Answer (d) We know that, Breaking stress $=\frac{\text { Breaking force }}{\text { Area of cross-section }}$

When length of the wire changes, area of cross-section remains same.

Hence, breaking force will be same when length changes.

Answer (d) We know that with increase in temperature length of a wire changes as

$$ L_{t}=L_{o}(1+\alpha \Delta T) $$

where $\Delta T$ is change in the temperature, $L_0$ is original length, $\alpha$ is coefficient of linear expansion and $L_{t}$ is Length at temperature $T$.

Now we can write

$$ \Delta L=L_{t}-L_0=L_0 \alpha \Delta T $$

where $\alpha$ is coefficient of linear expansion.

Young’s modulus $(Y)=\frac{\text { Stress }}{\text { Strain }}=\frac{F L_0}{A \times \Delta L}=\frac{F L_0}{A L_0 \alpha \Delta T} \propto \frac{1}{\Delta T}$

As,

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

When temperature increases $\Delta T$ increases, hence, $Y$ decreases.

Answer (c) Consider the diagram where a spring is stretched by applying a load to its free end. Clearly the length and shape of the spring changes.

The change in length corresponds to longitudinal strain and change in shape corresponds to shearing strain.

Answer (b) We know that Young’s modulus

$$ \begin{aligned} & Y=\frac{\text { Stress }}{\text { Strain }}=\frac{F / A}{\Delta L / L}=F / A \times \frac{L}{\Delta L} \\ &=\frac{F}{\pi(D / 2)^{2}} \times \frac{L}{\Delta L}=\frac{4 F L}{\pi D^{2} \Delta L} \\ & \Rightarrow \quad D^{2}=\frac{4 F L}{\pi \Delta L Y} \Rightarrow D=\sqrt{\frac{4 F L}{\pi \Delta L Y}} \\ & \text { As } F \text { and } \frac{L}{\Delta L} \text { are constants. } \end{aligned} $$

Hence,

$$ D \propto \sqrt{\frac{1}{Y}} $$

Now, we can find ratio as $\frac{D_{\text {copper }}}{D_{\text {iron }}}=\sqrt{\frac{Y_{\text {iron }}}{Y_{\text {copper }}}}$

Thinking Process

We will assume the vertical displacement $x$ to be very small compared to $L$. . Change in the length will be calculated by difference of final total length and initial length $2 L$.

Answer (a) Consider the diagram below

Hence, change in length

$$ \begin{matrix} \Delta L & =B O+O C-(B D+D C) & \\ & =2 B O-2 B D \\ & =2[B O-B D] \\ & =2[(x^{2}+L^{2})^{1 / 2}-L] \\ & =2 L[(1+\frac{x^{2}}{L^{2}})^{1 / 2}-1] \\ & \approx 2 L[1+\frac{1}{2} \frac{x^{2}}{L^{2}}-1]=\frac{x^{2}}{L} & \\ \therefore \quad \text { Strain } & =\frac{\Delta L}{2 L}=\frac{x^{2} / L}{2 L}=\frac{x^{2}}{2 L^{2}} & {[\because x«L]} \end{matrix} $$

Answer (c) Consider the FBD diagram of the rectangular frame

Balancing vertical forces $2 T \sin \theta-m g=0 $

$2 T \sin \theta=m g $

Total horizontal force $=T \cos \theta-T \cos \theta=0$

Now from Eq. (i), $\quad T=\frac{m g}{2 \sin \theta}$

As $m g$ is constant

$$ \begin{aligned} \Rightarrow \quad T \propto \frac{1}{\sin \theta} & \Rightarrow T_{\max }=\frac{m g}{2 \sin \theta_{\text {min }}} \\ \sin \theta_{\text {min }}=0 & \Rightarrow \theta_{\text {min }}=0 \end{aligned} $$

No option matches with $\theta=0^{\circ}$

$$ \begin{aligned} & T_{\min }=\frac{m g}{2 \sin \theta_{\max }} \\ & \sin \theta_{\max }=1 \Rightarrow \theta=90^{\circ} \\ & (\text { since, } \sin \theta_{\max }=1) \end{aligned} $$

Matches with option (b)

Hence, tension is least for the case (b).

Note We should be careful when measuring the angle, it must be in the direction as given in the diagram.

