Find the excess pressure inside a mercury drop of radius 2.0 mm. The surface tension of mercury = 0.464 N/m

928 N/m²

No worries! We‘ve got your back. Try BYJU‘S free classes today!

464 N/m²

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

464 N/m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

928 N/m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
464 N/m²

The excess pressure inside the drops is $P_{2} - P_{1} = \frac{2 S}{r}$
= $\frac{2 \times 0.464}{2 \times 10^{- 3} m} \frac{N}{m} = 464 \frac{N}{m^{2}}$

Suggest Corrections

Similar questions

Find the excess pressure inside a mercury drop of radius 2.0 mm. The surface tension of mercury = 0.464 N/m

Q. Surface tension of mercury is 0.465

N m^{- 1}

. The excess pressure inside a mercury drop of diameter 6 mm is then

Q. Find the excess pressure inside a drop of mercury of radius 2 mm, a soap bubble of radius 4 mm and an air bubble of radius 4 mm formed inside a tank of water. surface tension of mercury is 0.465 n/m and soap solution and water are 0.03 N/m and 0.076 N/m respectively.

Find the excess pressure inside (a) a drop of mercury of radius 2 mm (b) a soap bubble of radius 4 mm and (c) an air bubble of radius 4 mm formed inside a tank of water. Surface tension of mercury , soap solution and water are $0.46 N m^{- 1}, 0.03 N m^{- 1}$ and $0.076 N m^{- 1}$ respectively.

Join BYJU'S Learning Program

Find the excess pressure inside a mercury drop of radius 2.0 mm. The surface tension of mercury = 0.464 N/m

The correct option is B 464 N/m2 The excess pressure inside the drops is P2−P1=2Sr =2×0.4642×10−3mNm=464Nm2

The correct option is B
464 N/m²

The excess pressure inside the drops is $P_{2} - P_{1} = \frac{2 S}{r}$
= $\frac{2 \times 0.464}{2 \times 10^{- 3} m} \frac{N}{m} = 464 \frac{N}{m^{2}}$