A force of -P \hatk acts on the origin of the coordinate system. The torque about the point (2,-3) is P(a \hati+b \hatj) , The ratio of \fracab is \fracx2 . The value of x is -

Rotational Motion PYQ — JEE Main 2023 · 10 Apr · Shift 2 · Q57 with Solution

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Correct answer: 3

A force acts at the origin: $\vec F = -P\hat k$

We need the torque about the point $(2,-3)$ .

Since only two coordinates are given, the point is understood as $A=(2,-3,0)$

Position vector from the point about which torque is taken to the point of application of force

The force is applied at the origin $(0,0,0)$ .

So the position vector from $A$ to origin is $\vec r = (0-2)\hat i + (0-(-3))\hat j + (0-0)\hat k$ $\vec r = -2\hat i + 3\hat j$

Torque about point $A$ is $\vec \tau = \vec r \times \vec F$

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

Take out $-P$ : $\vec \tau = -P\left[(-2\hat i+3\hat j)\times \hat k\right]$

Now use cross products:

\qquad \hat j\times \hat k = \hat i$$ Hence, $$(-2\hat i+3\hat j)\times \hat k = -2(\hat i\times \hat k)+3(\hat j\times \hat k)$$ $$= -2(-\hat j)+3\hat i$$ $$= 3\hat i+2\hat j$$ Therefore, $$\vec \tau = -P(3\hat i+2\hat j)$$ $$\vec \tau = P(-3\hat i-2\hat j)$$ 4. **Compare with given form** Given torque is $$\vec \tau = P(a\hat i+b\hat j)$$ So, $$a=-3,\qquad b=-2$$ Thus, $$\frac{a}{b}=\frac{-3}{-2}=\frac{3}{2}$$ It is given that $$\frac{a}{b}=\frac{x}{2}$$ So, $$\frac{x}{2}=\frac{3}{2}$$ $$x=3$$ 5. **Final answer** $$\boxed{3}$$ The derived answer matches the stored correct answer.