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

Answer

$V=I R=1 \times(90+910)=1000 \mathrm{V}$

Given least count of voltmeter $=0.1 V$ 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 $=100$

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