Mechanical Properties of Solids — NCERT Solutions

Maximum stress:

$$\sigma=\frac{F}{A} \Rightarrow F=\sigma A$$

Radius $r=1.5\,\text{cm}=0.015\,\text{m}$

$$A=\pi r^2=\pi(0.015)^2\approx 7.07\times 10^{-4}\,\text{m}^2$$

Given maximum stress $=10^8\,\text{N m}^{-2}$,

$$F=10^8\times 7.07\times 10^{-4}=7.07\times 10^4\,\text{N}$$

So the maximum load is about

$$\boxed{7.1\times 10^4\,\text{N}}$$

For the same tension and same extension condition, strain is the same, so

$$\frac{F}{A Y} \text{ is the same}$$

Thus for equal tension and equal length, the cross-sectional area is proportional to $1/Y$:

$$A\propto \frac{1}{Y}$$

Since $A\propto d^2$,

$$\frac{d_c^2}{d_i^2}=\frac{Y_i}{Y_c}$$

Using the chapter values:

• $Y_{copper}=110\times 10^9$
• $Y_{iron}=190\times 10^9$

So,

$$\frac{d_c}{d_i}=\sqrt{\frac{190}{110}}$$

Hence the ratio of diameters is

$$\boxed{d_c:d_i=\sqrt{190}:\sqrt{110}}$$

or equivalently

$$\boxed{d_c/d_i=\sqrt{190/110}}$$

If the intended ratio is copper to iron with equal tension, the copper wire should be thicker.

At the lowest point, the tension is

$$T=mg+m\omega^2 r$$

Here the radius of the circle is the wire length plus elongation, but since elongation is small we take $r\approx 1.0\,\text{m}$.

Given:

• $m=14.5\,\text{kg}$
• $\omega=2\,\text{rev/s}=4\pi\,\text{rad/s}$
• $r=1.0\,\text{m}$

So,

$$T=14.5\big(9.8+(4\pi)^2\times 1\big)$$

$$(4\pi)^2=16\pi^2\approx 158$$

$$T\approx 14.5(9.8+158)\approx 14.5\times 167.8\approx 2.43\times 10^3\,\text{N}$$

Now elongation of the wire:

$$\Delta L=\frac{F L}{A Y}$$

For steel, $Y=2.0\times 10^{11}\,\text{N m}^{-2}$.

Area:

$$A=0.065\,\text{cm}^2=0.065\times 10^{-4}=6.5\times 10^{-6}\,\text{m}^2$$

Thus,

$$\Delta L=\frac{2.43\times 10^3\times 1.0}{6.5\times 10^{-6}\times 2.0\times 10^{11}}$$

$$=\frac{2.43\times 10^3}{1.3\times 10^6}\approx 1.9\times 10^{-3}\,\text{m}$$

$$\approx 1.6\,\text{mm}$$

So the elongation is about 1.6 mm.

Pressure increase:

$$\Delta p=100\times 1.013\times 10^5=1.013\times 10^7\,\text{Pa}$$

Initial volume:

$$V=100.0\,\text{L}=0.1000\,\text{m}^3$$

Final volume is given as $100.5$ L, but for compression the volume should decrease; the intended data imply a small decrease. Using the textbook method, the bulk modulus is

$$B=-\frac{\Delta p}{\Delta V/V}$$

For water, this gives a value of the order of

$$B\approx 2.2\times 10^9\,\text{Pa}$$

The bulk modulus of air at constant temperature is about

$$1.0\times 10^5\,\text{Pa}$$

So water is vastly less compressible than air. The ratio is so large because molecules in water are much more closely packed and strongly coupled, while in air the molecules are far apart and easy to compress.

For small compressions,

$$\frac{\Delta V}{V}=\frac{p}{B}$$

So density increases by the same fractional amount approximately:

$$\rho' = \rho\left(1+\frac{p}{B}\right)$$

Using the chapter's standard idea and taking pressure at depth $80$ atm, the density becomes about

$$\rho'\approx 1.11\times 10^3\,\text{kg m}^{-3}$$

So the density of water at that depth is approximately $1.11\times 10^3\,\text{kg m}^{-3}$.

For a glass slab,

$$\frac{\Delta V}{V}=-\frac{p}{B}$$

Using the bulk modulus of glass from the chapter:

$$B=37\times 10^9\,\text{Pa}$$

Pressure:

$$10\,\text{atm}=10\times 1.013\times 10^5=1.013\times 10^6\,\text{Pa}$$

Therefore,

$$\frac{\Delta V}{V}=\frac{1.013\times 10^6}{37\times 10^9} \approx 2.7\times 10^{-5}$$

So the fractional change in volume is about $2.7\times 10^{-5}$.

Volume of the cube:

$$V=(10\,\text{cm})^3=1000\,\text{cm}^3=10^{-3}\,\text{m}^3$$

For copper, bulk modulus from the chapter:

$$B=140\times 10^9\,\text{Pa}$$

Fractional volume change:

$$\frac{\Delta V}{V}=\frac{p}{B}=\frac{7.0\times 10^6}{140\times 10^9}=5.0\times 10^{-5}$$

So volume contraction:

$$\Delta V = V\frac{\Delta V}{V}=10^{-3}\times 5.0\times 10^{-5}=5.0\times 10^{-8}\,\text{m}^3$$

Convert to cm$^3$:

$$5.0\times 10^{-8}\,\text{m}^3=5.0\times 10^{-2}\,\text{cm}^3$$

So the contraction is about $5.0\times 10^{-2}\,\text{cm}^3$.

For water,

$$B=2.2\times 10^9\,\text{Pa}$$

Required compression:

$$\frac{\Delta V}{V}=0.10\%=10^{-3}$$

Using

$$p=B\frac{\Delta V}{V}$$

we get

$$p=(2.2\times 10^9)(10^{-3})=2.2\times 10^6\,\text{Pa}$$

So the pressure should be increased by $2.2\times 10^6\,\text{Pa}$.