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A stone of mass 1 kg tied to a string 4 ...

A stone of mass 1 kg tied to a string 4 m long and is rotated at constant speed of 40 m/s a vertical circle. The ratio of the tension at the top and the bottom, is `(g=10" m"//"s"^(2))`

A

`11 : 12`

B

`39 : 41`

C

`41 : 39`

D

`12 : 41`

Text Solution

Verified by Experts

The correct Answer is:
B
`T_("top")=("mv"^(2))/( r )-mg" "…(i)`

`T_("bottom")=("mv"^(2))/("r")+"mg"" "…(ii)`
By Eqs. (i) and (ii), we get
`(T_("top"))/(T_("bottom"))=(((v^(2))/(r)-g))/(((v^(2))/(r)+g))=(((40xx40)/(4)+10))/(((40xx40)/(4)+10))`
`=(400-10)/(400+10)=(390)/(410)=(39)/(41)`
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Knowledge Check

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