If the kinetic energy of a body is directly p | vedclass

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Question

$\mathrm{K} \propto \mathrm{t} \Rightarrow \mathrm{K}=\lambda \mathrm{t}, \lambda$ constant

$\frac{1}{2} \mathrm{mv}^{2}=\lambda \mathrm{t} \Righta

Vedclass · Accepted Answer

$(ii),\,(iv)$

Vedclass · Answer

$\mathrm{K} \propto \mathrm{t} \Rightarrow \mathrm{K}=\lambda \mathrm{t}, \lambda$ constant

$\frac{1}{2} \mathrm{mv}^{2}=\lambda \mathrm{t} \Rightarrow \mathrm{v}=\sqrt{\frac{2 \lambda}{\mathrm{m}} \mathrm{t}^{1 / 2}}$

$\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\sqrt{\frac{2 \lambda}{\mathrm{m}}} \frac{1}{2} \mathrm{t}^{-1 / 2}$

$\mathrm{F}=\mathrm{ma}=\sqrt{\frac{\lambda \mathrm{m}}{2}} \frac{1}{\mathrm{t}^{1 / 2}} \Rightarrow \mathrm{F} \propto \frac{1}{\sqrt{\mathrm{t}}},$ $(ii)$ is $\mathrm{O} . \mathrm{K}$

$\mathrm{F}=\sqrt{\frac{\lambda \mathrm{m}}{2}} \frac{1}{\mathrm{t}^{1 / 2}}=\sqrt{\frac{\lambda \mathrm{m}}{2}} 	imes \frac{1}{\mathrm{v}} \sqrt{\frac{2 \lambda}{\mathrm{m}}}$

$=\Rightarrow \mathrm{F} \propto \frac{1}{\mathrm{v}},(\mathrm{iv}) 	ext { is } \mathrm{O} \cdot \mathrm{K}$

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