The momentum of a moving particle given by $p = t (ln t)$ . Net force acting on this particle is defined by equation $F = \frac{d p}{d t}$ . The net force acting on the particle is zero at time:

t = 0

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t = \frac{1}{e}

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t = \frac{1}{e^{2}}

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t = 1

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Solution

The correct option is A $t = \frac{1}{e}$
$∵ p = t ℓ n t$
$∴ F = \frac{d p}{d t} = \frac{d}{d t} (t ℓ n t) = (1) ℓ n t + (t) (\begin{matrix} \frac{1}{t} \end{matrix}) = 1 + ℓ n t$
$F = 0 \Rightarrow 1 + ℓ n t = 0 \Rightarrow ℓ n t = - 1 \Rightarrow t = e^{- 1} = \frac{1}{e}$

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The momentum of a moving particle given by p=t(lnt). Net force acting on this particle is defined by equation F=dpdt. The net force acting on the particle is zero at time:

The correct option is A t=1e∵p=tℓnt∴F=dpdt=ddt(tℓnt)=(1)ℓnt+(t)(1t)=1+ℓntF=0⇒1+ℓnt=0⇒ℓnt=−1⇒t=e−1=1e

The momentum of a moving particle given by $p = t (ln t)$ . Net force acting on this particle is defined by equation $F = \frac{d p}{d t}$ . The net force acting on the particle is zero at time:

The correct option is A $t = \frac{1}{e}$
$∵ p = t ℓ n t$
$∴ F = \frac{d p}{d t} = \frac{d}{d t} (t ℓ n t) = (1) ℓ n t + (t) (\begin{matrix} \frac{1}{t} \end{matrix}) = 1 + ℓ n t$
$F = 0 \Rightarrow 1 + ℓ n t = 0 \Rightarrow ℓ n t = - 1 \Rightarrow t = e^{- 1} = \frac{1}{e}$