First slide

Pressure in a fluid-mechanical properties of fluids

Question

The tension in a string holding a solid block below the surface of a liquid (where $ρ_{liquid} > ρ_{block}$ ) as in shown in the figure is T₀ when the system is at rest. Then what will be the tension in the string if the system has upward acceleration a?

Moderate

A

$T_{0} (1 - \frac{a}{g})$
B

$T_{0} (1 + \frac{a}{g})$
C

$T_{0} (\frac{a}{g} - 1)$
D

$\frac{a}{g} T_{0}$

Solution

Let v be the volume of the solid block of density $ρ$ . Let $ρ_{1}$ be the density of water. Weight of body = v $ρ$ g. When the body is immersed in water,
Tension in the string = Upward thrust - Weight of the body
$\Rightarrow T = v ρ_{1} g - v ρ g = v g (ρ_{1} - ρ)$
When the lift is moving upwards with acceleration a, the tension in the string is $T = v (ρ_{1} - ρ) (g + a)$
$From Eqs. (i) and (ii), T = T_{0} (1 + a / g)$

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