Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 73

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Weekly Quiz Competition

Question 1
1 / -0

$$32 g$$ of $$O_{2}$$ is contained in a cubical container of side $$1 m$$ and maintained at a temperature of $$127 ^{0} C$$. The isothermal bulk modulus of elasticity of the gas in terms of universal gas constant $$R$$ is

A
$$127 R$$

B
$$400 R$$

C
$$200 R$$

D
$$560 R$$

Solution

Bulk modulus of elasticity,
$$ B = \dfrac{\Delta P}{ \Delta V / V} $$

"B" is directly proportional to "p"

Now, Ideal gas equation

$$PV = n RT$$

$$P = nRT / V$$

$$B = P = nRT /V$$

$$ B = \dfrac{1 \times R \times 400}{ 1^{3} } $$

$$ B =400 R $$
Question 2
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A uniform cylindrical wire is subjected to a longitudinal tensile stress of $$5\times 10^{7} N/m^{2}$$. Young's modulus of the material of the wire is $$2\times 10^{11} N/m^{2}$$. The volume change in the wire is $$0.02\%$$. The fractional change in the radius is

A
$$0.25\times 10^{-4}$$

B
$$0.5\times 10^{-4}$$

C
$$0.1\times 10^{-4}$$

D
$$1.5\times 10^{-4}$$

Solution

$$Y=\dfrac{Stress}{\Delta L/L}$$

$$\dfrac{\Delta L}{L}=\dfrac{stress}{Y}=\dfrac{5\times 10^7}{2\times 10^{11}}=2.5\times 10^{-4}$$

Now $$V=\pi r^2L$$

$$\dfrac{\Delta V}{V}=\dfrac{\pi\Delta (r^2L)}{\pi r^2L}$$$$=\dfrac{r^2\Delta L+L\times 2r\Delta r}{r^2L}$$

$$\dfrac{\Delta L}{L}+2\dfrac{\Delta r}{r}$$

$$2\dfrac{\Delta r}{r}=\dfrac{\Delta V}{V}-\dfrac{\Delta L}{L}$$$$=\dfrac{0.02}{100}-2.5\times 10^{-4}$$

$$\dfrac{\Delta r}{r}=1\times 10^{-4}-\dfrac{2.5}{2}\times 10^{-4}=-0.25\times 10^{-4}$$
Question 3
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Find how the volume density of the elastic deformation energy is distributed in a steel rod depending on the distance $$r$$ from its axis. The length of the rod is equal to $$l$$, the torsion angle to $$\varphi$$.

A
$$\displaystyle u=\frac{1}{2}\frac{G\varphi^2r^2}{l^2}$$

B
$$\displaystyle u=\frac{3}{2}\frac{G\varphi^2r^2}{l^2}$$

C
$$\displaystyle u=\frac{5}{2}\frac{G\varphi^4r^2}{l^2}$$

D
None of these

Solution

We know that $$E=\dfrac{G\pi r^4\varphi^2}{4l}$$

$$\dfrac{dE}{dr}=\dfrac{4G\pi r^3\varphi^2}{4l}$$

$$E=\dfrac{4G\pi r^3\varphi^2dr}{4l}$$

Also $$dV=2\pi lrdr$$ , where $$V$$ is volume.

$$\dfrac{dE}{dV}=\dfrac{G r^2\varphi^2}{2l^2}$$
Question 4
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A copper rod of length $$l$$ is suspended from the ceiling by one of its ends. Find the elongation $$\Delta l$$ of the rod due to its own weight.

A
$$\displaystyle\Delta l=\frac{1}{2}\frac{\rho gl^2}{E}$$

B
$$\displaystyle\Delta l=\frac{1}{3}\frac{\rho gl^2}{E}$$

C
$$\displaystyle\Delta l=\frac{1}{4}\frac{\rho gl^2}{E}$$

D
$$\displaystyle\Delta l=\frac{1}{5}\frac{\rho gl^2}{E}$$

Solution

Let the tension acting on $$dx $$ element of rod be $$T_x$$ and due to this the elongation in that element be $$\delta x$$
$$T_x = M_x g$$
$$T_x = \rho (S x)g$$ ...........(1) Where $$S$$ is the cross sectional area of the rod

From Hooke's law, $$E = \dfrac{T_x dx}{S \delta x}$$ $$\implies \delta x = \dfrac{T_x dx}{SE}$$

$$\therefore$$ Total elongation in the rod $$\int \delta x = \int_0^l \dfrac{T_x }{SE} dx$$
$$\therefore$$ $$\Delta l = \int_0^l \dfrac{\rho g}{E} xdx $$

OR $$\Delta l = \dfrac{\rho g}{E} \times \dfrac{x^2}{2}\bigg|_0^l$$ $$\implies$$ $$\Delta l = \dfrac{\rho g l^2}{2 E}$$
Question 5
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The length of a metal is $$l_1$$ when the tension in it is $$T_1$$ and is $$l_2$$ when the tension is $$T_2$$. The original length of the wire is :

