Physics Q8 – NEET-UG 2024

Home Physics Mechanical Properties of FluidsQuestion #15

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 :

Clue 1 of 3: Surface Tension Force

The force due to surface tension acts along the line of contact between the liquid and the object.

🧠 Full Solution Path

Step 1: Identify Given Parameters and Formula✦ Active

The radius of the circular disc is $r = 4.5 cm = 0.045 m$ . The surface tension of water is $σ = 0.07 {Nm}^{- 1}$ . The force required to lift the disc from the surface is due to surface tension acting along the circumference of the disc. The formula for this force is $F = σ L$ , where $L$ is the length of contact.

Step 2: Calculate the Length of Contact○ Expand

For a circular disc, the length of contact $L$ is its circumference:

L = 2 π r

Substitute the radius:

L = 2 \times π \times 0.045 m = 0.09 π m

💡 Teacher's Secret Hint

Ensure all units are converted to SI (meters for length) before calculation to avoid errors.

Step 3: Calculate the Force and Match with Options○ Expand

Now, calculate the force $F$ :

F = σ L = 0.07 {Nm}^{- 1} \times 0.09 π m

Using $π \approx 3.14159$ :

F = 0.07 \times 0.09 \times 3.14159 N \approx 0.0063 \times 3.14159 N \approx 0.019792 N

Convert the force to millinewtons (mN):

F \approx 0.019792 \times 1000 mN \approx 19.792 mN

Rounding to one decimal place, $F \approx 19.8 mN$ . This matches option (4).

💡 Teacher's Secret Hint

Pay close attention to unit conversions, especially between Newtons and milliNewtons, as options often differ by powers of ten.

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Physics Question 8 – NEET-UG 2024

Clue 1 of 3: Surface Tension Force

🧠 Full Solution Path