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A particle of mass 'm' moves along a circle of radius R with a normal acceleration varying with time as a=b*t^2 where b is a constant.The time dependence of the power developed by all the forces acting on the particle is:a) 2mRbt b)mRbt c)4mRbt d)3mRat
normal accln: an = v2/R = bt2 so v2 = bt2R Kinetic Energy(E) = 1/2 mv2 = 1/2*mbt2R power P =dE/dt = d(1/2*mbt2R) / dt P =mbt2R -- Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Naga Ramesh IIT Kgp - 2005 batch
normal accln: an = v2/R = bt2
so v2 = bt2R
Kinetic Energy(E) = 1/2 mv2 = 1/2*mbt2R
power P =dE/dt = d(1/2*mbt2R) / dt
P =mbt2R
--
Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Naga Ramesh IIT Kgp - 2005 batch
Hi I would like to make some corrections to the reply posted by Mr Ramesh. I think after differentiation of "E" we would be getting P = mRbt which is the answer i. e. the answer is the option (b).
Hi
I would like to make some corrections to the reply posted by Mr Ramesh. I think after differentiation of "E" we would be getting P = mRbt which is the answer i. e. the answer is the option (b).
Hi The differentiation needs to be corrected and accordingly the answer is (b) JAGABANDHU PANDA IIT Delhi
The differentiation needs to be corrected and accordingly the answer is (b)
JAGABANDHU PANDA
IIT Delhi
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