In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell ?

Answer

`e = 0.84, M = 15.21 kg`

Explanation

For body of mass m from A to B `u = 10 m//s` (given)
`a = -[(mg sin theta + f)/(m)] = -[(mg sin theta + mumg cos theta)/(m)]`
`= -[g sin theta + mug cos theta] = - g [ sin theta + mu cos theta]`
`= - 10 [0.05 + 0.25 xx 0.99]`
`= - 2.99 m//s^(2)`
`v^(2) - u^(2) = 2as`
`rArr v = sqrt(100 + 2 - (-2.99) xx 6) = 8 m//s`

After collision
Let `v_(1)` be the velocity of mass m after collision. Body `v_(2)` be the velocity of mass M after collision. Body of mass M moving from B to C and Coming to rest.
`u = v_(2), v = 0 " " a = - 2.99 m//s^(2)`
and `s = 0.5 " " v^(2) - u^(2) = 2as`
`rArr (0)^(2) = v_(2)^(2) = 2(-2.99) xx 0.5`
`rArr v_(2)^(2) = 1.73 m//s`
Body of mass m mocing from B to A after collision
`u = v_(1), v = +1 m//s`
`(K.E. + P.E.)_("initial") = (K.E. + P.E.)_("final") + W_("frication")`
`(1)/(2)mv_(1)^(2) + mgh = (1)/(2)mv^(2) + 0 + mugs`
`(1)/(2)v_(1)^(2) + 10 xx (6 xx 0.05) = (1)/(2) (1)^(2) + 0.25 xx 10 xx 6`
`v_(1) = -5 m//s`
`sin theta = (h)/(6)`
`h = 6 sin theta = 6 xx 0.05`

`:.` Coefficient of restituation
`e = |("Relative velocity of sepaeration")/("Relative velocity of approach")|`
`= |(-5 - 1.73)/(8 - 0)| = 0.84`