Physics M.C.Q (1 Marks)

Stress $=$ Elastic limit

$\delta_{\text {max }}=\frac{\sigma_{\text {elastlc }} \times L}{\text { Young's modulus }}=\frac{8 \times 10^8 \times 1}{2 \times 10^{11}}  =4 \times 10^{-3}$

$=4 \mathrm{~mm}$

i.e. maximum elongation is $4 \mathrm{~mm}$

Stress $=\frac{ W }{ A }$

(Here $A$ Cross-sectional Area)

$Y _{\text {copper }}< Y _{\text {stoel }}$

$=\frac{1}{2}(M g) \ell=\frac{1}{2} M g \ell$

Since initial volume of wires are same

$\therefore $ Their areas of cross sections are $A$ and $3A$ and lengths are $3l$ and $l$ respectively.

For Wire $1,$

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

For wire $2$, let $F'$ force is applied

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

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

From eqns $(i)$ and $(ii),$ 

$\left( {\frac{F}{{AY}}} \right)3l = \left( {\frac{{F'}}{{3AY}}} \right)l \Rightarrow F' = 9F$

Thus, $\Delta V =-\frac{ pV }{ B }$ or $V ^{\prime}- V =-\frac{ pV }{ B }$ Or, $V^{\prime}=V\left(1-\frac{p}{B}\right)$

Now, $\rho^{\prime}=\frac{ m }{ V ^{\prime}}=\frac{ m }{ V \left(1-\frac{ P }{ B }\right)}=\frac{\rho}{\left(1-\frac{ p }{ B }\right)}$

$\frac{\rho^{\prime}}{\rho}=\frac{1}{\left(1-\frac{ p }{ B }\right)}$

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

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

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

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

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}}$

$⇒\ m{\omega ^2}r = $ breaking stress  $\times$ cross sectional area

$⇒\ m{\omega ^2}r = p \times A$

$⇒\ \omega = \sqrt {\frac{{p \times A}}{{mr}}} = \sqrt {\frac{{4.8 \times {{10}^7} \times {{10}^{ - 6}}}}{{10 \times 0.3}}} $

$\omega = 4\,rad/\sec $

$\therefore$ ${\rm{Stress}} \propto \frac{{\rm{1}}}{{\pi {{\rm{r}}^{\rm{2}}}}}$

$\frac{{{S_B}}}{{{S_A}}} = {\left( {\frac{{{r_A}}}{{{r_B}}}} \right)^2} = {(2)^2} \Rightarrow {S_B} = 4{S_A}$

$\frac{{{l_A}}}{{{l_B}}} = {\left( {\frac{{{r_B}}}{{{r_A}}}} \right)^2}$

$\frac{{{l_A}}}{{{l_B}}}  = {\left( {\frac{{{r_B}}}{{2{r_B}}}} \right)^2} = \frac{1}{4}$

$⇒$ ${l_B} = 4{l_A}$

Ratio of strain =$\frac{{{A_2}}}{{{A_1}}} = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^2} = {\left( {\frac{4}{1}} \right)^2} = \frac{{16}}{1}$

E are:

Glass: $50 - 90\,GPa$

Rubber:$0.01 - 0.1\,GPa$

Steel:$200\,GPa$

Copper: $117\,GPa$

$\frac{{{E_2}}}{{{E_1}}} = \frac{{{P_2}}}{{{P_1}}} = \frac{{{V_1}}}{{{V_2}}} \times \frac{{{T_2}}}{{{T_1}}} = \left( {\frac{1}{4}} \right) \times \left( {\frac{{400}}{{300}}} \right) = \frac{1}{3}$ $⇒$ ${E_2} = \frac{{{E_1}}}{3}$

i.e. elasticity will become $\frac{1}{3}$ times.

$\therefore$ Breaking force $\propto$ $r^2$ (Breaking distance is constant)

If radius becomes doubled then breaking force will become $4$ times i.e. $40 \times 4 = 160\, kg\, wt$

If $\eta$ and $F$ are constant then $x \propto \frac{L}{A}$

For maximum displacement area at which force applied should be minimum and vertical side should be maximum, this is given in the $R$ position of rectangular block.

$\Rightarrow$ Breaking stress $=\frac{\mathrm{W}}{\mathrm{A}}=\frac{\mathrm{W}^{\prime}}{2 \mathrm{A}}$

to break new wire load $W^{\prime}=2 W$

on applying $W,$ new wire will not break.

$\frac{\mathrm{F}_{1}}{\mathrm{A}_{1}}=\frac{\mathrm{F}_{2}}{\mathrm{A}_{2}} \Rightarrow \mathrm{F}_{2}=\frac{\mathrm{A}_{2}}{\mathrm{A}_{1}} \times \mathrm{F}_{1}=\frac{\pi \mathrm{r}_{2}^{2}}{\pi \mathrm{r}_{1}^{2}} \mathrm{F}_{1}$

$\Rightarrow \mathrm{F}_{2}=\frac{(2)^{2}}{(1)^{2}} \mathrm{F}_{1} \Rightarrow \mathrm{F}_{2}=4 \mathrm{F}_{1}=80 \mathrm{kg}-\mathrm{wt}$

$\frac{(\mathrm{A} \ell) \rho \mathrm{g}}{\mathrm{A}}=\sigma$

$\ell=\frac{\sigma}{\rho g}$

$\Rightarrow \mathrm{F}_{2}=4 \mathrm{F}_{1}=80 \mathrm{kg}-\mathrm{wt}$

Area of this surface

$=4 \times\left(2 \times 10^{-2}\right) \times\left(0.2 \times 10^{-2}\right) \mathrm{m}^{2}=1.6 \times 10^{-4} \mathrm{m}^{2}$

$S_{\min }=3.5 \times 10^{8}=\frac{\mathrm{F}}{1.6 \times 10^{-4}} \Rightarrow \mathrm{F}=5.6 \times 10^{4} \mathrm{N}$

$\frac{F_{1}}{F_{2}}=\left(\frac{r_{1}}{r_{2}}\right)^{2}$

$\frac{200}{\mathrm{F}_{2}}=\left(\frac{\mathrm{r}}{2 \mathrm{r}}\right)^{2} \Rightarrow \mathrm{F}_{2}=800 \mathrm{N}$

Breaking stress $=\rho \times g \times L$

Substitute values from the question

Breaking stress $=7.5 \times 10^7 \,Nm ^{-2}$

$\left\{\begin{array}{l}\rho=\text { Density of material } \\ g=\text { Acceleration due to gravity } \\ L=\text { Length of wire that can hang without breaking }\end{array}\right\}$

