Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

The length of an iron wire is $$L$$ and area of cross-section is $$A$$. The increase in length is $$l$$ on applying the force $$F$$ on its two ends. Which of the statement is correct?

A
Increase in length is inversely proportional to its length

B
Increase in length is proportional to area of cross-section

C
Increase in length is inversely proportional to area of cross-section

D
Increase in length is proportional to Young's modulus.

Solution

$$l=\displaystyle\frac{FL}{YA}\Rightarrow l\propto \frac{1}{A}$$
Question 2
1 / -0

From a steel wire of density $$\rho$$ is suspended a brass block of density $$\rho_b$$. The extension of steel wire comes to $$e$$. If the brass block is now fully immersed in a liquid of density $$\rho_{\gamma}$$ the extension becomes $$e'$$. The ratio $$\dfrac{e}{e^{'}}$$ will be

A
$$\displaystyle\dfrac{\rho_b}{\rho_b-\rho_l}$$

B
$$\displaystyle\dfrac{\rho_b-\rho_l}{\rho_b}$$

C
$$\displaystyle\dfrac{\rho_b-\rho}{\rho_l-\rho}$$

D
$$\displaystyle\dfrac{\rho_l}{\rho_b-\rho_l}$$

Solution

Since everything else $$(L, A, Y)$$ is kept constant, extension will be directly proportional to the Force, which is mg or $${\rho}_{b} Vg$$
After immersing, tension will reduce due to buoyant force and will become $${\rho}_{b} Vg-{\rho}_{\gamma}Vg$$
Thus, $$\dfrac{e}{e'}=\dfrac{{\rho}_{b} Vg}{{\rho}_{b} Vg-{\rho}_{\gamma}Vg}=\dfrac{{\rho}_{b}}{{\rho}_{b}-{\rho}_{\gamma}}$$
Question 3
1 / -0

If a rubber ball is taken at the depth of $$200m$$ in a pool, its voulme decreases by $$0.1\%$$. If the density of the water is $$1\times 10^3kg/m^3$$ and $$g=10m/s^2$$, then the volume elasticity in $$N/m^2$$ will be

A
$$10^8$$

B
$$2\times 10^8$$

C
$$10^9$$

D
$$2\times 10^9$$

Solution

$$K=\displaystyle\frac{\Delta P}{\Delta V/V}=\frac{hpg}{\Delta V/V}=\displaystyle\frac{200\times 10^3\times 10}{0.1\times 100}=2\times 10^9$$
Question 4
1 / -0

A cube at temperature $$0^oC$$ is compressed equally from all sides by an external pressure $$P$$. By what amount should its temperature be raised to bring it back to the size it had before the external pressure was applied. The bulk modulus of the material of the cube is $$B$$ and the coefficient of linear expansion is $$\alpha$$.

A
$$P/B\alpha$$

B
$$P/3B\alpha$$

C
$$3\pi \alpha/ B$$

D
$$3B/P$$

Solution

Bulk modulus $$B=\displaystyle\frac{-P}{(\Delta V/V)}=\frac{-PV}{\Delta V}$$ .......(1)
and $$\Delta V= \gamma V\Delta T=3\alpha V. T$$ or $$\displaystyle\frac{-V}{\Delta V}=\frac{1}{3\alpha. T.}$$ .....(2)
From eqs. $$(1)$$ and $$(2)$$, $$B=P/(3\alpha . T)$$ or $$\displaystyle T=\frac{P}{3\alpha B}$$
Question 5
1 / -0

$$A$$ and $$B$$ are two wires. The radius of $$A$$ is twice that of $$B$$. They are stretched by the same load. Then the stress on $$B$$ is

A
equal to that on $$A$$

B
four times that on $$A$$

C
two times that on $$A$$

D
half that on $$A$$

Solution

$$stress=\dfrac{Force}{Area}$$
$$\therefore \displaystyle stress \propto \dfrac{1}{\pi r^2}$$
Thus, $$\dfrac{S_B}{S_A}=\left[\dfrac{r_A}{r_B}\right]^2=(2)^2\Rightarrow S_B=4S_A$$
Question 6
1 / -0

When a pressure of $$100$$ atmosphere is applied on a spherical ball, then its volume reduces to $$0.01\%$$. The bulk modulus of the material of the rubber in $$dyne/cm^2$$ is :

A
$$10\times 10^{12}$$

B
$$100\times 10^{12}$$

C
$$1\times 10^{12}$$

D
$$1000\times 10^{12}$$

Solution

$$B=-V\dfrac{dP}{dV}=\dfrac{100}{0.01/100}=10^6\ atm$$
which is $$10^{11}\ N/{m}^{2}$$
Now, $$1N={10}^{5}\ dynes$$ and $$1{m}^{2}={10}^{4}\ {cm}^{2}$$
Thus, we get, $$B={10}^{12}\ dyne/{cm}^{2}$$
Question 7
1 / -0