Answer (d) Consider the diagram.

A mass $M$ is attached at the centre. As the mass is attached to both the rods, both rod will be elongated, but due to different elastic properties of material rubber changes shape also.

Thinking Process

We have to compare ultimate tensile strength for each material. Material having more ultimate tensile strength will be elastic over larger region.

Answer (c, $d)$

It is clear from the two graphs, the ultimate tensile strength for material (ii) is greater, hence material(ii) is elastic over larger region as compared to material (i). For material(ii) fracture point is nearer, hence it is more brittle.

Answer $(a, d)$

As shown in the diagram

Clearly, forces at each cross-section is $F$.

Now applying formula,

$$ \begin{aligned} \text { Stress } & =\frac{\text { Tension }}{\text { Area }}=\frac{F}{A} \\ \text { Tension } & =\text { Applied force }=F \end{aligned} $$

Answer $(b, d)$

Let the mass is placed at $x$ from the end $B$.

Let $T_{A}$ and $T_{B}$ be the tensions in wire $A$ and wire $B$ respectively.

For the rotational equilibrium of the system,

$$ \begin{aligned} & \sum \tau=0 \\ & \Rightarrow \quad T_{B} x-T_{A}(l-x)=0 \\ & \Rightarrow \quad \frac{T_{B}}{T_{A}}=\frac{l-x}{x} \\ & \text { Stress in wire } A=S_{A}=\frac{T_{A}}{a_{A}} \\ & \text { Stress in wire } B=S_{B}=\frac{T_{B}}{a_{B}} \end{aligned} $$

where $a_{A}$ and $a_{B}$ are cross-sectional areas of wire $A$ and $B$ respectively.

By question $a_{B}=2 a_{A}$

Now, for equal stress

$$ \begin{aligned} & \Rightarrow \quad \frac{T_{A}}{a_{A}}=\frac{T_{B}}{a_{B}} \Rightarrow \frac{T_{B}}{T_{A}}=\frac{a_{B}}{a_{A}}=2 \\ & \Rightarrow \quad \frac{l-x}{x}=2 \Rightarrow \frac{l}{x}-1=2 \\ & \Rightarrow \quad x=\frac{l}{3} \Rightarrow l-x=l-l / 3=\frac{2 l}{3} \end{aligned} $$

Hence, mass $m$ should be placed closer to $B$.

For equal strain, $\quad( strain )_A $= $( strain)_B $

$ \Rightarrow \quad \frac{(Y_{A})}{S_{A}}=\frac{Y_{B}}{S_{B}} \quad \text { (where } Y_{A} \text { and } Y_{B} \text { are Young modulii) } $

$ \Rightarrow \quad \frac{Y_{\text {steel }}}{T_{A} / a_{A}}=\frac{Y_{A l}}{T_{B} / a_{B}} $

$ \Rightarrow \quad \frac{Y_{\text {steel }}}{Y_{\text {Al }}}=\frac{T_{A}}{T_{B}} \times \frac{a_{B}}{a_{A}}=(\frac{x}{l-x})(\frac{2 a_{A}}{a_{A}}) $

$ \Rightarrow \quad \frac{200 \times 10^{9}}{70 \times 10^{9}}=\frac{2 x}{l-x} \Rightarrow \frac{20}{7}=\frac{2 x}{l-x} $

$ \Rightarrow \quad \frac{10}{7}=\frac{x}{l-x} \Rightarrow 10 l-10 x=7 x $

$ \Rightarrow \quad 17 x=10 l \quad \Rightarrow \quad x=\frac{10 l}{17} $

$ l-x=l-\frac{10 l}{17}=\frac{7 l}{17}$

Hence, mass $m$ should be placed closer to wire $A$.

Answer $(a, d)$

As an ideal liquid is not compressible.

Hence, change in volume, $\quad \Delta V=0$

$$ \begin{aligned} & \text { Bulk modulus } B=\frac{\text { Stress }}{\text { Volume strain }}=\frac{F / A}{\Delta V / V}=\frac{F}{A} \times \frac{V}{\Delta V}=\infty \\ & \text { Compressibility, } K=\frac{1}{B}=\frac{1}{\infty}=0 \end{aligned} $$

As there is no tangential (viscous) force exists in case of an ideal fluid, hence, shear modulus $=0$.

