Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 32

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Question 1
1 / -0

A spiral spring is stretched to 20.5 cm gradation on a metre scale when loaded with a 100 g load and to the 23.3 cm gradation by 200 g load. The spring is used to support a lump of metal in air and the reading now is 24.0 cm. The mass of metal lump is :

A
250 gm

B
225 gm

C
145 gm

D
750 gm

Solution

Let original length of spring be $$L$$.

Thus, $$20.5-L=\dfrac{0.1kg\times g\times L}{AY}$$ or $$\dfrac{L}{AY}=20.5-L$$

Also, from second loading, we get similarly, $$\dfrac{0.2\times g\times L}{AY}=23.3-L$$ or, $$\dfrac{L}{AY}=\dfrac{23.3-L}{2}$$

Solve these to get, $$41-2L=23.3-L$$ or $$L=17.7$$

Also, $$\dfrac{L}{AY}=20.5-17.7=2.8$$

Thus, for $$24\ cm$$, $$\dfrac{mgL}{AY}=24-L$$ or $$mg=\dfrac{(24-17.7)}{2.8}=2.25$$ or $$m=225\ g$$
Question 2
1 / -0

A stone of mass m tied to one end of a thread of length l. The diameter of the thread is d and it is suspended vertically. The stone is now rotated in a horizontal plane and makes an angle $$\theta$$ with the vertical. Find the increase in length of the wire. Young's modulus of the wire is Y.

A
$$\displaystyle \frac{4mgl}{\pi d^{2}Y\cos\theta}$$

B
$$\displaystyle \frac{4mgl}{\pi d^{2}Y\sin\theta}$$

C
$$\displaystyle \frac{4mgl}{\pi d^{2}Y}$$

D
$$\displaystyle \frac{4mgl}{\pi d^{2}Y\sec\theta}$$

Solution

The Tension generated in string can be calculated as $$Tcos\theta =mg$$
$$Thus\quad T=\dfrac { mg }{ cos\theta } $$
Now, $$\Delta l=\dfrac { Tl }{ AY } =\dfrac { mg\times l }{ cos\theta \times \dfrac { \pi { d }^{ 2 } }{ 4 } \times Y } =\dfrac { 4mgl }{ \pi { d }^{ 2 }Ycos\theta } $$
Question 3
1 / -0

A steel cylindrical rod of length $$l$$ and radius $$r$$ is suspended by its end from the ceiling. Find the elastic deformation energy $$U$$ of the rod.

A
$$\displaystyle U=\frac{1}{6}\pi r^2\rho^2 g^2\frac{l^3}{E}$$

B
$$\displaystyle U=\frac{5}{6}\pi r^2\rho^2 g^2\frac{l^3}{E}$$

C
$$\displaystyle U=\frac{1}{6}\pi r^2\rho^2 g^2\frac{2l^3}{E}$$

D
$$\displaystyle U=\frac{5}{6}\pi r^2\rho^2 g^2\frac{2l^3}{E}$$

Solution

Let's consider,

$$g=$$ acceleration due to gravity

$$l=$$ length of the rod

$$r=$$ radius of the road

$$x=$$ change in length

$$\rho=$$ density

$$E=$$ modulus

From the given figure,

$$Stress=\dfrac{\rho.\pi r^2(l-x)g}{\pi r^2}=\rho g(l-x)$$

$$Strain, \dfrac{d\Delta x}{dx}=\dfrac{Stress}{E}=\dfrac{\rho g(l-x)}{E}$$

The elastic deformation energy is given by

$$dU=\dfrac{1}{2}\times stress\times strain\times volume$$

$$dU=\dfrac{1}{2}\times \rho g(l-x)\times\dfrac{\rho g(l-x)}{E}\times \pi r^2 dx $$

$$\int _0^UdU=\dfrac{\pi r^2 \rho^2 g^2}{2E} \int _0^l(l-x)^2dx$$

$$U=\dfrac{\pi r^2 \rho^2 g^2l^3}{6E}$$

The correct option is A.
Question 4
1 / -0

A uniform heavy rod of length $$L$$, weight $$W$$ and cross-sectional area $$A$$ is hanging from a fixed support. Young's modulus of the material is $$Y$$. Find the elongation of the rod.

