Physics

$M=ml$

$\Delta I=2\frac{l}{2}sin30^{?}$

$=\frac{l}{2}$

$M^{\prime }=mI/2$

$=M/2$

Power Consumed $=P=\frac{V^2}{R}$

$\frac{P_A}{P_B}=\frac{R_B}{R_A}$

$R_A=2R_B$

For Series Combination

$P_S=\frac{V^2}{3R_B}$

For Parallel Combination

$P_P=\frac{3V^2}{2R_B}$

$\frac{P_S}{P_P}=\frac{2}{9}$

From principle of homogeneity

$\left[F\right]=\left[\alpha t^2\right]=\left[\beta t\right]$

$\left[\alpha \right]=\frac{\left[F\right]}{\left[t^2\right]} and \left[\beta \right]=\frac{\left[F\right]}{\left[t\right]}$

$\therefore \left[\alpha \right]\left[t\right]=\left[\beta \right]$

$\therefore \frac{\alpha t}{\beta }= dimensionless$

In option (1),

$\frac{10}{15}=\frac{10}{5+R_D}$

The diode can conduct and have resistance  $R_D=10\Omega$  because diode have dynamic resistance. In that case bridge will be balanced.

The EM waves originate from an accelerating charge. The charge moving with uniform velocity produces steady state magnetic field.

Apply energy conservation,

$U_i+K_i=U_f+K_f$

$\Rightarrow -\frac{GMm}{R}+K_i=-\frac{GMm}{3R}+\frac{1}{2}m{v}^2$

$\Rightarrow -\frac{GMm}{R}+K_i=-\frac{GMm}{3R}+\frac{1}{2}*m*\frac{GM}{3R}$

$\Rightarrow K_i=-\frac{1}{6}\frac{GMm}{R}+\frac{GMm}{R}$

$K_i=\frac{5}{6}\frac{GMm}{R}$

At same temperature, curve with higher volume corresponds to lower pressure.

$V_3>V_2>V_1$

$\Rightarrow P_1>P_2>P_3$

(We draw a straight line parallel to volume axis to get this)

$T^{\prime }=2\pi \sqrt[]{\frac{{?}^{\prime }}{g}}$ where $t^{\prime }=\frac{\mathcal{l}}{2}$

$T=2\pi \sqrt[]{\frac{t}{g}}$

$T^{\prime }=\frac{x}{2}T$

$2\pi \sqrt[]{\frac{t}{2g}}=\frac{x}{2}2\pi \sqrt[]{\frac{t}{g}}$

$\frac{1}{\sqrt[]{2}}=\frac{x}{2}\Rightarrow x=\sqrt[]{2}$

Capacitive Reactance $X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}=\frac{1}{2*3.14*50*10*10^{-6}}$

$=\frac{1000}{3.14}$

$V_{rms}=210 V$

$i_{rms}=\frac{V_{rms}}{X_C}=\frac{210}{X_C}$

$Peak current =\sqrt[]{2}{i}_{ms}=\sqrt[]{2}*\frac{210}{1000}*3.14=0.932$

$?0.93 A$

According to modified Ampere's law

$?B.dI={\mu }_0\left(I_C+I_D\right)$

For Loop  $L_1{I}_C\ne 0$ and $I_D=0$

For Loop  $L_2{I}_C=0$ and $I_D\ne 0$

Due to  $KCL{I}_C=I_D$