A block of mass m is placed on a smooth sphere of radius R. It slides when pushed slightly. At what vertical distance h, from the top, will it leave the sphere?

A block of mass m is placed on a smooth sphere of radius R. It slides

1.	A block of mass m is placed on a smooth sphere of radius R. It slides when pushed slightly. At what vertical distance h, from the top, will it leave the sphere?
Answer» <html><body><p>`(R)/(4)`<br/>`(R)/(3)`<br/>`(R )/(2) `<br/> `R`</p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/MTG_NEET_GID_PHY_XI_C03_E01_142_S01.png" width="80%"/> <br/> Let the block leave the sphere at point B, which is at a distance h from the top of the sphere. <br/>`<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>(mv^2)/( R ) =mg <a href="https://interviewquestions.tuteehub.com/tag/costheta-2550146" style="font-weight:bold;" target="_blank" title="Click to know more about COSTHETA">COSTHETA</a> -N` <br/>where N is the normal reaction and v is the velocity of block at point B. <br/>When the block leaves the sphere at point B, the normal reaction N becomes <a href="https://interviewquestions.tuteehub.com/tag/zero-751093" style="font-weight:bold;" target="_blank" title="Click to know more about ZERO">ZERO</a>. <br/>`therefore(mv^2)/( R )= mgcos thetaorcos theta =(v^2)/(Rg)` <br/> From figure,` cos theta = (R-h)/( R)` <br/>` therefore(R-h)/(R ) = (2 gh)/(Rg) "" [:.v= sqrt(2 gh)]` <br/>` orR-h =2hor 3h=Rorh=(R )/(3 )`</body></html>