Correct Option is (B)

$P=Fv$

$P=m\frac{dv}{dt}v$

$\Rightarrow$ $vdv=\frac{P}{m}dt$

Integrating both sides, we get

$V^2=k^{\prime }t$

$\Rightarrow$ V = k"$\sqrt{t}$

$\Rightarrow$ $\frac{ds}{dt}$ = k"$\sqrt{t}$

$\Rightarrow$ $\int ds=\int k^{″}\sqrt{t}dt$

$\Rightarrow$ s $\propto$ t^3/2

Correct Option is (D)

Net force on motor will be

F_m = [920 + 68(10)]g + 6000 = 22000 N

So, required power for motor

P = F_m.V = 22000$\times$3 = 66000 W

Correct Option is (D)

F = mg + f

F = 20000 + 4000 = 24000 N

We know, Power(P) = Fv

$\Rightarrow$ v = $\frac{P}{F}$ = $\frac{60\times 746}{24000}$

$\Rightarrow$ v $\approx$ 1.9 m/s

Correct answer is (18)

Let s be the required distance.

From Work - Energy theorem,

Work = Change in kinetic energy

$\Rightarrow$ Power $\times$ Time = $\mathrm{\Delta }$K

i.e., Pt = $\mathrm{\Delta }$K $\Rightarrow$ Pt = $\frac{1}{2}$mv² ..... (i)

Given, P = 1 Js^$-$1, t = 9 s, m = 2 kg

Substituting all the given values in eq. (i), we get

1 $\times$ 9 = $\frac{1}{2}$(2) v²

v² = 9 $\Rightarrow$ v = 3 m/s (at t = 9 s)

As, Fv = P $\Rightarrow$ (ma)v = P [$\because$ F = ma]

$\Rightarrow m\left[\frac{dv}{dt}\right]v=P\Rightarrow m\left[\frac{ds}{dt}\frac{dv}{ds}\right]v=P$

$\Rightarrow m\left[v\frac{dv}{ds}\right]v=P$

$\Rightarrow 2v^{2}dv=ds$ {$\because$ P = 1 J/s and m = 2 kg}

Integrating both sides,

$\underset{0}{\overset{3}{\int }}2v^{2}dv=\underset{0}{\overset{s}{\int }}ds\Rightarrow \frac{2}{3}[v^3{]}_0^3=8$

$\frac{2}{3}[27-0]=s\Rightarrow s=18$ m

Hence, after 9 s, the body has moved a distance of 18 m.

Correct Option is (A)

We know,

centripetal acceleration = $\frac{V^2}{R}$

$\therefore$   According to question,

$\frac{V^2}{R}$ = $n^{2}R{t}^2$

$\Rightarrow$   V² = n² R² t²

$\Rightarrow$   V = nRt

$\Rightarrow$   $\frac{dV}{dt}$ = nR

Power (P) = Force (F) $\times$ Velocity (V)

= M $\frac{dV}{dt}$(V)

= M (nR) (nRt)

= Mn²R²t