Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 26

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Weekly Quiz Competition

Question 1
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A wire suspended from one end carries a sphere at its other end. The elongation in the wire reduces from $$2 mm$$ to $$1.6 mm$$ on completely immersing the sphere in water. The density of the material of the sphere is

A
3200 kg/m$$^{3}$$

B
800 kg/m$$^{3}$$

C
1250 kg/m$$^{3}$$

D
5000 kg/m$$^{3}$$

Solution

$$ Y = \dfrac{F/A}{\triangle L/L}$$
$$ Y = \dfrac{F_{1}/A}{2/L} = \dfrac{F_{2}/A}{1.6/L}$$
$$\dfrac { 4 }{ 5 } F_{1} = F_{2}$$
Bouyant force $$= F_{1} -F_{2} = \dfrac{F_{1}}{5}$$
$$ F_{1} = \rho_{s} V_{s} g$$
Bouyant force $$= \rho_{w} V_{s} g = \dfrac{F_{1}}{5}$$
$$ \Rightarrow 5\rho_{w} V_{s} g = \rho_{s} V_{s} g$$
$$ \Rightarrow \rho_{s} = 5\rho_{w}$$
$$ \Rightarrow \rho_{s} = 5000 kg / m^{3} (\because \rho_{w} = 1000 kg/m^{3})$$
So, density of material is $$5000 kg / m^{3}$$
Question 2
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If the force constant of a wire is $$K$$, the work done in increasing the length $$l$$ of the wire by $$l$$ is

A
$$3Kl$$

B
$$2Kl$$

C
$$\dfrac{Kl^{2}}{2}$$

D
$$\dfrac{Kl^{2}}{3}$$

Solution

$$ Work done = \int _{0}^{l} F dx$$
$$\displaystyle = \int _{0}^{l} K x d x (\because force (f) = Kx)$$
where $$x$$ is elongation in the wire
$$ = \dfrac{1}{2} Kx^{2} |_{0}^{l}$$
$$ = \dfrac{1}{2} Kl^{2} - \frac{1}{2} k(0)^{2}$$
$$ =\dfrac{1}{2} Kl^{2}$$
Question 3
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A 8m long string of rubber, having density 1.5 x 10$$^{3}$$ kg/m$$^{3}$$ and young's modulus $$5\times10^{6}$$ N/m$$^{2}$$ is suspended from the ceiling of a room. The increase in its length due to its own weight will be (g$$=$$10m/s$$^{2}$$)

A
$$9.6\times 10^{-2}\ \text{m}$$

B
$$19.2\times 10^{-5}\ \text{m}$$

C
$$9.6\times 10^{-3}\ \text{m}$$

D
$$9.6\ \text{m}$$

Solution

Mass of small dx element, $$\displaystyle dm = \lambda Adx$$

Also $$\displaystyle \frac{stress}{strain}=Y$$

=> $$\displaystyle \frac{F/A}{\frac{\triangle L}{L}} = Y$$

where $$F \rightarrow (x)(A)\lambda .g$$ , gravitational force experienced by the part below to $$dm$$ mass

Strain on $$dm$$ mass = $$\dfrac{d(\triangle L)}{dx}$$

=> $$\displaystyle \frac{x\, A\lambda g}{A\lambda}=\frac{d(\triangle L)}{dx}$$

$$\displaystyle \Rightarrow\int_{o}^{\triangle L} d(\triangle L) = \frac{\lambda g}{y}\int_{o}^{L}xdx$$

=>$$\displaystyle \triangle L = \frac{\lambda gL^2}{2y}=9.6\times 10^{-2}$$
Question 4
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Two wires of different materials each of length $$l$$ and cross sectional area ‘$$A$$’ are joined in series to form a composite wire. If their Young’s modulii are $$Y$$ and $$2Y$$, the total elongation produced by applying a force $$F$$ to stretch the composite wire.