$\rho=2.7 \times 10^3 \,kg m ^{-3}$

$g=9.8 \,m / s$

$7.5 \times 10^7=2.7 \times 10^3 \times 9.8 \times L$

$\frac{\text { Lateral strain }}{\text { Longitudinal strain }}=\eta$

$\frac{\Delta r / r}{\Delta l / I}=0.5$

Substitute $\Delta r / r=2 / 100$

$\frac{\Delta l}{l}=\frac{4}{100}$

$\therefore\% \text { increase }=\frac{\Delta l}{l} \times 100=4 \%$

$\because A \propto r^2$

So $\frac{\Delta A}{A}=\frac{2 \Delta r}{r}$

$\frac{4}{100}=2 \times \frac{\Delta r}{r}$

$\frac{2}{100}=\frac{\Delta r}{r}$

Tensile stress $=\frac{\text { Tension at point }}{\text { Area }}$

Tension at distance $x$ from top would be the amount of force acting due to all the weight below it

$=$ Mass per unit length of rod $\times$ length of rod $+M g$ $=m \times(I-x) g+M g$

So Tensile stress $=\frac{m(I-x) g+M g}{A}$

 $l \propto \frac{L}{{{d^2}}}$ [As F and Y are constant]

The ratio of $\frac{L}{{{d^2}}}$is maximum for case $(d)$

$ = \frac{1}{2} \times {\left( {\frac{1}{2}} \right)^2} = \frac{1}{8}$

$\delta = \frac{{W{L^3}}}{{4Yb{d^3}}}$

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

$\frac{{{l_2}}}{{{l_1}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = {(2)^2} = 4$

$ \Rightarrow {l_2} = 4{l_1} = 4cm$

$ = 2 \times {10^{11}} \times 3 \times {10^{ - 10}} = 60\;N/m$

 $\frac{{{l_s}}}{{{l_{cu}}}} = \frac{{1.2 \times {{10}^{11}}}}{{2 \times {{10}^{11}}}} = \frac{3}{5}$

$\frac{{{l_2}}}{{{l_1}}} = \frac{{{L_2}}}{{{L_1}}}{\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = 2 \times {\left( {\frac{1}{2}} \right)^2} = \frac{1}{2}$==>${l_2} = \frac{{{l_1}}}{2} = \frac{l}{2} = 0.5mm.$

$ = {10^{11}} \times {10^{ - 4}} \times {10^{ - 5}} \times 100 = {10^4}N$

= $\frac{1}{{14}} \times {10^{ - 2}}$$ = 7.1 \times {10^{ - 4}}{m^2}$

Longitudinal strain $\frac{l}{L} = \frac{{{\rm{stress}}}}{Y} = \frac{{{{10}^7}}}{{{{10}^{11}}}} = {10^{ - 4}}$

Percentage increase in length $ = {10^{ - 4}} \times 100 = 0.01\% $

i.e. force constant of a wire depends on young's modules (nature of the material), radius of the wire and length of the wire.

$ = 8 \times {10^{ - 6}} \times 2.2 \times {10^{11}} \times {10^{ - 2}} \times {10^{ - 4}} \times 5 = 8.8\;N$

$ Y$$ = 2 \times {10^{11}}N/{m^2}$

$l = \frac{{FL}}{{AY}} = \frac{{2 \times L}}{{2 \times {{10}^{ - 4}} \times {{10}^4}}} = L$

 Final length $=$ initial length $+$ increment $= 2L$

$\frac{{{l_2}}}{{{l_1}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = {\left( {\frac{1}{2}} \right)^2} \Rightarrow {l_2} = \frac{{{l_1}}}{4} = \frac{{2.4}}{4}$$ \Rightarrow {l_2} = 0.6\;cm$

If diameter is made four times then force required will be $16$ times. i.e. $16 \times 10^3 N$

$\frac{{{F_1}}}{{{F_2}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\left( {\frac{{{L_2}}}{{{L_1}}}} \right) = {\left( {\frac{2}{1}} \right)^2}\left( {\frac{1}{4}} \right) = 1$ $⇒$ $\frac{{{F_1}}}{{{F_2}}} = 1:1$

If volume is fixed then $l \propto {L^2}$

$l \propto \frac{L}{{{r^2}}}$ $(Y$ and $F$ are constant$)$

$\frac{{{l_2}}}{{{l_1}}} = \frac{{{L_2}}}{{{L_1}}} \times {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = (2) \times {\left( {\frac{1}{2}} \right)^2} = \frac{1}{2}$

==> ${l_2} = \frac{{{l_1}}}{2} = \frac{{0.01m}}{2} = 0.005m$

$\frac{{{l_2}}}{{{l_1}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = {(2)^2} \Rightarrow {l_2} = 4{l_1} = 4 \times 3 = 12mm$

Maximum extension takes place in that wire for which the ratio of $\frac{L}{{{r^2}}}$ will be maximum.

$\frac{{{l_1}}}{{{l_2}}} = \frac{{{F_2}}}{{{F_1}}}{\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = (4) \times {\left( {\frac{1}{2}} \right)^2} = 1$

${l_2} = {l_1} = 1mm$

$\frac{{{A_2}}}{{{A_1}}} = \frac{{{l_1}}}{{{l_2}}} \Rightarrow {A_2} = {A_1}\left( {\frac{{0.1}}{{0.05}}} \right)$= $2{A_1} = 2 \times 4 = 8m{m^2}$

As the length of wire get doubled therefore strain $= 1$

$Y =$ strain $= 20 \times 10^8 N/ m^2$

$\frac{{{l_2}}}{{{l_1}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = {(2)^2} \Rightarrow {l_2} = 4{l_1} = 4 \times 2 = 8\;mm.$

$\frac{{{F_A}}}{{{F_B}}} = {\left( {\frac{{{r_A}}}{{{r_B}}}} \right)^2} \times \left( {\frac{{{L_B}}}{{{L_A}}}} \right) = {\left( {\frac{2}{1}} \right)^2} \times \left( {\frac{2}{1}} \right) = \frac{8}{1}$

$l \propto {L^2}$ If volume of the wire remains constant

$\frac{{{l_2}}}{{{l_1}}} = {\left( {\frac{{{L_2}}}{{{L_1}}}} \right)^2} = {\left( {\frac{8}{2}} \right)^2} = 16$

${l_2} = 16 \times {l_1} = 16 \times 2 = 32mm = 3.2cm$

$l \propto \frac{1}{A}$                   $(F,L$ and $Y$ are constant$)$

$\frac{{{l_2}}}{{{l_1}}} = \frac{{{A_1}}}{{{A_2}}} = \frac{4}{8} = \frac{1}{2}$$ \Rightarrow {l_2} = \frac{{{l_1}}}{2} = \frac{{0.1}}{2} = 0.05mm$

Ratio of $\frac{L}{{{r^2}}}$ is maximum for wire in option $(b).$

 

$\frac{{{l_1}}}{{{l_2}}} = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^2} = {(2)^2} = 4$

Lateral strain due to the other two forces acting on perpendicular faces$ = \frac{{ - 2\sigma F}}{Y}$

Total increase in length $ = (1 - 2\sigma )\frac{F}{Y}$

Also ${l_c} - {l_s} = 0.5$…(ii)

On solving (i) and (ii) ${l_c} = 1.25\;cm$ and ${l_s} = 0.75\;cm$.

