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A particle of mass m is dropped inside a liquid medium which applies a bouncy force v and a drag force  k.v2. If acceleration due to gravity g is constant. What is the terminal velocity attended by the particle?

Option: 1

\sqrt{\frac{m g+B}{k}}


Option: 2

\sqrt{\frac{2 m g+B}{k}}


Option: 3

\sqrt{\frac{m g-B}{k}}


Option: 4

\sqrt{\frac{m g+B}{k}}


Answers (1)

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\text { At, } v=v_{t}

From figure, m g=B+k v_{t}^{2}

\begin{aligned} & m g-B=k v_{t}^{2} \\ & v_{t}^{2}=\frac{m g-B}{k} \\ & v_{t}=\sqrt{\frac{m g-B}{k}} \quad \text { Ans. } \end{aligned}

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Divya Prakash Singh

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