A spring balance A reads 2 ,kg with a block o | vedclass

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(C) When the block is immersed in the liquid,it experiences an upward buoyant force $F_b$ due to the liquid.For balance $A$,the reading corresponds to

Vedclass · Accepted Answer

the balance $A$ will read less than $2 \,kg$ and $B$ will read between $3 \,kg$ and $5 \,kg$

Vedclass · Answer

(C) When the block is immersed in the liquid,it experiences an upward buoyant force $F_b$ due to the liquid.For balance $A$,the reading corresponds to the tension $T$ in the spring. Initially,$T = mg = 2 \,kg 	imes g$. When immersed,$T' = mg - F_b$. Since $F_b > 0$,the new reading $T'$ will be less than $2 \,kg$.For balance $B$,the initial reading is the weight of the beaker plus the liquid. When the block is immersed,by Newton's third law,the block exerts an equal and opposite downward force $F_b$ on the liquid. Therefore,the new reading on balance $B$ will be the initial weight plus the buoyant force $F_b$. Since $F_b > 0$,the new reading will be greater than $3 \,kg$. However,the total weight of the system (beaker + liquid + block) is $2 \,kg + 3 \,kg = 5 \,kg$. Since the block is not resting on the bottom of the beaker but is supported by spring $A$,the reading on $B$ will be less than the total weight of $5 \,kg$.Thus,balance $A$ reads less than $2 \,kg$ and balance $B$ reads between $3 \,kg$ and $5 \,kg$.

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