A voltmeter of resistance 400 ohm is used to measure emf of a cell of internal resistance 2 ohm. The percentage error in reading of voltmeter is nearly equal to

1 %

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0.5 %

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1.25 %

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0.8 %

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Solution

The correct option is B 0.5 %

$R_{e q} = r + R = 402 Ω$

Reading of voltmeter is potential difference across cell

$= V = E - I r = E - \frac{E r}{r + R}$

$⟹ V = \frac{R}{R + r} E$

$⟹ \frac{V}{E} = \frac{R}{R + r}$

Error $\frac{Δ E}{E} = \frac{E - V}{E} = 1 - \frac{V}{E}$

$⟹ \frac{Δ E}{E} = \frac{r}{R + r} = \frac{2}{402} \times 100$

$= 0.5 %$

Answer-(B)

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A voltmeter of resistance 400 ohm is used to measure emf of a cell of internal resistance 2 ohm. The percentage error in reading of voltmeter is nearly equal to

The correct option is B 0.5 %Req=r+R=402ΩReading of voltmeter is potential difference across cell=V=E−Ir=E−Err+R⟹V=RR+rE⟹VE=RR+rError ΔEE=E−VE=1−VE⟹ΔEE=rR+r=2402×100=0.5%Answer-(B)