Consider the diagram,

Let the log be pressed and let the vertical displacement at the equilibrium position be $x_0$.

At equilibrium,

$$m g=\text { buoyant force }=\left(\rho A x_0\right) g \quad\left[\because m=V \rho=\left(A x_o\right) \rho\right]$$

When it is displaced by a further displacement $x$, the buoyant force is $A\left(x_0+x\right) \rho g$

$$\begin{aligned} \therefore \quad \text { Net restoring force } & =\text { Buoyant force }- \text { Weight } \\ & =A\left(x_0+x\right) \rho g-m g \end{aligned}$$

$$=(A \rho g) x \quad\left[\because m g=\rho A x_0 g\right]$$

As displacement $x$ is downward and restoring force is upward, we can write

$$\begin{aligned} F_{\text {restoring }} & =-(A \rho g) x \\ & =-k x\end{aligned}$$

where $$k=\text { constant }=A \rho g$$

So, the motion is SHM

($\because F \propto -x$)

$$\text { Now, } \quad \text { Acceleration } a=\frac{F_{\text {restoring }}}{m}=-\frac{k}{m} x$$

$$\begin{aligned} \text{Comparing with}\quad a & =-\omega^2 x \\ \Rightarrow \quad \omega^2 & =\frac{k}{m} \Rightarrow \omega=\sqrt{\frac{k}{m}} \\ \Rightarrow \quad \frac{2 \pi}{T} & =\sqrt{\frac{k}{m}} \Rightarrow T=2 \pi \sqrt{\frac{m}{k}} \\ \Rightarrow \quad T & =2 \pi \sqrt{\frac{m}{A \rho g}} \end{aligned}$$

Consider the diagram shown below

Let us consider an infinitesimal liquid column of length $d x$ at a height $x$ from the horizontal line.

If $\rho=$ density of the liquid

$$A=\text { cross-sectional area of } V \text {-tube }$$

PE of element $d x$ will be given as

$$P E=d m g x=(A p d x) g x \quad[\because d m=p V=p A d x]$$

where $$\quad A p d x=d m=\text { mass of element } d x$$

$\therefore$ Total PE of the left column

$$\begin{aligned} & =\int_0^{h_1} A \rho g x d x \\ & =A \rho g \int_0^{h_1} x d x \\ & =A \rho g\left|\frac{x^2}{2}\right|_0^{h_1}=A \rho g \frac{h_1^2}{2} \end{aligned}$$

$$\begin{aligned} &\text { But, }\\ &h_1=l \sin 45^{\circ}\\ &\therefore \quad P E=\frac{A \rho g}{2} l^2 \sin ^2 45^{\circ} \quad \text{... (i)} \end{aligned}$$

In a similar way,

$$ \begin{aligned} \text { PE of right column } & =\frac{A \rho g}{2} l^2 \sin ^2 45^{\circ} \\ \text { Total PE } & =\frac{A \rho g}{2} l^2 \sin ^2 45^{\circ}+\frac{A \rho g}{2} l^2 \sin ^2 45^{\circ} \quad \text{... (ii)}\\ & =2 \times \frac{1}{2} A \rho g l^2\left(\frac{1}{\sqrt{2}}\right)^2=\frac{A \rho g l^2}{2}\quad \text{... (iii)} \end{aligned}$$

If due to pressure difference created y element of left side moves on the right side, then

liquid present in the left arm $=l-y$

liquid present in the right arm $=l+y$

$$\begin{array}{ll} \therefore \quad & \text { Total PE }=A \rho g(l-y)^2 \sin ^2 45^{\circ}+A \rho g(l+y)^2 \sin ^2 45^{\circ} \\ & \text { Changes in } \mathrm{PE}=(\mathrm{PE})_{\text {tinal }}-(\mathrm{PE})_{\text {initial }} \end{array}$$

$$\begin{aligned} \text{or}\quad \Delta \mathrm{PE} & =\frac{A \rho g}{2}\left[(l-y)^2+(l+y)^2-l^2\right] \\ & =\frac{A \rho g}{2}\left[l^2+y^2-2 l y+l^2+y^2+2 l y-l^2\right] \\ & =\frac{A \rho g}{2}\left[2\left(l^2+y^2\right)\right] \\ & =A \rho g\left(l^2+y^2\right)\quad \text{... (iv)} \end{aligned}$$

