Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 76

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Question 1
1 / -0

A uniform pressure P is exerted on all sides of a solid cube at temperature t $$^oC$$. By what amount should the temperature of the cube be raised in order to bring its volume back to the value it had before the pressure was applied? The coefficient of volume expansion of cube is $$\alpha$$ and the bulk modulus is K.

A
$$\dfrac{P}{\alpha K}$$

B
$$\dfrac{P\alpha}{K}$$

C
$$\dfrac{PK}{\alpha}$$

D
$$\dfrac{2K}{P}$$

Solution

$$K=\dfrac{-PV}{\Delta V_1}$$

$$-\Delta V_1=\dfrac{PV}{K}$$

Let us suppose change in volume be $$\Delta V_2$$
when change in temperature is $$\Delta T$$

$$\Delta V_2=V\alpha\Delta T$$

For no change in volume after raising temperature,

$$\Delta V_1+\Delta V_2=0$$

$$\dfrac{PV}{K}=V\alpha\Delta T$$

$$\Delta T=\dfrac{P}{\alpha K}$$
Question 2
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Two wires are made of the same material and have the same volume. However $$wire\ 1$$ has cross-sectional area $$A$$ and $$wire\ 2$$ has cross-sectional area $$3A.$$ If the length of $$wire\ 1$$ is increases by $$\Delta x$$ on applying force $$F,$$ how much force is needed to stretch $$wire\ 2$$ by the same amount?

A
$$6F$$

B
$$9F$$

C
$$F$$

D
$$4F$$

Solution

$$\begin{array}{l}\text { Wire 1: Cross sectional area = A } \\\text { Volume }=V \\\text { Wire 2: cross sectional area = } 3 A \\\text { Volume }=V\end{array}$$

$$\begin{array}{l}\text { Since both wires are made of same } \\\text { material, so young's modulus for both } \\\text { wires is same. } \\\qquad \begin{array}{ll}A_{1} L_{1} & =A_{2} L_{2} \quad(\text { as volume is sane })\\A_{1} L_{1} & =\left(3 A_{1}\right)\left(L_{2}\right) \\L_{2} & =\frac{L_{1}}{3}...........(1)\end{array}\end{array}$$

$$\begin{array}{l}F=Y A_{1}\left(\frac{\Delta x}{L_{1}}\right) \\F^{\prime}=y\left(3 A_{1}\right)\left(\frac{3 \Delta x}{L_{1}}\right)={9 F} \\\text { Force required to stretch wire } 2 \text { by a }\\\text { distance } \triangle x \text { is } 9 \mathrm{~F}\text { . }\end{array}$$
Question 3
1 / -0

The ice storm in the state of Jammu strained many wires to the breaking point. In a particular situation, the transmission towers are separated by $$500\ m$$ of wire. The top grounding wire $$15^{o}$$ from horizontal at the towers, and has a diameter of $$1.5cm$$. The steel wire has a density of $$7860\ kg\ m^{-3}$$. When ice (density $$900\ kg\ m^{-3}$$) built upon the wire to a diameter $$10.0\ cm$$, the wire snapped. What was the breaking stress (force/ unit area) in $$N\ m^{-2}$$ in the wire at the breaking point? You may assume the ice has no strength.

A
$$7.4\ \times 10^{7}\ N\ m^{-2}$$

B
$$4.5\ \times 10^{8}\ N\ m^{-2}$$

C
$$2.6\ \times 10^{6}\ N\ m^{-2}$$

D
$$1.15\ \times 10^{7}\ N\ m^{-2}$$
Question 4
1 / -0

A wire suspended vertically from one of its ends stretched by attaching weight of $$200$$ N to the lower end. The weight stretches the wire by$$ 1 $$mm. Then the elastic energy stored in the wire is

A
$$0.2 J$$

B
$$10J $$

C
$$20 J$$

D
$$0.1 J$$

Solution

$$\begin{aligned}\text { Given, } & F=200 \mathrm{~N} \\\Delta l &=1 \mathrm{~mm}\end{aligned}$$

$$\begin{array}{l}\text { Using formula for elastic energy stored } \\\qquad U=\frac{1}{2}\left(\frac{y A}{L}\right)(\Delta L)^{2}=\frac{1}{2} \times F x^{\prime}(\Delta L)=0.1 J \\\text { Elastic Energy stored in the wire is } 0.1\mathrm{~J} \text { . }\end{array}$$
Question 5
1 / -0

Two wires of the same radius and material and having length in the ratio $$8.9:7.6$$ are stretched by the same force. The strains produced in the two cases will be in the ratio:

