A force of -P k acts on the… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Mechanics

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A force of $-P\hat{k}$ acts on the origin of the coordinate system. The torque about the point $(2,-3)$ is $P(a\hat{i}+b\hat{j})$ , The ratio of $\frac{a}{b}$ is $\frac{x}{2}$ . The value of $x$ is

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Held on 10 Apr 2023 · Verified 6 Jul 2026.

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Solution

It is given that the force $\vec{F}=-P\hat{k}$

The coordinates of the origin is $(0,0)$ .

The coordinates of the point given is $(2,-3)$

The displacement vector is

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

Hence, the torque is

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

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

So, it can be written

$a=-3$ , $b=-2$

$\therefore \frac{a}{b}=(\frac{3}{2})\Rightarrow$ so $x=3$

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Solution

It is given that the force $\vec{F}=-P\hat{k}$

The coordinates of the origin is $(0,0)$ .

The coordinates of the point given is $(2,-3)$

The displacement vector is

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

Hence, the torque is

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

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

So, it can be written

$a=-3$ , $b=-2$

$\therefore \frac{a}{b}=(\frac{3}{2})\Rightarrow$ so $x=3$

Did you get this right?

Solution

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Did you get this right?

Solution

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JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

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