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Work Power and Energy question

2005 · Shift 0 · Q179
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Work Power and Energy question

2005 · Shift 0 · Q179

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1
A body of mass mmm is accelerated uniformly from rest to a speed vvv in a time T.T.T. The instantaneous power delivered to the body as a function of time is given by
  1. A
    mv2T2.t2{{m{v^2}} \over {{T^2}}}.{t^2}T2mv2​.t2
  2. B
    mv2T2.t{{m{v^2}} \over {{T^2}}}.tT2mv2​.t
  3. C
    12mv2T2.t2{1 \over 2}{{m{v^2}} \over {{T^2}}}.{t^2}21​T2mv2​.t2
  4. D
    12mv2T2.t{1 \over 2}{{m{v^2}} \over {{T^2}}}.t21​T2mv2​.t
View written solutionFree

Correct answer: B

  1. Find the uniform acceleration

Since the body starts from rest and reaches speed vvv in time TTT under uniform acceleration,

a=v−0T=vT.a = \frac{v-0}{T} = \frac{v}{T}.a=Tv−0​=Tv​.

  1. Velocity as a function of time

For motion starting from rest with constant acceleration,

v(t)=at=vTt.v(t) = at = \frac{v}{T}t.v(t)=at=Tv​t.

  1. Instantaneous power

Instantaneous power is

P=F⃗⋅v⃗.P = \vec F \cdot \vec v.P=F⋅v.

Here force and velocity are in the same direction, so

P=Fv(t).P = Fv(t).P=Fv(t).

Also,

F=ma=mvT.F = ma = m\frac{v}{T}.F=ma=mTv​.

Therefore,

P(t)=(mvT)(vTt).P(t) = \left(m\frac{v}{T}\right)\left(\frac{v}{T}t\right).P(t)=(mTv​)(Tv​t).

  1. Simplify

P(t)=mv2T2t.P(t) = \frac{mv^2}{T^2}t.P(t)=T2mv2​t.

So the instantaneous power varies linearly with time.

  1. Compare with options
  • A: mv2T2t2\frac{mv^2}{T^2}t^2T2mv2​t2 ❌
  • B: mv2T2t\frac{mv^2}{T^2}tT2mv2​t ✅
  • C: 12mv2T2t2\frac{1}{2}\frac{mv^2}{T^2}t^221​T2mv2​t2 ❌
  • D: 12mv2T2t\frac{1}{2}\frac{mv^2}{T^2}t21​T2mv2​t ❌

Hence, the correct option is B.

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