Answer $(a, d)$

Consider the diagram where a deforming force $F$ is applied to the combination.

where $F$ is tension in each wire and $A$ is cross-section area of each wires.

As $F$ and $A$ are same for both the wire, hence, stress will be same for both the wire.

As,

$( Strain )_{steel } $ = $\frac{ Stress }{Y _ {steel }} $,

$(Strain ) _ {copper} $= $\frac{ Stress }{Y _ {copper }} $

$Y_{\text {steel }} \neq Y_{\text {copper }}$

Hence, the two wires will have differents strain.

Answer Young’s modulus $(Y)=\frac{\text { Stress }}{\text { Longitudinal strain }}$

For same longitudinal strain,

But

$\therefore$

$Y \propto \text { stress }$

$\frac{Y _ {\text {steel }}}{Y _ {\text {rubber }}} =\frac{(\text { stress }) _ {\text {steel }}}{(\text { stress }) _ {\text {rubber }}}$

Therefore, from Eq. (i),

$ \frac{(\text { stress }) _ {\text {steel }}}{(\text { stress }) _ {\text {rubber }}}>1 $ $ \Rightarrow \quad(\text { stress }) _ {\text {steel }}>\text { (stress) } _ {\text {rubber }}$

Answer Stress $=\frac{\text { Magnitude of internal reaction force }}{\text { Area of cross }- \text { section }}$

Therefore, stress is a scalar quantity not a vector quantity.

Answer Work done in stretching a wire is given by $W=\frac{1}{2} F \times \Delta l$

[where $F$ is applicd force and $\Delta l$ is extension in the wire] As springs of steel and copper are equally stretched. Therefore, for same force $(F)$,

$$ W \propto \Delta l $$

$$ \text { Young’s modulus }(Y)=\frac{F}{A} \times \frac{l}{\Delta l} \Rightarrow \Delta l=\frac{F}{A} \times \frac{l}{Y} $$

As both springs are identical, $\quad \Delta l \propto \frac{1}{Y}$

From Eqs. (i) and (ii), we get $\quad W \propto \frac{1}{Y}$

$$ \begin{matrix} \therefore & \frac{W_{\text {steel }}}{W_{\text {copper }}}=\frac{Y_{\text {copper }}}{Y_{\text {steel }}}<1 & \text { (As, } .Y_{\text {steel }}>Y_{\text {copper }}) \\ \Rightarrow & W_{\text {steel }}<W_{\text {copper }} \end{matrix} $$

Therefore, more work will be done for stretching copper spring.

Answer Young’s modulus $(Y)=\frac{F}{A} \times \frac{l}{\Delta l}$

For a perfectly rigid body, change in length $\Delta l=0$

$$ Y=\frac{F}{A} \times \frac{l}{0}=\infty $$

Therefore, Young’s modulus for a perfectly rigid body is infinite( $(\infty)$.

Answer Bulk modulus $(K)=\frac{p}{\Delta V / V}=\frac{p V}{\Delta V}$

For perfectly rigid body, change in volume $\Delta V=0$

$\therefore \quad K=\frac{p V}{0}=\infty$

Therefore, bulk modulus for a perfectly rigid body is infinity $(\infty)$.

Thinking Process

In this problem, we have to apply Hooke’s law and then elongation in each wire will be compared.

Answer The situation is shown in the diagram.

Now, $\quad$ Young’s modulus $(Y)=\frac{f}{A} \times \frac{L}{l}$

For first wire,

$Y=\frac{f}{\pi r^{2}} \times \frac{L}{l}$

For second wire,

$Y=\frac{2 f}{\pi(2 r)^{2}} \times \frac{2 L}{l^{\prime}}$

$=\frac{f}{\pi r^{2}} \times \frac{L}{l^{\prime}}$

From Eqs. (i) and (ii),

$$ \frac{f}{\pi r^{2}} \times \frac{L}{l}=\frac{f}{\pi r^{2}} \times \frac{L}{l^{\prime}} $$ $$ l=l^{\prime} $$

( $\because$ Both wires are of same material, hence, Young’s modulus will be same.)

Thinking Process

Due to increase in temperature of the rod, length, increases. The equation of thermal expansion for linear expansion will be applied.

