Find the maximum possible mass of a greased needle floating on water surface.

Options

A. `m_(max)=(2Tl)/(g)`
B. `m_(max)=(g)/(2Tl)`
C. `m_(max)=(2Tg)/(l)`
D. `m_(max)=(Tl)/(g)`

Answer

A

Explanation

Let the mass of the neddle be m. As the liquid surface is distorted, the surface tensionforces acing on both sides of the needle make an angle `theta`, say with vertical since the forces acting on the needle are F, F and mg, resolving the forces vertically for its equilibrium, we have
`sumF_(y)=Fcostheta+Fcostheta-mg=0`
Ths gives `m=(2Fcostheta)/(g)`
where `F=Tl`
Then `m=(2Tlcostheta)/(g)`
For m to be maximum `costheta=1`
Hence, `m_(max)=(2Tl)/(g)`