Class 11th

In the equation $v=\sqrt{\frac{\gamma P}{\rho }}$ ……(i)

$\rho$ Density = $\frac{Mass}{Volume}$ = $\frac{M}{V}$ where M = molecular weight of the gas, V = Volume of the gas, so we can write

$v=\sqrt{\frac{\gamma PV}{M}}$ …….(ii)

For ideal gas equation, PV = nRT, n = 1 so PV = RT

For constant T, PV = constant

In equation (ii), since PV = constant, $\gamma$ and M constant, v is also constant. Hence, at a constant temperature, the speed of sound in a gaseous medium is independent of the change in the pressure of the gas.

From equation (i) $v=\sqrt{\frac{\gamma P}{\rho }}$

For 1 mole of an ideal gas, the gas equation can be written as PV = RT or P = $\frac{RT}{V}$

Substituting in equation (i), we get $v=\sqrt{\frac{\gamma RT}{\rho V}}$ = $\sqrt{\frac{\gamma RT}{M}}$

Since $\gamma$ , R and M are constant, we get v $\propto \surd T$

Hence, the speed of sound in a gas is directly proportional to the square root of the temperature of the gaseous medium. i.e. the speed of sound increase with an increase in the temperature of the gaseous medium and vice versa.

Let $v_m$ and $v_d$ be the speed of sound in moist air and dry air respectively and ${\rho }_m$ and ${\rho }_d$ be the corresponding densities

From equation (i) $v=\sqrt{\frac{\gamma P}{\rho }}$ , we get $v_m\mathrm{}=\sqrt{\frac{\gamma P}{{\rho }_m}}$ and $v_d\mathrm{}=\sqrt{\frac{\gamma P}{{\rho }_d}}$

$\frac{v_m}{v_d}$ = $\sqrt{\frac{{\rho }_d}{{\rho }_m}}$

However, the presence of water vapour reduces the density of air, i.e. ${\rho }_d<$ ${\rho }_m$ , so $v_m<$ $v_d$

Length of the mild steel wire, l = 1.0 m

Area of cross-section, A = 0.5 $*{10}^{-2}$ ${cm}^2$ = 0.5 $*{10}^{-6}{m}^2$

A mass of 100 gm is suspended at the midpoint.

m = 100 gm= 0.1 kg

Due to the weight, the wire dips, as shown in the figure.

Original length = XZ, depression = l

The final length of the wire after it dips = XO + OZ

Increase in length of the wire, Δl = (XO + OZ) – XZ ……(i)

From Pythagoras theorem

XO = OZ = $\sqrt{{0.5}^2+l^2}$

From equation (i)

Δl = 2 $*\sqrt{{0.5}^2+l^2}$ - 1.0 = 2 $*0.5*\sqrt{1+{\left(\frac{l}{0.5}\right)}^2}$ - 1.0 = $\sqrt{1+{\left(\frac{l}{0.5}\right)}^2}$ - 1.0

Neglecting the smaller terms, we can write, Δl = $\frac{l^2}{0.5}$

We know, Strain = $\frac{Increaseinlength}{Originallength}$

Let T be the tension in the wire, then

mg = 2T $\cos ?\theta$

From the figure

$\cos ?\theta$ = $\frac{OY}{OX}$ = $\frac{0.5}{\sqrt{{0.5}^2+l^2}}$ = $\frac{0.5}{0.5*\sqrt{1+{\left(\frac{l}{0.5}\right)}^2}}$

Expanding the expression and eliminating the higher terms:

$\cos ?\theta$ = $\frac{l}{0.5*\sqrt{1+{\left(\frac{1}{2}\right)\left(\frac{l}{0.5}\right)}^2}}$

Since ( 1+ $\frac{l^2}{0.5}$ ) = 1 for small l, $\cos ?\theta$ = $\frac{1}{0.5}$

Then T = $\frac{mg}{2\left(\frac{l}{0.5}\right)}$ = $\frac{mg*0.5}{2l}$ = $\frac{mg}{4l}$

Stress = $\frac{Tension}{Area}$ = $\frac{mg}{4l*A}$

Young's modulus = $\frac{Stress}{Strain}$

Y = $\frac{mg*0.5}{?l*A*l^2}$

l= $\sqrt{\frac{mg*0.5}{4YA}}$

For steel, Y = 2 $*{10}^{11}$ Pa, then

l = $\sqrt{\frac{0.1*9.8*0.5}{4*2*{10}^{11}*0.5*{10}^{-6}}}$ = 0.0106 m

Hence, the depression at the midpoint is 0.0106 m

Cross-sectional area of wire A, $a_1$ = 1 ${mm}^2$ = 1 $*{10}^{-6}{m}^2$

Cross-sectional area of wire B, $a_2$ = 2 ${mm}^2$ = 2 $*{10}^{-6}{m}^2$

Young's modulus for steel, $Y_1$ = 2 $*{10}^{11}$ N/ $m^2$

Young's modulus for aluminium, $Y_2$ = 7 $*{10}^{10}$ N/ $m^2$

Stress in the wire = $\frac{Force}{area}$ = $\frac{F}{a}$

If the two wires have equal stresses, then

$\frac{F_1}{a_1}$ = $\frac{F_2}{a_2}$ or $\frac{F_1}{F_2}$ = $\frac{a_1}{a_2}$ = $\frac{1}{2}$ ………(i)

Where $F_1$ is the force exerted on steel wire and $F_2$ is the force exerted on aluminium wire

Taking a moment around the point of suspension, we get

$F_{1}y$ = $F_2(1.05-y)$

$\frac{F_1}{F_2}$ = $\frac{(1.05-y)\mathrm{}}{y}$ ……(ii)

Using equation (i) and (ii), we can write

$\frac{1}{2}=\frac{(1.05-y)\mathrm{}}{y}$ or y = 2.1 – 2y or y = 2.1/3 = 0.7 m

We know Young's modulus Y = Stress / Strain

Y = (F/A)/ Strain or Strain = (F/A)/Y

In case of Steel wire, ${Strain}_{steel}$ = $\frac{F_1}{a_1*Y_1}$

In case of Aluminium wire, ${Strain}_{alu}$ = $\frac{F_2}{a_2*Y_2}$

Since both the strains are equal, we get

$\frac{F_1}{F_2}$ = $\frac{a_1*Y_1}{a_2*Y_2}$ = $\frac{1*{10}^{-6}*2*{10}^{11}}{2*{10}^{-6}*7*{10}^{10}}$ = 1.429

From (ii) we get

$\frac{(1.05-y)\mathrm{}}{y}$ = 1.429

y = 0.432m

In order to produce an equal strain in the two wires. The mass should be suspended at a distance of 0.432m from the end A