A
$$\displaystyle \dfrac{WL}{AY}$$

B
$$\displaystyle \dfrac{WL}{2AY}$$

C
$$\displaystyle \dfrac{WL}{4AY}$$

D
$$\displaystyle \dfrac{WL}{3AY}$$

Solution

$$\displaystyle T=(L-x)\frac{W}{L};\:elongation=\frac{Tdx}{AY}$$
$$\displaystyle =\frac{(L-x)Wdx}{LAY};\:Total\:elongation=\frac{W}{LAY}\int_{0}^{L}(L-x)dx$$
$$\displaystyle =\frac{WL}{2AY}$$
Question 5
1 / -0

Poisson's ratio cannot exceed

A
0.25

B
1.0

C
0.75

D
0.5

Solution

Poisson's ratio = Lateral strain/Longitudinal strain
$$Y=3K(1-2\mu)\Rightarrow \mu=0.5-Y/6K$$
$$Y$$ is young's modulus.
$$\mu$$ is poisson ratio
$$K$$ is compressibility of the substance which is inverse of Bulk's modulus. Maximum value of $$K$$ is $$\infty$$
So maximum value of Poisson's ratio $$\mu=0.5$$
Question 6
1 / -0

The length of a wire is $$l_1$$ when tension is $$T_1$$ and $$l_2$$ when tension is $$T_2$$. The natural length of the wire is

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

B
$$\sqrt{l_1l_2}$$

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

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

E
$$\displaystyle \frac{l_1T_1-l_2T_2}{T_1-T_2}$$

Solution

$$\displaystyle \frac{T_1}{T_2}=\frac{l_1-l}{l_2-l}\:or\:T_1(l_2-l)=T_2(l_1-l)\:or\:l=\frac{l_1T_2-l_2T_1}{T_2-T_1}$$
Question 7
1 / -0

A sphere of mass M kg is suspended by a metal wire of length L and diameter d. When in equilibrium, there is a gap of $$\Delta l$$ between the sphere and the floor. The sphere is gently pushed aside so that it makes an angle $$\theta$$ with the vertical. Find $$\theta_{max}$$ so that sphere fails to rub the Floor, Young's modulus of the wire is Y.

A
$$\displaystyle \sin^{-1}\left(1-\frac{Y\pi d^{2}\Delta l}{8MgL}\right)$$

B
$$\displaystyle \tan^{-1}\left(1-\frac{Y\pi d^{2}\Delta l}{8MgL}\right)$$

C
$$\displaystyle \cos^{-1}\left(1-\frac{Y\pi d^{2}\Delta l}{8MgL}\right)$$

D
none

Solution

(c) $$\displaystyle Y=\dfrac{Fl}{A\Delta l}=\dfrac{2Mg(1-\cos\theta)L}{\pi\dfrac{d^{2}}{4}\Delta l}$$
or $$\displaystyle 1-\cos\theta=\dfrac{Y\pi d^{2}\Delta l}{8MgL}$$ or $$\displaystyle \cos\theta=1-\dfrac{Y\pi d^{2}\Delta l}{8MgL}$$
Here we have used the energy conservation
$$KE=PE$$
$$\displaystyle \dfrac{mv^{2}}{2}=mgl(1-\cos\theta)$$
or $$\displaystyle \dfrac{mv^{2}}{l}=2mg(1-\cos\theta)$$
Thus
$$\displaystyle \theta=\cos^{-1}\left(1-\dfrac{Y\pi d^{2}\Delta l}{8MgL}\right)$$
Question 8
1 / -0

A copper wire of cross-section A is under a tension T. Find the decrease in the cross-section area. Young's modulus is Y and Poisson's ratio is $$\sigma$$

A
$$\displaystyle \frac{\sigma T}{2AY}$$

B
$$\displaystyle \frac{\sigma T}{AY}$$

C
$$\displaystyle \frac{2\sigma T}{AY}$$

D
$$\displaystyle \frac{4\sigma T}{AY}$$

Solution

$$\dfrac { \Delta d }{ d } =-\sigma \dfrac { \Delta l }{ l } =-\sigma \dfrac { T }{ AY } \\ Thus\quad \Delta d=-\sigma d\dfrac { T }{ AY } ,\quad hence\quad \Delta A=\Delta (\pi /4{ d }^{ 2 })=2.\pi /4{ .d }\Delta d\\ Hence\quad \Delta A=2.\pi /4{ .d }.d.(\dfrac { -\sigma T }{ AY } )=2\dfrac { -\sigma T }{ Y } \\ or\quad Decrease\quad =\quad -\dfrac { \Delta A }{ A } =\dfrac { 2\sigma T }{ AY } $$
Question 9
1 / -0

Each wire in Fig. has cross-section area $$5\times10^{-3}cm^{2}$$ and $$Y=2\times10^{11}Pa$$. P, Q and R have mass $$3 kg$$ each. Find the strain developed in A. Assume surface to be smooth.

