Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 59

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Question 1
1 / -0

If in case $$A$$, elongation in wire of length $$L$$ is $$l$$, then for same wire elongation in case $$B$$ will be:

A
$$4l$$

B
$$2l$$

C
$$l$$

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

Solution

Guven, when weight 'W is hanged by wire of length 'L' Elongation in it is 'l
$$ w=\operatorname{yA}\left(\frac{1}{L}\right) \Rightarrow l=\frac{w L}{y_{A}} $$
Now, in case 2
$$ \begin{array}{l} \text { Net elongation }=\Delta L_{1}+\Delta L_{2} \\ W=Y A\left(\frac{\Delta L_{1}}{L / 2}\right) \\ W=Y A\left(\frac{\Delta L_{2}}{4 / 2}\right) \end{array} $$
$$ \begin{array}{l} \text { Adding beth: } \quad 2 W=\frac{2 y A}{L}\left(\Delta L_{1}+\Delta L_{2}\right) \\ \Delta L_{1}+\Delta L_{2}=\frac{W L}{Y A}=l \\ \text { Hence, net econgation in second case is } L \\ \text { Option } c \text { is correct option: } \end{array} $$
Question 2
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A thick rope of density $$\rho$$ and length $$L$$ is hung from a rigid support. The Young's modulus of the material of rope is $$Y$$. The increase in length of the rope due to its own weight is

A
$$\rho g{L}^{2}/4Y$$

B
$$\rho g{L}^{2}/2Y$$

C
$$\rho g{L}^{2}/Y$$

D
$$\rho gL/Y$$

Solution

Take a small section $$\delta y$$ at distance $$Y$$ from top as shown in the figure.Let area of rope be $$A$$
Stress due to this section $$=\rho Ag\dfrac{\delta y}{A}=\rho d \delta y$$
Strain on section$$Y=\dfrac{\delta l}{y}$$
Using stress$$=Y\times $$ strain
$$\Rightarrow \rho g\delta y=Y\times\dfrac{\delta l}{y}$$
$$\Rightarrow \Delta l=\dfrac{\rho g}{y}\displaystyle\int_{0}^{L}{y\delta y}$$
$$\Rightarrow \Delta l=\dfrac{\rho g{L}^{2}}{2Y}$$
Question 3
1 / -0

When a wire is stretched, its length increases by 0.3% and the diameter decreases by 0.1%. Poisson's ratio of the material of the wire is about

A
0.03

B
0.333

C
0.15

D
0.015

Solution

$$\begin{array}{l}\text { Given, increase in length of wire }=0.3\%\\\text { increase in diameter }=0.1 \% \\\text { We know that, } \\\text { } \quad \text { } \quad d=2r \\\qquad \ln (d)=\ln (2)+\ln (r) \\\text{Differentiating,}\qquad \frac{\Delta d}{d}=\frac{\Delta r}{r}\end{array}$$

$$\begin{aligned}&\therefore \Delta r=\frac{-0.1}{100}(r)=\frac{-r}{1000}\left(\begin{array}{l}\text { radius is } \\\text { decreasing })\end{array}\right.\\&\begin{aligned}\Delta l=\frac{0.3}{100}(l) &=\frac{3 l}{1000} \\\text { Now, poisson's ratio }(\sigma)&=\frac{\Delta r / r}{\Delta l / l} \\&=\frac{1 / 1000}{3 / 1000}=\frac{1}{3} \\&=0.333\end{aligned}\end{aligned}$$
Question 4
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A wire is stretched by $$0.01\ m$$ by a certain force $$F$$. Another wire of same material whose diameter and lengths are doubled to the original wire is stretched by the same force. Then its elongation will be:

A
$$0.005\ m$$

B
$$0.01\ m$$

C
$$0.02\ m$$

D
$$0.08\ m$$

Solution

Given that,

$$\Delta {{l}_{1}}=0.01\,m$$

Length $${{l}_{1}}=l$$

Length $${{l}_{2}}=2l$$

Radius $${{r}_{1}}=r$$

Radius $${{r}_{2}}=2r$$

We know that,

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

$$ \Delta l=\dfrac{Fl}{\pi {{r}^{2}}Y} $$

$$ \Delta l\propto \dfrac{l}{{{r}^{2}}} $$

Now, the diameter and lengths are doubled to the original wire is stretched by the same force

So,

$$ \dfrac{\Delta {{l}_{2}}}{\Delta {{l}_{1}}}=\dfrac{2l}{l}\times \dfrac{{{r}^{2}}}{4{{r}^{2}}} $$

$$ \dfrac{\Delta {{l}_{2}}}{\Delta {{l}_{1}}}=2\times \dfrac{1}{4} $$

$$ \Delta {{l}_{2}}=\dfrac{0.01}{2} $$

$$ \Delta {{l}_{2}}=0.005\,m $$

Hence, the elongation is $$0.005\ m$$
Question 5
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The maximum load a wire can withstand without breaking, when its length is reduced to half of its original length, will

A
be double

B
be half

C
be four times

D
remain same

Solution

Hint: The stress is the ratio of force and the area.

