Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 48

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Question 1
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A uniform rod of length $$'L'$$ and density $$'\rho'$$ is being pulled along a smooth floor with horizontal acceleration $$\alpha$$ as shown in the figure. The magnitude of the stress at the transverse cross-section through the mid-point of the rod is _________.

A
$$\dfrac {\rho l \alpha}{4}$$

B
$$4\rho l\alpha$$

C
$$2\rho l\alpha$$

D
$$\dfrac {\rho l\alpha}{2}$$

Solution

Given:
The length of the rod is $$L$$.
The density of the rod is $$\rho$$.
The angular acceleration of the rod is $$\alpha$$.

The mid-point will experience the stress due to the half length of the rod.

The mass of the half length of the rod is given as:
$$Mass=\rho\times V$$

The volume of the half of the rod is given as:
$$V=area\times \dfrac L2$$

So, $$Mass=\rho\times A\times \dfrac L2$$

The force experienced by the rod is given as:
$$F=m\times a$$
$$F=\rho\times A\times \dfrac L2\times\alpha$$
$$=\dfrac{\rho AL\alpha}{2}$$

Now, the stress experienced by the mid-point of the rod is :
$$Stress=\dfrac{Force}{area}$$

$$=\dfrac{\rho AL\alpha}{2A}$$

$$=\dfrac{\rho L\alpha}{2}$$
Question 2
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Which of the following statements are correct?

A
Poisson's ratio can be greater than $$0.5$$

B
Poisson's ratio is a characteristic property of the material of the body

C
Poisson's ratio of a body depends upon its shape and size

D
None of these

Solution

Suppose a uniform rod is pulled by a tension due to which a longitudinal stress $$S$$ is developed in its material. Then tensile strain in the material of the rod, $$\epsilon_{1} = \dfrac {S}{Y}$$. If its Poisson's ratio of $$\sigma$$, then lateral strain will be equal to $$\sigma \epsilon_{1}$$. Lateral strain has opposite nature to that of the longitudinal strain. We also know that volumetric strain $$\epsilon_{v} = \epsilon_{x} + \epsilon_{y} + \epsilon_{z}$$
$$\therefore \epsilon_{v} = \epsilon_{1} + 2(-\sigma \epsilon_{1}) = (1 - 2\sigma)\epsilon_{1}$$
Since a tension is applied on the material, therefore, its volume cannot decrease. Hence, $$\epsilon_{v}$$ cannot be negative. Therefore $$(1-2\sigma)$$ cannot be less than $$0$$ or $$\sigma$$ cannot be greater than $$0.5$$. Poisson's ratio is the property of the material of the body. (b) is correct.
Question 3
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Take, bulk modulus of water $$B = 2100\ MPa$$.
What increase in pressure is required to decrease the volume of $$200\ litres$$ of water by $$0.004$$ percent?

A
$$210\ kPa$$

B
$$840\ kPa$$

C
$$8400\ kPa$$

D
$$84\ kPa$$

Solution

Given,
bulk modulus of water $$(B) = 2100\ MPa$$
$$= 2100\times 10^{6} Pa$$
We know that,
$$\triangle P = B\left (\dfrac {-\triangle V}{V}\right )$$
Here, $$\triangle V = 0.004$$
and $$V = 100$$
$$= 2100\times 10^{6} \times \left (\dfrac {0.004}{100}\right )$$
$$= \dfrac {2100\times 10^{6}\times 4}{1000\times 100}$$
$$= \dfrac {8400\times 10^{6}}{10^{5}}$$
$$= 84000\ Pa$$
$$= 84\ kPa$$.
Question 4
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The compressibility of water $$4\times 10^{-5}$$ per unit atmospheric pressure. The decrease in volume of $$100\ cubic$$ centimeter of water under a pressure of $$100$$ atmosphere will be:

A
$$0.4\ cc$$

B
$$4\times 10^{-5} cc$$

C
$$0.025\ cc$$

D
$$0.004\ cc$$

Solution

Compressibility$$(\beta)=\cfrac{\cfrac{\triangle V}{V}}{P}$$
Here,
$$\beta=4 \times {10}^{-5}/atm$$
$$V=100\quad cc$$
$$P=100\quad atm$$
$$\therefore \triangle V=\beta PV$$
$$=4 \times {10}^{-5} \times 100 \times 100$$
$$\therefore \triangle V=0.4cc$$
Question 5
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The load versus elongation graph for four wires of the same material and same length is shown in the figure. The thinnest wire is represented by the line.

