
d=2Rcosθ
∴ Pressure difference across a liquid surface is
ΔP=T(R11+R21)
∵R1=RandR2=R
ΔP=T(R1+R1)
ΔP=R2T
∴ Pressure is more in the concave side hence pressure in water between the plates is lower by R2T
If two glass plates have water between them and are separated by very small distance (see figure), it is very difficult to pull them apart. It is because the water in between forms cylindrical surface on the side that gives rise to lower pressure in the water in comparison to atmosphere. If the radius of the cylindrical surface is R and surface tension of water is T then the pressure in water between the plates is lower by:

Held on 10 Apr 2015 · Verified 6 Jul 2026.
RT
4RT
R4T
R2T
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
A string $A$ of length $0.314$ m and Young's modulus $2 \times 10^{10}$ N/m$^2$ is connected to another string $B$ of length and Young's modulus both twice of those of $A$. This series combination of strings is then suspended from a rigid support and its free end is fixed to a load of mass $0.8$ kg. The net change in length of the combination is _____ mm. (radius of both the strings is $0.2$ mm and acceleration due to gravity $= 10$ m/s$^2$) (Mass of both strings is to be neglected as compared to the mass of load)
A particle of mass $m$ falls from rest through a resistive medium having resistive force, $F=-k v$, where $v$ is the velocity of the particle and $k$ is a constant. Which of the following graphs represents velocity ($v$) versus time ($t$)?
A spring of force constant $15 \mathrm{~N} / \mathrm{m}$ is cut into two pieces. If the ratio of their length is $1: 3$, then the force constant of smaller piece is $\_\_\_\_$ $\mathrm{N} / \mathrm{m}$.
Initially a satellite of 100 kg is in a circular orbit of radius $1.5 \mathrm{R}_{\mathrm{E}}$. This satellite can be moved to a circular orbit of radius $3 R_{E}$ by supplying $\alpha \times 10^{6} \mathrm{~J}$ of energy. The value of $\alpha$ is $\_\_\_\_$. (Take Radius of Earth $R_{E}=6 \times 10^{6} \mathrm{~m}$ and $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$)
A liquid drop of diameter $2$ mm breaks into $512$ droplets. The change in surface energy is $\alpha \times 10^{-6}$ J. The value of $\alpha$ is _______. (Take surface tension of liquid $= 0.08$ N/m)
Work through every JEE Main Mechanics PYQ, year by year.