A cubical block of wood having a mass of 160 | vedclass

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(B) Volume of the wood block $V_w = \frac{	ext{Mass}}{	ext{Density}} = \frac{160}{0.8 	imes 1} = 200 \,cm^3$. Let the mass of the metal piece be $x

Vedclass · Accepted Answer

$44.4$

Vedclass · Answer

(B) Volume of the wood block $V_w = \frac{	ext{Mass}}{	ext{Density}} = \frac{160}{0.8 	imes 1} = 200 \,cm^3$. Let the mass of the metal piece be $x \,g$. Volume of the metal piece $V_m = \frac{	ext{Mass}}{	ext{Density}} = \frac{x}{10 	imes 1} = \frac{x}{10} \,cm^3$. For the system to float in water, the total weight of the system must be equal to the buoyant force (upthrust) exerted by the water when the entire system is submerged. Total weight $W = (m_w + m_m)g = (160 + x)g$. Total volume of the system $V_{total} = V_w + V_m = (200 + \frac{x}{10}) \,cm^3$. Buoyant force $F_B = V_{total} 	imes ho_{water} 	imes g = (200 + \frac{x}{10}) 	imes 1 	imes g$. Equating weight and buoyant force: $(160 + x)g = (200 + \frac{x}{10})g$. $160 + x = 200 + 0.1x$. $0.9x = 40$. $x = \frac{400}{9} \approx 44.4 \,g$.

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