Two small metal spheres having equal charge and mass are suspended from some point on the ceiling of a damp room with silk threads of equal length. Let centre to centre distance between sphere be x, x << l, l is length of silk thread. Due to ionization of medium, charge leaks off from each sphere and they keep on coming closer to each other at a constant rate.Let their approach velocity y varies as v∝x−1/2If mass of each sphere is m then the rate at which charge varies with respect to time is dqdt∝N22πε0mgl. The value of N is _________.
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
answer is 3.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
At equilibrium, q24πε0x2cosθ=mgsinθ For small θ,cosθ→1 and sinθ≈θ=x/2l⇒q24πε0x2=mgx2l⇒q=x3/2⋅2πε0mgl⇒ dqq=32x3/2⋅2πε0mgl⋅dxdt Put v=dxdt∝x1/2⇒dqdt∝322πε0mgl Hence, N=3