question_answer

           

A jumper of mass m and length t is placed on two parabolic rails in x-y plane. Shape of the  rails can be described by

\[~Rail\text{ 1}:y={{x}^{2}}\text{ }\left( and\text{ }z=0 \right)\]

\[Rail\,2:y={{x}^{2}}(and=\ell )\]

If x is horizontal and y is vertical direction and magnetic field in the space is\[{{b}_{0}}j,\] the jumper can remain in equilibrium when y coordinate of its ends is (\[i\]= current in jumper)

A) \[\frac{i{{B}_{0}}\ell }{2mg}\]              

B) \[\frac{i{{B}_{0}}\ell }{mg}\]

C) \[{{\left( \frac{i{{B}_{0}}\ell }{mg} \right)}^{2}}\]

D) \[{{\left( \frac{i{{B}_{0}}\ell }{2mg} \right)}^{2}}\]

Correct Answer: D

Solution :

F= \[{{B}_{0}}i\ell \],y=\[{{x}^{2}}\]

For equilibrium, \[F\cos \theta =mgsin\theta \]

Or \[\tan \theta =\frac{F}{mg}=\frac{{{B}_{0}}i\ell }{mg}\]

Or \[\frac{dy}{dx}=\frac{{{B}_{0}}i\ell }{mg}\] Or \[2x=\frac{{{B}_{0}}i\ell }{mg}\]

Or \[x=\frac{{{B}_{0}}i\ell }{2mg}\] Or \[\sqrt{y}=\frac{{{B}_{0}}i\ell }{2mg}\] Or \[y={{\left( \frac{{{B}_{0}}i\ell }{2mg} \right)}^{2}}\]