Question

A needle is $7\text{ cm}$ long. Assuming that the needle is not wetted by water, what is the weight of the needle, so that it should float on water surface? [Surface tension of water = $70\text{ dyn/cm}$, $g = 980\text{ cm/s}^2$]

• A. $980\text{ dyn}$
• B. $490\text{ dyn}$
• C. $70\text{ dyn}$
• D. $68600\text{ dyn}$

Hint:

The most common mistake in needle floating problems is forgetting the multiplier of 2. A floating needle always creates two free surface contact lines with the liquid film (one on the left side and one on the right side), so the supporting force is always doubled: $F = 2TL$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A non-wetting needle of length $L = 7\text{ cm}$ rests on the surface of water. We need to calculate the maximum weight $W$ of the needle such that the upward surface tension force can keep it floating on the surface.

Step 2: Key Formula or Approach:
When an elongated needle floats on a liquid surface, it creates a surface indentation line along both of its long sides. Therefore, the liquid surface film contacts the needle along two separate boundary lines, each of length $L$. The total upward force due to surface tension $T$ is: $$F_T = 2 \cdot (T \cdot L)$$ For the needle to float in equilibrium, this upward surface tension force must perfectly balance its downward gravitational weight: $W = 2TL$.

Step 3: Detailed Explanation:
Let's list the given parameters, which are already in the standard CGS unit system: $$\text{Length, } L = 7\text{ cm}$$ $$\text{Surface Tension, } T = 70\text{ dyn/cm}$$ Substitute these values into the equilibrium weight equation: $$W = 2 \times T \times L$$ $$W = 2 \times 70 \times 7$$ $$W = 140 \times 7 = 980\text{ dyn}$$

Step 4: Final Answer:
The weight of the needle must be $980\text{ dyn}$, which corresponds to option (A).