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JEE Mains · Physics · STD 11 - Newtons laws of motion
  • JEE Mains 2023
  • Medium
  • Numerical

At any instant the velocity of a particle of mass \(500\,g\) is \(\left(2 t \hat{ i }+3 t ^2 \hat{ j }\right)\,ms ^{-1}\). If the force acting on the particle at \(t=1 s\) is \((\hat{i}+x \hat{j}) N\). Then the value of \(x\) will be:

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Correct Answer

(A) \(3\)

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\(\overrightarrow{ v }=2 t \hat{ i }+3 t ^2 \hat{ j }\)

\(\overrightarrow{ a }=2 \hat{i}+6 t \hat{j}\)

at \(t=1, \vec{a}=2 \hat{i}+6 \hat{j}\)

\(\overrightarrow{ F }= ma =0.5(2 \hat{i}+6 \hat{j})=\hat{i}+3 \hat{j}\)

\(\overrightarrow{ F }=\hat{i}+x \hat{j} \quad \text { Hence } x=3\)

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