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(B) The volume of the stone is $V = \frac{m}{\rho} = \frac{0.5}{500} = 10^{-3} \, m^3$.The buoyant force exerted by the water on the stone is $F_B = \

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$2.5$

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(B) The volume of the stone is $V = \frac{m}{ho} = \frac{0.5}{500} = 10^{-3} \, m^3$.The buoyant force exerted by the water on the stone is $F_B = ho_{water} V g = 1000 	imes 10^{-3} 	imes 10 = 10 \, N$.According to Newton's third law,the stone exerts an equal and opposite reaction force on the water,which increases the downward force on the spring balance.The initial reading of the balance is $1.5 \, kg$,which corresponds to a weight of $1.5 	imes 10 = 15 \, N$.The new reading on the spring balance will be the sum of the initial weight and the reaction force from the stone: $F_{total} = 15 \, N + 10 \, N = 25 \, N$.Converting this back to mass units,the balance reading is $\frac{25 \, N}{10 \, m/s^2} = 2.5 \, kg$.

$A$ beaker containing water is placed on the platform of a spring balance. The balance reads $1.5 \, kg$. $A$ stone of mass $0.5 \, kg$ and density $500 \, kg/m^3$ is immersed in water without touching the walls of the beaker. What will be the balance reading now?

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