**1**
Samarth Kashyap
**·**2009-04-01 17:13:59
is the answer for the second one MÏ‰^{2}(l^{2}-x^{2})/2l ??????

**66**
kaymant
**·**2009-04-01 19:01:06
2. The temperature increment of B will be greater. When we heat the spheres, they will expand. When the sphere A expands, its center of mass must rise up. Accordingly, a part of heat given is used to increase the gravitational potential energy. The remaining energy is used to increase the temperature. On the other hand, B's center of mass goes down. So it actually gets some extra energy apart from the supplied one. As such that temperature increment of B will be higher.

**66**
kaymant
**·**2009-04-01 19:10:12
1. The vertical force applied must balance (i) the force of gravity acting on the capillary, and (ii) the net vertically downward force arising due to surface tension. This second force is the combined effect of the surface tension at the lower end and the surface tension at the height upto which the liquid has climbed. Accordingly, the required net force is

F = mg + 2S(2\pi r)

Plugging in the value we get F â‰ˆ 34 mN.

**106**
Asish Mahapatra
**·**2009-04-01 19:18:49
at a distance x from the axis. The mass of the remaining rod = (L-x)M/L

now the tension there provides the necessary centripetal force for the remaining mass of rod..

So, if u take a elemental mass of length dr at a distance r from the axis, then the centrifugal force = rÏ‰^{2}dm

= rÏ‰^{2}(M/L)dr .... (1)

total centrifugal force on the (L-x)M/L is got by integrating (1) from x to L

centrifugal force on that remaining rod = MÏ‰^{2}/L*[r^{2}/2]_{x}^{L} = MÏ‰^{2}(L^{2}-x^{2})/2L

this is the tension

**1**
king_khan
**·**2009-04-02 01:39:08
kaymant sir...

plzz can u explain once again y u considered two liq. free surfaces..

y will the surface at the bottom be a free surface??

i feel it will only e the top one naa...

**1**
king_khan
**·**2009-04-02 01:39:34
cipher 1729...can u post the ans also ?

**1**
cipher1729
**·**2009-04-02 03:55:05
The answers are :

1. 10.8Ï€mN â‰ˆ 34mN

2.B's temp is higher

3. MÏ‰^{2}(L^{2}-x^{2})/2L

Thanks to everyone who solved(or tried to solve my problems)

**1**
cipher1729
**·**2009-04-02 04:02:14
Just wanted to know something, what is wrong in the following approach??

liq in capillary rises to a height h=2T/RÏg

weight of liquid= AhgÏ = 2TA/R........................1

weight of tube = Ï€g/1000 ( as it is in gram)...........2

so, total force reqd = total weight downwards =(1+2)

but the answer doesn't come this way. What's wrong???

**Anonymous**
**·**2018-02-24 01:48:21
Net force on capillary tube

= mg + 2F

F=force applied by capillary tube

mg = Ï€10^-2 N

2F = 2 x ( T x 2Ï€r) = 8 x 10^-4N

Required force = 10.8mN