Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 77

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Question 1
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A mass m is hanging from a wire of cross sectional are A and length L. Y is young's modulus of wire. An external force F is applied on he wire which is then slowly further pulled down by $$\triangle x$$ from its equilibrium position. Find the work done by the force F that the wire exerts on the mass:

A
$$mg\triangle x+\dfrac{YA}{2L}(\triangle x)^2$$

B
$$mg\triangle x+\dfrac{YA(\triangle x)^2}{L}$$

C
$$mg\dfrac{\triangle x }{2}+\dfrac{YA(\triangle x)^2}{L}$$

D
$$mg\dfrac{\Delta x}{2}+\dfrac{YA(\Delta x)^2}{2L}$$

Solution

When external force is applied, work done is given by
Work done $$W=mg\Delta x$$

And
$$ {{W}_{2}}=\dfrac{1}{2}\times stress\times strain\times volume $$

$$ \dfrac{1}{2}\times Y\times {{(strain)}^{2}}\times V $$

$$ \dfrac{1}{2}\times Y\times {{(\dfrac{\Delta x}{L})}^{2}}\times AL $$

$$ \dfrac{Y\Delta {{x}^{2}}A}{2L} $$

So, total work done is

$$W=mg\Delta x+\dfrac{Y\Delta {{x}^{2}}A}{2L}$$
Question 2
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A mild steel wire of length $$2l$$ meter cross-sectional area $$A\,{m^2}$$ is fixed horizontally between two pillars. A small mass $$m \ kg$$ is suspended from the mid point of the wire. If extension in wire are within elastic limit. Then depression at the mid point of wire will be

A
$${\left( {\frac{{Mg}}{{YA}}} \right)^{1/3}}$$

B
$${\left( {\frac{{Mg}}{{IA}}} \right)^{1/3}}$$

C
$${\left( {\frac{{Mg{l^3}}}{{YA}}} \right)^{1/3}}$$

D
$$\frac{{Mg}}{{2YA}}$$

Solution
Question 3
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A uniform rod of length L, has a mass per unit length $$\lambda$$ and area of cross-section A. The elongation in the rod is l due to its own weight, if it is suspended from the ceiling of a room.The Young's modulus of the rod is

A
$$\dfrac{ \lambda\,g L^2}{ 2 Al}$$

B
$$\dfrac{3 \lambda\,g L^2}{Al}$$

C
$$\dfrac{2 \lambda\,g L}{Al}$$

D
$$\dfrac{ \lambda\,g L^2}{Al}$$

Solution

Given, a ned of length $$=L$$
area of cross - sction $$=A$$
Yourng's modulues $$=y$$
Now , we can comert this
Sugstem into an equivalent spring - systerm of sprung equivalint $$K$$.
$$ \begin{array}{l} \text { Let } d x \text { be the elomgation in } \\ \text { sinall element } d z \text { . } \\ T_{1}=\left(\frac{M}{L}\right)(z)(g) \\ T_{1}=k(d x) \\ T_{1}=\left(\frac{y A}{d z}\right)(d x) \\ \frac{M g}{L} \int_{0}^{t} z d z=y A \int_{0}^{x} d x \end{array} $$
$$ \frac{M g L}{2 A Y}=x $$
$$ \text { } \begin{aligned} \text { Now , } & d m=\lambda d x \\ M &=\lambda \int_{0}^{0} d x \\ M &=\lambda L \end{aligned} $$
$$ \begin{array}{l} \therefore \text { Total elongation } \\ \qquad \begin{aligned} l &=\frac{\lambda g L^{2}}{2 A Y} \\ \Rightarrow & y=\frac{\lambda g L^{2}}{2 A l} \end{aligned} \end{array} $$
Question 4
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A metallic ring of radius $$2cm$$ and cross sectional area $$4cm^{2}$$ is fitted into a wooden circular disc of radius $$4cm$$.If the Young's module of the material of the ring is $$2 \times 10^{11}N/m^2$$, the metal ring expand is:

A
$$2 \times 10^7N$$

B
$$8 \times 10^7N$$

C
$$4 \times 10^7N$$

D
$$6 \times 10^7N$$

Solution

$$ \begin{array}{l} \text { Given, metal ring of radius }=2\mathrm{~cm} \\ \text { cross - sectional area }=4\mathrm{~cm}^{2} \\ \text { Young's modulus of ring }=2\times 10^{11} \frac{\mathrm{N}}{\mathrm{m}^{\prime}}\end{array} $$
$$ \begin{array}{l} l_{i}=2 \pi r \\ l_{+}=2 \pi R\\\frac{\Delta l_{i}}{l_{i}}=\frac{R-r}{\gamma}=\frac{4-2}{2}=1 \\ \therefore \quad \text { Stress }=\quad Y\left(\frac{\Delta l}{l}\right)=2 \times 10^{\prime\prime}\\\therefore \quad \text { Force }\quad=2\times 10^{\prime \prime} \times 4 \times 10^{-4} \\ =8 \times 10^{7} \mathrm{~N} \end{array} $$
Question 5
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Two wires of equal length and cross-section are suspended as shown in figure. Their young's modulus are $$Y_1$$ and $$Y_2$$ respectively. Their equivalent young's modulus of elasticity is :

