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:
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(3\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ v }=2 t \hat{ i }+3 t ^2 \hat{ j }\)
\(\overrightarrow{ a }=2 \hat{i}+6 t \hat{j}\)
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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