A galvanometer having `30` divisions has a current sensitivity of `20muA//di vision`. It has a resistance of `25Omega`. How will you convert it into an ammeter upto 1 ampere? How will you convert this ammeter into a voltmeter up to 1 volt?

Answer

(B, C, D)

Explanation

`m_(A) u = M_(A) v_(A) + M_(B)v_(B)` and `e = 1 = (v_(B) - v_(A))/(u) rArr v_(B) = (2m_(A)u)/(m_(A) + m_(B))`
For `m_(A) gt gt M_(B), v_(B) = 2u`
For `m_(A) = m_(B), v_(B) = u`
For `m_(A) lt lt m_(B), v_(B) = 0`
kinetic energy `K_(B) = (1)/(2)m_(B)v_(B)^(2) = (2m_(B)u^(2))/((1 + (m_(B))/(m_(A)))^(2)`