Class 11th

Work done = $\Delta KE$

$\Rightarrow -f=-x$

$\Rightarrow f=\frac{x}{y}$

Time (minimum) taken to cross the river

$t=\frac{W}{V_{s/r}}=\frac{200*{10}^{-3}}{2}$

$100*{10}^{-3}h$

$=0.1h=6minutes$

(b)  L₁₌  Length of rod =20.123cm

L₂₌  Length of rod = 18.1cm

Difference in length  

= 20.123-18.1

 =2.0 cm

$\begin{array}{l}{f}_l=\mu N=0.6*mgcos\theta \\ f_l=0.6*2*10*\frac{4}{5}=9.6N\end{array}$

Component of weight along incline.

$=mgsin\theta =2*10*\frac{3}{5}=12N$

$\Rightarrow mgsin\theta >f_{l}soT=2.4N$

Elastic potential energy stored in the spring

$U=\frac{1}{2}k{x}^2$

$=\frac{1}{2}Fx=\frac{F^2}{2k}$

$Hence\frac{U_1}{U_2}=\frac{k_2}{k_1}\left. [Fissameforbothsprings\right]$

$=\frac{500}{1200}=\frac{5}{12}$

$K.E.\propto t^2$

$\begin{array}{l}\Rightarrow \frac{1}{2}m{v}^2\propto t^2\\ \Rightarrow v\propto t\\ \Rightarrow \frac{dv}{dt}=constant\\ \Rightarrow F=ma=\frac{mdv}{dt}=constant\end{array}$

From work energy theorem

$W_{sp}+W_E=?K$

$-\frac{1}{2}k{x}_0^2+QE.x_0=0$

$?x_0=\frac{2QE}{k}$

$\Delta V_L=V_{0\gamma \Delta T}$

$\Rightarrow \Delta V_s=V_0*3*\frac{\gamma }{3}*\Delta T$

$\Rightarrow \Delta V_L-\Delta V_S=0$

Applying Bernoulli's theorem

$\Rightarrow P_0+\rho g2h+2\rho gh+0=P_0+\frac{1}{2}2\rho v^2$

$\Rightarrow V^2=4gh$

$\Rightarrow v=2\sqrt[]{gh}$

Since the water removed from the pond will be equal to the water displacement by the (man + boat) system, the level of water in the pond remains the same.