Question

When a part of a straight capillary tube is placed vertically in a liquid, the liquid rises upto certain height \( h \). If the inner radius of the capillary tube, density of the liquid and surface tension of the liquid decrease by \(1%\) each, then the height of the liquid in the tube will change by ________ %.

• A. \(-1\)
• B. \(-3\)
• C. \(+1\)
• D. \(+3\)

Hint:

For capillary rise problems, remember \( h \propto \dfrac{T}{\rho r} \). Use logarithmic differentiation to find percentage changes quickly.

Answer 1

The Correct Option is C

The height of rise of liquid in a capillary tube is given by \[ h = \frac{2T \cos\theta}{\rho g r}, \] where \( T \) = surface tension, \( \rho \) = density of the liquid, \( r \) = radius of the capillary tube.

Step 1: Write the relation for fractional change.
From the formula, \[ h \propto \frac{T}{\rho r}. \] Taking fractional change, \[ \frac{\Delta h}{h} = \frac{\Delta T}{T} - \frac{\Delta \rho}{\rho} - \frac{\Delta r}{r}. \]
Step 2: Substitute the given percentage changes.
Given: \[ \frac{\Delta T}{T} = -1%, \quad \frac{\Delta \rho}{\rho} = -1%, \quad \frac{\Delta r}{r} = -1%. \] \[ \frac{\Delta h}{h} = (-1) - (-1) - (-1) = +1%. \]
Step 3: Interpret the result.
The height of the liquid column increases by \(1%\).

Final Answer: \[ \boxed{+1%} \]