Question

A galvanometer coil has a resistance $80\Omega$ and full scale deflection current $10\text{mA}$. $920\Omega$ resistance is connected with the galvanometer to make it a voltmeter. If the least count of the voltmeter is $0.2\text{V}$, the number of divisions on the scale is

• A. $40$
• B. $50$
• C. $60$
• D. $30$

Answer 1

The correct option is B $50$

$G=80\Omega ;R=920\Omega \text{in series}$

$I_g=10\times 10\times {10}^{-3}\text{A}$

$V=I_g(G+R)$

$=10\times {10}^{-3}(80+920)$

$\Rightarrow V=10\text{V}$

According to question, $0.2\text{V}\to 1\text{div}$

Let there be $d$ divisions for $10\text{V}$

$\therefore d=\frac{1}{0.2}\times 10=\frac{100}{2}=50\text{div}$

Hence, option (b) is the correct answer.