A
$$\displaystyle\frac{l_1+l_2}{2}$$

B
$$\displaystyle\frac{l_1T_2+l_2T_1}{T_1+T_2}$$

C
$$\displaystyle\frac{l_1T_2-l_2T_1}{T_2-T_1}$$

D
$$\displaystyle \sqrt{T_1T_2l_1l_2}$$

Solution

If l is the original length of wire, then change in length of first wire, $$\displaystyle\displaystyle\displaystyle \Delta l_1=(l_1-l)$$
Change in length of second wire, $$\Delta l_{ 2 }=(l_{ 2 }-l)$$
Now, $$Y=\displaystyle\frac{T_1}{A}\times \frac{l}{\Delta l_1}=\frac{T_2}{A}\times \frac{l}{\Delta l_2}$$
or, $$\displaystyle\frac{T_1}{\Delta l_1}=\frac{T_2}{\Delta l_2}$$, or, $$\displaystyle\displaystyle\frac{T_1}{l_2-l}=\frac{T_2}{l_2-l}$$

Thus, $$T_{ 1 }l_{ 2 }-T_{ 1 }l=T_{ 2 }l_{ 1 }-lT_{ 2 }$$
$$\Rightarrow l=\displaystyle\frac{T_2l_1-T_1l_2}{T_2-T_1}$$
Question 6
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In Searle's experiment to find Young's modulus the diameter of wire is measured as $$d=0.05cm$$, length of wire is $$l=125cm$$ and when a weight,$$m=20.0kg$$ is put, extension in wire was found to be $$0.100cm$$. Find the maximum permissible error in Young's modulus $$(Y)$$. Use:$$Y=\displaystyle\frac{mgl}{(\pi/4)d^2x}$$.

A
$$6.3\%$$

B
$$5.3\%$$

C
$$2.3\%$$

D
$$1\%$$

Solution

$$Y=\displaystyle\frac{mgl}{(\pi/4)d^2x}$$ .....$$(1)$$
$$\displaystyle\left[\frac{dY}{Y}\right]_{max}$$ $$=\displaystyle \frac{\Delta m}{m}+\frac{\Delta l}{l}+2\frac{\Delta d}{d}+\frac{\Delta x}{x}$$
$$m=20.0kg\Rightarrow \Delta m=0.1kg$$
$$l=125cm\Rightarrow \Delta l=1cm$$
$$d=0.050cm\Rightarrow \Delta d=0.001cm$$
$$x=0.100cm\Rightarrow \Delta x=0.001cm$$
$$\displaystyle\left[\frac{dY}{Y}\right]_{max}=\displaystyle\left[\frac{0.1kg}{20.0kg}+\frac{1cm}{125cm}+2\times \frac{0.001cm}{0.05cm}+\frac{0.001cm}{0.100cm}\right]\times 100\%=6.3\%$$
Question 7
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Young's modulis of brass and steel are $$10 \times 10^{10}\ N/m$$ and $$2\times 10^{11}\ N/m^2$$, respectively. A brass wire and a steel wire of the same length are extended by $$1mm$$ under the same force. The radii of brass and steel wires are $$R_B$$ and $$R_S$$ respectively. Then

A
$$R_S=\sqrt 2R_B$$

B
$$R_S=\dfrac{R_B}{\sqrt 2}$$

C
$$R_S=4R_B$$

D
$$R_S=\dfrac{R_B}{4}$$

Solution

We know that $$Y=\dfrac{FL}{\pi r^2l}$$ or $$r^2=\dfrac{FL}{(Y\pi l)}$$

$$\therefore R^2_B=\dfrac{FL}{(Y_B\pi l)}$$ and $$R^2_S=\dfrac{FL}{(Y_S\pi l)}$$

$$\dfrac{R^2_B}{R^2_S}=\dfrac{Y_S}{Y_B}=\dfrac{2\times 10^{10}}{10^{10}}=2$$

$$R^2_B=2R^2_S$$

Thus, we get $$R_S=\dfrac{R_B}{\sqrt{2}}$$
Question 8
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If the ratio of lengths, radii and Young's modulus of steel and brass wires shown in the figure are a, b and c respectively, the ratio between the increase in lengths of brass and steel wires would be :

A
$$\dfrac{{ b }^{ 2 }}{2c}$$

B
$$\dfrac{bc}{{ 2a }^{ 2 }}$$

C
$$\dfrac{{ ba }^{ 2 }}{2c}$$

D
$$\dfrac{{ 2b }^{ 2 }c}{a}$$

Solution

$$\Delta L=\dfrac{FL}{AY}$$

Force on the Steel wire $$=20N$$

Elongation in the steel wire $$=\dfrac{20\times l_s}{A_s Y_s}$$

Force on the brass wire $$=20N+20N=40N$$

Elongation in the brass wire $$=\dfrac{40\times l_b}{A_b Y_b}$$

$$\Rightarrow \dfrac{\Delta L_b}{\Delta L_s}=\dfrac{40}{20}\dfrac{L_b}{L_s}\dfrac{r_s^2}{r_b^2}\dfrac{Y_s}{Y_b}=\dfrac{2b^2c}{a}$$
Question 9
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A tension of $$20\ N$$ is applied to a copper wire of cross sectional area $$0.01 cm^2$$, Young's Modulus of copper is $$1.1\times 10^{11} N/m^2$$ and Poisson's ratio is 0.32. The decrease in cross sectional area of the wire is:

