JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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?
- A \(\mathrm{P} \propto \sqrt{v}\)
- B \(\mathrm{P} \propto v\)
- C \(\mathrm{P}^2 \propto v^5\)
- D \(\mathrm{P}^2 \propto v^3\)
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, \(P^2 \propto V^5\)
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