A galvanometer gives full scale deflection with 0.006 A current. By connecting in to a `4990 Omega` resistance, it can be converted into a voltmeter of range 0-30V. If connected to a `(2n)/(249)Omega` resistance, it becomes an ammeter of range `0-1.5A`. The value of n is

Answer

`(5)`

Explanation

Here, `I_g=0*006A`, `R=4990Omega`, `V=30V`. Let G be the resistance of galvanometer, then
`V=I_g(G+R)`
or `G=V/I_g-R=(30)/(0*006)-4990=10Omega`
When galvanometer works as an ammeter, then shunt resistance required is
`S=(I_gG)/(I-I_g)=(0*006xx10)/((1*5-*006))=(6)/(149*4)Omega`
As per question,
`(2n)/(249)=(6)/(149*4)` or `n=(6xx249)/(2xx149*4)=5`