Question 22: Work, Energy and Power & Circular Motion

Question

A sand dropper drops sand of mass m(t) on a conveyer belt at a rate proportional to the square root of speed (v) of the belt, i.e., $\frac{dm}{dt} \propto \sqrt{v}$. If P is the power delivered to run the belt at constant speed then which of the following relationship is true?

Accepted Answer

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.

Answer

P $\propto \sqrt{v}$

Answer

P $\propto v$

Answer

$P^2 \propto v^3$

Answer

$P^2 \propto v^5$

Work, Energy and Power & Circular Motion

A sand dropper drops sand of mass m(t) on a conveyer belt at a rate proportional to the square root of speed (v) of the belt, i.e., \frac{dm}{dt} \propto \sqrt{v}. If P is the power delivered to run the belt at constant speed then which of the following relationship is true?

Choose the correct answer: