JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2025
- Hard
- MCQ
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. \(\frac{\mathrm{dm}}{\mathrm{dt}} \propto \sqrt{v}\). If P is the power delivered to run the belt at constant speed then which of the following relationship is true?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(\mathrm{P}^2 \propto v^5\)
Step-by-step Solution
Detailed explanation
Power \(=\vec{F} \cdot \vec{V}\)
and \(F=\frac{d p}{d t}=\) Rate of change of linear momentum
\(F=V \cdot \frac{d m}{d t}=K_1 V^{\frac{3}{2}}, K\) is constant
Power \((P)=\left(K V^{\frac{3}{2}}\right) \cdot(V)\)
\(=K V^{\frac{5}{2}}\)
So, \(P^2 \propto V^5\)
and \(F=\frac{d p}{d t}=\) Rate of change of linear momentum
\(F=V \cdot \frac{d m}{d t}=K_1 V^{\frac{3}{2}}, K\) is constant
Power \((P)=\left(K V^{\frac{3}{2}}\right) \cdot(V)\)
\(=K V^{\frac{5}{2}}\)
So, \(P^2 \propto V^5\)
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