { { \Delta V } }{ V } =2.1\times { 10^{ 9 } }\times { 10^{ -3 } }=2.1\times { 10^{ 6 } } \end{array}$$
Question 2
1 / -0

A plank of mass $$M$$ is suspended horizontally by using two wires as shown in the figure.They have same length $$(L)$$ and same cross sectional area $$(A)$$. Their Young's modulus are $$Y_1$$ and $$Y_2$$ respectively. The elastic potential energy of the system will be

A
$$\dfrac{2M^2g^2L}{A(Y_1+Y_2)}$$

B
$$\dfrac{m^2g^2L}{A(Y_1+Y_2)}$$

C
$$\dfrac{m^2g^2L}{2A(Y_1+Y_2)}$$

D
$$\dfrac{M^2g^2L(Y_1+Y_2)}{2AY_1Y_2}$$

Solution

In both wires tension =$$\left(\dfrac{m_2}{2}\right)=F$$
Total strain energy $$=S_1+S_2$$
$$=\dfrac{2F^2l}{Y_1}+\dfrac{2F^2l}{y_2}$$
$$=\dfrac{2M^2g^2l}{AY_1}+\dfrac{2M^2g^2l}{YA}$$
$$=\dfrac{M^2y^2l}{2A(Y_1+Y_2)}$$
Question 3
1 / -0

The compressibility of water is $$ 5 \times 10^{-10}m^2/N$$ and subjected to a pressure of 15 MPa.The fractional decrease in volume will be

A
$$7.5 \times 10^{-3}$$

B
$$5 \times 10^{-3}$$

C
$$2.5 \times 10^{-3}$$

D
$$1.25 \times 10^{-3}$$

Solution

$$\begin{array}{l}\text { Compressibily }(\gamma)=\dfrac{1}{\text { Bulk Modulus }(B)}=5\times10^{-10} \\P=15 \mathrm{MPa} \\\text { We know ,} \\P=-B \dfrac{\Delta V}{V}\end{array}$$
$$\begin{aligned}&\begin{aligned}\Rightarrow\dfrac{\Delta V}{V}=-\dfrac{P}{B}=-15 \times 10^{6} \times 5\times 10^{-10} &=-75 \times 10^{-4} \\&=-7.5 \times 10^{-3}\end{aligned}\\&\text { Answer: }\quad(A)\end{aligned}$$
Question 4
1 / -0

A copper wire is having length $$2m$$ and area of cross -section $$2mm^2$$. Then amount of work done (in joule ) in increasing its length by $$0.1mm$$ will be (Young's modulus o elasticity for copper $$Y=1.2\times 10^{11}Nm^{-2}$$

A
$$6\times 10^4J$$

B
$$6\times 10^{-3}J$$

C
$$6\times 10^{3}J$$

D
$$6\times 10^{-4}J$$

Solution

$$W=\dfrac{1}{2}\times $$stress $$\times $$ volume
$$stress=y\times strain$$
$$W=\dfrac{1}{2}(strain)^2\times Vol$$
Strain $$=\dfrac{\Delta L}{L}=\left(\dfrac{10-4}{2}\right)$$
$$vol=A\times L=2\times 10^{-6}\times 2$$
$$=4\times 10^{-6}$$
$$W=\dfrac{1.2\times 10^{11}}{2}\times \dfrac{10^{-8}}{4}\times 4\times 10^{-6}$$
$$=0.6\times 10^{-3}$$
$$W=6\times 10^{-4}J$$
Question 5
1 / -0

If the temperature of a wire of length $$2m$$ area of cross-section $$1{cm}^2$$ is increased from $${0^ \circ}C$$ to $${80^ \circ}C$$ and is not allowed to increase in length, then force required for it is {$$Y = 10^{10}N/m^2, \alpha = 10^{ - 6}/^oC$$}

