If $\mathop {\rm{F}}\limits^ \to = 2\,\widehat {\rm{i}} - 3\widehat {\rm{j}}\,\,{\rm{N}}$ and $\mathop r\limits^ \to = 3\,\widehat {\rm{i}} + 2\widehat {\rm{j}}\,\,m$ then torque $\overrightarrow \tau $ is

JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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Application +4 | -1

\mathop \mathrm{F}\limits^ \to = 2\,\widehat \mathrm{i} - 3\widehat \mathrm{j}\,\,\mathrm{N}

and

\mathop r\limits^ \to = 3\,\widehat \mathrm{i} + 2\widehat \mathrm{j}\,\,m

then torque

\overrightarrow \tau

View answer and explanation

Correct option: D

- 13\widehat \mathrm{k}

Option D (

- 13\widehat \mathrm{k}

) is correct because it is the torque calculated as the cross product of the position vector and the force vector.

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\mathrm{AB}

\mathrm{L}

\mu(x)=a+\frac{b x}{L}

x

\mathrm{A}

\mathrm{CM}

\left(\frac{7}{12}\right) \mathrm{L}

\mathrm{A}

\mathrm{b}

1.0 \mathrm{~g}

8 \mathrm{~cm}

\omega = {t^4} + 2{t^3}

t = 1\,sec

\left( {rad\,{s^{ - 2}}} \right)