Only option 'radius $3 \mathrm{mm},$ length $2 \mathrm{m}$ ' is satisfying the above relation.

$\frac{\ell}{\mathrm{r}^{2}} \rightarrow \max$ for $\mathrm{L}=100 \mathrm{cm}$

$\& r=0.2 \mathrm{mm}$

$[\mathrm{Y}]=\frac{\mathrm{MLT}^{-2}}{\mathrm{L}^{2}}=\mathrm{ML}^{-1} \mathrm{T}^{-2}$

Conversion factor from $CGS$ to $MKS$ is $0.1 .$

$\frac{\Delta \mathrm{l}_{2}}{\Delta \mathrm{l}_{1}}=\frac{\mathrm{D}_{1}^{2}}{\mathrm{D}_{2}^{2}}=4$ or $\Delta \mathrm{l}_{2}=4 \Delta \mathrm{l}_{1}=4 \mathrm{cm}$

$Y=5 \times 10$

$\frac{\Delta \ell}{\ell}=\frac{\text { Stress }}{\mathrm{Y}}$

$\frac{\Delta \ell}{\ell}=\frac{1.5 \times 10^{6} \times 10}{5 \times 10^{11}} \times 100$

$\frac{\Delta \ell}{\ell}=3 \times 10^{-3} \%$

$A=0.1 \mathrm{cm}^{2}=0.1 \times 10^{-4} \mathrm{m}^{2}$

$Y=2 \times 10^{11} \mathrm{N} / \mathrm{m}^{-2}$

$\frac{\Delta \mathrm{L}}{\mathrm{L}}=0.1 \%=\frac{0.1}{100}=0.1 \times 10^{-2}$

$\mathrm{As} \mathrm{Y}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \mathrm{L} / \mathrm{L}}$

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

$=2 \times 10^{11} \mathrm{Nm}^{-2} \times 0.1 \times 10^{-2} \times 0.1 \times 10^{-4} \mathrm{m}^{2}$

$=2 \times 10^{3} \mathrm{N}=2000 \mathrm{N}$

$A_{B}=2 c m^{2} A_{s}=1 \mathrm{cm}^{2}$

$\Delta \ell_{\mathrm{B}}=\Delta \ell_{\mathrm{s}}$

$\frac{\mathrm{F}}{\mathrm{A}_{\mathrm{B}}} \frac{\ell_{\mathrm{B}}}{\mathrm{Y}_{\mathrm{B}}}=\frac{\mathrm{F}}{\mathrm{A}_{\mathrm{S}}} \frac{\ell_{\mathrm{S}}}{\mathrm{Y}_{\mathrm{S}}}$

$\mathrm{L}=\frac{\mathrm{A}_{\mathrm{s}} \mathrm{Y}_{\mathrm{s}}}{\mathrm{A}_{\mathrm{B}} \mathrm{Y}_{\mathrm{B}}} \ell_{\mathrm{B}}=\frac{1}{2} \times \frac{2 \times 10^{11}}{10^{11}} \times 2=2$

$\mathrm{y}_{1}=\frac{4 \times 10^{8}}{0.1 / 100}=4 \times 10^{11}$

$\mathrm{y}_{2}=\frac{6 \times 10^{8}}{0.3 / 100}=2 \times 10^{11}$

$\Rightarrow \mathrm{y}_{1}=2 \mathrm{y}_{2}$

or $\Delta \mathrm{L}=\frac{\mathrm{FL}}{\mathrm{AY}}\left\{\mathrm{A}=\pi \mathrm{r}^{2}\right\}$

$\Rightarrow$ More $L/A,$ more elongation

$\frac{1_{2}}{\mathrm{l}_{1}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{2}=(\mathrm{n})^{2} \Rightarrow 1_{2}=\mathrm{n}^{2} \mathrm{l}_{1}$

$F \propto A^{2}$

$F=\mathrm{Ay} \cdot \frac{\Delta \ell}{\ell}$

$\Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{Ay}}=\frac{4 \mathrm{F} \ell}{\pi \mathrm{d} 2 \mathrm{y}}$

$\Delta \ell \propto \frac{\ell}{\mathrm{d}^{2}}$

$\Delta \mathrm{L}=\frac{(100)(6)}{10^{11}\left(10^{-5}\right)}=0.6 \times 10^{-3} \mathrm{m}$

We know

$\frac{\text { Force } \times \text { Length }}{\text { Area } \times \text { young's modulus }}=$ elongation $\quad\left\{\frac{F L}{A y}=\Delta x\right\}$

$\Rightarrow F=\left(\frac{\Delta x \cdot y}{L}\right) A$

$F=\left(\frac{\Delta x \cdot y}{L} \cdot \frac{\pi}{4}\right) d^2$

We can say $F \propto d^2$

So we can use

$\frac{F_1}{F_2}=\frac{d_1^2}{d_2^2}$  $\left\{\begin{array}{l}F_1=4 \times 10^5 N \\ d_1=2 mm \\ F_2=? \\ d_2=1.5 mm \end{array}\right\}$

Substituting values

$\frac{4 \times 10^5}{F_2}=\frac{(2)^2}{(1.5)^2}$

$F_2=2.3 \times 10^5 \,N$

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

Since $A, Y$ remain constant in given case We can say

$F L \propto \Delta x$

or $\frac{F_1 L_1}{F_2 L_2}=\frac{\Delta x_1}{\Delta x_2}$  $\left\{\begin{array}{l}F_1=200 \,N \\ \Delta x_1=1 \,mm \\ \Delta x_2=10.02 \,m -10 \,m =0.02 \,m =20 \,mm \\ L_1=1 \,m \\ L_2=10 \,m \end{array}\right.$

Substitute the values

$\frac{200 \times 1}{F_2 \times 10}=\frac{1 \,mm }{20 \,mm }$

$F_2=400 \,N$

$Y=\frac{F L}{A \Delta L} \Rightarrow \text { Percentage increase } \frac{\Delta L}{L} \times 100=\frac{F}{A Y} \times 100$

$\text { Diameter }=2.5 \,mm$

$d=\frac{2.5}{1000} \,m$

Area $=\frac{\pi d^2}{4} \quad Y=12.5 \times 10^{11} \,dyne / cm ^2$  $\left\{\frac{1 \text { dyne }}{ cm ^2}=\frac{0.1 \,N }{ m ^2}\right\}$

$\Rightarrow A=\frac{\pi \times(2.5)^2}{4} \quad F=100 \times 10=1000 \,N$

$\Rightarrow \frac{1000 \times L}{\frac{3.14 \times(2.5)^2}{4 \times(1000)^2} \times 12.5 \times 10^{11} \times 0.1}=\frac{\Delta L}{L} \times 100$

