Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 62

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Weekly Quiz Competition

Question 1
1 / -0

The Poisson's ratio of the material of a wire is$$0.25 .$$ If it is stretched by a force F, the longitudinal strain produced in the wire is $$5 \times 10 ^ { - 4 } .$$ What is the percentage increase in its volume?

A
$$0.2$$

B
$$2.5 \times 10 ^ { - 2 }$$

C
Zero

D
$$1.25 \times 10 ^ { - 6 }$$

Solution

$$\sigma =.25$$
$$\Rightarrow \ \dfrac {-\Delta R1R}{\Delta l 10}$$
$$\Rightarrow \ \dfrac {\Delta R}{R}=-.25\dfrac {\Delta l}{l}$$
$$=-.25\times 5\times 16^4$$
$$v=\pi R^2 l$$
$$\Rightarrow \ \dfrac {\Delta v}{v}\times 100 \left (2\dfrac {\Delta R}{R} + \dfrac {\Delta l}{l}\right)\times 100$$
$$=(2\times (-.25\times 5\times 10^{-4})+5\times 10^{-4})\times 150$$
$$=2.5\times 10^{-2}=.025\%$$
Question 2
1 / -0

The length of a steel wire is $$l_{1}$$ when the stretching force is $$T_{1}$$ and $$l_{2}$$ when the stretching force is $$T_{2}$$ The natural length of the wire is

A
$$\dfrac{l_{1}T_{1}+l_{2}T_{2}}{T_{1}+T_{2}}$$

B
$$\dfrac{l_{2}T_{1}+l_{2}T_{2}}{T_{1}+T_{2}}$$

C
$$\dfrac{l_{2}T_{1}+l_{2}T_{2}}{T_{1}-T_{2}}$$

D
$$\dfrac{l_{2}T_{1}-l_{1}T_{2}}{T_{1}-T_{2}}$$

Solution

assume natural length be $$l$$.
Hook's law :- $$T=K\triangle x$$ (T= Tension, $$\triangle x$$= elongation)
in case $$1 $$, $$T=T_{1}$$ and $$\triangle x= l_{1}-l$$
so, $$T_{1}=K(l_{1}-l)\rightarrow (1)$$
similarly for case 2, $$T_{2}=K(l_{2}-l)\rightarrow (2)$$
Divide $$(1)\div (2)$$,
$$\frac{T_{1}=K(l_{1}-l)}{T_{2}\, K(l_{2}-l)}\Rightarrow T_{1}(l_{2}-l)=T_{2}(l_{1}-l)$$
$$\Rightarrow l=\frac{T_{1}l_{2}-T_{2}l_{1}}{T_{1}-T_{2}}$$
Question 3
1 / -0

Determine the pressure required to reduce the given volume of water by 2%. Bulk modulus of water is $$2.2\times 10^{4}Nm^{-2}$$

A
$$4.4\times 10^{7}Nm^{-2}$$

B
$$2.2\times 10^{7}Nm^{-2}$$

C
$$3.3\times 10^{7}Nm^{-2}$$

D
$$1.1\times 10^{7}Nm^{-2}$$

Solution

Bulk modulus (B) is defined as :-
$$ B = \frac{-P}{\triangle v/v} $$ when, $$ \frac{\triangle v}{v} = 0.02 (2\%) $$
$$ \Rightarrow P = B\times \frac{Av}{v} = 2.2\times 10^{4}\times 0.02 = 4.4\times 10^{2} Atm $$
$$ = 4.4\times 10^{7}pa $$ $$ [1atm = 10^{5}pa] $$
Question 4
1 / -0

The pressure of a fixed mass of air is $$1.01\times10^5Pa.$$ It is compressed slowly, keeping its temperature constant. Then bulk modulus of air is,

A
$$1.414\times10^5$$pa.

B
$$1.67\times10^5$$pa.

C
$$51.2\times10^5$$pa.

D
$$1.01\times10^5$$pa.

Solution

$$\begin{array}{l}P=1.01 \times 10^{5} \mathrm{~Pa} \\\text { Temperature is constant } \Rightarrow \text { isothermal process }\end{array}$$

$$\begin{aligned}& P V=\text { constant }\\\Rightarrow&P(\Delta V)+(\Delta P)(V)=0\\\Rightarrow&\frac{\Delta V}{V}=-\frac{\Delta P}{P}\longrightarrow(1) \\B &=\frac{-\Delta P}{\Delta V / V}\longrightarrow \text { (2) }\end{aligned}$$

$$\begin{array}{l}\Rightarrow B=P \quad \longrightarrow\text{ using }(1) \\\Rightarrow B=1.01\times10^{5}\mathrm{~Pa}\\\text { Ans : }(D)\end{array}$$
Question 5
1 / -0

A uniform cyclinder rod of length $$L$$, cross -sectional area $$A$$ and youngs modules $$Y$$ is acted upon by the forces shown in fig. The elongation of the rod is

A
$$\dfrac{3FL}{5AY}$$

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

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

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

Solution

$$\begin{array}{l}\text {Using } \dfrac{F}{A}=Y \dfrac{\Delta l}{\ell} \\\text { (1) } \Delta l_{1}=\dfrac{F_{1} l_{1}}{A{Y}} \\\Rightarrow \Delta l_{1}=\dfrac{3 F(2 L / 3)}{A Y}\\\Rightarrow \Delta l_{1}=\dfrac{2 F L}{A Y}\end{array}$$

$$\begin{array}{l}\text { (2) } \Delta l_{2}=\dfrac{F_{2} l_{2}}{A Y} \\\Rightarrow \Delta l_{2}=\dfrac{2 F(L/3)}{A Y}\\\Rightarrow \Delta l_{2}=\dfrac{2 F L}{3 A Y}\end{array}$$

$$\begin{aligned}\text { Total elongation } &=\dfrac{2 F L}{A Y}+\dfrac{2 F L}{3 A Y}=\dfrac{8 F L}{3 A Y}\end{aligned}$$
Question 6
1 / -0

Which of the following is perfectly plastic?

