Question

The force required to take away a flat circular plate of radius 2 cm from the surface of water is [Surface tension of water = $70 \times 10^{-3}\ \text{Nm}^{-1}$, $\pi = \frac{22}{7}$]

• A. $4.4 \times 10^{-4}$ N
• B. $8.8 \times 10^{-3}$ N
• C. $6.6 \times 10^{-4}$ N
• D. $11 \times 10^{-3}$ N

Hint:

Canceling out the $7$ in the denominator with the $70$ from the surface tension gives a clean multiplier of $10$. Keeping track of pure powers of 10 separately ensures your decimal placement stays completely accurate!

Answer 1

The Correct Option is B

When a flat circular plate is lifted from a liquid surface, the pull force ($F$) required to overcome surface tension acts along the entire boundary perimeter of the plate touching the water. The total perimeter length of a circular plate of radius $r$ is $L = 2\pi r$. Therefore, the surface tension force is: $$F = T \cdot L = 2\pi r T$$ Given parameters: * Radius, $r = 2\ \text{cm} = 2 \times 10^{-2}\ \text{m}$ * Surface tension, $T = 70 \times 10^{-3}\ \text{Nm}^{-1}$ Substituting these values into the equation: $$F = 2 \times \frac{22}{7} \times (2 \times 10^{-2}\ \text{m}) \times (70 \times 10^{-3}\ \text{Nm}^{-1})$$ $$F = 2 \times 22 \times 2 \times 10^{-2} \times 10 \times 10^{-3}$$ $$F = 88 \times 10^{-4} = 8.8 \times 10^{-3}\ \text{N}$$
Final Answer:
The force required to pull the plate away from the surface is $8.8 \times 10^{-3}$ N, matching option (B).