Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 43

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Question 1
1 / -0

A metal wire of length l, area of cross-section A and Young's modulus Y behaves as a spring of spring constant k given by.

A
$k=\displaystyle\frac{YA}{l}$

B
$k=\displaystyle\frac{2YA}{l}$

C
$k=\displaystyle\frac{YA}{2l}$

D
$k=\displaystyle\frac{Yl}{A}$

Solution
Question 2
1 / -0

The increase in pressure required in $kPa$ , to decrease the $200$ litres volume of a liquid by $0.004$ % is (bulk modulus of the liquid $= 2100\ MPa$ )

A
$8.4$

B
$84$

C
$92.4$

D
$168$

Solution

Bulk modulus of elasticity is defined as the ratio of normal stress to the volumetric strain, within the elastic limit.
Thus,
$B = \dfrac {\text {normal stress}}{\text {volumetric strain}}$
$B = \dfrac {\triangle p}{-\triangle V/V}$
Here, negative sign shows that volume is decreased, when pressure is increased.
Here, $B = 2100\times 10^{6} Pa$
$V = 200\ L$
$\triangle V = 200\times \dfrac {0.004}{100} = 0.008\ L$
$\therefore 2100\times 10^{6} = \dfrac {\triangle p}{\left (\dfrac {0.008}{200}\right )}$
$\therefore \triangle p = 84 kPa$
Compressibility (C) of a material is the reciprocal of its bulk modulus of elasticity (B) ie,
$C = \dfrac {1}{B}$
Question 3
1 / -0

Four wires of the same material are stretched by the same load. Which one of them will elongate most if their dimensions are as follows

A
L = 100 cm, r = 1 mm

B
L = 200 cm, r = 3 mm

C
L = 300 cm, r = 3 mm

D
L = 400 cm, r = 4 mm

Solution

$\begin{array}{l}\text { L=length of the wire } \\r=\text { radius of the wine } \\Y \text { is same for all as made up of } \\\text { same material }\end{array}$

$Y=\frac{F / A}{\frac{d l}{l}} \Rightarrow d l=\frac{F l}{A Y}=\left(\frac{f}{\gamma}\right) \times \frac{\lambda}{\pi r^{2}}$

$\begin{aligned}\therefore d l &=\frac{c l}{\pi r^{2}} \\\therefore\left(\frac{e}{r^{2}}\right) \uparrow \Rightarrow d l \uparrow & \\\therefore \text { checking options we get }(L=100 \mathrm{~cm} \& r=\operatorname{lm} m) \text { wire elongate most} \\\text { }\end{aligned}$
Hence option A is correct
Question 4
1 / -0

The energy stored per unit volume in copper wire, which produces longitudinal strain of $0.1$ % is: $\left( Y=1.1\times { 10 }^{ 11 }{ N }/{ { m }^{ 2 } } \right)$

A
$11\times { 10 }^{ 3 }{ J }/{ { m }^{ 3 } }$

B
$5.5\times { 10 }^{ 3 }{ J }/{ { m }^{ 3 } }$

C
$5.5\times { 10 }^{ 4 }{ J }/{ { m }^{ 3 } }$

D
$11\times { 10 }^{ 4 }{ J }/{ { m }^{ 3 } }$

Solution

When a wire is stretched, some work is done against the internal restoring forces, acting between particles of the wire. This work done appears as elastic potential energy in the wire. Elastic potential energy $\left( U \right)$ is given by
$U=\dfrac { 1 }{ 2 } F\times \Delta l$
$=\dfrac { 1 }{ 2 } \times \dfrac { F }{ a } \times \dfrac { \Delta l }{ l } \times al$ ......(i)
where $l$ is length of wire, $a$ is area of cross section of wire, $F$ is stretching force and $\Delta l$ is increase in length.
Equation (i), may be written as
$U = \dfrac{1}{2} \times$ stress $\times$ strain $\times$ volume of the wire
$\therefore$ Elastic potential energy per unit volume of the wire
$u = \dfrac{U}{al} = \dfrac{1}{2} \times$ stress $\times$ strain
$= \dfrac{1}{2} \times$ (Young's modulus $\times$ strain) $\times$ strain
$=\dfrac { 1 }{ 2 } \times \left( Y \right) \times { \left( strain \right) }^{ 2 }$
Hence, $u=\dfrac { 1 }{ 2 } \times 1.1\times { 10 }^{ 11 }\times { \left( \dfrac { 0.1 }{ 100 } \right) }^{ 2 }$
$=5.5\times { 10 }^{ 4 }{ J }/{ { m }^{ 3 } }$
Question 5
1 / -0