A
$$OA$$

B
$$OB$$

C
$$OC$$

D
$$OD$$

Solution

As slop of the load-elongation graph given, young's modulus. So, OA has least value of young's modulus.
And, Young's modulus = $$\dfrac { stress }{ strain } $$
As, Young's modulus of OA is the least, so, strain is maximum. As strain is maximum, so, it is the thinnest wire.
Question 6
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Identical springs of steel and copper $$\left( { Y }_{ steel }>\quad { Y }_{ copper } \right) $$ are equally stretched

A
Less work is done on copper spring

B
Less work is done on steel spring

C
Equal work is done on both the springs

D
Data is incomplete

Solution

$$\begin{array}{l}\text { Since, young's modulus of steel is greater than } \\\text { that of copper. Hence, in order to produce } \\\text { same extension, lange force will have to be applied } \\\text { on the steel spring than that of copper spring } \\\therefore W\text { (steel)}>\text { W(copper )}\end{array}$$
Hence option A is correct
Question 7
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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
$$\displaystyle \dfrac{3FL}{5AY}$$

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

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

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

Solution

Divide the rod into 2 parts of length $$2l/m$$ and $$l/m$$ where the force F is acting.
Now solve the problem for the 2 rods separately.
Force acting on both sides of $$2l/3$$ is $$3F$$.
So extension $$x_1$$ will be $$x_1=\dfrac{\left[3F\times \dfrac{2l}{3}\right]}{AY}$$
Force acting on both sides of $$l/3$$ is $$2F$$
So extension $$x2$$ will be $$x_2=\dfrac{\left[2F\times \dfrac{l}{3}\right]}{AY}$$
Now total extension x is $$x=x_1+x_2=8FL/3AY$$
Question 8
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The increase in energy of a metal bar of length 'L' and cross-sectional area 'A' when compressed with a load 'M' along its length is
(Y = Young's modulus of the material of metal bar)

A
$$\dfrac{FL}{2AY}$$

B
$$\dfrac{F^2L}{2AY}$$

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

D
$$\dfrac{F^2L^2}{2AY}$$

Solution

Energy,$$U=\dfrac{1}{2}\times stress\times strain\times volume$$

Now, $$stress=\dfrac{F}{A}$$

$$strain=\dfrac{stress}{Y}=\dfrac{F}{AY}$$

$$Volume=AL$$

Hence, $$U=\dfrac{1}{2}\left(\dfrac{F}{A}\right)\left(\dfrac{F}{AY}\right)AL$$

$$\implies U=\dfrac{F^2L}{2AY}$$

Hence, answer is option-(B).
Question 9
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Two wires of equal length and cross-sectional area are suspended as shown in figure. Their Young's modulii are Y$$_1$$ and Y$$_2$$ respectively. The equivalent Young's modulii will be:

A
$$\displaystyle Y_1 + Y_2$$

B
$$\displaystyle \frac{Y_1 Y_2}{Y_1 + Y_2}$$

C
$$\displaystyle \frac{Y_1 + Y_2}{2}$$

D
$$\displaystyle \sqrt{Y_1 Y_2}$$

Solution

$$\begin{array}{l}{Y_1} = \dfrac{{{F_1}/A}}{{\Delta {L_1}/L}}\\{Y_2} = \dfrac{{{F_2}/A}}{{\Delta {L_2}/L}}\\{F_1} = {F_2} = F\\\Delta {L_1} = \dfrac{{FL}}{{{Y_1}A}}\\\Delta {L_2} = \dfrac{{FL}}{{{Y_2}A}}\\\therefore {Y_{12}} = \dfrac{{F/A}}{{\Delta {L_{12}}/L}}\\\Delta {L_{12}} = \Delta {L_2} + \Delta {L_2}\\\therefore \dfrac{1}{Y} = \frac{1}{{{Y_1}}} + \frac{1}{{{Y_2}}}\\Y = \dfrac{{{Y_1}{Y_2}}}{{{Y_1} + {Y_2}}}\end{array}$$
Question 10
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The bulk modulus of water is $$2.0\times { 10 }^{ 9 }\ { N }/{ { m }^{ 2 } }$$ . The pressure required to increase the density of water $$0.1 \%$$ is

A
$$2.0\times { 10 }^{ 3 }\ { N }/{ { m }^{ 2 } }$$

B
$$2.0\times { 10 }^{ 6 }\ { N }/{ { m }^{ 2 } }$$

C
$$2.0\times { 10 }^{ 5 }\ { N }/{ { m }^{ 2 } }$$

D
$$2.0\times { 10 }^{ 7 }\ { N }/{ { m }^{ 2 } }$$

Solution

$$\rho \quad \alpha \quad V$$
$$\dfrac{\Delta V}{V}=\dfrac{0.1}{100}$$
$$B=\dfrac{-p}{\Delta V/V}$$
$$\left| B \right| =\dfrac { P }{ \Delta V/V } \\ $$
$$p=2\times 10^9\times 10^{-3}$$
$$p=2\times 10^6 N/m^2$$