A
$$1:1$$

B
$$8.9:1$$

C
$$1:7.6$$

D
$$1:3.2$$

Solution

Y=$$\dfrac{stress}{strain}=\dfrac{\dfrac{F}{A}}{strain}$$
As the materials are made up of same material Youngs modulus is same for both and as radius is same for both area is also same
Therefore strains are in the ratio of 1:1
Question 6
1 / -0

Breaking stress for steel is $$F/A$$, where $$A$$ is the area of cross-section of steel wire. A body of mass $$M$$ is tied at the end of the steel wire of length $$L$$ and whirled in a horizontal circle. The maximum number of revolution it can make per second is:

A
$$\sqrt { \dfrac { FA }{ { 4\pi }^{ 2 }ML } }$$

B
$$\sqrt { \dfrac { F }{ { 4\pi }^{ 2 }MLA } }$$

C
$$\sqrt { \dfrac { F }{ { 4\pi }^{ 2 }ML } }$$

D
$$\sqrt { \dfrac { { 4\pi }^{ 2 } }{ MLA } }$$

Solution

Breaking stress for steel is $$\frac{F}{A}$$. On rotating the mass $$M$$, $$ F=m \omega^{2} L \quad \Rightarrow \quad w=\sqrt{\frac{F}{M L}} $$ $$\therefore$$ Revolutions per second $$=f=\frac{w}{2 \pi}=\sqrt{\frac{F}{4 \pi^{2} M L}}$$
Question 7
1 / -0

A spherical ball contracts in volume by $$0.01\%$$ when subjected to a normal uniform pressure of $$100$$ atmospheres. The bulk modulus of its material in dynes$$/cm^2$$ is?

A
$$10\times 10^{12}$$

B
$$100\times 10^{12}$$

C
$$1\times 10^{12}$$

D
$$2\times 10$$

Solution

Here,
$$\dfrac{Delta V}{V}=\dfrac{-0.01}{100}$$

$$P=100\ atm=100\times 10^5\ atm$$

$$\beta=\dfrac{-100\times V}{\Delta V}$$

$$=100\times 10^5\times 10^4=10\times 10^10 N/m^2$$

In $$dynes/cm^2$$ multiply it by 10,

$$=10^{12}\ dynes/cm^2$$
Question 8
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The length of a wire increases by $$8mm$$ when a weight of $$5kg$$ is suspended from it. If other things remain the same but the radius of the wire is doubled, what will be the increase in its length?

A
$$5mm$$

B
$$2mm$$

C
$$7mm$$

D
$$9mm$$

Solution

Stress = $$Y$$*Strain
$$\frac{F}{A}$$ = $$Y$$ $$(\frac{\Delta L}{L})$$
Here, $$F$$ = $$5*g$$ $$N$$
= $$ 50$$ $$N$$
$$\frac{50}{A}$$ = $$Y$$ $$(\frac{8 }{L})$$ ....(1)
($$(\Delta L)$$ = $$8$$ mm)
Now, when the radius of the wire is doubled ,
The area of cross section becomes $$4A$$.
Let the extension now be $${\Delta L'}$$.
Again , applying $$\frac{F}{A}$$ = $$Y$$ $$(\frac{\Delta L}{L})$$
$$\frac{50}{4A}$$ = $$Y$$ $$(\frac{\Delta L'}{L})$$ ....(2)

Divding equation (1) by equation (2) :
$$4$$ = $$\frac{8}{\Delta L'}$$

$${\Delta L'}$$ = $$2$$ mm.
Question 9
1 / -0

A uniform rod of mass m and length $$l$$ is rotating with constant angular velocity $$\omega$$ about an axis which passes through its one end and perpendicular to the length of rod. The area of cross section of the rod is A and its young's modulus is Y. Neglect gravity. The strain at the mid point of the rod is :

A
$$\dfrac{m\omega^2l}{8AY}$$

B
$$\dfrac{3m\omega^2l}{8AY}$$

C
$$\dfrac{3m\omega^2l}{4AY}$$

D
$$\dfrac{m\omega^2l}{4AY}$$

Solution

$$\begin{array}{l}\text { Stress }=y(\text { strain }) \\\text { Strain }=\dfrac{T_{\text {mid-point }}}{A Y}\end{array}$$

$$\begin{array}{l}\text { Let us assume a small element of } \\\text { length } d x \text { and mass dm. } \\\text { Mass per unit length }=\frac{m}{L} \\\qquad \begin{array}{rl}d & m=\left(\frac{m}{L}\right)(d x) \\d T & =(d m)(l-x) \omega^{2} \\& =\left(\frac{m}{L}\right)(1-x)\omega^{2} d x\end{array}\end{array}$$

$$\begin{array}{l}\text { Now, centripetal force at mid point is }\\\qquad \begin{aligned}\int_{0}^{l / 2} d T &=\int_{0}^{l / 2}\frac{m}{L}(l-x) w^{2} d x \\&=\frac{m w^{2}}{L}\left[l x\frac{x^{2}}{2}\right]_{0}^{1 / 2}\end{aligned}\end{array}$$

$$\begin{array}{l}T_{\text {mid-point }}=\dfrac{M \omega^{2}}{L}\left(\dfrac{L^{2}}{2}-\dfrac{L^{2}}{8}\right)=\dfrac{3 M \omega^{2} L}{8} \\\text { Strain }=\dfrac{3 m \omega^{2} L}{8 A Y}\end{array}$$
Question 10
1 / -0

