some fiitjee open test qs

i still can't figure out the following questions . Please help...

1.A long capillary tube of mass Î 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 ω. Find tension in rod at distance x from the axis.Length of rod is L

10 Answers

Samarth Kashyap ·

is the answer for the second one Mω2(l2-x2)/2l ??????

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 ≈ 34 mN.

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 = rω2dm
= rω2(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 = Mω2/L*[r2/2]xL = Mω2(L2-x2)/2L

this is the tension

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 ?

cipher1729 ·

The answers are :

1. 10.8πmN ≈ 34mN
2.B's temp is higher
3. Mω2(L2-x2)/2L

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

cipher1729 ·

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

Net force on capillary tube

= mg + 2F

F=force applied by capillary tube

mg = π10^-2 N

2F = 2 x ( T x 2Ï€r) = 8 x 10^-4N

Required force = 10.8mN

Aditya Singh ·

why 2F =2*T*\Pi R

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