Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 47

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Weekly Quiz Competition

Question 1
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 $$2\times 10^5$$ dyne is applied along its length, then its final length becomes.

A
$$3L$$

B
$$4L$$

C
$$2L$$

D
None of these

Solution

Given, $$Y=10^4N/m^2, A=2cm^2=2\times 10^{-4}m^2$$
$$F=2\times 10^5$$ dyne $$=2$$N
$$\therefore$$ Initial length $$I=\displaystyle\frac{FL}{AY}$$
$$=\displaystyle\frac{2\times L}{2\times 10^{-4}\times 10^4}=L$$
$$\therefore$$ Final length $$=$$ Initial length $$+$$ Increment
$$=2L$$.
Question 2
1 / -0

A rubber cord catapult has cross-sectional area 25 mm$$^2$$ and initial length of rubber cord is 10 cm. It is stretched to 5 cm and then released to project a missile of mass 5 g. Taking $$Y_{rubber} = 5 \times 10^8 Nm^{-2}$$, velocity of projected missile is:

A
20 ms$$^{-1}$$

B
100 ms^{-1}$$

C
250 ms$$^{-1}$$

D
200 ms$$^{-1}$$

Solution

Potential energy stored in the rubber cord catault will be converted into kinetic energy of mass
$$\dfrac{1}{2} mv^2 = \dfrac{1}{2} \dfrac{YAl^2}{L}$$
where $$l = 5\times 10^{-2} \ $$m, $$A = 25\times 10^{-6} \ m^2,$$ $$L = 10\times 10^{-2} \ $$m, $$m = 5\times 10^{-3} \ kg$$
$$\Rightarrow v = \sqrt{\dfrac{YAl^2}{mL}}$$
$$v= \sqrt{\dfrac{5 \times 10^8 \times 25 \times 10^{-6} \times (5 \times 10^{-2})^2}{5 \times 10^{-3} \times 10 \times 10^{-2}}}$$
$$v= 250 ms^{-1}$$
Question 3
1 / -0

A ball falling-in a lake of depth 400 m has a decrease of $$0.2\%$$ in its volume at the bottom. The bulk modulus of the material of the ball is (in $$Nm^-{2}$$)

A
$$9.8 \times 10^9$$

B
$$9.8 \times 10^{10}$$

C
$$1.96 \times 10^{10}$$

D
$$9.8 \times 10^{11}$$

E
$$1.96 \times 10^9$$

Solution

Given, $$h=400m, \dfrac{\triangle V}{V}=\dfrac{0.2}{100}$$
We know that the bulk modulus of the material
$$B=\dfrac{hpg}{\begin{pmatrix}\dfrac{\triangle V}{V}\end{pmatrix}}$$
$$B=\dfrac{400\times 10^3\times 9.8}{\dfrac{0.2}{100}}$$
$$B=\dfrac{400\times 10^3\times 9.8\times 100}{0.2}$$
$$B=2000\times 10^3\times 9.8\times 100$$
$$B=1.96\times 10^9 Nm^{-2}$$
Question 4
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One end of a uniform bar of weight $$\displaystyle { w }_{ 1 }$$ is suspended from the roof and a weight $$\displaystyle { w }_{ 2 }$$ is suspended from the other end, the area of cross-section is A. What is the stress at the mid point of the rod?

A
$$\displaystyle \frac { { w }_{ 1 }+{ w }_{ 2 } }{ A } $$

B
$$\displaystyle \frac { { w }_{ 1 }-{ w }_{ 2 } }{ A } $$

C
$$\displaystyle \frac { \left( { { w }_{ 1 } }/{ 2 } \right) +{ w }_{ 2 } }{ A } $$

D
$$\displaystyle \frac { { { w }_{ 2 } }/{ 2 }+{ w }_{ 1 } }{ A } $$

Solution

The downward tension at mid-point of the bar is due to weight of lower half of the bar and the hanging weight.

The downward tension at mid-point is, $$(w_1)/2 + w_2$$
As, $$stress = force/area = \dfrac{(w_1)/2 + w_2}{A}$$

Option C is correct.
Question 5
1 / -0

When temperature of a gas is $$20^{\circ}C$$ and pressure is changed from $$p_{1} = 1.01\times 10^{5} Pa$$ to $$p_{2} = 1.165\times 10^{5} Pa$$, the volume changes by $$10\%$$. The bulk modulus is

A
$$1.55\times 10^{5}$$

B
$$0.155\times 10^{5}$$

C
$$1.4\times 10^{5}$$

D
$$1.01\times 10^{5}$$

Solution

Bulk modulus =$$ B$$
Change in pressure = $$\triangle p$$
Change in volume = $$\triangle v$$
$$B = -\dfrac {\triangle p}{\triangle v/V} = \dfrac {(1.165\times 10^{5} - 1.01\times 10^{5})}{0.1}$$
$$= 1.55\times 10^{5} Pa$$.
Question 6
1 / -0

The Poisson's ratio of a material is $$0.8$$. If a force is applied to a wire of this material decreases its cross-sectional area by $$4\%$$, then the percentage increase in its length will be.

