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(B) The volume of the cubical block is $V = l^3$. Let $h$ be the height of the block above the surface of the mercury. The depth of the block submerge

Vedclass · Accepted Answer

$l(1 - \frac{\rho_s}{\rho_m})$

Vedclass · Answer

(B) The volume of the cubical block is $V = l^3$. Let $h$ be the height of the block above the surface of the mercury. The depth of the block submerged in mercury is $(l - h)$. The volume of mercury displaced is $V_{disp} = (l - h)l^2$. According to the law of floatation,the weight of the displaced mercury must equal the weight of the steel block. Weight of displaced mercury = $(l - h)l^2 \rho_m g$. Weight of the steel block = $l^3 \rho_s g$. Equating the two: $(l - h)l^2 \rho_m g = l^3 \rho_s g$. Dividing both sides by $l^2 g$: $(l - h) \rho_m = l \rho_s$. $l - h = l \frac{\rho_s}{\rho_m}$. $h = l - l \frac{\rho_s}{\rho_m} = l(1 - \frac{\rho_s}{\rho_m})$.

$A$ cubical block of steel of each side equal to $l$ is floating on mercury in a vessel. The densities of steel and mercury are $\rho_s$ and $\rho_m$. The height of the block above the mercury level is given by:

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