A
$$80N$$

B
$$160N$$

C
$$400N$$

D
$$120N$$

Solution

80N

Formula,

$$F=YA\alpha \Delta T$$

$$=10^{10}\times 0.0001\times 10^{-6}\times (80-0)$$

$$\therefore F=80N$$
Question 6
1 / -0

A mild steel wire of length $$2L$$ and cross-sectional area $$A$$ is stretched, well within elastic limit, horizontally between two pillars. A mass $$m$$ is suspended from the mind point of the wire. Strain in the wire is

A
$$\dfrac{x^{2}}{2L^{2}}$$

B
$$\dfrac{x}{L}$$

C
$$\dfrac{x^{2}}{L}$$

D
$$\dfrac{x^{2}}{2L}$$

Solution

$$\begin{array}{l}\text { Increase in Length }=\Delta L=A B+B C - A C=2 A B-A C \\\qquad A B=\sqrt{x^{2}+L^{2}} \quad , A C=2 L\\\Rightarrow\quad \Delta L=2 \sqrt{x^{2}+L^{2}}-2 L\\\Rightarrow \Delta L=2 L\left(1+\left(\dfrac{x}{L}\right)^{2}\right)^{1 / 2}-2 L \\\Rightarrow\Delta L=2 L\left \{\sqrt{1+\dfrac{x^{2}}{L^{2}}}-1\right\}.\end{array}$$

$$\begin{array}{l}\text { Using Binomial theorem. }\\\left(1+\dfrac{x^{2}}{L^{2}}\right)^{\dfrac{1}{2}}=1+\dfrac{x^{2}}{2 L^{2}} \\\Delta L=2 L\left\{1+\dfrac{x^{2}}{2 L^{2}}-1\right\}=\dfrac{x^{2}}{L}\\\text { strain }=\dfrac{\Delta l}{2 L}=\dfrac{x^{2}}{2 L^{2}}\end{array}$$

$$\text { Ans } \rightarrow A$$
Question 7
1 / -0

A hydraulic press contains 250 litres of oil. Find the decrease in volume of the oil when its pressure increase to $$10^7 Pa$$. The bulk modulus of the oil is $$K=5 \times 10^9 Pa$$.

A
-0.8 lit

B
-0.5 lit

C
-0.6 lit

D
-0.9 lit

Solution

Given,

$$\Delta P=10^7Pa$$

$$V=250 $$ lit

$$K=5\times 10^9 Pa$$

The bulk modulus of the oil is given by,

$$K=-\dfrac{\Delta P}{\Delta V/V}$$

$$\Delta V=-\dfrac{V\Delta P }{K}$$

$$\Delta V=-\dfrac{250\times 10^7}{5\times 10^9}$$

$$\Delta V=-0.5 $$ lit

The correct option is B.
Question 8
1 / -0

One end of uniform wire of length L and of weight W is attached rigidly to a point in the roof and a weight $$ W_1 $$ is suspended from its lower end. If s is the area of cross section of the wire, the stress in the wire at a height $$\dfrac{L}{4}$$ from its lower end is -

A
$$ \dfrac {W_1} {s} $$

B
$$ \dfrac { \left[ { W }_{ 1 }+\dfrac { W }{ 4 } \right] }{ s } $$

C
$$ \dfrac { \left[ { W }_{ 1 }+\dfrac { 3W }{ 4 } \right] }{ s } $$

D
$$ \dfrac { { W }_{ 1 }+W }{ 4 } $$

Solution

Consider the FBD as:
Let x be the weight of wire part of $$l=l/4$$
As wire is uniform,
$$\dfrac{x}{\dfrac{l}{4}}=\dfrac{w}{l}$$
$$\Rightarrow x=w/4$$
For equilibrium,
$$F=w_1+x=w_1+w/4$$
So stress at point P$$=\dfrac{F}{s}=\dfrac{(w_1+w/4)}{s}$$
option B is correct.
Question 9
1 / -0

When a metal wire is stretched by a load, the fractional change in its volume $$\Delta V/V$$ is proportional to?