If $v$ is the change in velocity of the total liquid column, then change in KE

$$\begin{aligned} \Delta \mathrm{KE} & =\frac{1}{2} m v^2 \\ \text{But}\quad m & =A \rho(2 l) \\ \therefore \quad \Delta \mathrm{KE} & =\frac{1}{2} A \rho 2 l v^2=A \rho l v^2\quad \text{... (v)} \end{aligned}$$

From Eqs. (iv) and (v),

$$\Delta \mathrm{PE}+\Delta \mathrm{KE}=A \rho g\left(l^2+y^2\right)+A \rho l v^2\quad \text{... (vi)}$$

System being conservative.

$\therefore$ Change in total energy $=0$

From Eq. (vi), $$\quad A \rho g\left(l^2+y^2\right)+A \rho l v^2=0$$

Differentiating both sides with respect to time $(t)$, we get

$$\begin{aligned} & A \rho g\left[0+2 y \frac{d y}{d t}\right]+A \rho l(2 v) \frac{d v}{d t}=0 \\ \text{But,}\quad & \frac{d y}{d t}=v \text { and } \frac{d v}{d t}=a \quad \text{[acceleration]} \end{aligned}$$

$$\begin{array}{l} \Rightarrow \quad A \rho g(2 y v)+A \rho l(2 v) a=0 \\ \Rightarrow \quad (g y+l a) 2 A \rho v=0 \\ 2 A \rho v=\text { constant and } 2 A \rho v \neq 0 \end{array}$$

$$ \begin{aligned} \therefore \quad l a+g y & =0 \\ a+\left(\frac{g}{l}\right) y & =0 \\ \text{or}\quad \frac{d^2 y}{d t^2}+\left(\frac{g}{l}\right) y & =0 \end{aligned}$$

$$\begin{aligned} &\text { This is the standard differential equation for SHM of the form }\\ &\begin{aligned} \frac{d^2 y}{d t^2}+\omega^2 y & =0 \\ \therefore\quad \omega & =\sqrt{\frac{g}{l}} \\ \therefore\quad T=\frac{2 \pi}{\omega} & =2 \pi \sqrt{\frac{l}{g}} \end{aligned} \end{aligned}$$

Consider the situation shown in the diagram.

The gravitational force on the particle at a distance $r$ from the centre of the earth arises entirely from that portion of matter of the earth in shells internal to the position of the particle. The external shells exert no force on the particle.

More clearly, let $g^{\prime}$ be the acceleration at $P$.

So, $$g^{\prime}=g\left(1-\frac{d}{R}\right)=g\left(\frac{R-d}{R}\right)$$

$$\begin{aligned} &\begin{aligned} &\text { From figure, }\quad R-d =y \\ \Rightarrow \quad g^{\prime} & =g \frac{y}{R} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { Force on body at } p \text {, }\\ &F=-m g^{\prime}=\frac{-m g}{R} y\quad \text{... (i)} \end{aligned}$$

$$\begin{aligned} &\Rightarrow \quad F \propto-y \quad \text { [where, } y \text { is distance from the centre] }\\ &\text { So, motion is SHM. } \end{aligned}$$

For time period, we can write Eq. (i)

As $$m a=-\frac{M g}{R} y \Rightarrow a=-\frac{g}{R} y$$

Comparing with $a=-\omega^2 y$

$$\begin{aligned} \omega^2 & =\frac{g}{R} \\ \Rightarrow \quad\left(\frac{2 \pi}{T}\right) & =\frac{g}{R} \Rightarrow T=2 \pi \sqrt{\frac{R}{g}} \end{aligned}$$

Consider the diagram,

Let us assume $t=0$ when $\theta=\theta_0$, then $\theta=\theta_0 \cos \omega t$

Given a seconds pendulum $\omega=2 \pi \Rightarrow \theta=\theta_0 \cos 2 \pi t\quad \text{... (i)}$

