Verified Solution • 2025–26 Session hard

QJEEM_2025_6: JEE Main 2025 Physics - Class 11 JEEM_2025_6 Solution: A force is applied on a particle, and it undergoes a…

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

Multiple Choice Options

  • A 13\displaystyle \frac{1}{3}
  • B 12\displaystyle \frac{1}{2}
  • C 0
  • D 2

Step-by-Step Verified Solution

To determine the value of bb such that the work done on the particle is zero, we use the definition of work in terms of vector dot products.

1. Physical Principle

Work done (WW) by a constant force F⃗\vec{F} over a displacement d⃗\vec{d} is given by the dot product:

W=F⃗⋅d⃗W = \vec{F} \cdot \vec{d}

2. Given Vectors

  1. Force: F⃗=2i^+bj^+k^\vec{F} = 2\hat{i} + b\hat{j} + \hat{k}
  2. Displacement: d⃗=i^−2j^−k^\vec{d} = \hat{i} - 2\hat{j} - \hat{k}

3. Dot Product Calculation

The dot product is calculated by multiplying the corresponding components (i^,j^,k^\hat{i}, \hat{j}, \hat{k}) and summing them:

F⃗⋅d⃗=(2)(1)+(b)(−2)+(1)(−1)\vec{F} \cdot \vec{d} = (2)(1) + (b)(-2) + (1)(-1)

Simplify the expression:

F⃗⋅d⃗=2−2b−1\vec{F} \cdot \vec{d} = 2 - 2b - 1

F⃗⋅d⃗=1−2b\vec{F} \cdot \vec{d} = 1 - 2b

4. Solve for bb

Since the problem states the work done is zero (W=0W = 0):

1−2b=01 - 2b = 0

2b=12b = 1

b=12b = \frac{1}{2}

Conclusion

The value of bb is 12\displaystyle \frac{1}{2}, which corresponds to Option (B).

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