The diameter of rain-drop is $0.02 \,cm$. If surface tension of water be $72 \times {10^{ - 3}}\,newton$ per metre, then the pressure difference of external and internal surfaces of the drop will be
$1.44 \times {10^4}\,dyne - c{m^{ - 2}}$
$1.44 \times {10^4}\,newton - {m^{ - 2}}$
$1.44 \times {10^3}\,dyne - c{m^{ - 2}}$
$1.44 \times {10^5}\,newton - {m^{ - 2}}$
The excess pressure inside a soap bubble is thrice the excess pressure inside a second soap bubble. The ratio between the volume of the first and the second bubble is:
A $U-$ tube with limbs of diameters $5\, mm$ and $2\, mm$ contains water of surface tension $7 \times 10^{-2}$ newton per metre, angle of contact is zero and density $10^3\, kg/m^3$. If $g$ is $10 \,m/s^2$, then the difference in level of two limbs is :-
A cylinder with a movable piston contains air under a pressure $p_1$ and a soap bubble of radius $'r'$ . The pressure $p_2$ to which the air should be compressed by slowly pushing the piston into the cylinder for the soap bubble to reduce its size by half will be: (The surface tension is $\sigma $ , and the temperature $T$ is maintained constant)
If the excess pressure inside a soap bubble is balanced by oil column of height $2\; mm$, then the surface tension of soap solution will be ($r = 1 \,cm$ and density $d = 0.8\, gm/cc$)
A bubble of $8$ mm diameter is formed in the air. The surface tension of soap solution is $30$ dynes/cm. The excess pressure inside the bubble is ........ $dynes/cm^2$