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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 A. `2g`, `pi//2omega`B. `g`, `3pi//omega`C. `2g`, `3pi//2`D. `g`, `pi//omega` |
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Answer» (`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)` |
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