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A spherical drop of coater has radius 1 mm if surface tension of contex is $70 \times 10^ {-3} N / m $ difference of pressures between inside and outside of the spherical drop is
$ \triangle p = { 2T \over R } $
In capilley pressure below the curved surface at water will be
Two bubbles A and B (A>B) are joined through a narrow tube than
$ r_A \gt r_B and p \alpha {1 \over r } $ $ So P_A \lt P_B $ So air will flow from B to A i.e. size ofA will increase
A capillary tube at radius R is immersed in water and water rises in it to a height H. Mass of water in the capillary tube is M. If the radius of the tube is doubled. Mass of water that will rise in the capillary tube will now be
Mass of liquid in capillary tube $ M = \pi R^2 - \rho $ $ M \alpha R^2 \times (1 / R } $ $ M \alpha R $ If radius become double then mass will become twice.
A vessel whose bottom has round holes with diametre of 0.1 mm is filled with water. The maximum height to which the water can be filled without leakage is $ ( S.T of water = 75 dyne /cm , g = 1000 m/s^2 ) $
The correct relation is
The correct relation for capillary rise is $r = rac{2T ext{ cos } heta}{hdg}$. Here, $r$ is the radius, $T$ is the surface tension, $ heta$ is the contact angle, $h$ is the height of the liquid column, $d$ is the density of the liquid, and $g$ is the acceleration due to gravity. Hence, option o1 is correct.
In a capillary tube water rises by 1.2 mm. The height of water that will rise in another capillary tube having half the radius of the first is
$ h = { 2T \over rdg} $ $ \therefore h \alpha { 1 \over r }$ $ \therefore r_1 h_1 = r_2 h_2 $
Water rises in a vertical capillary tube upto a beight of 2.0 cm. If tube is inclined at an angle of $ 60^ \circ $ with the verticalthen the what length the water will rise in the tube.
$l = { h \over cos \theta } = { \alpha \over cos 60 } = 4.0 cm $
The lower end of a glass capillary tube is dipped in water rises to a height of 8 cm the tube is then broken at a height of 6 cm. The height of water column and engled as contact will be
When a capillary tube is broken at a height of 6 cm the height of water column will be 6 cm. $ As h = { 25 cos \theta \over \rho rg } or { h \over cos \theta } = constant $ $ { 8 \over cos 0 ^\circ } = { 6 \over cos \theta } or cos \theta = { 6 cos \theta \over 8 } = {3 \over 4} $ $ \therefore \theta = cos ^ {-1} \left( {3 \over 4} \right) $
A large number of water drops each of rdius r combine to have a drop of radius R. If the surface tension is T and the mechanical equivalent at heat is J then the rise in tempreature will be
Rise in tempreture $ \triangle \theta = { 3T \over Jsd } \left( {1 \over r } - { 1 \over R} \right) $ $ \triangle \theta = { 3T \over J} \left( {1 \over r } - { 1 \over R} \right) $ (For water S = 1 and d = 1)
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