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JEE Main 202522 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Evening Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2025 · 22 January · Evening Shift · Physics · Work, Power and Energy

A force F =2 i +b j + k is applied on a particle and it undergoes a...

A force F→=2i^+bj^+k^\overrightarrow{\mathrm{F}}=2 \hat{i}+\mathrm{b} \hat{j}+\hat{k} is applied on a particle and it undergoes a displacement i^−2j^−k^\hat{i}-2 \hat{j}-\hat{k}. What will be the value of bb, if work done on the particle is zero.

Options

  1. A0
  2. B12\frac{1}{2}
  3. C2
  4. D13\frac{1}{3}

Correct answer

B. 12\frac{1}{2}

Step-by-step solution

w=0∴F⃗⋅S=0(2i^+bj^+k^)⋅(i^−2j^−k^)=02−2b−1=0b=12\begin{aligned} & w=0 \\ & \therefore \quad \vec{F} \cdot S=0 \\ & (2 \hat{i}+b \hat{j}+\hat{k}) \cdot(\hat{i}-2 \hat{j}-\hat{k})=0 \\ & 2-2 b-1=0 \\ & b=\frac{1}{2}\end{aligned}
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