Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 80

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Question 1
1 / -0

An elastic string carrying a body of mass 'm' extends by 'e'. The body rotates in a vertical circle with critical velocity. The extension in the string at the lowest position is

A
$$2$$ e

B
$$4$$ e

C
$$6$$ e

D
$$8$$ e

Solution
Question 2
1 / -0

Y,K,n represent the Young's modulus, bulk modulus and rigidity modulus of a body respectively. If rigidity modulus is twice the bulk modulus, then

A
$$Y=5K/18$$

B
$$Y=5 n/9$$

C
$$Y=9K/5$$

D
$$Y=18K/5$$
Question 3
1 / -0

A wire is stretched under a force. If the wire suddenly snaps, the temperature of the wire

A
Remains the same

B
Decreases

C
Increases

D
First decreases then increases

Solution
Question 4
1 / -0

In a Young's double slit experiment with sodium light, slits are 0.589 m apart. The angular separation of the maximum from the central maximum will be (given $$\lambda =589$$nm,):

A
$${ sin }^{ -1 }\left( { 0.33\times 10 }^{ 8 } \right) $$

B
$${ sin }^{ -1 }\left( { 0.33\times 10 }^{ 6 } \right) $$

C
$${ sin }^{ -1 }\left( { 3\times 10 }^{ 8 } \right) $$

D
$${ sin }^{ -1 }\left( { 3\times 10 }^{ -6 } \right) $$

Solution

Using relation, $$d \sin \theta =n \lambda \Rightarrow \sin \theta =\dfrac {n \lambda }{d}$$
For $$n=3. \sin \theta =\dfrac {3\lambda }{d}=\dfrac {3\times 589 \times 10^{-9}}{0.589}$$
$$=3\times 10^{-6}$$ or $$\theta =\sin^{-1}(3\times 10^{-6})$$
Question 5
1 / -0

A rod of length $$L$$ with square cross-section $$(a \times a)$$ is bend to form a circular ring.
Then the value of stress developed at points $$E$$ and $$F$$ respectively, ($$Y$$ is Young's modulus of metal rod)

A
Zero in both

B
$$\dfrac{2\pi a Y}{L}$$ compression, $$\dfrac{2\pi a Y}{L}$$ tension

C
$$\dfrac{\pi a Y}{L}$$ compression, $$\dfrac{\pi a Y}{L}$$ tension

D
$$2\dfrac{\pi a Y}{L}$$ compression in both.
Question 6
1 / -0

A uniform slender rod of length L, cross-sectional area A and Young's modulus Y is acted upon by the forces shown in the figure. The elongation of the rod is

A
$$\dfrac{3FL}{5AY}$$

B
$$\dfrac{2FL}{5AY}$$

C
$$\dfrac{3FL}{8AY}$$

D
$$\dfrac{8FL}{3AY}$$

Solution

$$ \begin{array}{l} \text { Cross section area }(A)=A \\ \text { if can be break into two pices at } 0 \end{array} $$
$$ \begin{array}{l} \Delta L \text { (net elongation) }=\Delta L_{1}+\Delta L_{2} \\ \text { we know } \\ \qquad \begin{array}{l} y=\frac{F L}{A \Delta L} \\ \Delta L=\frac{F L}{A Y} \end{array} \end{array} $$
$$ \begin{aligned} \Delta L_{1} &=\frac{F_{A C} L_{A_{C}}}{A Y} \\ \Delta L_{1} &=\frac{B F \times \frac{2 L}{X}}{A Y} \\ \Delta L_{1} &=\frac{2 F L}{A Y} \end{aligned} $$
$$ \begin{aligned} \text { Similarly } & \\ \qquad \Delta L_{2}=& \frac{F_{C B} L_{C B}}{A Y} \\ \Delta L_{2} &=\frac{2 F \times \frac{L}{3}}{A Y} \\ &=\frac{2 F L}{3 A Y} \end{aligned} $$
$$ \begin{aligned} \Delta L_{\text {net }} &=\Delta L_{1}+\Delta L_{2} \\ &=\frac{2 F L}{A Y}+\frac{2 F L}{3 A Y} \\ \Delta L_{\text {net }} &\left.=\frac{F L}{A Y} \times \frac{8}{3}\right] \end{aligned} $$
Question 7
1 / -0

Poisson ratio for a material is 0.3 and its bulk modulus is $$1.5\times { 10 }^{ 11 }N/{ m }^{ 2 }$$. Young's modulus of the material is-

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

B
$$1.85\times { 10 }^{ 11 }N/{ m }^{ 2 }$$

C
$$1.8\times { 10 }^{ 11 }N/{ m }^{ 2 }$$

D
$$1.4\times { 10 }^{ 11 }N/{ m }^{ 2 }$$

Solution

$$ \begin{array}{l} \text { Given ithat } \\ \text { Bulk modulus }(\beta)=1.5 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2} \\ \text { Poisson ratio }(\sigma)=0.3 \\ \text { Young modulus }(y)=? \\ \text { we know velation between } \beta, \sigma, y \end{array} $$
$$\beta=\frac{y}{3(1-2 \sigma)}$$
$$y=3 \beta(1-2 \sigma)$$

Putting value we get \begin{aligned} y &=3 \times 1.5 \times 10^{11}(1-0.6) \\ y &=1.8 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2} \end{aligned}
Question 8
1 / -0

