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Rotational Motion question

2006 · Shift 0 · Q110

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JEE Main
Physics
Rotational Motion
2006 · Shift 0 · Q110

Rotational Motion question

2006 · Shift 0 · Q110

JEE MainPhysicsRotational MotionMCQ+4 / −1

A force of

- F\widehat k

acts on

O,

the origin of the coordinate system. The torque about the point

(1, -1)

is

AIEEE 2006 Physics - Rotational Motion Question 212 English

A
$F\left( {\widehat i - \widehat j} \right)$
B
$- F\left( {\widehat i + \widehat j} \right)$
C
$F\left( {\widehat i + \widehat j} \right)$
D
$- F\left( {\widehat i - \widehat j} \right)$

View written solutionFree

Correct answer: B

Given data

Force acting at the origin $O(0,0,0)$ is $\vec F = -F\hat k$
We need torque about the point $(1,-1)$ .

Since the force is applied at the origin, the position vector from the point $(1,-1,0)$ to the origin is

\vec r = (0-1)\hat i + (0-(-1))\hat j = -\hat i + \hat j

Torque formula

Torque about the given point is

\vec \tau = \vec r \times \vec F

So,

\vec \tau = (-\hat i + \hat j) \times (-F\hat k)

Factor out $-F$ :

\vec \tau = -F[(-\hat i + \hat j) \times \hat k]

Evaluate the cross product

Using

\hat i \times \hat k = -\hat j, \qquad \hat j \times \hat k = \hat i

we get

(-\hat i + \hat j) \times \hat k = - (\hat i \times \hat k) + (\hat j \times \hat k)

= -(-\hat j) + \hat i = \hat j + \hat i

= \hat i + \hat j

Therefore,

\vec \tau = -F(\hat i + \hat j)

Match with options

This corresponds to:

\boxed{\text{B: } -F(\hat i + \hat j)}

Comparison with stored answer

Derived answer: B
Stored correct answer: C

These do not match. The stored answer appears to have the wrong sign.

The likely reason: torque must be computed using the position vector from the reference point $(1,-1,0)$ to the point of application $(0,0,0)$ , i.e.

\vec r = -\hat i + \hat j

not the reverse vector.

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