A beaker containing an ideal fluid executes plane `SHM` in a horizontal plane according to the equation `x=(sqrt(3)g)/(omega^(2))sinomegat`, `O` being the mean position. A bob is suspended at `S` through a string of length `L` as shown in the figure. The line `SO` is vertical. Assuming `L gtgt (sqrt(3)g)/(omega^(2))`.
The magnitude of maximum buoyant force and the time with it occurs for the second time are, respectively

Options

A. `2g`, `pi//2omega`
B. `g`, `3pi//omega`
C. `2g`, `3pi//2`
D. `g`, `pi//omega`

Explanation

(`C`) `2rhoVg`, `3pi//2omega`
Magnitude of component of Buoyant force `=rhova`
Where a is acceleration of beaker, `a=-sqrt(3) g sin omega t`
Net buoyant force `=sqrt(rho^(2)v^(2)g^(2)+rho^(2)v^(2)a^(2))`
`=rhovgsqrt(1+3sin^(2)omega t)`
Magnitude of maximum buoyant force `=2rhovg`
`sin^(2)omegat=1` at `t=(pi)/(2omega)`, `(3pi)/(2omega)`
`:. 2rhovg`, `(3pi)/(2omega)`