Correct answer is 216

${\mathrm{B}}_0$ is the magnetic field at origin

$\begin{array}{rl}& \frac{dB}{dx}=-\frac{{10}^{-3}}{{10}^{-2}}\\ \\ & {\int }_{B_0}^{B}dB=-{\int }_0^x{10}^{-1}dx\\ \\ & B-B_0=-{10}^{-1}x\\ \\ & B=(B_0-\frac{x}{10})\end{array}$

Motional emf in $\mathrm{AB}=0$

Motional emf in $\mathrm{CD}=0$

Motional emf in $\mathrm{AD}={\epsilon }_1={\mathrm{B}}_0\ell \mathrm{v}$

Magnetic field on rod BC B

$=({\mathrm{B}}_0-\frac{(-12\times {10}^{-2})}{10})$

Motional emf in $\mathrm{BC}={\epsilon }_2=({\mathrm{B}}_0+\frac{12\times {10}^{-2}}{10})\ell \times \mathrm{v}$

${\epsilon }_{\mathrm{eq}}={\epsilon }_2-{\epsilon }_1=300\times {10}^{-7}\mathrm{V}$

For time variation

$\begin{array}{rl}& {\left({\epsilon }_{\text{eq }}\right)}^{\prime }=\mathrm{A}\frac{\mathrm{dB}}{\mathrm{dt}}=60\times {10}^{-7}\mathrm{V}\\ \\ & {\left({\epsilon }_{\mathrm{eq}}\right)}_{\text{net }}={\epsilon }_{\mathrm{eq}}+{\left({\epsilon }_{\mathrm{eq}}\right)}^{\prime }=360\times {10}^{-7}\mathrm{V}\\ \\ & \text{ Power }=\frac{{\left({\epsilon }_{\text{eq }}\right)}_{\text{net }}^2}{\mathrm{R}}=216\times {10}^{-9}\mathrm{W}\end{array}$

Correct answer is 32

$\begin{array}{rl}& B_v=B\sin 30=\frac{1}{4}\times {10}^{-4}\\ & \omega =2\pi \times f=\frac{2\pi }{60}\times 1200\mathrm{rad}/\mathrm{s}\\ & \epsilon =\frac{1}{2}{B}_V\omega {\ell }^2\\ & =32\pi \times {10}^{-5}\mathrm{V}\end{array}$

Correct answer is 3

$\begin{array}{rl}& {\mathrm{B}}_{\mathrm{H}}=0.60\times {10}^{-4}\mathrm{Wb}/{\mathrm{m}}^2\\ & \text{ Induced emf e}={\mathrm{B}}_{\mathrm{H}}\mathrm{v}\ell \\ & =0.60\times {10}^{-4}\times 10\times 5\\ & =3\times {10}^{-3}\mathrm{V}\end{array}$