First slide

Fluid statics

Question

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

Easy

A

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

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

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

$\frac{a}{g} T$

Solution

here buoyant force is greater than weight of the block $T = {vd}_{l} g - {vd}_{s} g$
When it moves up with acceleration ‘a’.
$T^{'} = ({vd}_{l} - {vd}_{s}) (g + a)$
$= ({vd}_{l} - {vd}_{s}) g (1 + \frac{a}{g}) = T (1 + \frac{a}{g})$

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