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?
Solution
<p>Given, ${{dm} \over {dt}} \propto \sqrt v$</p>
<p>So, ${{dm} \over {dt}} = k\sqrt v$ .... (i)</p>
<p>where, k = constant</p>
<p>We know, Power = $P = FV$</p>
<p>$\Rightarrow P = {{dp} \over {dt}}v$ (As $F = {{dp} \over {dt}}$ where, p = linear momentum)</p>
<p>$\Rightarrow P = {d \over {dt}}(mv)v$</p>
<p>$\Rightarrow P = v{{dm} \over {dt}}\,.\,v$ (As speed is constant i.e. v = constant)</p>
<p>$\Rightarrow P = {v^2}(k\sqrt v )$ [From (i)]</p>
<p>$\Rightarrow P = k{v^{5/2}}$</p>
<p>by squaring both sides,</p>
<p>$\Rightarrow {P^2} = {k^2}{v^5}$</p>
<p>$\Rightarrow {P^2} \propto {v^5}$ (As $\mathrm{k^2}$ = constant)</p>
<p>Hence, option 4 is correct.</p>
About this question
Subject: Physics · Chapter: Work, Energy and Power · Topic: Work Done by a Force
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