Answer Given, Young’s modulus of steel $Y=2.0 \times 10^{11} N / m^{2}$

Coefficient of thermal expansion $\alpha=10^{-5}{ }^{\circ} C^{-1}$

Length $l=1 m$

Area of cross-section $A=1 cm^{2}=1 \times 10^{-4} m^{2}$

Increase in temperature $\Delta t=200^{\circ} C-0^{\circ} C=200^{\circ} C$

Tension produced in steel rod $(F)=Y A \alpha \Delta t$

$$ \begin{aligned} & =2.0 \times 10^{11} \times 1 \times 10^{-4} \times 10^{-5} \times 200 \\ & =4 \times 10^{4} N \end{aligned} $$

Answer Given, Bulk modulus of rubber $(K)=9.8 \times 10^{8} N / m^{2}$

Density of sea water $(\rho)=10^{3} kg / m^{3}$

Percentage decrease in volume, $(\frac{\Delta V}{V} \times 100)=0.1 \Rightarrow \frac{\Delta V}{V}=\frac{0.1}{100}$

$\Rightarrow \quad \frac{\Delta V}{V}=\frac{1}{1000}$

Let the rubber ball be taken upto depth $h$.

$\therefore \quad$ Change in pressure $(p)=h \rho g$

$\therefore \quad$ Bulk modulus $(K)=|\frac{p}{\Delta V / V}|=\frac{h \rho g}{(\Delta V / V)}$

$\Rightarrow \quad h=\frac{K \times(\Delta V / V)}{\rho g}=\frac{9.8 \times 10^{8} \times \frac{1}{1000}}{10^{3} \times 9.8}=100 m$

Answer

$$ \begin{aligned} \text { Length of steel cable } l & =9.1 m \\ \text { Radius } r & =5 mm=5 \times 10^{-3} m \\ \text { Tension in the cable } F & =800 N \\ \text { Young’s modulus for steel } Y & =2 \times 10^{11} N / m^{2} \\ \text { Change in length } \Delta l & =? \\ \text { Young’s modulus }(Y) & =\frac{F}{A} \times \frac{l}{\Delta l} \Rightarrow \Delta l=\frac{F}{\pi r^{2}} \times \frac{l}{Y} \\ & =\frac{800}{3.14 \times(5 \times 10^{-3})^{2}} \times \frac{9.1}{2 \times 10^{11}} \\ & =4.64 \times 10^{-4} m \end{aligned} $$

Thinking Process

To solve this question, we have to resolve the force along and perpendicular to the inclined plane. Now, we can easily calculate the tensile and shearing stress.

Answer Consider the adjacent diagram.

Let the cross-sectional area of the bar be $A$. Consider the equilibrium of the plane aa’. A force $F$ must be acting on this plane making an angle $\frac{\pi}{2}-\theta$ with the normal $O N$. Resolving $F$ into components, along the plane $(F P)$ and normal to the plane.

$$ \begin{aligned} & F_{P}=F \cos \theta \\ & F_{N}=F \sin \theta \end{aligned} $$

Let the area of the face aa’ be $A^{\prime}$, then

$$ \begin{aligned} \frac{A}{A^{\prime}} & =\sin \theta \\ \therefore \quad \text { The tensile stress } & =\frac{A}{\sin \theta} \\ & =\frac{\text { Normal force }}{A / \sin \theta}=\frac{F \sin \theta}{A^{\prime}} \\ \text { Shearing stress } & =\frac{\text { Parallel force }}{\text { Area }} \\ & =\frac{F \cos \theta}{A / \sin \theta}=\frac{F}{A} \sin \theta \cdot \cos \theta \\ & =\frac{F}{2 A}(2 \sin \theta \cdot \cos \theta)=\frac{F}{2 A} \sin 2 \theta \end{aligned} $$

(a) For tensile stress to be maximum, $\sin ^{2} \theta=1$

$$ \begin{aligned} \Rightarrow & \sin \theta & =1 \\ \Rightarrow & \theta & =\frac{\pi}{2} \end{aligned} $$

(b) For shearing stress to be maximum,

$$ \begin{matrix} \Rightarrow & & \sin 2 \theta & =1 \\ \Rightarrow & 2 \theta & =\frac{\pi}{2} \\ \Rightarrow & & \theta & =\frac{\pi}{4} \end{matrix} $$

Note We must not apply the formula for stress directly, forces must be resolved.

Thinking Process

When the wire is hanging different part of the wire will be acted upon by different forces. To find the tension due to whole wire, consider a small element and find the tension there and then integrate for the whole wire.

Answer Consider the diagram when a small element of length $d x$ is considered at $x$ from the load $(x=0)$.

(a) Let $T(x)$ and $T(x+d x)$ are tensions on the two cross-sections a distance $d x$ apart, then-$-T(x+d x)+T(x)=d m g=\mu d x g$ (where $\mu$ is the mass/length).

$(\because d m=\mu d x)$

$$ \begin{matrix} \Rightarrow & & T(x) & =\mu g x+C \\ \text { At } x=0, T(0)=M g \Rightarrow & C & =M g \\ \therefore & & T(x) & =\mu g x+M g \end{matrix} $$

Let the length $d x$ at $x$ increase by $d r$, then

$$ \begin{aligned} \text { Young’s modulus } Y & =\frac{\text { Stress }}{\text { Strain }} \\ \frac{T(x) / A}{d r / d x} & =Y \\ \Rightarrow \quad \frac{d r}{d x} & =\frac{1}{Y A} T(x) \\ r & =\frac{1}{Y A} \int_0^{L}(\mu g x+M g) d x \\ & =\frac{1}{Y A}[\frac{\mu g x^{2}}{2}+M g x]_0^{L} \\ & =\frac{1}{Y A}[\frac{m g L^{2}}{2}+M g L] \end{aligned} $$

$$ \begin{matrix} {[\because d T=T(x+d x)-T(x)]} \\ \text { (on integrating) } \end{matrix} $$

( $m$ is the mass of the wire)

$$ A=\pi \times(10^{-3})^{2} m^{2} $$

$$ \begin{aligned} & Y=200 \times 10^{9} Nm^{-2} \\ & m=\pi \times(10^{-3})^{2} \times 10 \times 7860 kg \end{aligned} $$

$$ \begin{aligned} \therefore \quad r & =\frac{1}{2 \times 10^{11} \times \pi \times 10^{-6}} \quad[\frac{\pi \times 786 \times 10^{-3} \times 10 \times 10}{2}+25 \times 10 \times 10] \\ & =[196.5 \times 10^{-6}+3.98 \times 10^{-3}] \approx 4 \times 10^{-3} m \end{aligned} $$

(b) Clearly tension will be maximum at $x=L$

$$\therefore T =\mu g L+M g=(m+M) g$$

$$ \text{The yield force =(Yield strength Y) area} =250 \times 10^{6} \times \pi \times(10^{-3})^{2}=250 \times \pi N$$
$$ \text{At yield point} \quad[\because m=\mu L]$$
$$\Rightarrow \quad(m+M) g =250 \times \pi$$
$$\Rightarrow m =\pi \times(10^{-3})^{2} \times 10 \times 7860 \ll M$$
$$\therefore M g \approx 250 \times \pi$$
$$\text{Hence}, M =\frac{250 \times \pi}{10}=25 \times \pi \approx 75 kg$$.

Thinking Process

To solve this question, we have to consider a small element, find tension in the rod at this element and then calculate for the whole rod.

Answer Consider an element of width $d r$ at $r$ as shown in the diagram. Let $T(r)$ and $T(r+d r)$ be the tensions at $r$ and $r+d r$ respectively.

Net centrifugal force on the element $=\omega^{2} r d m$

(where $\omega$ is angular velocity of the rod)

$$ \begin{aligned} & =\omega^{2} r \mu d r \\ & \Rightarrow \quad T(r)-T(r+d r)=\mu \omega^{2} r d r \\ & \Rightarrow \quad-d T=\mu \omega^{2} r d r \\ & {[\because \text { Tension and centrifugal forces are opposite] }} \\ & \therefore \quad-\int_{T=0}^{T} d T=\int_{r=l}^{r=r} \mu \omega^{2} r d r \quad[\because T=0 \text { at } r=l] \\ & \Rightarrow \quad T(r)=\frac{\mu \omega^{2}}{2}(l^{2}-r^{2}) \end{aligned} $$

