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Work Power and Energy question

2003 · Shift 0 · Q157
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Work Power and Energy question

2003 · Shift 0 · Q157

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1
A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time ′t′'t'′t′ is proportional to
  1. A
    t3/4{t^{3/4}}t3/4
  2. B
    t3/2{t^{3/2}}t3/2
  3. C
    t1/4{t^{1/4}}t1/4
  4. D
    t1/2{t^{1/2}}t1/2
View written solutionFree

Correct answer: B

  1. Given: The machine delivers constant power PPP to move a body along a straight line.

  2. Use the definition of power: P=dWdtP = \frac{dW}{dt}P=dtdW​ Since power is constant, W=PtW = PtW=Pt

  3. Relate work to kinetic energy: Assuming the body starts from rest and all the work goes into increasing kinetic energy, Pt=12mv2Pt = \frac{1}{2}mv^2Pt=21​mv2

  4. Find how velocity depends on time: v2∝t⇒v∝t1/2v^2 \propto t \quad \Rightarrow \quad v \propto t^{1/2}v2∝t⇒v∝t1/2

  5. Now relate distance and velocity: v=dxdt∝t1/2v = \frac{dx}{dt} \propto t^{1/2}v=dtdx​∝t1/2 So, dx∝t1/2dtdx \propto t^{1/2}dtdx∝t1/2dt

  6. Integrate: x∝∫t1/2dt=23t3/2x \propto \int t^{1/2}dt = \frac{2}{3}t^{3/2}x∝∫t1/2dt=32​t3/2

  7. Therefore: x∝t3/2x \propto t^{3/2}x∝t3/2

  8. Option check:

    • A: t3/4t^{3/4}t3/4 — incorrect
    • B: t3/2t^{3/2}t3/2 — correct
    • C: t1/4t^{1/4}t1/4 — incorrect
    • D: t1/2t^{1/2}t1/2 — incorrect

Hence, the distance moved is proportional to t3/2\boxed{t^{3/2}}t3/2​.

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