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  • A thin uniform rod of mass m and length is in equilibrium as shown. P is pivot when a current I flows along the rod, it rotates from its place about point P. Assume that there is no effect of curre

A thin uniform rod of mass m and length is in equilibrium as shown. P is pivot when a current I flows along the rod, it rotates from its place about point P. Assume that there is no effect of current on the spring and the spring remain horizontal. The torque experienced by the rod about P and the spring force, when the rod makes an angle \theta with the horizontal, would be:

Option: 1

\mathrm{\frac{\mathrm{iB} \ell^2}{2}+\frac{\mathrm{mgl}}{2} \cos \theta ; \mathrm{K} \ell \cos \theta }


Option: 2

\mathrm{\frac{\mathrm{iB} \ell^2}{2}-\mathrm{K} \ell^2 \sin \theta \cos \theta, \mathrm{K} \ell \sin \theta }


Option: 3

\frac{\mathrm{iB} \ell^2}{2}+\frac{\mathrm{mg} \ell}{2} \cos \theta-\mathrm{K} \ell^2 \sin \theta \cos \theta ; \mathrm{K} \ell \cos \theta


Option: 4

0 ; \mathrm{K} \ell \cos \theta


Answers (1)

best_answer

\mathrm{x}=\ell \cos \theta ; \mathrm{F}=\mathrm{k} \ell \cos \theta \
\mathrm{ \tau=\int_0^{\ell}\left(\text { Bidy.y) }-\mathrm{kx} \ell \sin \theta+\frac{\mathrm{mg} \ell}{2} \cos \theta\right. }
\mathrm{ \tau=\frac{\mathrm{iB} \ell^2}{2}+\frac{\mathrm{mg} \ell}{2} \cos \theta-\mathrm{k} \ell^2 \sin \theta \cos \theta}

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manish painkra

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