Explanation

The height of rise of liquid in a capillary tube is given by the relation

$$ h = \frac{2T \cos\theta}{\rho g r} $$

Here, $T$ is the surface tension of the liquid, $\rho$ is the density of the liquid, $r$ is the inner radius of the capillary tube and $g$ is acceleration due to gravity.

Since $\cos\theta$ and $g$ are constants, we can write

$$ h \propto \frac{T}{\rho r} $$

Taking percentage change on both sides,

$$ \frac{\Delta h}{h} = \frac{\Delta T}{T} - \frac{\Delta \rho}{\rho} - \frac{\Delta r}{r} $$

Given that surface tension decreases by $1\%$, density decreases by $1\%$ and radius decreases by $1\%$,

$$ \frac{\Delta T}{T} = -1\%, \quad \frac{\Delta \rho}{\rho} = -1\%, \quad \frac{\Delta r}{r} = -1\% $$

Substitute these values,

$$ \frac{\Delta h}{h} = (-1) - (-1) - (-1) $$

$$ \frac{\Delta h}{h} = -1 + 1 + 1 = +1\% $$

Thus, the height of the liquid in the tube increases by

1%

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