A sand droper 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., . If P is the power delivered to run the belt at constant speed then which of the following relationship is true? [

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., $\frac{d m}{d t} \propto \sqrt{v}$ . If P is the power delivered to run the belt at constant speed then which of the following relationship is true? [2025]

Smart Achievers · Accepted Answer

(4)

Power = $\vec{F} \cdot \vec{v} and F = \frac{d p}{d t} = Rate of change of linear momentum$

$F = v \cdot \frac{d m}{d t} = K v^{3 / 2}, K is constant$

Power $(P) = {(K v)}^{3 / 2} \cdot (v) = K v^{5 / 2}$

So, $P^{2} \propto v^{5}$

Solution

Q.
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., $\frac{d m}{d t} \propto \sqrt{v}$ . If P is the power delivered to run the belt at constant speed then which of the following relationship is true [2025]

1 $P^{2} \propto v^{3}$

2 $P \propto \sqrt{v}$

3 $P \propto v$

4 $P^{2} \propto v^{5}$

Ans.
(4)

Power = $\vec{F} \cdot \vec{v} and F = \frac{d p}{d t} = Rate of change of linear momentum$

$F = v \cdot \frac{d m}{d t} = K v^{3 / 2}, K is constant$

Power $(P) = {(K v)}^{3 / 2} \cdot (v) = K v^{5 / 2}$

So, $P^{2} \propto v^{5}$

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Solution

Q. 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., dmdt∝v. If P is the power delivered to run the belt at constant speed then which of the following relationship is true [2025]

1 P2∝v3

2 P∝v

3 P∝v

4 P2∝v5

Ans. (4) Power = F→·v→ and F=dpdt=Rate of change of linear momentum F=v·dmdt=Kv3/2, K is constant Power (P)=(Kv)3/2·(v)=Kv5/2 So, P2∝v5