Question

Two particles of equal mass are connected to a
rope AB of negligible mass such that one is at
end A and other dividing the length of rope in the
ratio 1:2 from B. The rope is rotated about end B
in a horizontal plane. Ratio of tensions in the
smaller part to the other is (ignore effect of
gravity)
Critical
Thinking
(a) 4:3 (b) 1:4 (0) 1:2 (d) 1:3​

Answer 1

Answer:

(c) 1:2

Explanation:

Given,

Let the two particles of equal mass be 'm'.

The length of rope is 'AB'.

The mass 'm' is attached to one end 'A'.

The other mass 'm' is attached to a point 'C' at a distance that makes a ratio in the length of rope 1:2 from 'B'.

Now,

When the string is rotated with respect to 'B', tension is produced in the string.

           T = mg + ma  Newton

where T - Tension produced

           m - mass of particle

           a - acceleration produced in the string due to applied force

Since the effect of gravity is neglected

            g = 0

Therefore the equation becomes

             T = ma  Newton

The applied force is experienced in the form of rotation

             F = m$v^{2}$/r  newton

Let 'r' and 'R' be the distance of mass from end 'B' in the ratio 1:2 and 'ω' be the angular velocity of rotation. Then,

Force experienced by the particle at 'C'

            f = mrω$^{2$           (Since ω = v/r)

Similarly,

           F = mRω$^{2$

The ratio of tension produced in the string due to two particles is given by

          f : F

Substituting the value of 'f' and 'F' from the above equation and solving, we get

         f : F  = r : R

Since, r : R is in the ratio 1:2,

          f : F should be in the ratio 1:2

So, the ratio of the tensions in the smaller part to the other is (c) 1:2