JEE Main 2025 · 29 January · evening · Single correct · +4 / −1
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., dm/dt ∝ √v. If P is the power delivered to run the belt at constant speed then which of the following relationship is true?
- (a) P ∝ √v
- (b) P ∝ v
- (c) P² ∝ v³
- (d) P² ∝ v⁵
Answer and solution
Answer (d) P² ∝ v⁵
Solution
Physics sir · 3 steps
The belt keeps giving fresh sand momentum, so find the force first, then multiply by v for power.
- Force on the belt. Sand lands at rest and must be brought to speed v. Each second the belt gives momentum v dm/dt to new sand, so it needs a force F = v dm/dt to keep its speed constant.
P = Fv = v² dm/dt
- Put in the rate. With dm/dt ∝ √v, the power becomes P ∝ v² × v^(1/2) = v^(5/2).
- So. Squaring both sides gives P² ∝ v⁵, which is option (d).
The idea to keep. When mass is added to a moving belt, F = v dm/dt and P = v² dm/dt. Put in how dm/dt depends on v and count the powers of v carefully.
Answer from the NTA final answer key. Solution by PrepWiser.