$=0.16 \%$

Let us say that elongation in copper $=x$

Than elongation in steel $=2-x$

We know

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

$\because F, A, L$ are same only material is different We can say

$\frac{1}{Y} \propto \Delta x$

$\frac{Y_2}{Y_1}=\frac{\Delta x_1}{\Delta x_2}$   $\left\{\begin{array}{l}\text { Where } \\ Y_2=Y_{\text {steel }} \\ Y_1=Y_{\text {copper }} \\ \Delta x_1=\text { elongation in copper }=x \\ \Delta x_2=2-x\end{array}\right.$

Substituting values

$\frac{20 \times 10^{11}}{12 \times 10^{11}}=\frac{x}{2-x}$

$x=1.25 \,cm$

So $\Delta x_{\text {copper }}=1.25 \,cm , \Delta x_{\text {sleel }}=0.75 \,cm$

$\text { Strain }=0.32 \%$

$\Rightarrow \frac{\Delta L}{L} \times 100=0.32$

$\Rightarrow \frac{\Delta L}{L}=\frac{0.32}{100}$

$A=\pi r^2=3.14 \times\left(\frac{10}{1000}\right)^2$

$Y=2 \times 10^{11} \,Nm ^2$

We know

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

$F=\left(\frac{\Delta L}{L}\right) \times A \times Y$

Substituting values

$F=\frac{0.32}{100} \times 3.14 \times\left(\frac{10}{1000}\right)^2 \times 2 \times 10^{11}$

$F=201 \,kN$

At proportional limit

Stress $\propto$ strain

Stress $=Y \times$ strain   $\{Y=$ Young's Modulus $\}$

Stress $=Y \times \frac{\Delta L}{L}$  $\left\{\begin{array}{l}\text { Stress }=8 \times 10^8 \,N / m ^2 \\ Y=2 \times 10^{11} \,N / m ^2 \\ L=1 \,m \end{array}\right.$

Substituting values

$\frac{8 \times 10^8 \times 1}{2 \times 10^{11}}=\Delta L$

$4 \,mm=\Delta L$

For the same load wire with maximum elongation has minimum cross-section area

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

$F, L, Y$ are fixed so $\frac{1}{A} \propto \Delta x$

$\Rightarrow O A$ is the thinnest.

Since tension in both cases is same and all other parametrs $(Y, A, L)$ are also same $\Rightarrow$ Elongation will be same in both cases.

At same value of load

$A$ has less elongation than $B \quad \ldots (1)$

$\frac{F L}{A Y}=\Delta L$    $\left\{\begin{array}{l}\because L, A \text { are same } \\ F \text { - Load is also taken same }\end{array}\right\}$

So $\frac{1}{Y} \propto \Delta L \quad \ldots (2)$

Using conditions $(1)$ and $(2)$

We can say

$Y_A > Y_B$    $\{$ Young's modulus of $A$ greater than $B\}$

From the graph we can see young's modulus is less for $T_1$ as compared to $T_2$ ( $Y=$ slope of stress-strain curve)

As $T$ increases $Y$ decreases

So $T_1 > T_2$

Total mass can be assumed to be concentrated at center of mass at distance $\frac{L}{2}$ from top

$\frac{M}{L}=\lambda$

$M=\lambda L$

$\Delta x=\frac{F L / 2}{A Y}=\frac{1}{2} \times \frac{\lambda L^2 g}{A Y}$

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

For wire $A$

$\Delta L_A=\frac{F \cdot L_A}{\pi r_A^2 \cdot Y_A} \ldots (1)$

For wire $B$

$\Delta L_B=\frac{F \cdot L_B}{\pi r_B^2 \cdot Y_B} \ldots (2)$

Divide $(1)$ by $(2)$

$\frac{\Delta L_A}{\Delta L_B}=\frac{F \cdot L_A}{\pi r_A^2 \cdot Y_A} \times \frac{\pi r_B^2 \cdot Y_B}{F \times L_B}=\frac{L_A}{L_B} \times\left(\frac{r_B}{r_A}\right)^2 \times \frac{Y_B}{Y_A}$

Substituting the value of ratio's

$\frac{\Delta L_A}{\Delta L_B}=\frac{4}{1} \times\left(\frac{1}{3}\right)^2 \times \frac{2}{1}=\frac{8}{9}$

Tension is same (given)

From free body diagram

$3 T=150 \,N$

$T=50 \,N$

Since the bar has to be supported symmetrically Therefore extension in each wire will be same

We know $\Delta x=\frac{F L}{A Y}$

Compare $1$ copper wire with another steel wire

$\frac{F L}{A_C Y_C}=\frac{F L}{A_S Y_S}$

$\Rightarrow \frac{A_S}{A_C}=\frac{Y_C}{Y_S}$  $\left\{\begin{array}{l}\text { Where, } \\ A_C-\text { Area of copper wire } \\ Y_C-\text { Young's modulus copper } \\ A_S-\text { Area of steel wire } \\ Y_S-\text { Young's modulus steel }\end{array}\right.$

Substitutuing value of $Y_C$ and $Y_s$

$\frac{\pi d_S^2}{4 \times \frac{\pi}{4} \times d_C^2}=\frac{110 \times 10^9}{190 \times 10^9}$

$\frac{d_S}{d_C}=\sqrt{\frac{11}{19}}$  $\left\{\begin{array}{l}d_S \text {-diameter of steel wire } \\ d_C \text {-diameter of copper wire }\end{array}\right.$

$\frac{d_C}{d_S}=\sqrt{\frac{19}{11}}$

For elongation of rod under its own weight

We know $\Delta x=\frac{\rho g x^2}{2 Y}$ $\left\{\begin{array}{l}\text { Where, } \\ \Delta x=\text { Elongation } \\ \rho=\text { Density of rod } \\ Y=\text { Young's modulus } \\ L=\text { Length } \\ g=\text { Acceleration due to gravity } \\ x=\text { Distance of point from lower end }\end{array}\right.$

We can clearly see that elongation $\propto\left(x^2\right)$

So graph of $\Delta x$ vs $x$ should be a upward parabola.