A
$$1.16\times 10^{-6}cm^2$$

B
$$1.16\times 10^{-5}m^2$$

C
$$1.16\times 10^{-4}m^2$$

D
$$1.16\times 10^{-3}cm^2$$

Solution

In case Poisson ratio ($$\nu $$) is given, the relation between stress and strain contains factor of 2,

Hence, $$\dfrac { \Delta A }{ A } =\dfrac { 2 }{ Y } \dfrac { F }{ A } \nu $$

$$\Delta A=\dfrac { 2F }{ Y } \nu =2\dfrac { 20N }{ 1.1 \times { 10 }^{ 11 }N/{ m }^{ 2 } } (0.32)=1.16 \times { 10 }^{ -6 }{ cm }^{ 2 }$$
Question 10
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The length of a metal wire is $$l_1$$ when the tension in it is $$F_1$$ and $$l_2$$ when the tension is $$F_2$$. Then original length of the wire is:

A
$$\dfrac{l_1F_1+l_2F_2}{F_1+F_2}$$

B
$$\dfrac{l_2-l_1}{F_2-F_1}$$

C
$$\dfrac{l_1F_2-l_2F_1}{F_1-F_2}$$

D
$$\dfrac{l_1F_1-l_2F_2}{F_2-F_1}$$

Solution

$$F\propto \Delta l$$

$$F_1\propto l_1-l$$
Where $$l$$, is the original length of the wire

$$F_2\propto l_2-l$$

$$\dfrac{F_1}{F_2}=\dfrac{l_1-l}{l_2-l}$$

$$F_1l_2-F_1l=F_2l_1-F_2l$$

$$(F_2-F_1)l=F_2l_1-F_1l_2$$

$$l=\dfrac{F_2l_1-F_1l_2}{F_2-F_1}$$

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 73

Result

Mechanical Properties of Solids Test - 73

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Rank

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SHARING IS CARING

Question 1

$$127 R$$

$$400 R$$

$$200 R$$

$$560 R$$

Solution

Question 2

$$0.25\times 10^{-4}$$

$$0.5\times 10^{-4}$$

$$0.1\times 10^{-4}$$

$$1.5\times 10^{-4}$$

Solution

Question 3

$$\displaystyle u=\frac{1}{2}\frac{G\varphi^2r^2}{l^2}$$

$$\displaystyle u=\frac{3}{2}\frac{G\varphi^2r^2}{l^2}$$

$$\displaystyle u=\frac{5}{2}\frac{G\varphi^4r^2}{l^2}$$

None of these

Solution

Question 4

$$\displaystyle\Delta l=\frac{1}{2}\frac{\rho gl^2}{E}$$

$$\displaystyle\Delta l=\frac{1}{3}\frac{\rho gl^2}{E}$$

$$\displaystyle\Delta l=\frac{1}{4}\frac{\rho gl^2}{E}$$

$$\displaystyle\Delta l=\frac{1}{5}\frac{\rho gl^2}{E}$$

Solution

Question 5

$$\displaystyle\frac{l_1+l_2}{2}$$

$$\displaystyle\frac{l_1T_2+l_2T_1}{T_1+T_2}$$

$$\displaystyle\frac{l_1T_2-l_2T_1}{T_2-T_1}$$

$$\displaystyle \sqrt{T_1T_2l_1l_2}$$

Solution

Question 6

$$6.3\%$$

$$5.3\%$$

$$2.3\%$$

$$1\%$$

Solution

Question 7

$$R_S=\sqrt 2R_B$$

$$R_S=\dfrac{R_B}{\sqrt 2}$$

$$R_S=4R_B$$

$$R_S=\dfrac{R_B}{4}$$

Solution

Question 8

$$\dfrac{{ b }^{ 2 }}{2c}$$

$$\dfrac{bc}{{ 2a }^{ 2 }}$$

$$\dfrac{{ ba }^{ 2 }}{2c}$$

$$\dfrac{{ 2b }^{ 2 }c}{a}$$

Solution

Question 9

$$1.16\times 10^{-6}cm^2$$

$$1.16\times 10^{-5}m^2$$

$$1.16\times 10^{-4}m^2$$

$$1.16\times 10^{-3}cm^2$$

Solution

Question 10

$$\dfrac{l_1F_1+l_2F_2}{F_1+F_2}$$

$$\dfrac{l_2-l_1}{F_2-F_1}$$

$$\dfrac{l_1F_2-l_2F_1}{F_1-F_2}$$

$$\dfrac{l_1F_1-l_2F_2}{F_2-F_1}$$

Solution

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