Rotational Motion

JEE Main level, numerical answer, medium

110 of 280

Question 110 of 280

A force of −Pk^-P\hat{k} acts on the origin of the coordinate system. The torque about the point (2,−3)(2,-3) is P(ai^+bj^)P(a\hat{i}+b\hat{j}), The ratio of ab\dfrac{a}{b} is x2\dfrac{x}{2}. The value of xx is

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Answer: 3

It is given that the force F⃗=−Pk^\vec{F}=-P\hat{k}
The coordinates of the origin is (0,0)(0,0).
The coordinates of the point given is (2,−3)(2,-3)
The displacement vector is
r⃗=(0−2)i^+(0+3)j^=−2i^+3j^\vec{r}=(0-2)\hat{i}+(0+3)\hat{j}=-2\hat{i}+3\hat{j}
Hence, the torque is
τ⃗=r⃗×F⃗=(−2i^+3j^)×(−Pk^)\vec{\tau }=\vec{r}\times \vec{F}=(-2\hat{i}+3\hat{j})\times (-P\hat{k})
⇒τ⃗=−2Pj^−3Pi^=P(ai^+bj^)\Rightarrow \vec{\tau }=-2P\hat{j}-3P\hat{i}=P(a\hat{i}+b\hat{j})
So, it can be written
a=−3a=-3, b=−2b=-2
∴ab=(32)⇒∴\dfrac{a}{b}=\left(\dfrac{3}{2}\right)\Rightarrow so x=3x=3

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