Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 68

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Question 1
1 / -0

A metallic beam having Young's modulus $Y$ is supported at the two ends. It is loaded at the centre. The depression of the centre is proportional to

A
$Y$

B
$1/Y$

C
$Y^{2}$

D
$1/Y^{2}$

Solution

Depression in the beam is given by $\delta=\dfrac{W{L}^{3}}{4Yb{d}^{3}}$
where $W$ is the weight, $L$ is length, $b$ is breadth, $Y$ is young's modulus and $d$ is the depth and $\delta$ is the depression in the beam.
Here, it is clear that,
Depression in the beam is inversely proportional to Young's modulus
$\delta\propto \dfrac{1}{Y}$
Question 2
1 / -0

If rigidity modulus is 2.6 times of youngs modulus then the value of poission's ratio is

A
0.2

B
0.3

C
0.5

D
0.1

Solution

$\begin{array}{l}\text{Young's Modulus}= 2 \times\text { Rigidity Modulus } \times \text { (1+ poisson's Ratio)}\\\text { Given , } \text{ Young's Modulus }=2.6 \times \text {Rigidity Modulus}\\ 2.6 = 2 \times (1+ \sigma)\\ 1.3 = 1 + \sigma \\ \sigma =0.3 \end{array}$
Question 3
1 / -0

A steel wire of mass 3.16 Kg is stretched to a tensile strain of 1 x ${ 10 }^{ -3 }$ . What is the elastic deformation energy if density p=7.9g/cc and Y=2x $10^{ 11 }N/m^{ 2 }$

A
4 KJ

B
0.4 KJ

C
0.04 KJ

D
40 J

Solution

$\begin{array}{l}\text { Given } \\\text { mass }=3.16 \mathrm{~kg} \\\text { Strain }\left(\frac{\Delta L}{L}\right)=1\times10^{-3}\end{array}$
$\rho(\text { density })=7.8 \mathrm{~g} /\mathrm{cc}$
$Y(\text { young modulus })=2\times 10^{11}$
$f=\frac{7.9 \mathrm{gram}}{\mathrm{cm}^{3}}=\frac{7.9\times 10^{-3}\mathrm{~kg}}{10^{-6}\mathrm{~m}^{3}}$
$=7.9 \times 10^{3} \mathrm{~kg}\mathrm{m}^{3}$
$\begin{array}{l}\text { Now, } \\\text { Energy stored }=\frac{1}{2} \times \text {stress }\times \text { strain } \times \text { volume }\end{array}$
$\begin{array}{c}\therefore \quad Y=\frac{\text { stress }}{\text { strain }} \\\text { Stress }=Y\text { strain }\end{array}$
$E=\frac{1}{2} \times Y\times(\operatorname{strain})^{2}\times\frac{(\rho V)}{\rho}$
$\therefore \rho V=m$
$=\frac{1 \times 2 \times 10^{11} \times\left(1\times 10^{-3}\right)^{2}\times 3.16}{2 \times7.9\times 10^{+3}}$
$\begin{array}{l}=0.4 \times 10^{2} \\E=40 \mathrm{~J}\end{array}$
Question 4
1 / -0

Which of the following substances has the highest value of the young's modulus?

A
Steel

B
Rubber

C
wood

D
Plastic

Solution

$\begin{array}{l}\text { we know, } \\\text { young's modulus }=\frac{\text { Stress }}{\text { strain }}\end{array}$
$\begin{array}{l}Y=\frac{F L}{A \Delta L} \\Y\alpha \frac{1}{\Delta L}\end{array}$
$\begin{array}{l}\text { By observation, }\Delta \mathrm{L} \text { in case of steel is }\\\text { smallest. So } Y \text { (young's modulus) will be } \\\text { highest for steel. }\end{array}$
Question 5
1 / -0

A telephone wire between the two poles is 50 m long and 1 mm in radius.When it is stretched by a load of 65 kg,its length becomes 50.12 m.Calculate Young's modulus of elasticity.

A
$4.22\times10^{10} Nm^{-2}$

B
$8.44\times10^{10} Nm^{-2}$

C
$4.22\times10^{9} Nm^{-2}$

D
$8.44\times10^{9} Nm^{-2}$

Solution

$\begin{array}{l} y=\frac{F l}{A \Delta l}=\frac{650 \times 50}{\pi \times 10^{-6} \times 0.12} \\ y=\frac{65 \times 5}{\pi \times 12} \times 10^{4}=8.44 \times 10^{4} \end{array}$
Question 6
1 / -0

Two wires of equal length and cross-section area suspended as shown in figure. Their Young's modulus are $Y_1$ and $Y_2$ respectively. The equivalent Young's modulus will be

A
$Y_1 + Y_2$

B
$\dfrac{Y_1 + Y_2}{2}$

C
$\dfrac{Y_1Y_2}{Y_1 + Y_2}$

D
$\sqrt{Y_1Y_2}$

Solution

We know that

$K = \dfrac{F}{ \triangle l}$ = $\dfrac{YA}{l}$

and, $K_{eq}=k_1+k_2$

$\Rightarrow$ $\dfrac{Y_{eq}2A}{l}$ = $\dfrac{Y_1A}{l}$ + $\dfrac{Y_2A}{l}$

$\Rightarrow$ $Y_{eq} = \dfrac{Y_1 + Y_2}{2}$
Question 7
1 / -0

A thin ring of radius R is made up of a material of density $\rho$ and young's modulus Y. If the ring is rotated about its centre in its own plane with an angular velocity $\omega$ then the small increase in radius $(\Delta R)$ of the ring is

