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QJEEM16Phb87f: JEE Main 2016 Physics - Class 11 JEEM16Phb87f Solution: Which of the following option correctly describes the variation …

Which of the following option correctly describes the variation of the speed v and acceleration ‘a’ of a point mass falling vertically in a viscous medium that applies a force F = − kv, where ‘k’ is a constant, on the body ? (Graphs are schematic and not drawn to scale)

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/eb13fa7f88ddd4c6adbd8642cdbcac5ed906c82f876562d6bf1f6c127d0a3b52.webp
  • B https://ik.imagekit.io/novatra/photos/600f73291a5acbea908f66ceca86ec7122d3b2f01acbd587dc124e360ab899e3.webp
  • C https://ik.imagekit.io/novatra/photos/54320e255edec275bc9262b84a8d05aa3b76aec99c9de8df0ff58aab894dcefa.webp
  • D https://ik.imagekit.io/novatra/photos/15dd83a4c860dda508a3526ef1f9f2667c1155e0fa659323fcabb40628a3eabe.webp

Step-by-Step Verified Solution

Equation of motion for the mass,

ma = mg −- kv

⇒\Rightarrow     dvdt=mg−kvm{{dv} \over {dt}} = {{mg - kv} \over m}

⇒\Rightarrow     ∫0vdvmg−kv=1m∫0tdt\int\limits_0^v {{{dv} \over {mg - kv}}} = {1 \over m}\int\limits_0^t {dt}

⇒\Rightarrow     −1k[ln⁡(mg−kv)]0v=tm- {1 \over k}\left[ {\ln \left( {mg - kv} \right)} \right]_0^v = {t \over m}

⇒\Rightarrow     ln⁡(mg−kvmg)=−ktm\ln \left( {{{mg - kv} \over {mg}}} \right) = - {{kt} \over m}

⇒\Rightarrow     1−kvmg=e−ktm1 - {{kv} \over {mg}} = {e^{ - {{kt} \over m}}}

⇒\Rightarrow     kvmg=1−e−ktm{{kv} \over {mg}} = 1 - {e^{ - {{kt} \over m}}}

⇒\Rightarrow     v=mgk(1−e−ktm)v = {{mg} \over k}\left( {1 - {e^{ - {{kt} \over m}}}} \right)

ma == mg −- k ×\times mgk{{mg} \over k} (1 −- e −ktm^{ - {{kt} \over m}} )

==  mg −- mg + mge −ktm^{ - {{kt} \over m}}

a == g e −ktm^{ - {{kt} \over m}}

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