$Y=\frac{m g / A}{\Delta \ell / \ell}=\frac{m g \ell}{A \Delta \ell}$

Also $\Delta \ell=\ell \alpha \Delta T$

From (1) and (2)

$Y=\frac{m g \ell}{A \ell \alpha \Delta T}=\frac{m g}{A \alpha \Delta T}$

$\therefore m=\frac{Y A \alpha \Delta T}{g}$

$=\frac{10^{11} \times \pi\left(10^{-3}\right)^2 \times 10^{-5} \times 10}{10}=\pi \approx 3$

$ = 4 \times {10^{ - 5}} \times 100 \times 100 = 0.4cc$

$\%$ change in volume $= 3 \times (\% $ change in length$)$

$= 3 \times 1\% = 3\%$    $\therefore$  Bulk strain $\frac{{\Delta V}}{V} = 0.03$

$ = 19.6 \times {10^8}N/{m^2}$

$ = 1.55 \times {10^5}pa$

$\frac{\Delta \rho}{\rho}=\frac{\mathrm{P}}{\mathrm{B}}=\frac{\mathrm{dgh}}{\mathrm{B}}=\frac{1.025 \times 10^{3} \times 10 \times 10^{3}}{1.6 \times 10^{6} \times 10^{3}}=0.64 \times 10^{-2}$

$\%$ change $=\frac{\Delta \rho}{\rho} \times 100=0.64 \%$

$9.1 \times 10^{8}=\frac{\mathrm{h} \times 10^{3} \times 10}{\left(\frac{0.1}{100}\right)} \quad \Rightarrow \mathrm{h}=91 \mathrm{m}$

$\Rightarrow \Delta \mathrm{P}=\mathrm{B}\left(-\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)$

$\Rightarrow \Delta \mathrm{P}=\left(2100 \times 10^{6}\right)\left(\frac{0.004}{100}\right)$

$\left(\text { As }-\frac{\Delta \mathrm{V}}{\mathrm{V}} \times 100=0.004\right)$

$\Rightarrow \Delta \mathrm{P}=84 \times 10^{3} \mathrm{P}_{\mathrm{a}}$

$\frac{10^{5}}{50 \times 10^{-6}}=\frac{\rho \mathrm{gh} \times 100}{1}$

$h=2 \mathrm{km}$

$\Rightarrow \Delta \mathrm{P}=\mathrm{B}\left(-\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)$

$\Rightarrow \Delta \mathrm{P}=\left(2100 \times 10^{6}\right)\left(\frac{0.004}{100}\right)$

$\Rightarrow\left(\text { As }-\frac{\Delta \mathrm{V}}{\mathrm{V}} \times 100=0.004\right)$

$\Rightarrow \Delta \mathrm{P}=84 \times 10^{3} \mathrm{P_a}$

$=1.05 \times 10^{9} \mathrm{Pa}$

$\mathrm{B}=-\frac{\mathrm{dpV}}{\Delta \mathrm{V}}$

Here $\mathrm{p}=$ Pressure (stress)

$-\frac{\Delta \mathrm{V}}{\mathrm{V}}=\text { Volume strain }$

But liquid is incompressible, so

${\Delta \mathrm{V}=0} $

Hence,  ${\mathrm{B}=-\frac{\mathrm{pV}}{0}=\infty}$

or $\mathrm{B}=\infty$ (infinity)

Fractional change in volume, $\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{B}}$

Here, $P=10$ atm $=10 \times 1 \times 10^{5} \mathrm{N} \mathrm{m}^{-2}$

$\mathrm{B}=37 \times 10^{9} \mathrm{N} \mathrm{m}^{-2}$

$\therefore \quad \frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{1 \times 10^{6} \mathrm{Nm}^{-2}}{37 \times 10^{9} \mathrm{Nm}^{-2}}=0.027 \times 10^{-3}$

$=2.7 \times 10^{-5}$

$\Rightarrow \Delta \mathrm{P}=\mathrm{B}\left(-\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)$

$\Rightarrow \Delta \mathrm{P}=\left(2100 \times 10^{6}\right)\left(\frac{0.004}{100}\right)$

$\Rightarrow\left(\text { As }-\frac{\Delta \mathrm{V}}{\mathrm{V}} \times 100=0.004\right)$

$\Rightarrow \Delta \mathrm{P}=84 \times 10^{3} \mathrm{P}_{a}$

$h = 91\,m$

$=-\frac{2.5 \times 10^{7}}{1.25 \times 10^{11}} \times(60)^{3} \mathrm{mm}^{3}=-43.2 \mathrm{mm}^{3}$

$\frac{1}{\sigma}=\frac{\mathrm{PV}}{\Delta \mathrm{V}} $

$ \Rightarrow \Delta {\rm{V}} = \sigma {\rm{PV}}$

$p = h\rho g = 200 \times {10^3} \times 10N/{m^2}$

$k = 22000\,atm = 22000 \times {10^5}N/{m^2}$

$V = 1{m^3}$

$\Delta V = \frac{{200 \times {{10}^3} \times 10 \times 1}}{{22000 \times {{10}^5}}} = 9.1 \times {10^{ - 4}}{m^3}$

$\frac{\Delta V}{V}=\frac{-\Delta P}{B}$   $\left\{\begin{array}{l}\Delta V=10 \% \text { of } V \text { ( } \because \text { Pressure increases volume must } \\ \text { If } \begin{array}{l}\text { decreases by } 10 \% \text { so we will use a +ve sign) }\end{array} \\ \begin{array}{rl}\Rightarrow \Delta V & =100 cc \\ \Delta P & =P_2-P_1 \\ & =1.165 \times 10^5-1.01 \times 10^5\end{array}\end{array}\right.$

Substituting the values

$\frac{-10}{100}=\frac{-\left(1.165 \times 10^6-1.01 \times 10^6\right)}{B}$

$\frac{1}{10}=\frac{.155 \times 10^5}{B}$

$B=1.55 \times 10^5 \,Pa$

Volume $=(\text { side })^3$

$v=(a)^3$

So $\frac{\Delta V}{V}=\frac{3 \Delta a}{a}$  $\left\{\operatorname{given} \frac{\Delta a}{a}=-2 \%\right\}$

$\therefore \frac{\Delta v}{v}=3 \times-2 \quad \text { Side decreases solve used (-)ve sign }$

$=-6 \%$

We know

$\frac{\Delta v}{v}=-\frac{P}{B}$

Substituting value of $\Delta v / v$

$-\frac{6}{100}=-\frac{P}{B}$

So bulk strain produced is $0.06$

Pressure at the bottom of sea $=\rho_w g h=1000 \,kg / m ^3 \times 10 \,m / s ^2 \times 1400 \,m =14000000 \,N / m ^2$

Also we know

$\frac{\Delta v}{v}=-\frac{P}{B} \quad\left\{\frac{\Delta v}{v}=\frac{-2}{100}\right\}$

$\frac{-2}{100}=\frac{-14000000}{B}$

$B=7 \times 10^8 \,N / m ^2$

$\text { Modulus of rigidity }(G)=\frac{\text { Force } \times \text { Length }}{\text { Area } \times \text { Lateral displacement }}=\frac{F L}{A \times \Delta x}$

$F=10 \,kN =10 \times 10^3 \,N$

$L=4 \,cm =0.04 \,m$

$A=1 \,cm ^2=1 \times 10^{-4} \,m ^2$

$G=8 \times 10^{11}\,N / m ^2$

Substituting values

$8 \times 10^{11}=\frac{10 \times 10^3 \times 0.04}{1 \times 10^{-4} \times \Delta x}$

$\Delta x=\frac{10 \times 10^3 \times 0.04}{1 \times 10^{-4} \times 8 \times 10^{11}}=5 \times 10^{-6} \,m$

