Question

A piece of metal of specific gravity 7 floats in mercury of specific gravity 13.6 what fraction of its volume is under mercury

Answer 1

Answer:

The exact answer is 0.515

Explanation:

Answer 2

Answer:

The fraction of metal's volume which is under mercury is 0.5147.

Concept:

Specific Gravity: Specific gravity is defined as the ratio of the density of the body to the density of water at 4° C.

Density of water at 4° C, $\rho_{water}$ = 1000 kg/m³

Solution:

Specific gravity of metal, S₁ = 7

But Specific gravity = $\frac{Density\hspace{1mm} of \hspace{1mm} metal}{Density\hspace{1mm} of \hspace{1mm} water \hspace{1mm} at\hspace{1mm} 4\textdegree C}$

$S_1 = \frac{\rho_{metal}}{\rho_{water}}$

$\rho_{metal} = S_1\times \rho_{water}\\\\\rho_{metal} = 7\times 1000\\\\\rho_{metal} = 7000 kg/m^3$

Specific gravity of mercury, S₂ = 13.6

But Specific gravity = $\frac{Density\hspace{1mm} of \hspace{1mm} mercury}{Density\hspace{1mm} of \hspace{1mm} water \hspace{1mm} at\hspace{1mm} 4\textdegree C}$

$S_2 = \frac{\rho_{mercury}}{\rho_{water}}$

$\rho_{mercury} = S_2\times \rho_{water}\\\\\rho_{mercury} = 13.6\times 1000\\\\\rho_{mercury} = 13600 kg/m^3$

Now, applying equilibrium condition, we get

Weight of metal piece (W) = Byoyancy force (F)

mg = F

But $m = \rho_{metal} \times V_{metal}\\F = \rho_{mercury} \times V_{displaced} \times g$

Putting the above values in the equation, we get

$\rho_{metal} \times V_{metal}\times g = \rho_{mercury} \times V_{displaced}\times g \\\rho_{metal} \times V_{metal} = \rho_{mercury} \times V_{displaced}\\\\\frac{V_{displaced}}{V_{metal}} = \frac{\rho_{metal}}{\rho_{mercury}}$

              $= \frac{7000}{13600}\\\\= \frac{70}{136} \\\\= 0.5147$

Hence, the fraction of metal's volume which is under mercury is 0.5147.

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