Physics

${\displaystyle \int I.dt}={\displaystyle \underset{0}{\overset{10}{\int }}\left(20+3t\right)dt}$

$={\left(20t\right)}_0^{10}+3{\left(\frac{t^2}{2}\right)}_0^{10}=350C$

$\frac{1}{20}=\left(1.5-1\right)\left(\frac{2}{R}\right)$

R = 20cm

$\frac{1}{f\text{'}}=\left(\frac{1.5}{1.6}-1\right)\left(\frac{2}{R}\right)$            

$=\frac{-1}{16}*\frac{2}{20}$                  

f' = – 160cm

Work Area

->W = $\frac{1}{2}$ * 600 * 4

= 1200J

Equivalent circuit

$i=\frac{9}{4+1}=1.8\text{?}A$

$\frac{i}{2}=0.9\text{?}A$

(V_P– V_Q) = 0.9 * 6 – 0.9 * 2

V_C = 3.6V

U =  $\frac{1}{2}$ CV² =   $\frac{1}{2}$ * 4 * 3.6 * 3.6mJ

= 25.92mJ

$I=Q_{0.\omega }$            

$=\frac{CV}{\sqrt[]{LC}}=V\sqrt[]{\frac{C}{L}}$

$=12\sqrt[]{\frac{100*10^{-6}}{10*10^{-3}}}$

= 1.2 A

$a_c=\frac{v^2}{r}$ , $\frac{{\left(a_c\right)}_1}{{\left(a_c\right)}_2}={\left(\frac{v_1}{v_2}\right)}^2$

$\frac{3}{4}={\left(\frac{v_1}{v_2}\right)}^2\to \frac{v_1}{v_2}=\sqrt[]{3}:2$

$\frac{h}{p_1}=\frac{h}{p_2}$

$\sqrt[]{2m_1{k}_1}=\sqrt[]{2m_2{k}_2}$

$\frac{k_2}{k_1}=\frac{m_1}{m_2}=1835$

S₁ = $\frac{1}{2}$ a (p – 1)²

S₂ = $\frac{1}{2}$ ap²

S₁ + S₂ = $\frac{1}{2}$ a [ (p – 1)² + p²] = $\frac{1}{2}$ at²

$t=\sqrt[]{2p^2+1-2p}$

$h=\frac{2Scos\theta }{\rho gR}$

as T­, S¯

The correct answer is Option (2).

  $\frac{GMm}{{\left(R+x\right)}^2}=\frac{GMm\left(R-x\right)}{R^3}$

->R³ = (R + x)² (R – x)

->R³ = (R²– x²) (R + x)

->x² + Rx – R² = 0

$x=\frac{-R\pm \sqrt[]{R+4R^2}}{2}$

$x=\frac{\left(\sqrt[]{5R}-1\right)}{2}R$