A
$\cfrac { \rho { \omega }^{ 2 }{ R }^{ 3 } }{ Y }$

B
$\cfrac{\rho\omega^2}{YR^3}$

C
$\cfrac{Y}{\rho\omega^2R^3}$

D
None

Solution

Consider an element PQ of length dl. Let T be the tension and A the area of cross-section of the wire .
Mass of element $dm=volume \times density = A(dl) \times \rho$
The component of T, towards the centre provides the necessary centripetal force to portion PQ $F= 2T sin \left( \dfrac{d \theta}{2} \right) = (dm) R \omega^2$
For small angles $\dfrac{sin(d \theta)}{2} = \dfrac{d \theta}{2} = \dfrac{dl/R}{2}$
or $d \theta = \dfrac{d}{R}$
Substituting in Eq. (i), we have
$T \ times \dfrac{dl}{R} = A(dl)( \rho R \omega^2$
or $T= A \rho \omega^2 R^2$
Let $\Delta R$ be the increase in radius.
Longitudinal strain= $\dfrac{ \Delta l}{l} = \dfrac{ \Delta(2 \pi R)}{ 2 \pi R} = \dfrac{ \Delta R}{R}$
Now, $Y= \dfrac{T/A}{ \Delta R/ R}$
$\Delta R = \dfrac{T R}{AY} = \dfrac{A \rho \omega^2 R^2 ) R}{ AY}$ or $\Delta R= \dfrac{ \rho \omega^2 R^3}{Y}$
Question 8
1 / -0

The Young's modulus of a wire is numerically equal to the stress which will

A
not charge the length of the wire

B
double the length of the wire

C
increase the length by 50%

D
charge the radius of the wire to half

Solution

$\frac{\Delta l}{l}=1$ $l_{f i n a l}=l+\Delta l=2 l$ Therefere. length of wire will be doubled.
Question 9
1 / -0

What happens when the applied load increases and upto breaking stress in the experiment to determine the Young's modulus of elasticity?

A
The area of wire goes in decreasing and wire extends and breaks.

B
The area of wire goes in increasing and wire breaks.

C
The wire extends and area remains constant.

D
The area remains same and wire length is also same.

Solution

$\begin{array}{l}\text { In searl's experiment, when applied load } \\\text { increases upto breaking stress. then, area } \\\text { of wire goes on decreasing and length } \\\text { will extend and finally break. }\end{array}$
Hence option A is correct
Question 10
1 / -0

The SI unit of stress is same as the SI unit of

A
Strain

B
Modulus of elasticity

C
Pressure

D
Both (2) and (3)

Solution

The $SI$ unit of stress is same as the $SI$ unit of pressure
Stress $=\dfrac { F }{ A } =\dfrac { Force }{ acceleration }$
Similarly, Pressure $=\dfrac { Force }{ acceleration }$ .

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Mechanical Properties of Solids Test - 68

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Mechanical Properties of Solids Test - 68

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Question 1

YYY

1/Y1/Y1/Y

Y2Y^{2}Y2

1/Y21/Y^{2}1/Y2

Solution

Question 2

0.2

0.3

0.5

0.1

Solution

Question 3

4 KJ

0.4 KJ

0.04 KJ

40 J

Solution

Question 4

Steel

Rubber

wood

Plastic

Solution

Question 5

4.22×1010Nm−24.22\times10^{10} Nm^{-2}4.22×1010Nm−2

8.44×1010Nm−28.44\times10^{10} Nm^{-2}8.44×1010Nm−2

4.22×109Nm−24.22\times10^{9} Nm^{-2}4.22×109Nm−2

8.44×109Nm−28.44\times10^{9} Nm^{-2}8.44×109Nm−2

Solution

Question 6

Y1+Y2Y_1 + Y_2Y1​+Y2​

Y1+Y22\dfrac{Y_1 + Y_2}{2}2Y1​+Y2​​

Y1Y2Y1+Y2\dfrac{Y_1Y_2}{Y_1 + Y_2}Y1​+Y2​Y1​Y2​​

Y1Y2\sqrt{Y_1Y_2}Y1​Y2​​

Solution

Question 7

ρω2R3Y\cfrac { \rho { \omega }^{ 2 }{ R }^{ 3 } }{ Y } Yρω2R3​

ρω2YR3\cfrac{\rho\omega^2}{YR^3}YR3ρω2​

Yρω2R3\cfrac{Y}{\rho\omega^2R^3}ρω2R3Y​

None

Solution

Question 8

not charge the length of the wire

double the length of the wire

increase the length by 50%

charge the radius of the wire to half

Solution

Question 9

The area of wire goes in decreasing and wire extends and breaks.

The area of wire goes in increasing and wire breaks.

The wire extends and area remains constant.

The area remains same and wire length is also same.

Solution

Question 10

Strain

Modulus of elasticity

Pressure

Both (2) and (3)

Solution

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$Y$

$1/Y$

$Y^{2}$

$1/Y^{2}$

$4.22\times10^{10} Nm^{-2}$

$8.44\times10^{10} Nm^{-2}$

$4.22\times10^{9} Nm^{-2}$

$8.44\times10^{9} Nm^{-2}$

$Y_1 + Y_2$

$\dfrac{Y_1 + Y_2}{2}$

$\dfrac{Y_1Y_2}{Y_1 + Y_2}$

$\sqrt{Y_1Y_2}$

$\cfrac { \rho { \omega }^{ 2 }{ R }^{ 3 } }{ Y }$

$\cfrac{\rho\omega^2}{YR^3}$

$\cfrac{Y}{\rho\omega^2R^3}$