A wooden stick of length `L`, radius `R` and density `rho` has a small metal piece of mass `m` ( of negligible volume) attached to its one end. Find the minimum value for the mass `m` (in terms of given parameters) that would make the stick float vertically in equilibrium in a liquid of density `sigma(gtrho)`.

Options

A. `2piR^(2)Lrho(sqrt((sigma)/(rho))-1)`
B. `piR^(2)Lrho(sqrt((2sigma)/(rho))-1)`
C. `piR^(2)Lrho(sqrt((sigma)/(rho))-1)`
D. `piR^(2)Lrho(sqrt((sigma)/(rho))-1)`

Answer

C

Explanation

`(M+m)g=piR^(2)hsigmag` ..(i)
`M=piR^(2)Lrho`.
The heigth of centre of mass from bottom
`=((M) L//2+mxx0)/(m+M)=(ML)/((m+M))`
For rotatory equilibrium and for minimum m, this should be equal to `h//2`.
`because(h)/(2)=(ML)/(2(m+M))` .(ii)
`becauseh=(ML)/(2(m+M))`,
`m=piR^(2)Lsqrt(sigmarho)-piR^(2)Lrho=piR^(2)Lrho(sqrt((sigma)/(rho))-1)`