$r_1^4 \phi_1=r_2^4 \phi_2$

$\frac{\phi_A}{\phi_B}=\frac{\left(r_B\right)^4}{\left(r_A\right)^4}=\left(\frac{r_2}{r_1}\right)^4$

$\frac{{{U_2}}}{{{U_1}}} = {\left( {\frac{{{l_2}}}{{{l_1}}}} \right)^2} = {\left( {\frac{{10}}{2}} \right)^2} = 25$ $\Rightarrow$  ${U_2} = 25{U_1}$

i.e. potential energy of the spring will be $25 \,V$

 $\frac{{{W_1}}}{{{W_2}}} = \frac{{{l_1}}}{{{l_2}}} = \frac{l}{{2l}} = \frac{1}{2}$

$ = \frac{1}{2} \times 10 \times 10 \times 1 \times {10^{ - 1}} = 0.05\;J$

$=$$\frac{1}{2} \times \frac{{{F^2}L}}{{AY}} = \frac{1}{2} \times \frac{{{{(50)}^2} \times 0.2}}{{1 \times {{10}^{ - 4}} \times 1 \times {{10}^{11}}}}$$=$ $2.5 \times {10^{ - 5}}J$

$W = \frac{1}{2}Fl = \frac{1}{2} \times 10 \times 0.5 \times {10^{ - 3}}$= $2.5 \times {10^{ - 3}}J$

Work done to displace it through $1.5 \,mm$

$W = F \times l = 5 \times {10^{ - 3}}J$

The ratio of above two work $= 1 : 2$

$\frac{{{E_1}}}{{{E_2}}} = \frac{{{Y_2}}}{{{Y_1}}}$ (Stress is constant) 

$\frac{{{E_1}}}{{{E_2}}} = \frac{3}{2}$

$ = \frac{1}{2} \times \frac{{{{(50)}^2} \times 20 \times {{10}^{ - 2}}}}{{2 \times {{10}^{ - 4}} \times 1.4 \times {{10}^{11}}}}$$ = 8.57 \times {10^{ - 6}}J$

$ = 0.075\;J$

At extension ${l_2}$, the stored energy$ = \frac{1}{2}Kl_2^2$

Work done in increasing its extension from ${l_1}$ to ${l_2}$

                 $ = \frac{1}{2}K(l_2^2 - l_1^2)$

Work done$ = \frac{1}{2}K{x^2} = \frac{1}{2} \times (0.2) \times {(0.05)^2} = 2.5\;J$

$\therefore \quad W=U=\frac{1}{2} \times$ Stress $\times$ Strain $\times$ Volume

OR $\quad W=\frac{1}{2} \times Y(\text { Strain })^{2} \times$ Volume   $……(1)$

$(\text {Strain}=Y \times$Strain)

Strain $=\frac{x}{L}$and volume $=A L$

$\therefore W=\frac{1}{2} \times Y \times \frac{x^{2}}{L^{2}} \times A L \quad \Longrightarrow W=\frac{Y x^{2} A}{2 L}$

$\frac{{{U_A}}}{{{U_B}}} = \left( {\frac{{{L_A}}}{{{L_B}}}} \right) \times {\left( {\frac{{{r_A}}}{{{r_B}}}} \right)^2}$= $(3) \times {\left( {\frac{1}{2}} \right)^2} = \frac{3}{4}$

 

$F=200 N, x=1 m m=10^{-3} m$

$\therefore E=\frac{1}{2} \times 200 \times 1 \times 10^{-3}=0.1 J$

$=\frac{1}{2} \times \mathrm{Y} \times(\alpha \Delta \theta)^{2}$

$=\frac{1}{2} \times 10^{11} \times\left(12 \times 10^{-6} \times 20\right)^{2}=2880 \mathrm{J} / \mathrm{m}^{3}$

$=\frac{1}{2} \times Y \times(\alpha \Delta \theta)^{2}$

$=\frac{1}{2} \times 10^{11} \times\left(12 \times 10^{-6} \times 20\right)^{2}=2880 \mathrm{J} / \mathrm{m}^{3}$

$\left[\frac{\text { stress }}{Y}\right]$ (volume) $=\frac{1}{2}\left[\frac{\text { stress }^{2}}{Y}\right] \times$ volume

$=\frac{1}{2}\left[\frac{M g}{A}\right]^{2} \cdot\left[\frac{1}{Y}\right](A L)=\frac{1}{2} \frac{M^{2} g^{2} L}{A Y}$

$=\frac{1}{2} \times \frac{2 \times 10^{11} \times 1 \times 10^{-6}}{1} \times\left(10^{-3}\right)^{2}=0.1 \mathrm{J}$

Work done $=$ Force $\times$ elongation

$W=F \cdot \Delta x \ldots .$  $\left\{F=\Delta x \cdot \frac{A Y}{L}\right\}$

$\Rightarrow W=\frac{A Y}{L} \times \Delta x^2$

Multiply and divide by $L$

We get

$W=\frac{\text { Volume } \cdot Y}{L^2} \Delta x^2$

Cross multiply ( $L \cdot A)$ volume

$\frac{W}{L \cdot A}=Y \cdot\left(\frac{\Delta x}{L}\right)^2$

Work done per unit volume

Substitute values

$=8 \times 10^{10} \cdot\left(\frac{2}{100}\right)^2$

$=32 \,MJ / m ^3$

Shear modulus $=\frac{\text { Shear } \text { stress }}{\text { Shear } \text { stress }}$

$\eta=\frac{\text { Shear stress }}{\phi}$

$\eta \phi=$ Shear stress

Strain energy per unit volume $=\frac{1}{2} \times$ shear stress $\times$ shear strain $\Rightarrow \frac{\text { Strain energy }}{\text { Volume }}=\frac{1}{2} \times \eta \phi \times \phi \quad$ (Cross multiply volume)

Strain energy $=\frac{1}{2} \eta \phi^2 V$

$\frac{\text { Lateral strain }}{\text { Longitudinal strain }}=$ Poisson's ratio

$\frac{0.02 / 100}{\Delta l / l}=\frac{1}{4}$  $\left\{\begin{array}{l}Y=\text { (Young's modulus) } \\ \quad=8 \times 10^{10} \text { (given) } \\ \text { Poission's ratio }=\frac{1}{4} \text { (given) } \\ \text { Lateral strain }=0.02 \% \text { (given) }\end{array}\right.$

$\frac{\Delta l}{l}=\frac{0.08}{100}$

$\Delta U$ (Elastic potential energy per unit volume $=\frac{1}{2} \times Y \times($ Longitudinal strain $\left.)\right)$

Substituting values

$\Delta U=\frac{1}{2} \times 8 \times 10^{10} \times\left(\frac{0.08}{100}\right)^2$

$\Delta U=2.56 \times 10^4 \,J / m ^3$

$U=$ work done

We know

Work done $=\frac{\pi S r^4 \phi^2}{4 L}$ $\left\{\begin{array}{l}\text { Where, } \\ \phi=\text { Angle of twist }=\alpha \\ S=\text { Modulus of rigidity }=4\end{array}\right.$