A
$$1\%$$

B
$$2\%$$

C
$$2.5\%$$

D
$$4\%$$

Solution

The Poisson's ratio $$\sigma$$ is given by
$$\sigma =\displaystyle\frac{lateral strain}{longitudinal strain}=\frac{\Delta d/d}{\Delta l/l}$$ ........(i)
where, l$$=$$length of wire
d$$=$$diameter of wire
The cross-sectional area of wire is given by
$$A=\displaystyle \pi r^2=\frac{\pi d^2}{4}$$
$$\therefore \displaystyle\frac{\Delta A}{A}=\frac{2\Delta d}{d}$$
$$\Rightarrow \displaystyle\frac{\Delta d}{d}=\frac{\Delta A}{2A}$$ ........(ii)
From Eqs. (i) and (ii), we get
$$\displaystyle \sigma =\frac{\displaystyle\frac{\Delta A}{2A}}{\Delta l/l}$$
$$\Rightarrow \displaystyle\frac{\Delta l}{l}=\frac{\Delta A}{2\sigma \cdot A}=\dfrac{1}{2\times \sigma }\left(\displaystyle\frac{\Delta A}{A}\right)$$
$$=\displaystyle\frac{1}{2\times 0.8}\times 4\%$$
$$\displaystyle\frac{\Delta l}{l}=2.5\%$$.
Question 7
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In the shown figure, length of the rod is $$L$$, area of cross-section $$A$$, Young's modulus of the material of the rod is $$Y$$. Then, $$B$$ and $$A$$ is subjected to a tensile force $$F_{A}$$ while force applied at end $$B, F_{B}$$ is lesser than $$F_{A}$$. Total change in length of the rod will be

A
$$F_{A} \times \dfrac {L}{2AY}$$

B
$$F_{B} \times \dfrac {L}{2AY}$$

C
$$\dfrac {(F_{A} + F_{B})L}{2AY}$$

D
$$\dfrac {(F_{A} - F_{B})L}{2AY}$$

Solution

Tension at $$x$$,
$$T = F_{A} - [F_{A} - F_{B}] \dfrac {x}{L} = F_{A} \left (1 - \dfrac {x}{L}\right ) + F_{B}\left (\dfrac {x}{L}\right )$$
Change in length
$$\triangle (dx) = \dfrac {Tdx}{AY} = \left [F_{A}\left (1 - \dfrac {x}{L}\right ) + F_{B} \left (\dfrac {x}{L}\right )\right ] = \dfrac {dx}{AY}$$
Total change in length
$$= \dfrac {1}{AY} \left [\int_{0}^{1} F_{A} (1 - x/L) dx + \int_{0}^{1} F_{B}\left (\dfrac {x}{L}\right )dx \right ]$$
$$= \dfrac {1}{AY} \left [F_{A}\left (L - \dfrac {L}{2}\right ) + F_{B}\times \dfrac {L}{2}\right ]$$
$$= \dfrac {1}{AY} [F_{A} + F_{B}] \dfrac {L}{2} = \dfrac {L}{2AY} (F_{A} + F_{B})$$.
Question 8
1 / -0

An aluminium rod (Young's modulus $$= 7\times 10^{9} N/m^{2})$$ has a breaking strain of $$0.2$$%. The minimum cross-sectional area of the rod in order to support a load of $$10^{4}$$ Newton's is:

A
$$1\times 10^{-2}m^{2}$$

B
$$1.4\times 10^{-3}m^{2}$$

C
$$3.5\times 10^{-3}m^{2}$$

D
$$7.1\times 10^{-4}m^{2}$$

Solution

No correct option.