A uniform wire of length $$L$$ and radius $$r$$ is twisted by an angle $$\alpha$$. If modulus of rigidity of the wire is $$\eta$$, then the elastic potential energy stored in wire, is

A
$$\dfrac { \pi \eta { r }^{ 4 }\alpha }{ 2{ L }^{ 2 } }$$

B
$$\dfrac { \pi \eta { r }^{ 4 }{ \alpha }^{ 2 } }{ 4{ L } }$$

C
$$\dfrac { \pi \eta { r }^{ 4 }\alpha }{ 4{ L }^{ 2 } }$$

D
$$\dfrac { \pi \eta { r }^{ 4 }{ \alpha }^{ 2 } }{ 2{ L } }$$

Solution

$$\begin{array}{c}\text { Given, wire of length }=L \text { and } \\\text { radius }=r \\\text { medulus of rigidity }=\eta\end{array}$$
Due to twisting there will shear stress and torque will be generated.
$$\tau=\frac{\alpha \cdot \eta \cdot \gamma}{L}\quad\begin{array}{c}\text { where } \tau=\text { torque }\\n=\text { modulus of } \\\text { rigidity }\end{array}$$$$\alpha= angle twrited$$
$$\begin{array}{l}\text { Now, elastic potential energy stored } \\\qquad E=\frac{\tau^{2}}{4 n} \times \text { volume }\end{array}$$
$$\begin{array}{l}=\frac{\alpha^{2} \cdot \eta^{2} \cdot r^{2}}{L^{2} \cdot 4\eta} \times \pi r^{2} L \\E=\frac{\pi \eta r^{4}\alpha^{2}}{4 L}\end{array}$$

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 76

Result

Mechanical Properties of Solids Test - 76

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SHARING IS CARING

Question 1

$$\dfrac{P}{\alpha K}$$

$$\dfrac{P\alpha}{K}$$

$$\dfrac{PK}{\alpha}$$

$$\dfrac{2K}{P}$$

Solution

Question 2

$$6F$$

$$9F$$

$$F$$

$$4F$$

Solution

Question 3

$$7.4\ \times 10^{7}\ N\ m^{-2}$$

$$4.5\ \times 10^{8}\ N\ m^{-2}$$

$$2.6\ \times 10^{6}\ N\ m^{-2}$$

$$1.15\ \times 10^{7}\ N\ m^{-2}$$

Question 4

$$0.2 J$$

$$10J $$

$$20 J$$

$$0.1 J$$

Solution

Question 5

$$1:1$$

$$8.9:1$$

$$1:7.6$$

$$1:3.2$$

Solution

Question 6

$$\sqrt { \dfrac { FA }{ { 4\pi }^{ 2 }ML } }$$

$$\sqrt { \dfrac { F }{ { 4\pi }^{ 2 }MLA } }$$

$$\sqrt { \dfrac { F }{ { 4\pi }^{ 2 }ML } }$$

$$\sqrt { \dfrac { { 4\pi }^{ 2 } }{ MLA } }$$

Solution

Question 7

$$10\times 10^{12}$$

$$100\times 10^{12}$$

$$1\times 10^{12}$$

$$2\times 10$$

Solution

Question 8

$$5mm$$

$$2mm$$

$$7mm$$

$$9mm$$

Solution

Question 9

$$\dfrac{m\omega^2l}{8AY}$$

$$\dfrac{3m\omega^2l}{8AY}$$

$$\dfrac{3m\omega^2l}{4AY}$$

$$\dfrac{m\omega^2l}{4AY}$$

Solution

Question 10

$$\dfrac { \pi \eta { r }^{ 4 }\alpha }{ 2{ L }^{ 2 } }$$

$$\dfrac { \pi \eta { r }^{ 4 }{ \alpha }^{ 2 } }{ 4{ L } }$$

$$\dfrac { \pi \eta { r }^{ 4 }\alpha }{ 4{ L }^{ 2 } }$$

$$\dfrac { \pi \eta { r }^{ 4 }{ \alpha }^{ 2 } }{ 2{ L } }$$

Solution

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