Chapter 8: Mechanical Properties of Solids

Ans: In the above question it is given that:

Mass is \[m=14.5\text{ }kg\].

Length of the steel wire is \[l=1.0\text{ }m\]

Angular velocity is $\omega =2rev/s$

Cross-sectional area of the wire is \[a=0.065c{{m}^{2}}=0.065\times {{10}^{-4}}{{m}^{2}}\].

Consider the elongation of the wire when the mass is at the lowest point of its path to be $\Delta l$.

The total force on the mass when the mass is placed at the position of the vertical circle is given by:

$F=mg+ml{{\omega }^{2}}$

$\Rightarrow F=14.5\times 9.8+14.5\times 1\times {{2}^{4}}=200.1N$

Young’s modulus is given by:

 $Y=\frac{Stress}{Strain}$

$\Rightarrow Y=\frac{\left( \frac{F}{A} \right)}{\left( \frac{\Delta l}{l} \right)}$

$\Rightarrow \Delta l=\frac{Fl}{AY}$

We know that Young’s modulus for steel is $2\times {{10}^{11}}Pa$.

Therefore,

$\Delta l=\frac{220.1\times 1}{0.065\times {{10}^{-4}}\times 2\times {{10}^{11}}}=1.53\times {{10}^{-4}}m$

Thus, the elongation of the wire is $1.53\times {{10}^{-4}}m$.

Ans: In the above question it is given that:

Initial volume is ${{V}_{1}}=100.0l=100\times {{10}^{-3}}{{m}^{3}}$.

Final volume is ${{V}_{2}}=100.5l=100.5\times {{10}^{-3}}{{m}^{3}}$.

Thus, the increase in volume is ${{V}_{2}}-{{V}_{1}}=0.5\times {{10}^{-3}}{{m}^{3}}$.

Increase in pressure is $\Delta p=100atm=100\times 1.013\times {{10}^{5}}Pa$.

The formula for bulk modulus is 

\[\text{Bulk Modulus}=\frac{\Delta p}{\left( \frac{\Delta V}{{{V}_{1}}} \right)}=\frac{\Delta p{{V}_{1}}}{\Delta V}\]

\[\Rightarrow \text{Bulk Modulus=}\frac{100\times 1.013\times {{10}^{5}}\times 100\times {{10}^{-3}}}{0.5\times {{10}^{-3}}{{m}^{3}}}=2.026\times {{10}^{9}}Pa\]

We know that Bulk modulus of air is $1\times {{10}^{5}}Pa$.

$\Rightarrow \frac{Bulk\text{ }modulus\text{ }of\text{ }water}{Bulk\text{ }modulus\text{ }of\text{ }air}=\frac{2.026\times {{10}^{9}}}{1\times {{10}^{5}}}=2.026\times {{10}^{4}}$

This ratio is very high because air is more compressible than water.

Ans: In the above question it is given that:

Pressure at the given depth is $p=80.0atm=80\times 1.01\times {{10}^{5}}Pa$.

Consider the given depth to be h.

Density of water at the surface is ${{\rho }_{1}}=1.03\times {{10}^{3}}kg/{{m}^{3}}$

Consider ${{\rho }_{2}}$ to be the density of water at the depth h.

Consider ${{V}_{1}}$ to be the volume of water of mass m at the surface.

Consider ${{V}_{2}}$ to be the volume of water of mass m at the depth h.

Consider $\Delta V$ to be the change in volume.

$\Delta V={{V}_{1}}-{{V}_{2}}$

$\Rightarrow \Delta V=m\left[ \left( \frac{1}{{{\rho }_{1}}} \right)-\left( \frac{1}{{{\rho }_{2}}} \right) \right]$

Now,

$Volumetric\text{ }strain=m\left[ \left( \frac{1}{{{\rho }_{1}}} \right)-\left( \frac{1}{{{\rho }_{2}}} \right) \right]\times \left( \frac{{{\rho }_{1}}}{m} \right)$

$\Rightarrow \frac{\Delta V}{{{V}_{1}}}=1-\left( \frac{{{\rho }_{1}}}{{{\rho }_{2}}} \right)$ …… (1)