Let the increase in length of the element $d r$ be $\Delta r$

$$ \begin{aligned} & \text { So, } \quad \text { Young’s modulus } Y=\frac{\text { Stress }}{\text { Strain }}=\frac{T(r) / A}{\frac{\Delta r}{d r}} \\ & \therefore \quad \frac{\Delta r}{d r}=\frac{T(r)}{A}=\frac{\mu \omega^{2}}{2 Y A}(l^{2}-r^{2}) \\ & \therefore \quad \Delta r=\frac{1}{Y A} \frac{\mu \omega^{2}}{2}(l^{2}-r^{2}) d r \\ & \therefore \quad \Delta=\text { change in length in right part }=\frac{1}{Y A} \frac{\mu \omega^{2}}{2} \int_0^{l}(l^{2}-r^{2}) d r \\ & =(\frac{1}{Y A}) \frac{\mu \omega^{2}}{2}[l^{3}-\frac{l^{3}}{3}]=\frac{1}{3 Y A} \mu \omega^{2} L^{2} \\ & \therefore \quad \text { Total change in length }=2 \Delta=\frac{2}{3 Y A} \mu \omega^{2} l^{2} \end{aligned} $$

Thinking Process

Due to increase in temperature length of each side will change, hence, the angle corresponding to any vertex also changes.

Answer Consider the diagram shown

Let

$$ \begin{aligned} l_1 & =A B, l_2=A C, l_3=B C \\ \cos \theta & =\frac{l_3^{2}+l_1^{2}-l_2^{2}}{2 l_3 l_1} \end{aligned} $$

$$ \Rightarrow \quad 2 l_3 l_1 \cos \theta=l_3^{2}+l_1^{2}-l_2^{2} $$

Differentiating $2(l_3 d l_1+l_1 d l_3) \cos \theta-2 l_1 l_3 \sin \theta d \theta$

(assume $\angle A B C=\theta$ )

Now,

and

$$ =2 l_3 d l_1+2 l_1 d l_1-2 l_2 d l_2 $$

$d l_1=l_1 \alpha_1 \Delta t \quad$ (where, $\Delta t=$ change in temperature)

$d l_2=l_2 \alpha_1 \Delta t \Rightarrow d l_3=l_3 \alpha_2 \Delta t$

$l_1=l_2=l_3=l$

$(l^{2} \alpha_1 \Delta t+l^{2} \alpha_1 \Delta t) \cos \theta+l^{2} \sin \theta d \theta=l^{2} \alpha_1 \Delta t+l^{2} \alpha_1 \Delta t-l^{2} \alpha_2 \Delta t$

Putting

$$ \sin \theta d \theta=2 \alpha_1 \Delta t(1-\cos \theta)-\alpha_2 \Delta t $$

$$ \theta=60^{\circ} $$

(for equilateral triangle)

$$ d \theta \times \sin 60^{\circ}=2 \alpha_1 \Delta t(1-\cos 60^{\circ})-\alpha_2 \Delta t $$

$$ =2 \alpha_1 \Delta t \times \frac{1}{2}-\alpha_2 \Delta t=(\alpha_1-\alpha_2) \Delta t $$

$\Rightarrow \quad d \theta=$ change in the angle $\angle A B C$

$$ =\frac{(\alpha_1-\alpha_2) \Delta T}{\sin 60^{\circ}}=\frac{2(\alpha_1-\alpha_2) \Delta T}{\sqrt{3}} \quad(\because \Delta t=\Delta T \text { given }) $$

Thinking Process

In this question, the elastic torque is given we have to find the torque caused by the weight due to bending and equate with the given value.

Answer Consider the diagram according to the question, the bending torque on the trunk of radius $r$ of the tree $=\frac{Y \pi r^{4}}{4 R}$

where $R$ is the radius of curvature of the bent surface.

When the tree is about to buckle $W d=\frac{Y \pi r^{4}}{4 R}$

If $R \gg h$, then the centre of gravity is at a height $l \approx \frac{1}{2} h$ from the ground.

