JEE Physics 2023 Question with Solution

Given: A force $\vec{F} = -P\hat{k}$ acts at the origin. The torque is required about the point $(2,-3)$.

Find: The value of $x$ if $\frac{a}{b} = \frac{x}{2}$ and $\vec{\tau} = P(a\hat{i}+b\hat{j})$.

The torque is given by

$$\vec{\tau} = \vec{r} \times \vec{F}$$

The position vector from the point $(2,-3)$ to the origin is

$$\vec{r} = (0-2)\hat{i} + (0-(-3))\hat{j} = -2\hat{i} + 3\hat{j}$$

The force is

$$\vec{F} = -P\hat{k}$$

Now,

$$\vec{\tau} = (-2\hat{i} + 3\hat{j}) \times (-P\hat{k})$$

Using cross product,

$$\vec{\tau} = -P(3\hat{i} + 2\hat{j})$$

Comparing with $P(a\hat{i} + b\hat{j})$, we get

$$a = 3, \quad b = 2$$

So,

$$\frac{a}{b} = \frac{3}{2}$$

Given $\frac{a}{b} = \frac{x}{2}$, therefore

$$\frac{3}{2} = \frac{x}{2}$$

Hence,

$$x = 3$$

Therefore, the value of $x$ is $3$.

Given: $\vec{\tau} = P(a\hat{i}+b\hat{j})$ and force $\vec{F} = -P\hat{k}$.

Find: The numerical value of $x$.

First find the displacement vector from the reference point to the point of application of force:

$$\vec{r} = -2\hat{i} + 3\hat{j}$$

Then evaluate

$$\vec{r} \times \vec{F} = (-2\hat{i} + 3\hat{j}) \times (-P\hat{k})$$

Using $\hat{i} \times \hat{k} = -\hat{j}$ and $\hat{j} \times \hat{k} = \hat{i}$,

$$(-2\hat{i}) \times (-P\hat{k}) = -2P\hat{j}$$

and

$$(3\hat{j}) \times (-P\hat{k}) = -3P\hat{i}$$

So,

$$\vec{\tau} = -3P\hat{i} - 2P\hat{j}$$

The solution compares magnitudes of coefficients to write

$$a = 3, \quad b = 2$$

Thus,

$$\frac{a}{b} = \frac{3}{2} = \frac{x}{2}$$

Hence,

$$x = 3$$

Therefore, the required answer is $3$.