QJEEM25Ph7493: JEE Main 2025 Physics - Class 11 JEEM25Ph7493 Solution: A sand dropper drops sand of mass m(t) on a conveyer belt at a…

Given, ${{dm} \over {dt}} \propto \sqrt v$

So, ${{dm} \over {dt}} = k\sqrt v$ .... (i)

where, k = constant

We know, Power = $P = FV$

$\Rightarrow P = {{dp} \over {dt}}v$ (As $F = {{dp} \over {dt}}$ where, p = linear momentum)

$\Rightarrow P = {d \over {dt}}(mv)v$

$\Rightarrow P = v{{dm} \over {dt}}\,.\,v$ (As speed is constant i.e. v = constant)

$\Rightarrow P = {v^2}(k\sqrt v )$ [From (i)]

$\Rightarrow P = k{v^{5/2}}$

by squaring both sides,

$\Rightarrow {P^2} = {k^2}{v^5}$

$\Rightarrow {P^2} \propto {v^5}$ (As $\mathrm{k^2}$ = constant)

Hence, option 4 is correct.