The mass and length of a wire are M and L respectively. The density of the material of the wire is d. On applying the force F on the wire, the increase in length is l, then the Young's modulus of the material of the wire will be

A
$$Fdl/Ml$$

B
$$FL/Mdl$$

C
$$FMl/dl$$

D
$$Fd{ L }^{ 2 }/Ml$$

Solution

$$ \begin{array}{l} \text { Given, } \\ \text { Mass of wire }=M . \\ \text { length of wire }=L \\ \text { Force applied }=F \\ \text { increased length }=l \\ \text { density }=d \\ \text { we know } \\ \text { young modulus }(Y)=\frac{\text { Stress }}{\text { strain }} \end{array} $$
$$y=\frac{\left(\frac{F}{A}\right)}{\left(\frac{l}{L}\right)}$$ $$y=\frac{F L}{A}$$ $$\therefore d=\frac{M}{V}$$ $$\rightarrow d=\frac{M}{A L} \quad \rightarrow A L=\frac{M}{d}$$ $$y=\frac{F L x L}{(A \times L) \cdot l}$$ $$y=\frac{F L^{2} d}{M l}$$
Question 9
1 / -0

A rod AD consisting of three segments AB, BC and CD are joined together is A. The length of the three segment are respectively 0.1 m, 0.2 m and 0.15 m. The cross-section of the rod is uniformly $$10^{-4} m^2$$.A weight of 10 kg is hung from D. Calculate the displacement of point D. If [$$Y_{AB} = 2.5 \times 10^{10} N/{m^2}$$ , $$Y_{BC} = 4 \times 10^{10} N/{m^2}$$ and $$Y_{CD} = 1 \times 10^{10} N/{m^2}$$. Neglect the weight of the rod. Take g = 10 $$m/s^2$$]

A
4 $$\times { 10 }^{ -6 }$$ m

B
9 $$\times { 10 }^{ -6 }$$ m

C
24 $$\times { 10 }^{ -6 }$$ m

D
18 $$\times { 10 }^{ -6 }$$ m

Solution

$$ \begin{array}{l} \text { Eiven, } \\ \text { cross section of wire }(4)=10^{-4} \mathrm{~m}^{2} \\ Y_{A B}(\text { young modulus })=2.5 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2} \\ Y_{B C}=4 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2} \\ Y_{C D}=1 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2} \\ g=10 \mathrm{~m} / \mathrm{s}^{2} \end{array} $$
$$ \begin{aligned} & Y=\frac{F l}{A \Delta l} \\ & \Delta L=\frac{F L}{A Y} \\ \Delta L_{A B} &=\frac{10 \times 10}{10^{-4}} \times \frac{L_{A B}}{Y_{A B}} \\ &=\frac{10^{6} \times 0.1}{2.5 \times 10^{10}} \\ &=4 \times 10^{-6} \mathrm{~m} \end{aligned} $$
$$\begin{aligned} \Delta L_{B C} &=10^{6} \times \frac{L_{B C}}{Y_{B C}} \\ &=10^{6} \times \frac{0.2}{4 \times 10^{10}} \\ &=5 \times 10^{-6 \mathrm{~m}} \\ \Delta L_{C D} &=10^{6} \times \frac{L_{C P}}{Y_{C D}} \\ &=10^{6} \times \frac{0.15}{10^{10}} \\ &=15 \times 10^{-6} \mathrm{~m} \end{aligned}$$
$$\therefore $$displacement of $$D=\Delta L_{A B} + \Delta L_{B C}+\Delta_{C D}$$
$$=\left(4 \times 10^{-6}\right)+\left(5 \times 10^{-6}\right)+\left(15 \times 10^{-6}\right)$$
$$d=24 \times 10^{-6}$$
Question 10
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A rubber cord has a cross-sectional area 1 $$mm^{ 2 }$$ and total unstretched length 10cm. It is stretched to 12cm and then released to project a mass of 80g. The Young's modulus for rubber is 5$$\times 10^{ 8 }Nm^{ -2 }$$. Find the velocity of mass (in m/s)?

A
5

B
3

C
7

D
2

Solution

$$ \begin{array}{l} \text { Civen, } \\ \text { cross sectional Area }(A)=1 m m^{2}=10^{-6} m^{2} \\ \text { Unstreched length }(L)=10 c m=10 \times 10^{-2} \mathrm{~m} \\ \text { mass of missile } \\ \text { young modulus }(y)=5 \times 10^{8} \mathrm{Nm}^{-2} \\ \text {Streched length }(L+\Delta L)=12 \mathrm{~cm} \\ \text { Let us consider rubber cord as spring } \\ \text { then } \\ \qquad \begin{aligned} k &=\frac{y A}{L} \\ &=5 \times 10^{3} \mathrm{~N} / \mathrm{m} \end{aligned} \end{array} $$ we know, and this energy is transtered to $K E$ of missile $$ \begin{aligned} \frac{1}{2} k(\Delta L)^{2} &=\frac{1}{2} m v^{2} \\ v &=\sqrt{\frac{K}{m}} \times \Delta L \\ v &=\sqrt{\frac{5 \times 10^{3}}{80 \times 10^{-3}} \times 2 \times 10^{-2}} \\ & v=5 \mathrm{~m} / \mathrm{s} \end{aligned} $$