Verified Solution • 2025–26 Session moderate

QJEEM16Ph4d96: JEE Main 2016 Physics - Class 11 JEEM16Ph4d96 Solution: A particle of mass m is acted upon by a force F given by the…

A particle of mass m is acted upon by a force F given by the empirical law F = Rt2 v(t).{R \over {{t^2}}}\,v\left( t \right). If this law is to be tested experimentally by observing the motion starting from rest, the best way is to plot :

Multiple Choice Options

  • A υ\upsilon (t) against t2
  • B log υ\upsilon (t) against 1t2{1 \over {{t^2}}}
  • C log υ\upsilon (t) against t
  • D log υ\upsilon (t) against 1t{1 \over {{t}}}

Step-by-Step Verified Solution

Given,

F = Rt2{R \over {{t^2}}} v(t)

⇒\Rightarrow    m dvdt{{dv} \over {dt}} = Rt2{R \over {{t^2}}} (v)

⇒\Rightarrow     dvv{{dv} \over v} = Rm{R \over m} dtt2{{dt} \over {{t^2}}}

Intergrating both sides,

∫dvv=Rm∫dtt2\int {{{dv} \over v} = {R \over m}\int {{{dt} \over {{t^2}}}} }

⇒\Rightarrow    lnv = Rm{{R \over m}} ×\times (−1t)\left( { - {1 \over t}} \right) + C

⇒\Rightarrow    lnv = −- Rm{{R \over m}} (1t)\left( {{1 \over t}} \right) + C

Graph between lnv and 1t{{1 \over t}} will be straight line curve.

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