Young's modulus of the material of a wire is Y. If it is under a stress S, the energy stored per unit volume is given by:

A
$\dfrac{1}{2}\dfrac{S}{Y}$

B
$\dfrac{1}{2}\dfrac{S^2}{Y}$

C
$\dfrac{1}{2}\dfrac{S}{Y^2}$

D
$\dfrac{1}{2}\dfrac{S^2}{Y^2}$

Solution

Energy stored per unit volume can be given as:
$E =\dfrac{1}{2} \times stress \times strain$ -----------(1)
From Hooke's law :
Young's modulus, $Y = \dfrac{Stress}{Strain}$

$\implies$ $Strain = \dfrac{Stress}{Y} = \dfrac{S}{Y}$ -----------(2)

From equation (1) and (2):
$\therefore$ $E = \dfrac{1}{2} \times S \times \dfrac{S}{Y}$

$\Rightarrow E= \dfrac{1}{2} \dfrac{S^2}{Y}$
Hence, the correct option is $(B)$
Question 6
1 / -0

There is no change in volume of a wire due to change in its length of stretching. The Poisson's ratio of the material of the wire is:

A
0.50

B
- 0.50

C
0.25

D
- 0.25

Solution

Hint: The negative of the ratio of transverse strain to lateral or axial strain is Poisson's ratio.
Explanation:
Step 1: Concept used:

Volume of a wire of radius $r$ and length $l$ is,
$V=\pi{r}^{2}l$ .........................(1)
Differentiate the equation from both sides-
$\therefore dV=2\pi rldr + \pi{r}^{2}dl$
$(dV=0,$ as volume is unchanged $)$
$0=2\pi rldr + \pi{r}^{2}dl$
$\therefore 2rldr = -{r}^{2}dl$
$\cfrac{dr}{r}= -\cfrac{1}{2} \cfrac{dl}{l}$

Step 2: Calculating the Poisson's ratio:

“The ratio of transverse contraction strain to longitudinal extension strain in the direction of the stretching force,” as defined by Poisson.

$\eta = \frac{Transverse\ strain}{Longitudional\ strain}$
$\therefore \cfrac{\cfrac{dr}{r}}{\cfrac{dl}{l}}=-\cfrac{1}{2}$
$\therefore \sigma= -\cfrac{1}{2}$
As, here $-ve$ sign implies that if, length increased, radius decreased, so, we can write
$\sigma=\cfrac{1}{2}=0.5$

Option A is correct.
Question 7
1 / -0

The value of Poisson's ratio (theoretically) lies between

A
$-1$ to $\dfrac { 1 }{ 2 }$

B
$-\dfrac { 3 }{ 4 }$ to $-\dfrac { 1 }{ 2 }$

C
$-\dfrac { 1 }{ 2 }$ to $1$

D
$1$ to $2$

Solution
Question 8
1 / -0

Bulk modulus of water is $2\times 10^9N/m^2$ . The change in pressure required to increase the density of water by $0.1\%$ is.

A
$2\times 10^9N/m^2$

B
$2\times 10^8N/m^2$

C
$2\times 10^6N/m^2$

D
$2\times 10^4N/m^2$

Solution

$Mass=volume\times density=Vd=constant$
$V\Delta d+d\Delta V=0$

$\Rightarrow \displaystyle\frac{\Delta V}{V}=-\frac{\Delta d}{d}=-\frac{0.1}{100}=10^{-3}$

$\therefore K=-\displaystyle\frac{\Delta p}{\Delta V/V}$

$\Rightarrow \Delta p=-K\frac{\Delta V}{V}$

$\Delta p=2\times 10^9\times 10^{-3}$
$=2\times 10^6N/m^2$
Question 9
1 / -0

A wire of initial length $L$ and radius $r$ is stretched by a length $l$ . Another wire of same material but with initial length $2L$ and radius $2r$ is stretched by a length $2l$ . The ratio of stored elastic energy per unit volume in the first and second wire is:

A
$1:4$

B
$1:2$

C
$2:1$

D
$1:1$

Solution

Elastic energy per unit volume of wire
$\mu =\cfrac { 1 }{ 2 } \times Y \times { (Strain) }^{ 2 }$
$\therefore \cfrac { { \mu }_{ 1 } }{ { \mu }_{ 2 } } ={ \left\{ \cfrac { { (Strain) }_{ 1 } }{ { (Strain) }_{ 2 } } \right\} }^{ 2 }=\cfrac { { l }^{ 2 } }{ { L }^{ 2 } } \times \cfrac { 4{ L }^{ 2 } }{ { 4l }^{ 2 } } = 1:1$
Question 10
1 / -0

