JEE MAIN - Physics (2025 - 22nd January Evening Shift - No. 19)

A force $\overrightarrow{\mathrm{F}}=2 \hat{i}+\mathrm{b} \hat{j}+\hat{k}$ is applied on a particle and it undergoes a displacement $\hat{i}-2 \hat{j}-\hat{k}$ What will be the value of $b$, if work done on the particle is zero.
$\frac{1}{3}$
$\frac{1}{2}$
0
2

Explanation

To determine the value of $$b$$ such that the work done on the particle is zero, follow these steps:

The work done by a force on a displacement is given by the dot product:

$$\text{Work} = \overrightarrow{F} \cdot \overrightarrow{d}.$$

Given the force

$$\overrightarrow{F} = 2\hat{i} + b\hat{j} + \hat{k},$$

and the displacement

$$\overrightarrow{d} = \hat{i} - 2\hat{j} - \hat{k},$$

compute the dot product:

$$\overrightarrow{F} \cdot \overrightarrow{d} = (2)(1) + (b)(-2) + (1)(-1).$$

Simplify the expression:

$$\overrightarrow{F} \cdot \overrightarrow{d} = 2 - 2b - 1 = 1 - 2b.$$

Since the work done is zero:

$$1 - 2b = 0.$$

Solve for $$b$$:

$$2b = 1,$$

$$b = \frac{1}{2}.$$

Thus, the value of $$b$$ is $$\frac{1}{2}$$, which corresponds to Option B.

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