⇒Oscillations

Correct Option is (D)

Time period of simple pendulum $\mathrm{T}=2\pi \sqrt{\frac{l}{\mathrm{g}}}$

$\Rightarrow \mathrm{T}\propto \frac{1}{\sqrt{\mathrm{g}}}$

Net downward force acting on the bob inside the liquid $=$ Weight of bob $-$ Upthrust

$\Rightarrow \mathrm{Vpg}-\mathrm{V}\frac{\rho }{\rho }\mathrm{g}=\frac{7}{8}\mathrm{V}\rho \mathrm{g}$

The value of $g$ inside the liquid will be $\frac{g}{8}$

$\therefore$ Time period in liquid ${\mathrm{T}}_1=\frac{1}{2\pi }\sqrt{\frac{\frac{l}{7}}{8}}=\sqrt{\frac{8}{7}}\mathrm{T}$

Correct Option is (B)

Time period of pendulum: $T=2\pi \sqrt{\frac{L}{g}}$

When lift is accelerated downward with acceleration $\frac{\mathrm{g}}{4}$,

$\begin{array}{rl}& g=g-\frac{g}{4}\\ & g=\frac{3g}{4}\end{array}$

$\therefore$ New Time period will be

$\begin{array}{rl}& T_1=2\pi \sqrt{\frac{4L}{3g}}\\ & T_1=2\pi \frac{2}{\sqrt{3}}\sqrt{\frac{L}{g}}\\ & T_1=\frac{2}{\sqrt{3}}T\end{array}$

Correct Option is (C)

$\begin{array}{rl}\mathrm{v}& =\omega \sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}\\ & \text{ At }\mathrm{x}=\frac{\mathrm{a}}{2},\\ \therefore \mathrm{v}& =\omega \sqrt{{\mathrm{a}}^2-\frac{{\mathrm{a}}^2}{4}}\\ & =\omega \frac{\sqrt{3\mathrm{a}}}{2}\\ & =\frac{2\pi }{\mathrm{T}}\times \frac{\sqrt{3}\mathrm{a}}{2}.....(\because \omega =\frac{2\pi }{T})\\ \therefore \mathrm{v}& =\frac{\pi \mathrm{a}\sqrt{3}}{\mathrm{T}}\end{array}$

Correct Option is (B)

Time period of a simple pendulum is $\mathrm{T}=2\pi \sqrt{\left(\frac{l}{\mathrm{a}}\right)}$

In stationary lift, the value of acceleration is $\mathrm{a}=\mathrm{g}$

$\mathrm{T}=2\pi \sqrt{\left(\frac{l}{\mathrm{g}}\right)}$

When the lift is accelerating in an upward direction, there is a pseudo force acting in a downward direction.

$\begin{array}{rl}& \therefore \mathrm{ma}=\mathrm{mg}+\frac{\mathrm{mg}}{3}\\ & \therefore \mathrm{a}=\mathrm{g}+\frac{\mathrm{g}}{3}\\ & \therefore \mathrm{a}=\frac{4\mathrm{g}}{3}\end{array}$

$\therefore$ Period for a pendulum in accelerating lift is

$\begin{array}{rl}{\mathrm{T}}^{\prime }& =2\pi \sqrt{\frac{l}{\mathrm{a}}}=2\pi \sqrt{\frac{3l}{4\mathrm{g}}}\\ \therefore {\mathrm{T}}^{\prime }& =\frac{\sqrt{3}}{2}\left(2\pi \sqrt{\frac{l}{\mathrm{g}}}\right)\\ \therefore {\mathrm{T}}^{\prime }& =\frac{\sqrt{3}}{2}\mathrm{T}\end{array}$

Correct Option is (B)

$\begin{array}{rl}& \text{ Given }:\mathrm{K}\cdot \mathrm{E}=\mathrm{P}\cdot \mathrm{E}+\frac{25}{100}\cdot \mathrm{P}\cdot \mathrm{E}\\ & \mathrm{K}\cdot \mathrm{E}=\mathrm{P}\cdot \mathrm{E}+\frac{1}{4}\mathrm{P}\cdot \mathrm{E}\\ & \text{ We know K.E }=\frac{1}{2}\mathrm{m}{\omega }^2({\mathrm{A}}^2-{\mathrm{x}}^2)\text{ and P.E }=\frac{1}{2}\mathrm{m}{\omega }^2{\mathrm{x}}^2\\ & \therefore \mathrm{K}\cdot \mathrm{E}=\frac{5}{4}\mathrm{P}\cdot \mathrm{E}\\ & \frac{1}{2}\mathrm{m}{\omega }^2({\mathrm{A}}^2-{\mathrm{x}}^2)=\frac{5}{4}\left(\frac{1}{2}\mathrm{m}{\omega }^2{\mathrm{x}}^2\right)\\ & {\mathrm{A}}^2-{\mathrm{x}}^2=\frac{5}{4}{\mathrm{x}}^2\\ & {\mathrm{A}}^2=\frac{9}{4}{\mathrm{x}}^2\\ \therefore & \mathrm{A}=\frac{3}{2}\mathrm{x}\\ \therefore & \mathrm{x}=\mathrm{A}\times \frac{2}{3}=2\mathrm{cm}\end{array}$