A galvanometer coil has a resistance `90 Omega` and full scale deflection current `10 mA`. A `910 Omega` resistance is connected in series with the galvanometer to make a voltmeter. If the least count of the voltmeter is `0.1 V` the number of divisions on its scale is

Answer

`100`

Explanation

`V_("max")=(I_("max")R_(g)+R_(s))`
`=(10xx10^(-3))(1000)`
`=10V`
L.C `=0.1V`
`N(L.C.)=V_("max")`
`impliesN=10/0.1=100`