A ring of mass m and radius R rests in equilibrium on a smooth cone of semi-vertical angle 45^(@) as shown. The radius of the cone is 2R. the radius of circular crosssection of the ring is r(r lt lt R). What will be the tension in the ring?

A ring of mass m and radius R rests in equilibrium on a smooth cone of

1.	A ring of mass m and radius R rests in equilibrium on a smooth cone of semi-vertical angle 45^(@) as shown. The radius of the cone is 2R. the radius of circular crosssection of the ring is r(r lt lt R). What will be the tension in the ring?
Answer» `(MG)/(2sqrt(2)PI)` `(mg)/(2PI)` `(mg)/(sqrt(2)pi)` `(sqrt(2)mg)/(pi)` Solution :FBD of smallsection of ringsubtendingangle `d theta` at centre of ring is as shown. `(N)/(sqrt(2))=T d theta, (N)/(sqrt(2))=(m d theta)/(2pi)G` `rArr T=(mg)/(2pi)`

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A ring of mass m and radius R rests in equilibrium on a smooth cone of semi-vertical angle 45^(@) as shown. The radius of the cone is 2R. the radius of circular crosssection of the ring is r(r lt lt R). What will be the tension in the ring?

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