Substituting values

$U=\frac{\pi \eta r^4 \alpha^2}{4 L}$

$\sigma = \frac{{3K - Y}}{{6K}}$=$\frac{{3 \times 11 \times {{10}^{10}} - 7.25 \times {{10}^{10}}}}{{6 \times 11 \times {{10}^{10}}}}$$ \Rightarrow \sigma = 0.39$

$\frac{{dV}}{V} = 2 \times 2 \times {10^{ - 3}} = 4 \times {10^{ - 3}}$

 Percentage change in volume =$4 \times {10^{ - 1}} = 0.4\% $

$2.4\eta = 2\eta (1 + \sigma ) \Rightarrow 1.2 = 1 + \sigma \Rightarrow \sigma = 0.2$

$Y=3 K(1-2 \sigma)$

$\Rightarrow 1-2 \sigma=\frac{Y}{3 K}=0 \Rightarrow \sigma=\frac{1}{2}$

$\frac{\text { Lateral strain }}{\text { Longitudinal strain }}=\eta=0.5$

$\frac{-\Delta r / r}{\Delta l / l}=\frac{1}{2}$

$\frac{-2 \Delta r}{r}=\frac{\Delta l}{l}$

Magnitute wise both are equal but sign's would be different as both quantities cannot increase

Now volume $\propto$ area $\times$ length $v \propto r^2 \cdot L$

$\frac{\Delta V}{V}=\frac{2 \Delta r}{r}+\frac{\Delta L}{L}$

Substituting value of $\frac{\Delta L}{L}$

$\frac{\Delta V}{V}=0$

$\therefore$ No change in volume.

So we can say that $Y = \cot \theta = \frac{1}{{\tan \theta }} = \frac{1}{{{\rm{slope}}}}$

So elasticity of wire $P$ is minimum and of wire $R$ is maximum

Applied force $\propto$ extension

but the graph between extension and stored elastic energy will be parabolic in nature

As $U = 1/2\,\,k{x^2}$or $U \propto {x^2}$.

$Y = \frac{{{\rm{Stress}}}}{{{\rm{Strain}}}} = \,\frac{{8 \times {{10}^7}}}{{4 \times {{10}^{ - 4}}}} = 2 \times {10^{11}}\,\,N/{m^2}$

i.e. $\tan {\theta _A} > \tan {\theta _B} > \tan {\theta _C}$

or ${Y_A} > {Y_B} > {Y_C}$

$ A, B,$ and $C$ graph are for steel, brass and rubber respectively.
 

i.e. $\ell \propto\left(1 / r^{2}\right)$

i.e. for same load, thickest wire (radius is more) will show minimum elongation.

From the graph, minimum elongation is shown by graph $OD.$

Hence OD must be thickness wire's graph.

Let $r_1$ and $r_2$ be the radii of two wires, then

$\frac{r_1}{r_2}=\frac{2}{1}$

(given)

We know that,stress $=\frac{F}{A}$

As force applied is same, so

$\frac {(Stress)_1}{(Stress)_2}$=$\frac {A_2}{A_1}=\frac {\pi r_2^2}{\pi r_1^2}$

$=\left(\frac{r_2}{r_1}\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}$

As, thermal strain $=$ strain caused by

$\alpha \Delta \theta=\frac{\Delta l}{l}$

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

$\Rightarrow \Delta \theta=\frac{5000}{4 \times 10^{12} \times 10^{-4} \times 2.5 \times 10^{-6}}=5^{\circ} C$

Note $\Delta \theta$ in $K$ is same as $\Delta \theta$ in ${ }^{\circ} C$.

$\therefore 20-T=5^{\circ} C \text { or } T=15^{\circ} C$

As the steel rail is contrained from expansion, the expansion pressure causes stress in the steel rail.

Thermal stress depends upon coefficient of expansion $\alpha$ and rise of temperature $\Delta T$.

$\therefore$ 'Thermal stress, $\sigma \propto \alpha \Delta T$

$\Rightarrow \sigma=Y \cdot \alpha \cdot \Delta T$ (where, $Y=$ Young's modulus)

$\therefore \sigma =2 \times 10^{11} \times 11 \times 10^{-5} \times(40-25)$

$=3.3 \times 10^7 \,Pa$

Kinetic energy of block is converted into potential energy of stretched wire.

$\Rightarrow \quad \frac{1}{2} m v^2=\frac{1}{2} \text { Stress } \times \text { Strain } \times \text { Volume }$

$\Rightarrow m v^2=\frac{\text { Stress }}{\text { Strain }} \times(\text { Strain })^2 \times \text { Volume }$

$\Rightarrow m v^2=Y \times \frac{\Delta l^2}{L^2} \times A \times L \Rightarrow \Delta l^2=\frac{m v^2 L}{A Y}$

$\therefore \quad \Delta l=v \sqrt{\frac{m L}{A Y}}$

So, distance travelled by block in stretching the wire is $\Delta l=v \sqrt{\frac{m L}{A Y}}$.

Potential energy versus separation plot for two atoms in a solid is as shown below.

From above graph, we see that as temperature rises $\left(T_3 > T_2 > T_1\right)$, total energy of atoms also increases $\left(E_3 > E_2 > E_1\right)$

As a result, mean separation between atoms also increases $\left(r_3 > r_2 > r_1\right)$.

Hence, crystalline solids in general expands on heating.

Adiabatic bulk modulus for an ideal gas is

$\quad K=\frac{d p}{-d V / V}=-V\left(\frac{d p}{d V}\right)$

$\quad=\gamma P \quad \dots(i)$

Now given,

$\quad K \propto p^n \quad \dots(ii)$

$\therefore$ From Eqs. $(i)$ and $(ii)$, we have

$\Rightarrow \quad n=1$

$\mathrm{~A}=\pi \mathrm{r}^2=16 \pi \times 10^{-10} \mathrm{~m}^2$

$\text { Strain }=\frac{\Delta \ell}{\ell}=\frac{\mathrm{F}}{\mathrm{AY}}=\frac{\mathrm{T}}{\mathrm{AY}}$

$=\frac{80 / 3}{16 \pi \times 10^{-10} \times 2 \times 10^{11}}=\frac{1}{12 \pi}$

$\alpha=12$

$\sigma=\frac{\mathrm{T}}{\mathrm{A}}=\frac{\mathrm{mg}}{\mathrm{A}}$

$\frac{(\sigma \mathrm{A} \ell) \mathrm{g}}{\mathrm{A}}$

$\Rightarrow \ell=\frac{\sigma}{\rho \mathrm{g}}=\frac{1.2 \times 10^8 \times 3}{6 \times 10^4 \times 10}=600$