A
$$Y_1+Y_2$$

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

C
$$Y_1-Y_2$$

D
$$Y_1Y_2$$

Solution
Question 6
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The Young's modulus of brass and steel are respectively $$1.0\times {10}^{10}N/{m}^{2}$$ and $$2\times {10}^{10}N/{m}^{2}$$. A brass wire and a steel wire of the same length are extended by $$1\ mm$$ under the same force, the radii of brass and steel wires are $${R}_{B}$$ and $${R}_{S}$$ respectively. Then

A
$${R}_{S}=\sqrt{2}{R}_{B}$$

B
$${R}_{S}=\cfrac{{R}_{B}}{\sqrt{2}}$$

C
$${R}_{S}=4{R}_{B}$$

D
$${R}_{S}=\cfrac{{R}_{B}}{4}$$

Solution

$$\begin{array}{l}\text { Given, young modulus of brass }=10^{10}\mathrm{~N} / \mathrm{m}^{2} \\\text { Young modulus of steel }=2 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2} \\\text { Both wires have same length and } \\\text { extended by } 1 \mathrm{~mm} \text { under same force. }\end{array}$$

$$\begin{array}{l}F=Y_{B}\left(\pi R B^{2}\right)\left(\frac{\Delta L}{L}\right) \\F=Y_{S}\left(\pi R_{S}^{2}\right)\left(\frac{\Delta L}{L}\right)\\Y_{B}\left(R_{B}\right)^{2}=Y_{S}\left(R_{S}\right)^{2}\\R_{B}=\sqrt{2} R_{S} \quad \text { Hence, option } B \text { is}\\\text { the correct answer. }\end{array}$$
Question 7
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A solid sphere of radius R made of a material of bulk modulus B surrounded by a liquid in a cylindrical container.A massless piston of area A floats on the surface of the liquid. Find the fractional decreases in the radius of the sphere $$ \left( \frac { d R }{ R } \right) $$ when a mass M is placed on the piston to compress the liquid:

A
$$ \left( \dfrac { 3Mg }{ AB } \right) $$

B
$$ \left( \dfrac {2 Mg }{ AB } \right) $$

C
$$ \left( \dfrac { Mg }{ 3AB } \right) $$

D
$$ \left( \dfrac { Mg }{ 2AB } \right) $$

Solution

Given,

Radius of sphere, $$R$$

Mass placed on massless piston, $$M$$.

Area of piston, $$A$$

Change in pressure $$\Delta P=\dfrac{\Delta F}{A}=\dfrac{Mg-0}{A}=\dfrac{Mg}{A}$$

Volume of sphere, $$v=\dfrac{4}{3}\pi {{R}^{3}}$$

Small decrease in volume, $$-dv=d\left( \dfrac{4}{3}\pi {{R}^{^{3}}} \right)=4\pi {{R}^{2}}dR$$

Bulk modulus, $$B$$

$$ B=\dfrac{dp}{-\dfrac{dv}{v}}=\dfrac{\dfrac{Mg}{A}}{-\dfrac{4\pi {{R}^{2}}dR}{\dfrac{4}{3}\pi {{R}^{3}}}}=\dfrac{Mg}{-3A\dfrac{dR}{R}} $$

$$ -\dfrac{dR}{R}=\dfrac{Mg}{3AB} $$

Hence, fractional decrease in radius of sphere is $$\dfrac{Mg}{3AB}$$
Question 8
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A neutron moving with a velocity $$v$$ and kinetic energy $$E$$ collides perfectly elastically head on with the nucleus of an atom of mass number $$A$$ at rest. The energy received by the nucleus and the total energy of the system are related by

A
$$\cfrac { 4A }{ { \left( A+1 \right) }^{ 2 } } $$

B
$${ \left( \cfrac { A-1 }{ A+1 } \right) }^{ 2 }$$

C
$$\cfrac{(A+1)}{4{A}^{2}}$$

D
$${ \left( \cfrac { A+1 }{ A-1 } \right) }^{ 2 }$$
Question 9
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When a $$20g$$ mass hangs attached to one end of a light spring of length $$10cm$$, the spring stretches by $$2cm$$. The mass is pulled down unitl the total length of the spring is $$14cm$$. The elastic energy, in Joule stored in the spring is -

A
$$4\times {10}^{-2}$$

B
$$4\times {10}^{-3}$$

C
$$8\times {10}^{-2}$$

D
$$8\times {10}^{-3}$$

Solution
Question 10
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One end of a uniform rod of mass $$M$$ and cross-sectional area $$A$$ is suspended from the other end. The stress at the mid-point of the rod will be :

A
$$ \dfrac {2Mg}{A}$$

B
$$ \dfrac {3Mg}{2A}$$

C
$$ \dfrac {Mg}{A}$$

D
Zero

Solution

The weight of suspended mass is given as,

$${W_1} = mg$$

The weight of the rod acting at the midpoint is given as,

$${W_2} = \dfrac{{mg}}{2}$$

The stress at the midpoint is given as,

$$\sigma = \dfrac{{{W_1} + {W_2}}}{A}$$

$$\sigma = \dfrac{{mg + \dfrac{{mg}}{2}}}{A}$$

$$\sigma = \dfrac{{3mg}}{{2A}}$$