Question

In a capillary tube of radius 'R', a straight thin metal wire of radius 'r' is inserted symmetrically and one end of the combination is dipped vertically in water such that lower end of the capillary and thin wire are at same level. If 'T' is the surface tension of water and '\(ρ\)' is the density of water then the rise of water in the capillary is

• A. \(\frac{T}{(R-r)ρg}\)
• B. \(\frac{2T}{(R-r)ρg}\)
• C. \(\frac{4T}{(R+r)ρg}\)
• D. \(\frac{T}{(R+r)ρg}\)

Hint:

The wetted perimeter is 2 pi (R + r), and the water column cross-section is pi (R^2 - r^2).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The upward pull of surface tension acts along the wetted perimeter and balances the weight of the liquid column.

Step 2: Upward force:
Water touches the capillary wall (perimeter \(2\pi R\)) and the wire (perimeter \(2\pi r\)). Assuming zero contact angle, the total upward force is \(T\cdot2\pi(R+r)\).

Step 3: Weight of the column:
The area of cross-section of the water column is \(\pi(R^2-r^2)\). Weight \(=\pi(R^2-r^2)h\rho g\).

Step 4: Equate and solve:
\(T\cdot2\pi(R+r)=\pi(R^2-r^2)h\rho g\). Cancel \(\pi(R+r)\):
\[ h=\dfrac{2T}{(R-r)\rho g} \]
Option B.

Step 5: Why the other options are wrong.
Option A drops the factor of 2. Options C and D have \(R+r\) in the denominator, which comes from using the area \(\pi(R+r)^2\) or other wrong geometry.

Final Answer:
The rise is 2T / ((R - r) rho g). \[ \boxed{\text{(B) }\dfrac{2T}{(R-r)\rho g}} \]