The spring balance A reads 2 kg with a block | vedclass

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(C) When the block $m$ is suspended in air,the reading of spring balance $A$ is equal to the weight of the block,$T = mg = 2 \ kgf$.When the block is

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the balance $A$ will read less than $2 \ kg$ and $B$ will read more than $5 \ kg$

Vedclass · Answer

(C) When the block $m$ is suspended in air,the reading of spring balance $A$ is equal to the weight of the block,$T = mg = 2 \ kgf$.When the block is immersed in the liquid,an upward buoyant force $F_B$ acts on the block. The new tension in the spring $A$ becomes $T' = mg - F_B$. Since $F_B > 0$,$T' According to Newton's third law,the block exerts an equal and opposite downward force $F_B$ on the liquid. This force is transmitted to the beaker and consequently to the pan of balance $B$. Thus,the new reading of balance $B$ becomes $W_{beaker+liquid} + F_B$. Since $F_B > 0$,the reading of balance $B$ will be more than $5 \ kg$.

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