Physics

$f_0=140 cm$ and $f_e=5 cm$

For distant object,

$m=\frac{f_0}{f_e}=\frac{140}{5}=28$

Given  $V^{\prime }=V=$ Constant

 (i) $C^{\prime }=\frac{{\epsilon }_{0}A}{d^{\prime }},C=\frac{{\epsilon }_{0}A}{d}$

$d^{\prime }<d$

$C^{\prime }>C$
Hence, final capacitance greater than initial capacitance,

(ii)  $U^{\prime }=\frac{1}{2}{C}^{\prime }{V}^2$

$U=\frac{1}{2}C{V}^2$

$U^{\prime }>U$

Hence final energy is greater than initial energy

(iii)  $\frac{Q^{\prime }}{V^{\prime }}=C^{\prime }$ and $\frac{Q}{V}=C$

$\frac{Q^{\prime }}{V^{\prime }}\ne \frac{Q}{V}$

(iv) Product of charge and voltage

$X^{\prime }=Q^{\prime }V=C^{\prime }{V}^2$

$X=QV=C{V}^2$

$X^{\prime }>X$

Initially, the body has zero velocity and zero slope. Hence the acceleration would be zero initially.

After that, the slope of v-t curve is constant and positive.

After some time, velocity becomes constant and acceleration is zero.

After that, the slope of v-t curve is constant and negative.

Thermal strain $=$ Longitudinal strain $=\alpha \Delta T$

$\Rightarrow$ Longitudinal strain, $\delta =10^{-5}*10^2=10^{-3}$

$\Rightarrow$ Compressive stress $=\delta *$  Young's Modulus

$=10^{-3}*0.5*10^{11}$

$=0.5*10^8$

$\Rightarrow$ Compressive force $=0.5*10^8*10^{-3}=0.5*10^5$

$=5*10^4*\frac{10}{10}$

$=50*10^3 N$

$M=ml$

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

$=\frac{l}{2}$

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

$=M/2$

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,

$U_i+K_i=U_f+K_f$

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

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

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

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