A
$$10^{-5}$$

B
$$10^{-6}$$

C
$$10^{-4}$$

D
none

Solution

From the free body diagram, we can easily see,
$$3g-T_{ 1 }=3a\\ T_{ 1 }-T_{ 2 }=3a\\ T_{ 2 }=3a$$
Adding the three, we get:
$$a=\dfrac { g }{ 3 } ;\: T_{ 2 }=3a=g\\ \dfrac { \Delta l }{ l } =\dfrac { F/A }{ Y } =\dfrac { T }{ AY } =\dfrac { g }{ 5\times 10^{ -3 }\times 2\times 10^{ 11 }\times 10^{ -4 } } =10^{ -4 }$$
Question 10
1 / -0

Young's modulus of rubber is $$10^{4} N/m^{2}$$ and area of cross section is $$2\ cm^{-2}$$. If force of $$2\times 10^{5}$$ dyn is applied along its length, then its initial $$l$$ becomes.

A
$$3l$$

B
$$4l$$

C
$$2l$$

D
None of these

Solution

We have $$1\ dyne=10^{ -5 }N$$
$$\Delta l=\dfrac { F.L }{ A.Y } =\dfrac { 2\times 10^{ 5 }\times 10^{ -5 }\times l }{ 2\times { 10 }^{ -4 }\times { 10 }^{ 4 } } =l\\ Thus\quad new\quad length\quad =\quad l+\Delta l=2l$$

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 32

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Mechanical Properties of Solids Test - 32

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SHARING IS CARING

Question 1

250 gm

225 gm

145 gm

750 gm

Solution

Question 2

$$\displaystyle \frac{4mgl}{\pi d^{2}Y\cos\theta}$$

$$\displaystyle \frac{4mgl}{\pi d^{2}Y\sin\theta}$$

$$\displaystyle \frac{4mgl}{\pi d^{2}Y}$$

$$\displaystyle \frac{4mgl}{\pi d^{2}Y\sec\theta}$$

Solution

Question 3

$$\displaystyle U=\frac{1}{6}\pi r^2\rho^2 g^2\frac{l^3}{E}$$

$$\displaystyle U=\frac{5}{6}\pi r^2\rho^2 g^2\frac{l^3}{E}$$

$$\displaystyle U=\frac{1}{6}\pi r^2\rho^2 g^2\frac{2l^3}{E}$$

$$\displaystyle U=\frac{5}{6}\pi r^2\rho^2 g^2\frac{2l^3}{E}$$

Solution

Question 4

$$\displaystyle \dfrac{WL}{AY}$$

$$\displaystyle \dfrac{WL}{2AY}$$

$$\displaystyle \dfrac{WL}{4AY}$$

$$\displaystyle \dfrac{WL}{3AY}$$

Solution

Question 5

0.25

1.0

0.75

0.5

Solution

Question 6

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

$$\sqrt{l_1l_2}$$

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

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

$$\displaystyle \frac{l_1T_1-l_2T_2}{T_1-T_2}$$

Solution

Question 7

$$\displaystyle \sin^{-1}\left(1-\frac{Y\pi d^{2}\Delta l}{8MgL}\right)$$

$$\displaystyle \tan^{-1}\left(1-\frac{Y\pi d^{2}\Delta l}{8MgL}\right)$$

$$\displaystyle \cos^{-1}\left(1-\frac{Y\pi d^{2}\Delta l}{8MgL}\right)$$

none

Solution

Question 8

$$\displaystyle \frac{\sigma T}{2AY}$$

$$\displaystyle \frac{\sigma T}{AY}$$

$$\displaystyle \frac{2\sigma T}{AY}$$

$$\displaystyle \frac{4\sigma T}{AY}$$

Solution

Question 9

$$10^{-5}$$

$$10^{-6}$$

$$10^{-4}$$

none

Solution

Question 10

$$3l$$

$$4l$$

$$2l$$

None of these

Solution

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