A rigid insulated wire frame in the form of a right-angled triangle ABC is set in a vertical plane as shown in fig. two beads of equal masses m each and carrying charges `q_(1)` and `q_(2)` are connected by a cord of length l and can slide without friction on the wires.
Considering the case when the beads are stationary, determine
(i) the angel `alpha`
(ii) the tension in the cord, and
(iii) The normal reaction on the beads.
If the cord is now cut, what are the value of the charges for which the beads continue to remain stationary?

Answer

A, B, C

Explanation

Because of equilibrium of charge `q_1` `N_1=mgsin60^@+(T-F)sinalpha` …(i)
and `(T-F) cos alpha=mgcos 60^@` …(ii)


Because of equilibrium of charge `q_2`
`(T-F) sin alpha=mgcos 30^@` ...(iii)
and `N_2=(T-F) cos alpha+mg sin 30^@` ...(iv)

From (i) and (iii)
`N_1=mgsin60^@+mgcos30^@`
`=mg(sqrt3/2+sqrt3/2)=sqrt3mg`
From (ii) and (iv)
`N_2=mgcos60^@+mgsin30^@=mg(1/2+1/2)=mg`
Also, `F=k(q_1q_2)/(l^2)`
Now from eqn. (ii) and (iii), we get
`(T-F)^2cos^2alpha+(T-F)^2sin^2alpha`
`=m^2g^2cos^(2)60^@+m^2g^2cos^(2)30^@`
`implies (T-F)^2=m^2g^2[1/4+3/4]=m^2g^2`
`implies T-F=+-mg` ...(v)
`implies T=mg+F=mg+k(q_1q_2)/(l^2)` ...(vi)
[Taking positive sign]
From (ii) and (v)
`mgcosalpha=mgcos60^@impliescos alpha=cos 60^@`
`:.` `alpha=60^@`
When the string is cut, `T=0`
`:.` From (vi)
`mg=+-k(q_1q_2)/(l^2)impliesq_1q_2=+-(mgl^2)/(k)`
Now the charges should be unlike for equilibrium.