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JEE Main 20205 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
5 September
Shift
Morning Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2020 · 5 September · Morning Shift · Physics · Rotational Motion

A force → F = ( ^ i + 2 ^ j + 3 ^ k ) N acts at a point ( 4 ^ i + 3...

A force \overset{\to}{F} = ( \overset{^}{i} + 2\overset{^}{j} + 3\overset{^}{k} ) N acts at a point ( 4\overset{^}{i} + 3\overset{^}{j} - \overset{^}{k} ) m. Then the magnitude of torque about the point ( \overset{^}{i} + 2\overset{^}{j} + \overset{^}{k} ) m will be xN−m.\sqrt{x}N - m .The value of xx is..........

Correct answer

0

Step-by-step solution

\overset{\to}{r} = \left(4 - 1\right)\overset{^}{i} + \left(3 - 2\right)\overset{^}{j} + \left(- 1 - 1\right)\overset{^}{k}

= 3\overset{^}{i} + \overset{^}{j} - 2\overset{^}{k}

τ = \overset{\to}{r} \times \overset{\to}{F} = \left|\begin{cases}\overset{^}{i} & j & \overset{^}{k} \\ 3 & 1 & - 2 \\ 1 & 2 & 3\end{cases}\right|

= \overset{^}{i}\left(7\right) - \overset{^}{j}\left(11\right) + \overset{^}{k}\left(5\right) = 7\overset{^}{i} - 11\overset{^}{j} + 5\overset{^}{k}

=49+121+25=195= \sqrt{49 + 121 + 25} = \sqrt{195}

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