From $\quad \triangle A B C R^{2} \simeq(R-d)^{2}+(\frac{1}{2} h)^{2}$

if $d \ll R, \quad R^{2} \simeq R^{2}-2 R d+\frac{1}{4} h^{2}$

$\therefore \quad d=\frac{h^{2}}{8 R}$

If $\omega_0$ is the weight/volume

$$ \begin{aligned} & \qquad \frac{Y \pi r^{4}}{4 R}=\omega_0(\pi r^{2} h) \frac{h^{2}}{8 R} \quad[\because \text { Torque is caused by the weight }] \\ & \Rightarrow \quad h \simeq(\frac{2 Y}{\omega_0})^{1 / 3} r^{2 / 3} \\ & \text { Hence, critical height }=h=(\frac{2 Y}{\omega_0})^{1 / 3} r^{2 / 3} . \end{aligned} $$

Thinking Process

In this problem ,the given string is elastic and there is no dissipation of energy we can apply conservation of mechanical energy.

Answer Consider the diagram the stone is dropped from point $p$.

(a) Till the stone drops through a length $L$ it will be in free fall. After that the elasticity of the string will force it to a SHM. Let the stone come to rest instantaneously at $y$.

The loss in PE of the stone is the PE stored in the stretched string.

$$ m g y=\frac{1}{2} k(y-L)^{2} $$

$$ \begin{aligned} \Rightarrow \quad m g y & =\frac{1}{2} k y^{2}-k y L+\frac{1}{2} k L^{2} \Rightarrow \frac{1}{2} k y^{2}-(k L+m g) y+\frac{1}{2} k L^{2}=0 \\ y & =\frac{(k L+m g) \pm \sqrt{(k L+m g)^{2}-k^{2} L^{2}}}{k}=\frac{(k L+m g) \pm \sqrt{2 m g k L+m^{2} g^{2}}}{k} \end{aligned} $$

Retain the positive sign.

$$ \therefore \quad y=\frac{(k L+m g)+\sqrt{2 m g k L+m^{2} g^{2}}}{k} $$

(b) In SHM, the maximum velocity is attained when the body passes, through the “equilibrium, position” i.e., when the instantaneous acceleration is zero. That is $m g-k x=0$, where $x$ is the extension from $L$.

$\Rightarrow$

$$ m g=k x $$

Let the velocity be $v$. Then,

Now,

$$ \begin{aligned} & \frac{1}{2} m v^{2}+\frac{1}{2} k x^{2}=m g(L+x) \quad \text { (from conservation of energy) } \\ & \frac{1}{2} m v^{2}=m g(L+x)-\frac{1}{2} k x^{2} \end{aligned} $$

$$ \begin{aligned} \therefore \quad \frac{1}{2} m v^{2} & =m g(L+\frac{m g}{k})-\frac{1}{2} k \frac{m^{2} g^{2}}{k^{2}}=m g L+\frac{m^{2} g^{2}}{k}-\frac{1}{2} \frac{m^{2} g^{2}}{k} \\ \frac{1}{2} m v^{2} & =m g L+\frac{1}{2} \frac{m^{2} g^{2}}{k} \\ \therefore \quad v^{2} & =2 g L+m g^{2} / k \\ v & =(2 g L+m g^{2} / k)^{1 / 2} \end{aligned} $$

(c) When stone is at the lowest position i.e., at instantaneous distance $Y$ from $P$, then equation of motion of the stone is

$$ \frac{m d^{2} y}{d t^{2}}=m g-k(y-L) \Rightarrow \frac{d^{2} y}{d t^{2}}+\frac{k}{m}(y-L)-g=0 $$

Make a transformation of variables, $z=\frac{k}{m}(y-L)-g$

$$ \therefore \quad \frac{d^{2} z}{d t^{2}}+\frac{k}{m} z=0 $$

It is a differential equation of second order which represents SHM.

Comparing with equation $\frac{d^{2} z}{d t^{2}}+\omega^{2} z=0$

Angular frequency of harmonic motion $\omega=\sqrt{\frac{k}{m}}$

The solution of above equation will be of the type $z=A \cos (\omega t+\phi)$; where $\omega=\sqrt{\frac{k}{m}}$

$$ y=(L+\frac{m g}{k})+A^{\prime} \cos (\omega t+\phi) $$

Thus, the stone will perform SHM with angular frequency $\omega=\sqrt{k / m}$ about a point $y_0=L+\frac{m g}{k}$.