# some fiitjee open test qs

1.A long capillary tube of mass ╬ gram, radius 2mm and negligible thickness is partially immersed in a liq of surface tension 0.1N/m, The angle of contact =0 and NEGLECTING BUOYANT FORCE , The force required to hold tube vertically will be.

2.Consider two identical iron spheres A and B. A lies on a thermally insulating plate and B hangs from a thermally insulating thread.
If equal amount of heat is given to both, will the temperature in both A and B be same??

3.A rod is rotating about an axis perpendicular to it one end, with angular velocity Žē. Find tension in rod at distance x from the axis.Length of rod is L

• king_khan ·

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...

• king_khan ·

cipher 1729...can u post the ans also ?

• Samarth Kashyap ·

is the answer for the second one MŽē2(l2-x2)/2l ??????

• cipher1729 ·

1. 10.8ŽĆmN Ōēł 34mN
2.B's temp is higher
3. MŽē2(L2-x2)/2L

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

• kaymant ·

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.

• kaymant ·

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 Ōēł 34 mN.

• cipher1729 ·

Just wanted to know something, what is wrong in the following approach??
liq in capillary rises to a height h=2T/RŽüg
weight of liquid= AhgŽü = 2TA/R........................1
weight of tube = ŽĆg/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???

• Asish Mahapatra ·

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 = rŽē2dm
= rŽē2(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 = MŽē2/L*[r2/2]xL = MŽē2(L2-x2)/2L

this is the tension

•  Anonymous ·

Net force on capillary tube

= mg + 2F

F=force applied by capillary tube

mg = ŽĆ10^-2 N

2F = 2 x ( T x 2ŽĆr) = 8 x 10^-4N

Required force = 10.8mN

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