Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

A metal rod of Young's modulus $$2\times { 10 }^{ 10 }N{ m }^{ -2 }$$ undergoes an elastic strain of 0.02%. The energy per unit volume stored in the rod, in $$Joules$$ is ?

A
$$400$$

B
$$800$$

C
$$1200$$

D
$$1600$$

Solution

$$\text{Energy per unit voume = Energy density}$$
$$\begin{array}{l}=\dfrac{1}{2} \times \text { stress } \times \text { strain } \\=\dfrac{1}{2}\times\dfrac{F}{A} \times \dfrac{\Delta l}{l}\\=\dfrac{1}{2} \times\left(\dfrac{\gamma}{l}\right)\left(\dfrac{\Delta l}{l}\right) \\=\dfrac{1}{2} Y\left(\dfrac{\Delta l}{l}\right)^{2} \\=\dfrac{1}{2} \times 2 \times 10^{10} \times\left(\dfrac{0.02}{100}\right)^{2}\end{array}$$
$$=400 \mathrm{Joule}$$
$$\text{Ans: (A)}$$
Question 2
1 / -0

A wire elongates by $$l$$ mm when a load W is hanged from it. If the wire goes over a pulley and two weights W each are hung at the two ends, the elongation of the wire will be (In mm) -

A
$$l$$

B
$$2l$$

C
zero

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

Solution

$$\begin{array}{l}\text { Hence at equilibriun }\\\qquad \begin{array}{l}T=W \\\frac{W}{A}=Y\frac{l_{1}}{L} \\\Rightarrow l_{1}=\frac{W L}{A Y}\end{array}\end{array}$$

$$\begin{array}{l}\text { At equilibnumu } T=W \text{ for each } \\\text { side of wire }\rightarrow\\\frac{W}{A}=Y \frac{l_{2}}{L}\\\Rightarrow l_{2}=\frac{W L}{A Y}\end{array}$$

$$\text { hence elongation will be same }\Rightarrow\text { ans } \rightarrow(A)$$
Question 3
1 / -0

A metal wire of length $$2.5 m$$ and area of cross section $$1.5\times { 10 }^{ -6 }{ m }^{ -2 }$$, is stretched through $$2 mm$$. Calculate the work done during stretching. $$Y=1.2\times { 10 }^{ 11 }N{ m }^{ -2 }$$

A
$$0.15 J$$

B
$$0.51 J$$

C
$$1.51 J$$

D
$$5.1 J$$

Solution
Question 4
1 / -0

A brass wire of diameter $$1 mm$$ and of length $$2 m$$ is stretched by applying a force of $$20 N$$. If the increase in length is $$0.51 mm$$. Then the Young's modulus of the wire is

A
$$8.848 \times 10^{10} N/m^2$$

B
$$7.984 \times 10^{10} N/m^2$$

C
$$6.984 \times 10^{10} N/m^2$$

D
$$9.984 \times 10^{10} N/m^2$$

Solution

Given,
$$F=20N,l=2m,d=1mm,\Delta l=0.51mm$$
We know,

$$Y=\dfrac{Fl}{A\Delta l}$$

$$Y=\dfrac{20\times 2}{\pi 5^2 \times 10^{-8}\times 0.51\times 10^{-3}}$$

$$Y=\dfrac{40}{25\pi\times 0.51}\times 10^11$$

$$Y=0.9984\times 10^11=9.984\times 10^{10}$$

Option $$\textbf D$$ is the correct answer
Question 5
1 / -0

A force of $$15 N$$ increases the length ofa by $$1 \mathrm { mm }$$. The additional force require increase the length by $$2.5 \mathrm { mm }$$ in N is

A
$$2.25$$

B
$$22.5$$

C
$$37.5$$

D
$$3.75$$

Solution

We know,

$$Y=\dfrac{FL}{A\Delta L}$$

For a wire, A,L,Y are constant.

$$\dfrac{F}{\Delta L}=$$constant

Let extra Force needed be $$ F$$,

$$\dfrac{15}{1}=\dfrac{F+15}{2.5}$$

$$F+15=2.5\times 15$$

$$F=1.5\times 15=22.5$$

Option $$\textbf B$$ is the correct answer
Question 6
1 / -0

Ball A strikes with velocity u elastically with identical ball B at rest, inclined at an angle of with line joining the centers of two balls. What will be the speed of ball B after the collision:

A
u

B
$$\dfrac{u\sqrt{3}}{2}$$

C
$$\dfrac{u}{2}$$

D
$$\dfrac{u}{\sqrt{2}}$$

Solution

It is not possible for ball $$A$$ to collide with ball $$B$$ at an angle to the line joining their centers because it will always consider with the line joining their centers
$$\rightarrow $$ It will always collide along the line joining their centers.
As the collision is elastic the velocity of ball $$B$$ will be $$u$$.
Question 7
1 / -0

A material has poisson's ratio $$0.3$$. If a uniform rod of it suffers a longitudinal strain of $$25\times 10^{-3}$$, then the percentage increase in its volume is

