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A small block of wood of relative density 0.5 is submerged in water at a depth of 5 m When the block is released it starts moving upwards, the acceleration of the block is `(g=10ms^(-2)`)A. `5ms^(-2)`B. `10ms^(-2)`C. `7.5ms^(-2)`D. `15ms^(-2)` |
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Answer» `ma_("apparent")="force of buoyancy-weight of body"` `=Vd_(w)g-Vd_(B)g` `impliesa_("apparent")=(Vd_(w)g-Vd_(B)g)/(Vd_(B))=g((d_(w))/(d_(B))-1)` |
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