A
$$3FA/2Yl$$

B
$$2FA/3Yl$$

C
$$2FA/3AY$$

D
$$3Fl/2AY$$

Solution

Applying a force F to stretch is applied on both wires simultaneously so we can directly add the elongation of each wire with force F, so,
$$ Y = \dfrac{F/A}{\triangle L_{1} / L }$$ for wire with youngs modulus $$Y$$
$$2Y = \dfrac{F/A}{\triangle L_{2} / L } $$ for wire with youngs modulus $$2Y$$
$$ \triangle L_{1} = \dfrac{F L}{Y A }$$
$$ \triangle L_{2} = \dfrac{F L}{2YA}$$
$$\triangle L_{1} + \triangle L_{2} = \dfrac{F L}{Y A} + \dfrac{F L}{2YA} = \dfrac{3F L}{2YA}$$
Question 5
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A force of $$15N$$ increases the length of a wire by 1mm. The additional force required to increase the length by $$2.5mm$$ in N is

A
$$22.5$$

B
$$37.5$$

C
$$52.5$$

D
$$75$$

Solution

$$Y=\dfrac{F/A}{\Delta l/l}$$

$$\Delta l=\dfrac{Fl}{AY}$$

$$\Delta l_1=\dfrac{F_1l}{AY}$$-----(1)

$$\Delta l_2=\dfrac{F_2l}{AY}$$-----(2)

Dividing (1) by (2)

$$\dfrac{\Delta l_1}{\Delta l_2}=\dfrac{F-1}{F-2}$$

$$\dfrac{1}{2.5}=\dfrac{15}{F_2}$$

$$F_2=15\times 2.5=37.5$$ N

So, additional force$$=F_2-F_1$$

$$=37.5-15$$

$$=22.5$$N
Question 6
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The density of water at the surface of the ocean is $$\rho $$ . If the bulk modulus of water is B, what is the density of ocean water at a depth where the pressure is nP$$_{o}$$ , where P$$_{o}$$ is the atmosphreic pressure:

A
$$\dfrac{\rho B}{B-(n-1)P_{o}}$$

B
$$\dfrac{\rho B}{B+(n-1)P_{o}}$$

C
$$\dfrac{\rho B}{B-nP_{o}}$$

D
$$\dfrac{\rho B}{B+nP_{o}}$$

Solution

Let the volume of certain mass (M) of water at surface be $$V$$ $$\implies \rho =\dfrac{M}{V}$$
Pressure at a depth $$P' = n P_o$$
Thus change in pressure $$\Delta P = nP_o - P_o =(n-1)P_o$$
Bulk modulus $$B = \dfrac{\Delta P}{\Delta V/V}$$ $$\implies \Delta V = \dfrac{(n-1) P_o V}{B}$$

As at a depth, the pressure increases. Thus the volume of water will decrease at that depth.
$$\therefore$$ Volume of water of mass (M) at that depth $$V' = V - \Delta V$$
$$\therefore$$ $$V' = V - \dfrac{(n-1) P_o V}{B} = (B - (n-1) P_o) \dfrac{V}{B}$$ ..............(1)

Density at that depth $$\rho' = \dfrac{M}{V'} = \rho \dfrac{V}{V'}$$
$$\therefore$$ $$\rho' = \dfrac{\rho B}{B - (n-1) P_o}$$ (using 1)
Question 7
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When a mass is suspended from the end of a wire the top end of which is attached to the roof of the lift, the extension is $$e$$ when the lift is stationary. If the lift moves up with a constant acceleration $$g/2$$, the extension of the wire would be

A
$$\dfrac{2e}{3}$$

B
$$\dfrac{3e}{2}$$

C
$$2e$$

D
$$3e$$

Solution

$$ Y = \dfrac{mg/A}{e/L} = \dfrac{(3/2) mg/A}{x/L} = Y$$
$$ x = \dfrac{3e}{2}$$
Hence, new elongation is $$ \dfrac{3e}{2}$$
In this problem Young's modulus will be same because were is same. No change in area, natural length will be also same. But tension will change. In the frame of lift when it is moving up by acceleration g/2 a psuedo force of mg/2 units will add in the tension so elongation will increase to $$ \dfrac{3e}{2}$$
Question 8
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When a sphere of radius $$2 cm$$ is suspended at the end of a wire, elongation is $$'e$$'. When the same wire is loaded with a sphere of radius $$3cm$$ and made of the same material, the elongation would be :

A
$$\dfrac{8}{27}e$$

B
$$\dfrac{27}{8}e$$

C
$$\dfrac{4}{9}e$$

D
$$\dfrac{9}{4}e$$

Solution

$$Y=\dfrac{F/A}{\Delta l/l}$$

$$\Delta l=\dfrac{Fl}{YA}$$

So, $$\dfrac{\Delta l_1}{\Delta l_2}=\dfrac{p\frac{4}{3}\pi r{_{1}}^{3}g}{p\dfrac{4}{3}\pi r{_{2}}^{3}g}$$

$$\dfrac{\Delta l_1}{\Delta l_2}=\dfrac{(2)^3}{(3)^3}$$

$$\dfrac{\Delta l_1}{\Delta l_2}=\dfrac{8}{27}$$

$$\Delta l_2=\dfrac{27}{8}\Delta l_1$$

$$=\dfrac{27}{8}e$$.
Question 9
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Two bars $$A$$ and $$B$$ of circular section and of the same volume and made of the same material are subjected to tension. If the diameter of $$A$$ is half that of $$B$$ and if the force supplied to both the rods is the same and is within the elastic limit, the ratio of the extension of $$A$$ to that of $$B$$ will be