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

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

B
$$100ms^{-1}$$

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

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

Solution

Young's modulus of rubber, $$Y_{rubber} =\displaystyle\frac{F}{A}\times \frac{l}{\Delta l}\Rightarrow F=YA\cdot \frac{\Delta l}{l}$$
Thus, we get, $$F=\displaystyle\frac{5\times 10^8\times 25\times 10^{-6}\times 5\times 10^{-2}}{10\times 10^{-2}}$$
or, $$F=\displaystyle 25\times 25\times 10^{2-1}=6250N$$
Kinetic energy=Potential energy of rubber
Thus, $$\displaystyle\frac{1}{2}mv^2=\frac{1}{2}F\Delta l$$
or, $$\displaystyle v=\displaystyle\sqrt{\frac{F\Delta l}{m}}=\displaystyle\sqrt{\frac{6250\times 5\times 10^{-2}}{5\times 10^{-3}}}=\sqrt{62500}$$
Thus, $$v=250m/s$$
Question 8
1 / -0

Two wire $$A$$ and $$B$$ are of the same material. Their lengths are in the ratio of $$1:2$$ and the diameter are in the ratio $$2:1$$. If they are pulled by the same force, then increase in length will be in the ratio of

A
$$2:1$$

B
$$1:4$$

C
$$1:8$$

D
$$8:1$$

Solution

We know that Young's modulus $$Y=\displaystyle\dfrac{F}{\pi r^2}\times \dfrac{L}{l}$$
Since $$Y$$, $$F$$ are same for both the wires, we have,
$$\displaystyle\dfrac{1}{r^2_1}\dfrac{L_1}{l_1}=\dfrac{1}{r^2_2}\dfrac{L_2}{l_2}$$

or, $$\displaystyle\dfrac{l_1}{l_2}=\dfrac{r^2_2\times L_1}{r^2_1\times L_2}=\displaystyle\dfrac{(D_2/2)^2\times L_1}{(D_1/2)^2\times L_2}$$

or, $$\displaystyle\dfrac{l_1}{l_2}=\displaystyle\dfrac{D^2_2\times L_1}{D^2_1\times L_2}=\displaystyle\dfrac{D^2_2}{(2D_2)^2}\times \dfrac{L_2}{2L_2}=\dfrac{1}{8}$$

so, $$l_1:l_2=1:8$$
Question 9
1 / -0

When a $$4kg$$ mass is hung vertically on a light spring that obeys Hooke's law, the spring stretches by $$2cms$$. The work required to be done by an external agent in stretching this spring by $$5cms$$ will be $$(g=9.8m/sec^2)$$

A
$$4.900$$ joule

B
$$2.450$$ joule

C
$$0.495$$ joule

D
$$0.245$$ joule

Solution

Since the spring obeys Hooke's law, thus F=kx, or $$K=\displaystyle\dfrac{F}{x}=\dfrac{4\times 9.8}{2\times 10^{-2}}=19.6\times 10^2$$
Thus, work done can be written as stored potential energy=$$\dfrac{1}{2}k{x}^{2}$$
=$$\displaystyle\dfrac{1}{2}19.6\times 10^2\times (0.05)^2=2.45J$$
Question 10
1 / -0

A mass m is suspended from a wire. Change in length of the wire is $$\Delta l$$. Now the same wire is stretched to double its length and the same mass is suspended from the wire. The change in length in this case will become (it is assumed that elongation in the wire is within the proportional limit)

A
$$\Delta l$$

B
$$2\Delta l$$

C
$$4\Delta l$$

D
$$8\Delta l$$

Solution

Elongation is given by
$$\Delta l=\dfrac {Fl}{AY}=\dfrac {Fl}{\frac {V}{l}Y}=\dfrac {Fl^2}{VY}$$
SInce the force, volume and Y are constant
$$\Delta l\propto l^2$$
New length is double the original length, so the elongation would be
$$\dfrac {\Delta l_2}{\Delta l_1}=\left (\dfrac {l_2}{l_1}\right )^2$$
$$\Delta l_2=\Delta l_1\left (\dfrac {2l}{l}\right )^2=\Delta l_1(4)$$
Option C.

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 37

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Mechanical Properties of Solids Test - 37

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Question 1

Increase in length is inversely proportional to its length

Increase in length is proportional to area of cross-section

Increase in length is inversely proportional to area of cross-section

Increase in length is proportional to Young's modulus.

Solution

Question 2

$$\displaystyle\dfrac{\rho_b}{\rho_b-\rho_l}$$

$$\displaystyle\dfrac{\rho_b-\rho_l}{\rho_b}$$

$$\displaystyle\dfrac{\rho_b-\rho}{\rho_l-\rho}$$

$$\displaystyle\dfrac{\rho_l}{\rho_b-\rho_l}$$

Solution

Question 3

$$10^8$$

$$2\times 10^8$$

$$10^9$$

$$2\times 10^9$$

Solution

Question 4

$$P/B\alpha$$

$$P/3B\alpha$$

$$3\pi \alpha/ B$$

$$3B/P$$

Solution

Question 5

equal to that on $$A$$

four times that on $$A$$

two times that on $$A$$

half that on $$A$$

Solution

Question 6

$$10\times 10^{12}$$

$$100\times 10^{12}$$

$$1\times 10^{12}$$

$$1000\times 10^{12}$$

Solution

Question 7

$$20ms^{-1}$$

$$100ms^{-1}$$

$$250ms^{-1}$$

$$200ms^{-1}$$

Solution

Question 8

$$2:1$$

$$1:4$$

$$1:8$$

$$8:1$$

Solution

Question 9

$$4.900$$ joule

$$2.450$$ joule

$$0.495$$ joule

$$0.245$$ joule

Solution

Question 10

$$\Delta l$$

$$2\Delta l$$

$$4\Delta l$$

$$8\Delta l$$

Solution

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