There is a small hole in a hollow sphere. Wat | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For water to enter the hole,the pressure due to the depth of water must overcome the capillary pressure (excess pressure) at the hole.The pressure

Vedclass · Accepted Answer

$0.07$

Vedclass · Answer

(B) For water to enter the hole,the pressure due to the depth of water must overcome the capillary pressure (excess pressure) at the hole.The pressure due to depth is $P = h ho g$.The excess pressure due to surface tension for a hole of radius $r$ is $P_{ex} = \frac{2T}{r}$.Equating the two,we have $h ho g = \frac{2T}{r}$.Given: $h = 40 \, cm = 0.4 \, m$,$ho = 1000 \, kg/m^3$,$g = 10 \, m/s^2$,$T = 0.07 \, N/m$.Rearranging for $r$: $r = \frac{2T}{h ho g} = \frac{2 	imes 0.07}{0.4 	imes 1000 	imes 10} = \frac{0.14}{4000} = 0.000035 \, m = 0.035 \, mm$.The diameter $D = 2r = 2 	imes 0.035 \, mm = 0.07 \, mm$.

There is a small hole in a hollow sphere. Water enters it when it is taken to a depth of $40 \, cm$ under water. The surface tension of water is $0.07 \, N/m$. The diameter of the hole is ...... $mm$.

Similar Questions

The $EN$ (Electronegativity) of element $A$ is $E_1$ and its $IP$ (Ionization Potential) is $E_2$. Hence,its $EA$ (Electron Affinity) will be:

$A$ dipole having dipole moment $p$ is placed in front of a solid uncharged conducting sphere as shown in the diagram. The net potential at point $A$ lying on the surface of the sphere is:

The ratio of minimum wavelengths of Lyman and Balmer series will be

The radius of the circle $r = 12 \cos \theta + 5 \sin \theta$ is

On heating one end of a rod,the temperature of the whole rod will be uniform when

Vedclass Test Series

Exam Paper Generator

Online Exam Module