Question

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

• A. $90$
• B. $91$
• C. $100$
• D. $none$

Answer 1

The correct option is C $100$

We know,

$V=IR=1\times (90+910)=1000V$

Given,

least count of voltmeter $=0.1V$ i.e. 0.1 volts/division

hence, $V=\frac{\text{Total no. of divisions}}{\text{least count}}$

$1000=\frac{\text{Total no. of divisions}}{0.1}$

Total no. of divisions $=100$