A
$$16$$

B
$$8$$

C
$$4$$

D
$$2$$

Solution

Using
$$Y = \dfrac{F/A}{\triangle L/L }$$

$$ Y = \dfrac{F/ \triangle_{1}}{\triangle L_{1} / L_{1}} - (1)\text{For bar A} $$

$$ Y = \dfrac{F/ \triangle_{2}}{\triangle L_{2} / L_{2}} - (2)\text{For bar B} $$
Volume is same for A and B
$$So, R(\dfrac{d_{1}}{2})^{2} L_{1} = R (\dfrac{d_{2}}{2})^{2} L_{2}$$

$$ d_{1}^{2} L_{1} = 4d_{1}^{2} L_{2} (given d_{1} = \dfrac{1}{2}d_{2})$$

$$ \dfrac{L_{1}}{L_{2}} = 4$$

Dividing 1 and 2

$$ 1 = \dfrac{L_{1} \triangle L_{2} A_{2}}{L_{2} \triangle L_{1} A_{1}}$$

$$ \dfrac{\triangle L_{1} }{\triangle L_{2}} = 4\times 4 (\because \dfrac{A_{2}}{A_{1}} = (\dfrac{d_{2}}{d_{1}})^{2} = 4) $$

$$ \dfrac{\triangle L_{1} }{\triangle L_{2}} = 16$$
Question 10
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The density of a metal at normal pressure is $$\rho$$. It's density when it is subjected to an excess pressure p is $$\rho$$'. If B is the bulk modulus of the metal, the ratio $$\rho $$'/$$\rho$$ is :

A
$$\dfrac{1}{1-p/B}$$

B
$$1+\dfrac{P}{B}$$

C
$$\dfrac{1}{1-B/p}$$

D
$$1+B/p$$

Solution

$$B=-V\dfrac{dP}{dV}$$

Thus, $$\Delta V=-\dfrac{pV}{B}$$ or $$V'-V=-\dfrac{pV}{B}$$
Or, $$V'=V(1-\dfrac{p}{B})$$

Now, $$\rho '=\dfrac{m}{V'}=\dfrac{m}{V(1-\dfrac{p}{B})}=\dfrac{\rho}{(1-\dfrac{p}{B})}$$

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 26

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Mechanical Properties of Solids Test - 26

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SHARING IS CARING

Question 1

3200 kg/m$$^{3}$$

800 kg/m$$^{3}$$

1250 kg/m$$^{3}$$

5000 kg/m$$^{3}$$

Solution

Question 2

$$3Kl$$

$$2Kl$$

$$\dfrac{Kl^{2}}{2}$$

$$\dfrac{Kl^{2}}{3}$$

Solution

Question 3

$$9.6\times 10^{-2}\ \text{m}$$

$$19.2\times 10^{-5}\ \text{m}$$

$$9.6\times 10^{-3}\ \text{m}$$

$$9.6\ \text{m}$$

Solution

Question 4

$$3FA/2Yl$$

$$2FA/3Yl$$

$$2FA/3AY$$

$$3Fl/2AY$$

Solution

Question 5

$$22.5$$

$$37.5$$

$$52.5$$

$$75$$

Solution

Question 6

$$\dfrac{\rho B}{B-(n-1)P_{o}}$$

$$\dfrac{\rho B}{B+(n-1)P_{o}}$$

$$\dfrac{\rho B}{B-nP_{o}}$$

$$\dfrac{\rho B}{B+nP_{o}}$$

Solution

Question 7

$$\dfrac{2e}{3}$$

$$\dfrac{3e}{2}$$

$$2e$$

$$3e$$

Solution

Question 8

$$\dfrac{8}{27}e$$

$$\dfrac{27}{8}e$$

$$\dfrac{4}{9}e$$

$$\dfrac{9}{4}e$$

Solution

Question 9

$$16$$

$$8$$

$$4$$

$$2$$

Solution

Question 10

$$\dfrac{1}{1-p/B}$$

$$1+\dfrac{P}{B}$$

$$\dfrac{1}{1-B/p}$$

$$1+B/p$$

Solution

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