$ \mathrm{Y}=\frac{\frac{\mathrm{F}}{\mathrm{r}^2}}{\frac{\ell}{\mathrm{L}}} $

$ \mathrm{F}=\mathrm{Y} \pi \mathrm{r}^2 \times \frac{\ell}{\mathrm{L}} $  $...........(i)$

$ \mathrm{Y}=\frac{\frac{\pi \mathrm{r}^2 / 4}{\Delta \ell}}{\mathrm{L}} $

$ \mathrm{F}=\mathrm{Y} \frac{\Delta \ell}{\mathrm{L}} \times 2 \times \frac{\pi \mathrm{r}^2}{4} $

$ \text { From }(\mathrm{i}) $

$ \mathrm{Y} \pi \mathrm{r}^2 \frac{\ell}{\mathrm{L}}=\mathrm{Y} \frac{\Delta \ell}{\mathrm{L}} \frac{\pi \mathrm{r}^2}{2} $

$ \Delta \ell=2 \ell$

$ \Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{AY}} $

$ \mathrm{V}=\mathrm{A} \ell \Rightarrow \ell=\frac{\mathrm{V}}{\mathrm{A}} $

$ \Delta \ell=\frac{\mathrm{FV}}{\mathrm{A}^2 \mathrm{Y}}$

$Y$ & $V$ is same for both the wires

$ \Delta \ell \propto \frac{\mathrm{F}}{\mathrm{A}^2} $

$ \frac{\Delta \ell_1}{\Delta \ell_2}=\frac{\mathrm{F}_1}{\mathrm{~A}_1^2} \times \frac{\mathrm{A}_2^2}{\mathrm{~F}_2} $

$ \Delta \ell_1=\Delta \ell_2 $

$ \mathrm{~F}_1 \mathrm{~A}_2^2=\mathrm{F}_2 \mathrm{~A}_1^2 $

$ \frac{\mathrm{F}_1}{\mathrm{~F}_2}=\frac{\mathrm{A}_1^2}{\mathrm{~A}_2^2}=\left(\frac{4}{1}\right)^2=16$

$ 30-T_1=3 \times a $

$ T_1=20 \mathrm{~N} $

$ \text { strain }=\frac{\text { stress }}{Y} $

$ =2 \times 10^{-4}$

$\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{\mathrm{F}}{\mathrm{AY}}$

$\frac{\frac{\Delta \mathrm{L}_1}{\mathrm{~L}_1}}{\frac{\Delta \mathrm{L}_2}{\mathrm{~L}_2}}=\frac{\mathrm{F}_1}{\mathrm{~F}_2}=\frac{30}{10}=3$

using small angle approximation $\sin \theta=\theta$

$\theta=\frac{1}{100}$

$\therefore \quad \mathrm{T}=\frac{10}{\theta}$

$T=1000 \mathrm{~N}$

Change in length $\Delta \mathrm{L} \quad=2 \sqrt{\mathrm{x}^2+\mathrm{L}^2}-2 \mathrm{~L}$

$=2 \mathrm{~L}\left[1+\frac{\mathrm{x}^2}{2 \mathrm{~L}^2}-1\right]$

$\Delta \mathrm{L} =\frac{\mathrm{x}^2}{\mathrm{~L}}$

$\therefore$ Modulus of elasticity $=\frac{\text { stress }}{\text { strain }}$

$2 \times 10^{11}=\frac{10^3}{\mathrm{~A} \times \frac{\mathrm{x}^2}{\mathrm{~L}}} \times 2 \mathrm{~L}$

$\therefore \quad  \mathrm{A}=1 \times 10^{-4} \mathrm{~m}^2$

$=\frac{5}{500}+\frac{0.02}{2}=0.01+0.01$

$\frac{\Delta \mathrm{Y}}{\mathrm{Y}}  =0.02 \Rightarrow \% \frac{\Delta \mathrm{Y}}{\mathrm{Y}}=2 \%$

$\Delta \ell=\frac{200 \times 2}{2 \times 10^{-4} \times 10^{11}}=2 \times 10^{-5}=20 \mu$

$ -\left(\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)=\frac{\rho \mathrm{gh}}{\mathrm{B}}=\frac{1000 \times 10 \times 4000}{2 \times 10^9} $

$ =2 \times 10^{-2}[-\mathrm{ve} \text { sign represent compression }]$

$\Delta \mathrm{P}=-\beta \frac{\Delta \mathrm{V}}{\mathrm{V}}$

$\rho \mathrm{gh}=-\beta \frac{\Delta \mathrm{V}}{\mathrm{V}}$

$10^3 \times 10 \times \mathrm{h}=-9 \times 10^8 \times\left(-\frac{0.02}{100}\right)$

$\Rightarrow \mathrm{h}=18 \mathrm{~m}$

$\therefore m=\frac{\pi D^2 \sigma}{4 g}$

$=\frac{22}{7} \times \frac{\left(14 \times 10^{-3}\right)^2 \times 7 \times 10^5}{4 \times 9.8}$

$=11\,kg$

$\Rightarrow F =\frac{ YA }{\ell} \Delta \ell$

$\left(\frac{ A \Delta \ell}{\ell}\right)_1=\left(\frac{ A \Delta \ell}{\ell}\right)_2$ $\Rightarrow \frac{\Delta \ell_2}{\Delta \ell_1}=\frac{ A _1}{ A _2} \times \frac{\ell_2}{\ell_1}$

$\Rightarrow \frac{\Delta \ell_2}{0.2}=\frac{1}{2.4 \times 2.4} \times \frac{2}{1}$ $\Rightarrow \Delta \ell_2=6.9 \times 10^{-2}\,mm$

$=4 \times \frac{10}{4}=10\,N$

$\Delta L =\frac{ FL }{ AY }$

$=\frac{10 \times 6}{3 \times 10^{-6} \times 2 \times 10^{11}}$

$=10^{-4}\,m =0.1\,mm$

$\frac{\Delta L_{ A }}{\Delta L _{ B }}=\frac{ A _{ B }}{ A _{ A }} \frac{Y_{ B }}{Y_{ A }}=12$

When $T _1=100\,N$, Extension $=\ell_1-\ell_0$

When $T _2=120\,N$, Extension $=\ell_2-\ell_0$

Then $100= K \left(\ell_1-\ell_0\right)$

And $120= K \left(\ell_2-\ell_0\right)$

$\frac{1}{2} \Rightarrow \frac{5}{6}=\frac{\ell_1-\ell_0}{\ell_2-\ell_0}$

$5 \ell_2-5 \ell_0=6 \ell_1-6 \ell_0$

$\ell_0=6 \ell_1-5 \ell_2$

$\ell_0=6 \ell_1-5\left(\frac{11 \ell_1}{10}\right)$

$\ell_0=6 \ell_1-\frac{11 \ell_1}{2}$

$\ell_0=\frac{\ell_1}{2}$

$\therefore x=2$

$\Rightarrow \frac{2}{1}=\frac{2 \ell^{\prime}}{\ell} \Rightarrow \ell^{\prime}=\ell$