Given,

Young's modulus $$=7\times10^9N/m^2$$ Breaking strain $$=0.2\% f=10^4N$$

So from young's modulus formula.

$$y=\dfrac{fI'}{AI}$$ Where $$I'=\dfrac{2}{1000}$$

By putting the value then we get,

$$7\times10^9=10^4\times\dfrac{2}{1000}\times A\Rightarrow A=3.5\times10^8m^2$$
Question 9
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One end of a long metallic wire of length $$L$$ is tied to the ceiling. The other end is tied to massless spring of spring constant $$k$$. A mass $$m$$ hangs freely from the free end of the spring. The area of cross-section and Young's modulus of the wire are $$A$$ and $$Y$$ respectively. If the mass is slightly pulled down and released, it will oscillate with a time period $$T$$ equal to

A
$$2\pi \sqrt {m/k}$$

B
$$2\pi \sqrt {\dfrac {m(YA + kL)}{YAk}}$$

C
$$2\pi \sqrt {\dfrac {mYA}{kL}}$$

D
$$2\pi \sqrt {\dfrac {mL}{YA}}$$

Solution

young's modulus $$Y = \dfrac {F\times L}{A\times \triangle L}$$ or $$\dfrac {F}{\triangle L} = \dfrac {YA}{L}$$
Force constant of wire $$k_{1} = \dfrac {F}{\triangle L} = \dfrac {YA}{L}$$
Force constant of wire $$k_{1}$$ and spring of force constant $$k$$ are in series
$$\dfrac {1}{k_{2}} = \dfrac {1}{k} + \dfrac {1}{k_{1}}$$
or $$k_{2} = \dfrac {kk_{1}}{k + k_{1}} = \dfrac {k\left (\dfrac {YA}{L}\right )}{k + \dfrac {YA}{L}} = \dfrac {kYA}{kL + YA}$$
The time period of the combination is
$$T' = 2\pi \sqrt {\dfrac {m}{k_{2}}} = 2\pi \sqrt {\dfrac {m(kL + YA)}{kYA}}$$.
Question 10
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A cubical ball is taken to a depth of $$200 m$$ in a sea. The decrease in volume observed to be $$0.1$$%. The bulk modulus of the ball is - $$(g = 10/ m/s^{-2})$$

A
$$2\times 10^{7} Pa$$

B
$$2\times 10^{6} Pa$$

C
$$2\times 10^{9} Pa$$

D
$$1.2\times 10^{9} Pa$$

Solution

Let the bulk modulus be $$B$$. The initial volume be $$v$$ and final volume is $$(v)(0.1/100)=0.001\ v$$
Change in volume $$\rightarrow dv=0.001\ v-v=-0.99$$
Pressure on the body, $$P=\rho gh$$
$$=1000\times 10\times 200$$
$$=2000000$$
$$=2\times 10^6\ Pa$$

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 47

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Mechanical Properties of Solids Test - 47

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SHARING IS CARING

Question 1

$$3L$$

$$4L$$

$$2L$$

None of these

Solution

Question 2

20 ms$$^{-1}$$

100 ms^{-1}$$

250 ms$$^{-1}$$

200 ms$$^{-1}$$

Solution

Question 3

$$9.8 \times 10^9$$

$$9.8 \times 10^{10}$$

$$1.96 \times 10^{10}$$

$$9.8 \times 10^{11}$$

$$1.96 \times 10^9$$

Solution

Question 4

$$\displaystyle \frac { { w }_{ 1 }+{ w }_{ 2 } }{ A } $$

$$\displaystyle \frac { { w }_{ 1 }-{ w }_{ 2 } }{ A } $$

$$\displaystyle \frac { \left( { { w }_{ 1 } }/{ 2 } \right) +{ w }_{ 2 } }{ A } $$

$$\displaystyle \frac { { { w }_{ 2 } }/{ 2 }+{ w }_{ 1 } }{ A } $$

Solution

Question 5

$$1.55\times 10^{5}$$

$$0.155\times 10^{5}$$

$$1.4\times 10^{5}$$

$$1.01\times 10^{5}$$

Solution

Question 6

$$1\%$$

$$2\%$$

$$2.5\%$$

$$4\%$$

Solution

Question 7

$$F_{A} \times \dfrac {L}{2AY}$$

$$F_{B} \times \dfrac {L}{2AY}$$

$$\dfrac {(F_{A} + F_{B})L}{2AY}$$

$$\dfrac {(F_{A} - F_{B})L}{2AY}$$

Solution

Question 8

$$1\times 10^{-2}m^{2}$$

$$1.4\times 10^{-3}m^{2}$$

$$3.5\times 10^{-3}m^{2}$$

$$7.1\times 10^{-4}m^{2}$$

Solution

Question 9

$$2\pi \sqrt {m/k}$$

$$2\pi \sqrt {\dfrac {m(YA + kL)}{YAk}}$$

$$2\pi \sqrt {\dfrac {mYA}{kL}}$$

$$2\pi \sqrt {\dfrac {mL}{YA}}$$

Solution

Question 10

$$2\times 10^{7} Pa$$

$$2\times 10^{6} Pa$$

$$2\times 10^{9} Pa$$

$$1.2\times 10^{9} Pa$$

Solution

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