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 needle of length 7 cm floats on water with surface tension T = 70 dyn/cm; find the maximum weight it can have.

Step 2: Key Formula or Approach:
The water film contacts both long sides of the needle, so total upward force F = 2TL. At equilibrium, W = 2TL.

Step 3: Detailed Explanation:
W = 2 × 70 × 7 = 140 × 7 = 980 dyn.

Step 4: Final Answer:
Maximum weight is 980 dyn, matching option (A).