1.

A ball tied to a rope making an angle 30^(@) with the horizontal is held so that the rope is just If the ball is released, calculate the heat generated in joule when it crosses 30^(@) down the horizontal (dotted)Givenm=Kg,L=2m,g=10m//s^(2).

Answer»

SOLUTION :The speed just after JERKING
`v=sqrt(2gL) cos 30^(@)=sqrt((3gL)/(2))`
`therefore` heat=10 SEC in kinetic ENERGY
`=(1)/(2)m(2gL-(3gL)/(2))`
`=(MGL)/(4)=5J`


Discussion

No Comment Found

Related InterviewSolutions