Correct answer is 2

$R=8\mathrm{\Omega }$

$\varphi =\frac{2}{3}(9-t^2)$

At $t=3$, $\varphi =0$

$\epsilon =\left|-\frac{d\varphi }{dt}\right|=\frac{4}{3}t$

$H={\int }_0^3\frac{V^2}{R}dt={\int }_0^3\frac{1}{8}\times \frac{16}{9}{t}^{2}dt$

$=\frac{2}{9}\times {\left(\frac{t^3}{3}\right)}_0^3=\frac{2}{9\times 3}\times 27=2J$

Correct answer is 250

$R=20\mathrm{\Omega }$

$\varphi =8t^2-9t+5$

$\epsilon =\left|-\frac{d\varphi }{dt}\right|=|16t-9|=|16(0.25)-9|=5$

$i=\frac{\epsilon }{R}=\frac{5}{20}=0.25A=\frac{0.25}{{10}^3}\times {10}^{3}A=250mA$

Correct answer is 440

$V_{max}=NAB\omega$

$=1000\times 1\times 0.07\times (2\pi \times 1)$

$\simeq 440$ volts

Correct Option is (A)

$\overrightarrow{B}$ must not be parallel to the plane of coil for non zero flux and according to lenz law if B is outward it should be decreasing for anticlockwise induced current.

Correct Option is (C)

$\to$ When magnet passes through centre region of solenoid, no current / Emf is induced in loop.

$\to$ While entering flux increases so negative induced emf

$\to$ While leaving flux decreases so positive induced emf.