A
$$-\dfrac{\Delta l}{l}$$

B
$$\left(\dfrac{\Delta l}{l}\right)^2$$

C
$$\sqrt{\Delta l/l}$$

D
None of these

Solution

$$v=\dfrac { \pi { d }^{ 2 }l }{ 4 } $$

$$⟹\dfrac { ΔV }{ V } =\dfrac { 2Δd }{ d } +\dfrac { Δl }{ l } $$
$$⟹d\frac { ΔV }{ V } =\dfrac { (1−2σ)Δl }{ l } $$

$$(\dfrac { Δd }{ d } =\dfrac { −σΔl }{ l } )$$
where $$σ$$ is Poisson's ratio.
Question 10
1 / -0

A cubical ball is taken to a depth of $$200m$$ in a sea. The decrease in volume observed to be $$0.1\%$$. The bulk modulus of the ball is ($$10={ ms }^{ -2 }$$)

A
$$2\times { 10 }^{ 7} \quad Pa$$

B
$$2\times { 10 }^{ 6} \quad Pa$$

C
$$2.1\times { 10 }^{ 9} \quad Pa$$

D
$$1.2\times { 10 }^{ 9} \quad Pa$$

Solution

We know,

$$\beta=\dfrac{-PV}{\Delta V}$$

$$\beta=\dfrac{-2.1\times 10^6}{-10^{-3}}$$

$$\beta=2.1\times 10^9$$

Option $$\textbf C$$ is the correct answer

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 60

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Mechanical Properties of Solids Test - 60

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Question 1

$$2.1 \times 10^5 N/m^2$$

$$2.1 \times 10^3 N/m^2$$

$$2.1 \times 10^6 N/m^2$$

$$2.1 \times 10^7 N/m^2$$

Solution

Question 2

$$\dfrac{2M^2g^2L}{A(Y_1+Y_2)}$$

$$\dfrac{m^2g^2L}{A(Y_1+Y_2)}$$

$$\dfrac{m^2g^2L}{2A(Y_1+Y_2)}$$

$$\dfrac{M^2g^2L(Y_1+Y_2)}{2AY_1Y_2}$$

Solution

Question 3

$$7.5 \times 10^{-3}$$

$$5 \times 10^{-3}$$

$$2.5 \times 10^{-3}$$

$$1.25 \times 10^{-3}$$

Solution

Question 4

$$6\times 10^4J$$

$$6\times 10^{-3}J$$

$$6\times 10^{3}J$$

$$6\times 10^{-4}J$$

Solution

Question 5

$$80N$$

$$160N$$

$$400N$$

$$120N$$

Solution

Question 6

$$\dfrac{x^{2}}{2L^{2}}$$

$$\dfrac{x}{L}$$

$$\dfrac{x^{2}}{L}$$

$$\dfrac{x^{2}}{2L}$$

Solution

Question 7

-0.8 lit

-0.5 lit

-0.6 lit

-0.9 lit

Solution

Question 8

$$ \dfrac {W_1} {s} $$

$$ \dfrac { \left[ { W }_{ 1 }+\dfrac { W }{ 4 } \right] }{ s } $$

$$ \dfrac { \left[ { W }_{ 1 }+\dfrac { 3W }{ 4 } \right] }{ s } $$

$$ \dfrac { { W }_{ 1 }+W }{ 4 } $$

Solution

Question 9

$$-\dfrac{\Delta l}{l}$$

$$\left(\dfrac{\Delta l}{l}\right)^2$$

$$\sqrt{\Delta l/l}$$

None of these

Solution

Question 10

$$2\times { 10 }^{ 7} \quad Pa$$

$$2\times { 10 }^{ 6} \quad Pa$$

$$2.1\times { 10 }^{ 9} \quad Pa$$

$$1.2\times { 10 }^{ 9} \quad Pa$$

Solution

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