(1)

Here, $l=1$ $\text{m}$

Young’s modulus $=2\times {10}^{11} {\text{N m}}^{-2}$ and Stress $=8\times {10}^8 {\text{N m}}^{-2}$

$\Delta l=\frac{\text{Stress}}{Y}\times l=\frac{8\times {10}^8}{2\times {10}^{11}}\times 1=4\times {10}^{-3}=4$ $\text{mm}$

(4)

Longitudinal stress is the normal stress which changes the length of the body, it is given by $=\frac{W}{A}$

(3)

The stretching of a coiled spring is determined by its shear modulus because neither its length nor its volume changes but only the shape of the spring is changed. Elasticity of steel is more than copper.

(4)

Given : initial length = L, area of cross section = A

New length after mass M is suspended on the wire = L

$\therefore$ Change in length, $\Delta L=L_1-L.$

Now Young's modulus, $Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F}{A}\times \frac{L}{\Delta L}$

        $=\frac{mg}{A}\frac{L}{\Delta L} \text{or} \frac{MgL}{A(L_1-L)}$

(1)

Young's modulus, $Y=\frac{Fl}{A\Delta l}$

Since initial volume of wires are same and their areas of cross-sections are A and 3A, so lengths are $3l$ and $l$ respectively.

For wire 1,

          $\Delta l=\left(\frac{{\displaystyle F}}{{\displaystyle AY}}\right)3l$                                    ...(i)

For wire 2, let $F\text{'}$ force be applied, 

       $\frac{F\text{'}}{3A}=Y\frac{\Delta l}{l}$

$\Rightarrow \Delta l=\left(\frac{F\text{'}}{3AY}\right)l$                                    ...(ii)

From equations (i) and (ii), $\left(\frac{{\displaystyle F}}{{\displaystyle AY}}\right)3l=\left(\frac{{\displaystyle F\text{'}}}{{\displaystyle 3AY}}\right)l\Rightarrow F\text{'}=9F$

(3)

Bulk modulus $B$ is given as

  $B=\frac{-pV}{\Delta V}$                                 ...(i)

The volume of a spherical object of radius $r$ is given as

     $V=\frac{4}{3}\pi r^3, \Delta V=\frac{4}{3}\pi \left(3r^2\right)\Delta r$

$\therefore$   $-\frac{V}{\Delta V}=\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi 3r^2\Delta r} \text{or} -\frac{V}{\Delta V}=-\frac{r}{3\Delta r}$

Put this value in eqn. (i), we get $B=-\frac{pr}{3\Delta r}$

Fractional decrease in radius is $-\frac{\Delta r}{r}=\frac{p}{3B}$

(4)

Let $L$ and $A$ be the length and area of cross-section of each wire. In order to have the lower ends of the wires to be at the same level (i.e. same elongation is produced in both wires), let weights $W_s$ and $W_b$ are added to steel and brass wires respectively. Then, by definition of Young’s modulus, the elongation produced in the steel wire is

$\Delta L_s=\frac{W_{s}L}{Y_{s}A}$                                       $(\text{as }Y=\frac{{\displaystyle W/A}}{{\displaystyle \Delta L/L}})$

and that in the brass wire is $\Delta L_b=\frac{W_{b}L}{Y_{b}A}$

But $\Delta L_s=\Delta L_b$                                    (given)

$\therefore$   $\frac{W_{s}L}{Y_{s}A}=\frac{W_{b}L}{Y_{b}A} \text{or} \frac{W_s}{W_b}=\frac{Y_s}{Y_b}$

As   $\frac{Y_s}{Y_b}=2;$            $\therefore$   $\frac{W_s}{W_b}=\frac{2}{1}$

(1)

Depth of ocean, $d=2700\text{m}$

Density of water, $\rho ={10}^3\text{kg}$ ${\mathrm{m}}^{-3}$

Compressibility of water, $K=45.4\times {10}^{-11}{\text{Pa}}^{-1}$

         $\frac{\Delta V}{V}= ?$

Excess pressure at the bottom, $\Delta P=\rho gd$

        $={10}^3\times 10\times 2700=27\times {10}^6\text{Pa}$

We know, $B=\frac{\Delta P}{(\Delta V/V)}$

$\left(\frac{{\displaystyle \Delta V}}{{\displaystyle V}}\right)=\frac{\Delta P}{B}=K·\Delta P$                   $(\because K=\frac{1}{B})$

          $=45.4\times {10}^{-11}\times 27\times {10}^6=1.2\times {10}^{-2}$

(2)

As $V=Al$                                         ...(i)

where $A$ is the area of cross-section of the wire.

Young’s modulus,  $Y=\frac{(F/A)}{(\Delta l/l)}=\frac{Fl}{A\Delta l}$

         $\Delta l=\frac{Fl}{YA}=\frac{F{l}^2}{YV} \text{(Using (i))}$

       $\Delta l\propto l^2$

Hence, the graph between $\Delta l$ and $l^2$ is a straight line.