Bulk modulus is given by:

\[Bulk\text{ }modulus=\frac{p{{V}_{1}}}{\Delta V}\]

$\Rightarrow \frac{\Delta V}{{{V}_{1}}}=\frac{p}{B}$

Compressibility of water is given by:

$\frac{1}{B}=45.8\times {{10}^{-11}}P{{a}^{-1}}$

\[\Rightarrow \frac{\Delta V}{{{V}_{1}}}=80\times 1.013\times {{10}^{5}}\times 45.8\times {{10}^{-11}}=3.71\times {{10}^{-3}}\] …… (2)

From equations (1) and (2) we get:

$1-\left( \frac{{{\rho }_{1}}}{{{\rho }_{2}}} \right)=3.71\times {{10}^{-3}}$

$\Rightarrow {{\rho }_{2}}=\frac{1.03\times {{10}^{3}}}{\left[ 1-\left( 3.71\times {{10}^{-3}} \right) \right]}=1.034\times {{10}^{3}}kg{{m}^{-3}}$

Clearly, the density of water at the given depth (h) is $1.034\times {{10}^{3}}kg{{m}^{-3}}$.

Ans: In the above question it is given that:

The hydraulic pressure exerted on the glass slab is $p=10atm=10\times 1.013\times {{10}^{5}}Pa$.

Also, we know that the bulk modulus of glass is $B=37\times {{10}^{9}}N/{{m}^{2}}$.

Bulk modulus is given by the relation:

\[B=\frac{p}{\left( \frac{\Delta V}{V} \right)}\]

Where,

\[\frac{\Delta V}{V}\] is the fractional change in volume.

\[\Rightarrow \left( \frac{\Delta V}{V} \right)=\frac{p}{B}\]

\[\Rightarrow \left( \frac{\Delta V}{V} \right)=\frac{10\times 1.013\times {{10}^{5}}}{37\times {{10}^{9}}}=2.73\times {{10}^{-5}}\]

Clearly, the fractional change in the volume of the glass slab is \[2.73\times {{10}^{-5}}\].

Ans: In the above question it is given that:

The length of an edge of the solid copper cube is $l=10cm=0.1m$.

Hydraulic pressure is $p=7\times {{10}^{6}}Pa$.

Bulk modulus of copper is $B=140\times {{10}^{9}}Pa$.

Bulk modulus is given by the relation:

\[B=\frac{p}{\left( \frac{\Delta V}{V} \right)}\]

Where,

\[\frac{\Delta V}{V}\] is the volumetric strain

$\Delta V$ is the change in volume.

$V$ is the original volume.

\[\Rightarrow \Delta V=\frac{pV}{B}\]

The original volume of the cube is $V={{l}^{3}}$.

\[\Rightarrow \Delta V=\frac{p{{l}^{3}}}{B}\]

\[\Rightarrow \Delta V=\frac{7\times {{10}^{6}}\times {{\left( 0.1 \right)}^{3}}}{140\times {{10}^{9}}}=5\times {{10}^{-8}}{{m}^{3}}=5\times {{10}^{-2}}c{{m}^{3}}\]

Clearly, the volume contraction of the solid copper cube is \[5\times {{10}^{-2}}c{{m}^{3}}\].

Ans: In the above question it is given that:

Volume of water is $V=1L$.

The water is to be compressed by \[0.10%\].

$\therefore Fractional\text{ }change=\frac{\Delta V}{V}=\frac{0.1}{100\times 1}={{10}^{-3}}$ 

Bulk modulus is given by the relation:

\[B=\frac{p}{\left( \frac{\Delta V}{V} \right)}\]

\[\Rightarrow p=B\times \left( \frac{\Delta V}{V} \right)\]

We know that, bulk modulus of water is $B=2.2\times {{10}^{9}}N/{{m}^{2}}$

\[\Rightarrow p=2.2\times {{10}^{9}}\times {{10}^{-3}}=2.2\times {{10}^{6}}N/{{m}^{2}}\].

Clearly, the pressure on water should be \[2.2\times {{10}^{6}}N/{{m}^{2}}\].