A
Plasticine

B
Quartz fibre

C
steel

D
Rubber

Solution

$$\begin{array}{l}\text { Plasticine is a perfectly plastic body as it undergoes } \\\text { no work hardening after yield. } \\\text { Answer: } (A)\end{array}$$
Question 7
1 / -0

When a mass of 8 kg is suspended from a string, its length is $$l_1$$. If a mass 10 kg is suspended, its length is $$l_2$$. Length, when a mass of 16 kg is suspended from it, is given by

A
$$2l_2 - l_1$$

B
$$2l_2 + l_1$$

C
$$4l_2 + 3l_1$$

D
$$4l_2 - 3l_1$$

Solution

$$\begin{array}{l}m_{1}=8 \mathrm{~kg},\ l_{1}\\m_{2}=10\mathrm{~kg},\ l_{2} \\m_{3}=16 \mathrm{~kg} ,\ \mathrm{y}_{1}\end{array}$$

$$\begin{array}{l}\text { Let original length of string} =2\\\text { To find:}\ y_{1} \text { in terms of } L_1\text { and } L_2\end{array}$$

$$\begin{array}{l}\text { using } \dfrac{F}{A}=Y \dfrac{\Delta l}{l} \\\text { (1) } \dfrac{8 g}{A}=Y \dfrac{\ell_{1}-x}{x}\\\text { (2) } \dfrac{10 g}{A}=Y \dfrac{l_{2}-x}{x} \\\text { (3) } \dfrac{16 g}{A}=Y \dfrac{y_{1}-x}{x}\end{array}$$

$$\begin{array}{l}\text { Using }(1) \text { and }(2) \text { find } x \text { , and eliminate } x \text { in }(3) \text { , we get } \\y_{1}=2 l_{2}-l_{1}\quad \\\text { Ans }: \text { (A) }\end{array}$$
Question 8
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A mass of $$0.5\ kg$$ is suspended from wire, then length of wire increase by $$3\ mm$$ then find out work done.

A
$$4.5\times 10^{-3} J$$

B
$$7.5\times 10^{-3} J$$

C
$$9.3\times 10^{-2} J$$

D
$$2.5\times 10^{-2} J$$

Solution

$$\begin{aligned}\text { Work done in stretching of a wire }&=\frac{1}{2} F l \\&=\frac{1}{2} m g l \\&=\frac{1}{2}(0.5)(10)\times\left(3\times10^{-3}\right)\\&=7.5\times10^{-3}\\\text {Hence answer is } B &\end{aligned}$$
Question 9
1 / -0

A wire 2 m in length suspended vertically stretches by 10 mm when the mass of 10 kg is attached to the lower end. The elastic potential energy gain by the wire is? (take g = 10 $$m/s^2$$)

A
0.5 J

B
5 J

C
50 J

D
500 J

Solution

Formulae:

spring force by wire = weight of mass attached

$$k\Delta x=mg$$

$$k(0.01)=10 \times 9.8$$

$$k=9800N/m$$

elastic potential energy,

$$U=(0.5)k\Delta x^2$$

$$=(0.5)\times 9800 \times (0.01)^2$$

$$\therefore U=0.49\approx 0.5J$$
Question 10
1 / -0

What should be done if the gas cylinder at your home catches fire?

A
Water should be sprinkled.

B
Sand, soil should be put at it.

C
Cylinder should be covered with wet blanket.

D
One should run away

Solution

if the gas cylinder at your home catches fire the cylinder should be covered with wet cloth or dry powder extinguisher, and this the only possible way to stop the leak.
So the correct option is 'Cylinder should be covered with a wet blanket.'

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 62

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Mechanical Properties of Solids Test - 62

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Question 1

$$0.2$$

$$2.5 \times 10 ^ { - 2 }$$

Zero

$$1.25 \times 10 ^ { - 6 }$$

Solution

Question 2

$$\dfrac{l_{1}T_{1}+l_{2}T_{2}}{T_{1}+T_{2}}$$

$$\dfrac{l_{2}T_{1}+l_{2}T_{2}}{T_{1}+T_{2}}$$

$$\dfrac{l_{2}T_{1}+l_{2}T_{2}}{T_{1}-T_{2}}$$

$$\dfrac{l_{2}T_{1}-l_{1}T_{2}}{T_{1}-T_{2}}$$

Solution

Question 3

$$4.4\times 10^{7}Nm^{-2}$$

$$2.2\times 10^{7}Nm^{-2}$$

$$3.3\times 10^{7}Nm^{-2}$$

$$1.1\times 10^{7}Nm^{-2}$$

Solution

Question 4

$$1.414\times10^5$$pa.

$$1.67\times10^5$$pa.

$$51.2\times10^5$$pa.

$$1.01\times10^5$$pa.

Solution

Question 5

$$\dfrac{3FL}{5AY}$$

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

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

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

Solution

Question 6

Plasticine

Quartz fibre

steel

Rubber

Solution

Question 7

$$2l_2 - l_1$$

$$2l_2 + l_1$$

$$4l_2 + 3l_1$$

$$4l_2 - 3l_1$$

Solution

Question 8

$$4.5\times 10^{-3} J$$

$$7.5\times 10^{-3} J$$

$$9.3\times 10^{-2} J$$

$$2.5\times 10^{-2} J$$

Solution

Question 9

0.5 J

5 J

50 J

500 J

Solution

Question 10

Water should be sprinkled.

Sand, soil should be put at it.

Cylinder should be covered with wet blanket.

One should run away

Solution

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