In Young's double slit experiment, the ratio of intensities of bright and dark bands is $16$ which means

A
The ratio of their amplitudes is $5$

B
Intensities of individual sources are $25$ and $9$ units respectively

C
The ratio of their amplitudes is $4$

D
Intensities of individuals sources are $4$ and $3$ units respectively

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Mechanical Properties of Solids Test - 43

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Mechanical Properties of Solids Test - 43

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Question 1

k=YAlk=\displaystyle\frac{YA}{l}k=lYA​

k=2YAlk=\displaystyle\frac{2YA}{l}k=l2YA​

k=YA2lk=\displaystyle\frac{YA}{2l}k=2lYA​

k=YlAk=\displaystyle\frac{Yl}{A}k=AYl​

Solution

Question 2

8.48.48.4

848484

92.492.492.4

168168168

Solution

Question 3

L = 100 cm, r = 1 mm

L = 200 cm, r = 3 mm

L = 300 cm, r = 3 mm

L = 400 cm, r = 4 mm

Solution

Question 4

11×103J/m311\times { 10 }^{ 3 }{ J }/{ { m }^{ 3 } }11×103J/m3

5.5×103J/m35.5\times { 10 }^{ 3 }{ J }/{ { m }^{ 3 } }5.5×103J/m3

5.5×104J/m35.5\times { 10 }^{ 4 }{ J }/{ { m }^{ 3 } }5.5×104J/m3

11×104J/m311\times { 10 }^{ 4 }{ J }/{ { m }^{ 3 } }11×104J/m3

Solution

Question 5

12SY\dfrac{1}{2}\dfrac{S}{Y}21​YS​

12S2Y\dfrac{1}{2}\dfrac{S^2}{Y}21​YS2​

12SY2\dfrac{1}{2}\dfrac{S}{Y^2}21​Y2S​

12S2Y2\dfrac{1}{2}\dfrac{S^2}{Y^2}21​Y2S2​

Solution

Question 6

0.50

- 0.50

0.25

- 0.25

Solution

Question 7

−1-1−1 to 12\dfrac { 1 }{ 2 } 21​

−34-\dfrac { 3 }{ 4 } −43​ to −12-\dfrac { 1 }{ 2 } −21​

−12-\dfrac { 1 }{ 2 } −21​ to 111

111 to 222

Solution

Question 8

2×109N/m22\times 10^9N/m^22×109N/m2

2×108N/m22\times 10^8N/m^22×108N/m2

2×106N/m22\times 10^6N/m^22×106N/m2

2×104N/m22\times 10^4N/m^22×104N/m2

Solution

Question 9

1:41:41:4

1:21:21:2

2:12:12:1

1:11:11:1

Solution

Question 10

The ratio of their amplitudes is 555

Intensities of individual sources are 252525 and 999 units respectively

The ratio of their amplitudes is 444

Intensities of individuals sources are 444 and 333 units respectively

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$k=\displaystyle\frac{YA}{l}$

$k=\displaystyle\frac{2YA}{l}$

$k=\displaystyle\frac{YA}{2l}$

$k=\displaystyle\frac{Yl}{A}$

$8.4$

$84$

$92.4$

$168$

$11\times { 10 }^{ 3 }{ J }/{ { m }^{ 3 } }$

$5.5\times { 10 }^{ 3 }{ J }/{ { m }^{ 3 } }$

$5.5\times { 10 }^{ 4 }{ J }/{ { m }^{ 3 } }$

$11\times { 10 }^{ 4 }{ J }/{ { m }^{ 3 } }$

$\dfrac{1}{2}\dfrac{S}{Y}$

$\dfrac{1}{2}\dfrac{S^2}{Y}$

$\dfrac{1}{2}\dfrac{S}{Y^2}$

$\dfrac{1}{2}\dfrac{S^2}{Y^2}$

$-1$ to $\dfrac { 1 }{ 2 }$

$-\dfrac { 3 }{ 4 }$ to $-\dfrac { 1 }{ 2 }$

$-\dfrac { 1 }{ 2 }$ to $1$

$1$ to $2$

$2\times 10^9N/m^2$

$2\times 10^8N/m^2$

$2\times 10^6N/m^2$

$2\times 10^4N/m^2$

$1:4$

$1:2$

$2:1$

$1:1$

The ratio of their amplitudes is $5$

Intensities of individual sources are $25$ and $9$ units respectively

The ratio of their amplitudes is $4$

Intensities of individuals sources are $4$ and $3$ units respectively