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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2025
  • Easy
  • MCQ

A body of mass 4 kg is placed on a plane at a point P having coordinate \((3,4) \mathrm{m}\). Under the action of force \(\overrightarrow{\mathrm{F}}=(2 \hat{i}+3 \hat{j}) \mathrm{N}\), it moves to a new point Q having coordinates \((6,10) \mathrm{m}\) in 4 sec . The average power and instanteous power at the end of 4 sec are in the ratio of :

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Correct Answer

(D) \(6: 13\)

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\(\begin{aligned} & \langle\mathrm{p}\rangle=\frac{(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \cdot(3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}})}{4}=6 \\ & \overrightarrow{\mathrm{a}}=\left(\frac{\overrightarrow{\mathrm{F}}}{\mathrm{m}}=\frac{1}{2} \hat{\mathrm{i}}+\frac{3}{4} \hat{\mathrm{j}}\right) \\ & \overrightarrow{\mathrm{v}} \text { at } \mathrm{t}=4 \mathrm{sec}=\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{3}{4} \hat{\mathrm{j}}\right) \times 4=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \\ & P_{\text {ins }}=(2 \hat{\mathrm{i}}+3)(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}})=13 \\ & \frac{\langle\mathrm{P}\rangle}{P_{\text {ins }}}=\frac{6}{13}\end{aligned}\)
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