A smooth track is in the form of a quarter circle of radius 10 m lies in the vertical plane. A bead moves from A and B along the circular track under the action of forces `vec(F)_(1),vec(F)_(2)` and `vec(F)_(3)` Force `vec(F)_(1)` acts always towards point B and is always 20 N in magnitude Force `vec(F)_(2)` always acts horizontally and is always 25 N in magnitude. Force `vec(F)_(3)` always acts tangentially to the track and is of constant magnitude 10 N. Select the correct answer.

Options

A. Work done by `vec(F)_(1)` is 750 J on the bead
B. Work done by `vec(F)_(2)` is 20 J on the bead
C. Work done by `vec(F)_(3)` is `50 pi J` on the bead
D. Work done by `vec(F)_(3)` is `100 sqrt(2) J`on the bead

Answer

B, C

Explanation

Work done by `F_(1) = 20 xx 10 sqrt(2) = 200 sqrt(2) J`
Work done `F_(2) = 25 xx 10 = 250 J`
Work done by `F_(3)=10 xx(2pi R)/(4)=50 pi J`.