Question

A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is 0.07 Nm^-1, then the excess force required to take it away from the surface is :

• A. 19.8 mN
• B. 198 N
• C. 1.98 mN
• D. 99 N

Answer 1

The Correct Option is A

Step 1: Use the Formula for Force Due to Surface Tension

The force required to overcome surface tension is given by:

$$ F = 2\pi r \cdot T $$

Where:

\( F \) = Force due to surface tension

\( r \) = Radius of the liquid surface

\( T \) = Surface tension

Step 2: Convert Radius to Meters

Given:

$$ r = 4.5 \text{ cm} = 0.045 \text{ m} $$

Step 3: Substitute the Values

Given surface tension:

$$ T = 0.07 \text{ N/m} $$

Substituting the values into the formula:

$$ F = 2\pi (0.045) (0.07) $$

Step 4: Solve for \( F \)

Calculating:

$$ F = 0.0198 \text{ N} $$

Converting to milliNewtons:

$$ F = 19.8 \text{ mN} $$

Conclusion

The force required to overcome surface tension is 19.8 mN.

Answer 2

The excess force required to take the disc away from the water surface is calculated using the formula for surface tension:\(F = 2\pi r \cdot T\) where \(r\) is the radius of the disc and \(T\) is the surface tension of water. Given: \(r = 4.5 \text{ cm} = 0.045 \text{ m}\) and \(T = 0.07 \text{ Nm}^{-1}\).

Now, substitute the known values into the equation:

\(F = 2 \times \pi \times 0.045 \times 0.07\)

\(F \approx 0.0198 \text{ N}\).

To convert Newtons to milliNewtons, multiply by 1000:

\(F = 0.0198 \times 1000 = 19.8 \text{ mN}\).

Therefore, the correct answer is 19.8 mN.