At time $t_1$ let $\theta=\theta_0 / 2$

$$\begin{aligned} \therefore \quad \cos 2 \pi t_1 & =1 / 2 \Rightarrow t_1=\frac{1}{6} \quad\left[\because \cos 2 \pi t_1=\cos \frac{\pi}{3} \Rightarrow 2 \pi t_1=\frac{\pi}{3}\right] \\ \frac{d \theta}{d t} & =-\left(\theta_0 2 \pi\right) \sin 2 \pi t \quad \text{[from Eq. (i)]} \end{aligned}$$

$$\begin{aligned} \text { At } \quad t & =t_1=\frac{1}{6} \\ \frac{d \theta}{d t} & =-\theta_0 2 \pi \sin \frac{2 \pi}{6}=-\sqrt{3} \pi \theta_0 \end{aligned}$$

Negative sign shows that it is going left.

Thus, the linear velocity is

$$u=-\sqrt{3} \pi \theta_0 l \text { perpendicular to the string. }$$

The vertical component is

$$u_y=-\sqrt{3} \pi \theta_0 l \sin \left(\theta_0 / 2\right)$$

and the horizontal component is

$$u_x=-\sqrt{3} \pi \theta_0 l \cos \left(\theta_0 / 2\right)$$

At the time it snaps, the vertical height is

$$H^{\prime}=H+l\left(1-\cos \left(\theta_0 / 2\right)\right)\quad \text{... (ii)}$$

$$\begin{aligned} &\text { Let the time required for fall be } t \text {, then }\\ &H^{\prime}=u_y t+(1 / 2) g t^2 \quad \text { (notice } g \text { also in the negative direction) } \end{aligned}$$

or $$\frac{1}{2} g t^2+\sqrt{3} \pi \theta_0 l \sin \frac{\theta_o}{2} t-H^{\prime}=0$$

$\therefore$ $$t=\frac{-\sqrt{3} \pi \theta_0 l \sin \frac{\theta_0}{2} \pm \sqrt{3 \pi^2 \theta_0^2 l^2 \sin ^2 \frac{\theta_0}{2}+2 g H^{\prime}}}{g}$$

$$=\frac{-\sqrt{3} \pi l \frac{\theta_0^2}{2} \pm \sqrt{3 \pi^2\left(\frac{\theta_0{ }^4}{4}\right) l^2+2 g H^{\prime}}}{g}\left[\because \sin \frac{\theta_0}{2} \approx \frac{\theta_0}{2} \text { for small angle }\right]$$

$$\begin{aligned} &\text { Given that } \theta_0 \text { is small, hence neglecting terms of order } \theta_0^2 \text { and higher }\\ &t=\sqrt{\frac{2 H^{\prime}}{g}} \quad \text{[from Eq. (iii)]} \end{aligned}$$

Now, $$H^{\prime} \approx H+l(1-1) \quad\left[\therefore \cos \theta_0 / 2 \approx 1\right]$$

$$ \begin{aligned} &\begin{aligned} & =H \quad \text { [from Eq. (ii)] }\\ \Rightarrow \quad t & =\sqrt{\frac{2 H}{g}} \end{aligned}\\ \end{aligned}$$

$$\begin{aligned} &\text { The distance travelled in the } x \text {-direction is } u_x t \text { to the left of where the bob is snapped }\\ &\begin{gathered} X=U x t=\sqrt{3} \pi \theta_0 l \operatorname{Cos}\left(\frac{\theta_0}{2}\right) \sqrt{\frac{2 H}{g}} \mathrm{~s} \\ \text { As } \theta_0 \text { is small } \Rightarrow \cos \left(\frac{\theta_0}{2}\right) \approx 1 \\ X=\sqrt{3} \pi \theta_0 l \sqrt{\frac{2 H}{g}}=\sqrt{\frac{6 H}{g}} \theta_0 l \pi \end{gathered} \end{aligned}$$

At the time of snapping, the bob was at a horizontal distance of $l \sin \left(\theta_0 / 2\right) \approx l \frac{\theta_0}{2}$ from $A$.

Thus, the distance of bob from $A$ where it meets the ground is

$$\begin{aligned} & =\frac{l \theta_0}{2}-X=\frac{l \theta_0}{2}-\sqrt{\frac{6 H}{g}} \theta_0 l \pi \\ & =\theta_0 l\left(\frac{1}{2}-\pi \sqrt{\frac{6 H}{g}}\right) \end{aligned}$$