A pendulum bob has a speed of `3ms^(-1)` at its lowest position. The pendulum is `0.5` m long. The speed of the bob, when string makes an angle of `60^(@)` to the vertical is `("take, g"=10ms^(-1))`

Options

A. `(1)/(2) m//s`
B. `2 m//s`
C. `(1)/(3) m//s`
D. `3 m//s`

Answer

B

Explanation

`h=l-l "cos" 60^(@)=l(1-"cos" 60^(@))`
By u"sin"g the principle of conservation of energy
`(P.E.)_(A)+(K.E.)_(A)=(P.E.)_(B)+(K.E.)_(B)`
`therefore 0+(1)/(2)mv^(2)=mgl (1-"cos" 60^(@))+(1)/(2) mv_(B)^(2)`
`therefore (1)/(2)xx3^(2)=10xx0.5(1-(1)/(2))+(1)/(2)v_(B)^(2)`
`9=10xx(1)/(2)+v_(B)^(2)`
`therefore 9-5=v_(B)^(2) therefore v_(B)=2m//s`