JEE Physics 2025 Question with Solution

Given: $\mathbf{F} = 2\hat{i} + b\hat{j} + \hat{k}$ and $\mathbf{r} = \hat{i} - 2\hat{j} - \hat{k}$.

Find: The value of $b$ for which the work done is zero.

Work done is the dot product of force and displacement.

$$W = \mathbf{F} \cdot \mathbf{r}$$

Substituting the given vectors,

$$W = (2\hat{i} + b\hat{j} + \hat{k}) \cdot (\hat{i} - 2\hat{j} - \hat{k})$$

Using the dot product of corresponding components,

$$W = 2(1) + b(-2) + 1(-1)$$ $$W = 2 - 2b - 1 = 1 - 2b$$

Since the work done is zero,

$$1 - 2b = 0$$ $$b = \frac{1}{2}$$

Therefore, the value of $b$ is $\frac{1}{2}$. The solution concludes this value, although the answer key and listed option key disagree with the working.

Given: $\mathbf{F} = 2\hat{i} + b\hat{j} + \hat{k}$ and $\mathbf{r} = \hat{i} - 2\hat{j} - \hat{k}$.

Find: The value of $b$ such that $W = 0$.

1. Use the formula for work done by a constant force:

$$W = \mathbf{F} \cdot \mathbf{r}$$

2. Substitute the vectors:

$$W = (2\hat{i} + b\hat{j} + \hat{k}) \cdot (\hat{i} - 2\hat{j} - \hat{k})$$

3. Multiply corresponding components and add:

$$W = (2)(1) + (b)(-2) + (1)(-1)$$

4. Simplify:

$$W = 2 - 2b - 1$$ $$W = 1 - 2b$$

5. Apply the condition that work done is zero:

$$1 - 2b = 0$$

6. Solve for $b$:

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

Hence, the correct value is $\frac{1}{2}$, which corresponds to option A.