Class 11th

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

Let us assume the depth = h, pressure at depth, $P_h$ = 80 atm = 80 $*1.013*{10}^5$ Pa

Density of water at the surface, ${\rho }_s$ = 1.03 $*{10}^3$ kg/ $m^3$

Let density of water at depth h be ${\rho }_h$

Let $V_s$ be the volume at the surface and $V_h$ be the volume at depth h and ΔV be the change in volume. Let m be the mass of water.

ΔV = $V_s-V_h$ From the relation m = $\rho V,$ we get

ΔV = m ( $\frac{1}{{\rho }_s}$ - $\frac{1}{{\rho }_h}$ ) = ${\rho }_s{V}_s$ ( $\frac{1}{{\rho }_s}$ - $\frac{1}{{\rho }_h}$ )

$\frac{\mathrm{\Delta }\mathrm{V}}{V_s}$ = 1- $\frac{{\rho }_s}{{\rho }_h}$ ……(i)

Bulk modulus of water, k = $V_1\frac{\mathrm{\Delta }\mathrm{P}}{\mathrm{\Delta }\mathrm{V}}$ = $V_s\frac{\mathrm{\Delta }\mathrm{P}}{\mathrm{\Delta }\mathrm{V}}$

$\frac{\mathrm{\Delta }\mathrm{V}}{V_s}$ = $\frac{\mathrm{\Delta }\mathrm{P}}{\mathrm{k}}$ …….(ii)

Bulk modulus of water, k = 2.1 $*{10}^9$ Pa

Hence $\frac{\mathrm{\Delta }\mathrm{V}}{V_s}=\frac{80\mathrm{}*1.013*{10}^5}{2.1*{10}^9}$ = 3.86 $*{10}^{-3}$

From equation (i) $\frac{{\rho }_s}{{\rho }_h}$ = 0.99614

${\rho }_h=1.03*{10}^3/0.99614$ = 1.034 $*{10}^3$ kg/ $m^3$