A
$$1\%$$

B
$$2\%$$

C
$$3\%$$

D
$$4\%$$

Solution

$$\begin{aligned}\mu &=\frac{1}{2}-\frac{Y}{6 B}\\\Rightarrow& \frac{3}{10}=\frac{1}{2}-\frac{Y}{6 B}\\\Rightarrow & \frac{Y}{6 B}=\frac{1}{2}-\frac{3}{10}=\frac{1}{5} \\\Rightarrow & \frac{Y}{6 B}=\frac{1}{5}\\\Rightarrow & \frac{Y}{B}=\frac{6}{5}\end{aligned}$$

$$\begin{aligned}\Rightarrow \frac{(\Delta V / V)}{(\Delta l / l)} &=\frac{6}{5} \\\Rightarrow \quad(\Delta v / v)&=\frac{25 \times 10^{-3} \times 6}{5}=30\times10^{-3}\\\Rightarrow \quad \frac{\Delta V}{V}\%&=30 \times 10^{-3} \times 100=3 \% \\A n s: &(C)\end{aligned}$$
Question 8
1 / -0

When a weight of $$10 \,kg$$ is suspended from a copper wire of length $$3 \,m$$ and diameter $$0.4 \,mm$$. Its length increases by $$2.4 \,cm$$. If the diameter of the wire is doubled, then the extension is its length will be :

A
$$7.6 \,cm$$

B
$$4.8 \,cm$$

C
$$1.5 \,cm$$

D
$$0.6 \,cm$$

Solution

Mass $$= 10 \,kg$$ $$D = 0.4 \,mm$$
$$f = mxg = 100 N$$

$$A = \pi r^2 = \dfrac{\pi \times (0.4 \times 10^{-3})^2}{4} = \dfrac{\pi D^2}{4}$$
$$L = 3$$

$$T = \dfrac{fL}{A\Delta L}$$

$$Y = \dfrac{100 \times 3}{\dfrac{\pi \times (0.4 \times 10^{-3})^2}{4} \times \Delta L}$$

$$\Delta L = \dfrac{300 \times 4}{\pi (0.4 \times 10^{-3})^2 \times 4}$$

Now, $$Y$$ is same for the material.
$$Y = \dfrac{F \times L'}{A' \times \Delta L'}$$

$$\Delta L' = \dfrac{f \times \dfrac{L}{4}}{4 \times A \times Y}$$

$$\Delta L' = \dfrac{\dfrac{100 \times 3}{4}}{4 \times 1.25 \times 10^{-7} \times 1 \times 10^{11}}$$

$$\Delta L' = \dfrac{75}{4 \times 1.25 \times 10^4}$$

$$\Delta L' = 1.5 \times 10^{-3}m$$
$$\Delta L' = 1.5 \,cm$$
Question 9
1 / -0

The Young's modulus a rubber string 8 cm long and density 1.5 kg/ $$m^{3}$$ is $$5\times 10^{8}N/m^{2}$$, is suspended on the ceiling in a room. The increases in length due to its own weigth will be

A
$$9.6\times 10^{-5}$$m

B
$$9.6\times 10^{-11}$$m

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

D
9.6 m

Solution

Given that,

Length of rubber band, $$l=8\,cm$$

Density, $$d=1.5\,Kg/{{m}^{3}}$$

Young’s modulus, $$Y=5\times {{10}^{8}}\,N/{{m}^{2}}$$

The relation for increase in length and young’s modulus is as follows:

$$ l=\dfrac{{{L}^{2}}dg}{2Y} $$

$$ l=\dfrac{{{(8\times {{10}^{-2}})}^{2}}1.5\times 10}{2\times 5\times {{10}^{8}}} $$

$$ l=\dfrac{64\times {{10}^{-4}}\times 1.5\times 10}{{{10}^{9}}} $$

$$ l=96\times {{10}^{-12}} $$

$$ l=9.6\times {{10}^{-11}}\,m $$
Question 10
1 / -0

Write Copper, Steel, Glass and Rubber in order of increasing coefficient of elasticity

A
Steel, Rubber, Copper, Glass

B
Rubber. Copper, Glass, Steel

C
Rubber. Glass, Steel, Copper

D
Rubber. Glass, Copper, Steel

Solution

D.
rubber, glass, copper, steel

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 61

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Mechanical Properties of Solids Test - 61

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SHARING IS CARING

Question 1

$$400$$

$$800$$

$$1200$$

$$1600$$

Solution

Question 2

$$l$$

$$2l$$

zero

$$\dfrac{l}{2}$$

Solution

Question 3

$$0.15 J$$

$$0.51 J$$

$$1.51 J$$

$$5.1 J$$

Solution

Question 4

$$8.848 \times 10^{10} N/m^2$$

$$7.984 \times 10^{10} N/m^2$$

$$6.984 \times 10^{10} N/m^2$$

$$9.984 \times 10^{10} N/m^2$$

Solution

Question 5

$$2.25$$

$$22.5$$

$$37.5$$

$$3.75$$

Solution

Question 6

u

$$\dfrac{u\sqrt{3}}{2}$$

$$\dfrac{u}{2}$$

$$\dfrac{u}{\sqrt{2}}$$

Solution

Question 7

$$1\%$$

$$2\%$$

$$3\%$$

$$4\%$$

Solution

Question 8

$$7.6 \,cm$$

$$4.8 \,cm$$

$$1.5 \,cm$$

$$0.6 \,cm$$

Solution

Question 9

$$9.6\times 10^{-5}$$m

$$9.6\times 10^{-11}$$m

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

9.6 m

Solution

Question 10

Steel, Rubber, Copper, Glass

Rubber. Copper, Glass, Steel

Rubber. Glass, Steel, Copper

Rubber. Glass, Copper, Steel

Solution

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