First slide

Pressure in a fluid-mechanical properties of fluids

Question

A block of silver of mass 4 kg hanging from a string is immersed in a liquid of relative density 0.72.If relative density of silver is 10, then tension in the string will be

Moderate

A

37.12 N
B

42 N
C

73 N
D

21 N

Solution

Let $ρ_{S} and ρ_{L}$ be the densities of silver and liquid, respectively, and m and V be the mass and volume, respectively, of the silver block. Therefore,
Tension in the string = mg - buoyant force
$\begin{array}{l} \Rightarrow T = ρ_{S} V g - ρ_{L} V g = (ρ_{S} - ρ_{L}) V g \\ Also, V = \frac{m}{ρ_{S}} \\ \begin{aligned} ∴ T & = (\frac{ρ_{S} - ρ_{L}}{ρ_{S}}) m g \\ = \frac{(10 - 0.72) \times 10^{3}}{10 \times 10^{3}} \times 4 \times 10 = 37.12 N \end{aligned} \end{array}$

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