Step 1: Explanation,
The expression or the stress can be written as,
$$\begin{array}{l}\text { Stress }=\frac{F}{A}\\\qquad\Rightarrow \quad F=(\text { Stress }) \times(A) \\\text { On reducing the length of the wire } \\\text { to half , neither the stress change } \\\text { nor the area of cross- section. } \\\text { Maximum load wire can hold } \\\text { remains same. }\end{array}$$
Question 6
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One end of a uniform wire of length L and of weight W you is attached rigidly see through a point in the roof and weight W1 is suspended from its lower end. If S is true area of cross section of the wire the stress in the wire at a height (3L/4) from its lower end is :

A
$$[W_1 + (W/4)] / S$$

B
$$W_1/S$$

C
$$[W_1 + (3W/4)] / S$$

D
$$(W_1+W) / S$$

Solution
Question 7
1 / -0

The pressure of a medium is changed from $$1.01\times 10^{5}\ Pa$$ to $$1.165\times 10^{5}\ Pa$$ and change in volume is $$10\%$$ keeping temperature constant. The bulk modulus of the medium is

A
$$204.8\times 10^{5}\ Pa$$

B
$$102.4\times 10^{5}\ Pa$$

C
$$51.2\times 10^{5}\ Pa$$

D
$$1.55\times 10^{5}\ Pa$$

Solution

$$\textbf{Step 1 - Bulk modulus K}$$
Given, $$P_{i} = 1.01\times 10^{5} Pa, P_{f} = 1.165\times 10^{5} Pa$$

$$\dfrac {\triangle V}{V} = 10\% = \dfrac {10}{100}$$

$$K = \dfrac {\triangle P}{\left (\dfrac {\triangle V}{V}\right )} = \dfrac {(P_{f} - P_{i})}{\left (\dfrac {\triangle V}{V}\right )} = \dfrac {(1.165 \times 10^{5} Pa - 1.01\times 10^{5} Pa)}{\left (\dfrac {10}{100}\right )}$$

$$\Rightarrow K = 1.55\times 10^{5} Pa$$
Question 8
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The relationship between Young's modulus $$Y$$, Bulk modulus $$K$$ and modulus of rigidity $$\eta$$ is:-

A
$$Y = \dfrac{9 \eta K}{3 + k}$$

B
$$Y = \dfrac{9 \eta K}{n + 3}$$

C
$$Y=\dfrac{9 \eta K}{\eta+3 k}$$

D
$$Y = \dfrac{3 \eta K}{9 \eta + k}$$

Solution
Question 9
1 / -0

To wires $$A$$ and $$B$$ have the same length and area of cross section. But Young's modulus of $$A$$ is two times the Young's modulus of $$B$$. Then the ratio of force constant of $$A$$ to that of $$B$$ is

A
$$1$$

B
$$2$$

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

D
$$12\ 2$$

Solution

$$\begin{array}{l}\text { Given, } Y_{A}=2 Y_{B} \\\text { Same length and area of cross - section of wires } A \text { & } B \\\text { According to the spring equivalent concept ,}\end{array}$$

$$\begin{array}{l}F=k x \quad(x=\text { extension in spring })\\F=\left(\frac{Y A}{L}\right)(\Delta L) \\\text { From above equations, we conclude } \\\qquad k=\frac{Y A}{L}=\text { force constant } \\\therefore \quad \frac{Y_{A}}{Y_{B}}=\frac{k_a}{k_{b}}=2\quad(\text { area and length } \\\text { are same. })\end{array}$$
Question 10
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A rod of mass $$'M'$$ is subjected to force $$'t'$$ and $$'2f'$$ at both the ends as shown in the figure. If young modulus of its material is $$'y'$$ and its length is $$L$$ find total elongation of rod.

A
$$\dfrac{fl}{2Ay}$$

B
$$\dfrac{fl}{Ay}$$

C
$$\dfrac{3fl}{2Ay}$$

D
$$\dfrac{4fl}{2Ay}$$

Solution

To solve problem of these types we consider average force acting
$$\begin{array}{l} \Delta l=\dfrac { { FL } }{ { Ay } } =\left( { \dfrac { { { f_{ 1 } }+{ f_{ 2 } } } }{ 2 } } \right) \dfrac { L }{ { Ay } } \\ =\left( { \dfrac { { f+2f } }{ 2 } } \right) \dfrac { L }{ { Ay } } \\ =\dfrac { { 3fL } }{ { 2Ay } } \\ Total\, elongation\, =\Delta l=\dfrac { { 3fL } }{ { 2Ay } } \\ Hence,\, the\, option\, \; C\, is\, the\, correct\, answer. \end{array}$$