Physics

Conservation of Mechanical Energy is linked to newton's second law of motion (F = ma) in combination with work energy theorem. The work energy theorem states that total work done on an object equals to change in kinetic energy. This theorem is used to prove that the sum of kinetic energy and potential energy remains constant.

$\frac{GM}{{\left(3R/2\right)}^2}=\frac{GM}{R^3}*r$

$\Rightarrow OA=\frac{4R}{9}=r$

$AB=R-\frac{4R}{9}=\frac{5R}{9}\Rightarrow OA:AB=4:5=x:y\Rightarrow x=4$

With the help of conservation of volume, we can write

$27*\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3\Rightarrow R=3r.......\left(1\right)$

With the help of conservation of charge, we can write

Q = 27 q.(2)

Potential energy of single drop = U1 = $\frac{q^2}{8\pi {\epsilon }_{0}r}$ $\frac{{}^{}}{{}_{}}$

Potential energy of bigger drop = $U_2=\frac{Q^2}{8\pi {\epsilon }_{0}R}=\frac{27*27*q^2}{8\pi {\epsilon }_0\left(3r\right)}=243\left(\frac{q^2}{8\pi {\epsilon }_{0}r}\right)=243U_1$ ${}_{}\frac{{}^{}}{{}_{}}\frac{{}^{}}{{}_{}}\frac{{}^{}}{{}_{}}{}_{}$

$\Rightarrow \frac{U_2}{U_1}=243$

$l=\frac{90-30}{4000}=15mA$

$I_1=\frac{30}{5000}=6mA\&l_2=9mA$

$v=\omega \sqrt[]{A^2-s^2}\Rightarrow \frac{A\omega }{2}=\omega \sqrt[]{A^2-s^2}\Rightarrow s=\frac{\sqrt[]{3}A}{2}\Rightarrow x=3$

$\Delta Q=\Delta U+\Delta W$

^{$\Rightarrow Q=\Delta U+\frac{Q}{5}\Rightarrow \Delta U=\frac{4Q}{5}=n{C}_v\Delta T\Rightarrow \frac{4Q}{5}=\frac{5R}{2}\Delta T\Rightarrow \Delta T=\frac{8Q}{25R}$}

^{$Q=nc\Delta T=1*C*\frac{80}{25R}\Rightarrow C=\frac{25R}{8}\Rightarrow x=25$}

Band Width = 2 * n * the highest modulation frequency

$\Rightarrow n=\frac{90k{H}_z}{2*5kHz}=9$

Maximum Distance = 3 (x + x) = 6x = 3 * 2x = 3 * 50 = 150 cm

Case I :- $2?-?=\frac{1}{2}m{v}_1^2........\left(i\right)$ $\frac{}{}{}_{}^{}$

Case II :- $10?-?=\frac{1}{2}m{v}_2^2........\left(ii\right)$ $\frac{}{}{}_{}^{}$

$\Rightarrow \frac{1}{9}=\frac{v_1^2}{v_2^2}$

$\Rightarrow v_1:v_2=1:3$

x = 1

$V=k{T}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

$T{V}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=K$

$\Rightarrow \gamma -1=-3/2$

$\Rightarrow \gamma =-1/2$

Work done = $\frac{nR\Delta T}{-y+1}=\frac{1*R*90}{3/2}=6R$

n = 60