A) 1%
B) 0.8%
C) 0.1%
D) 8%
Correct Answer: B
Solution :
Let E = emf of cell. Current in the circuit is \[i=\frac{E}{R+r}=\frac{E}{500+4}\] |
Potential drop across terminals of voltmeter is \[V=iR=500\times \frac{E}{500+4}=\frac{500E}{504}\] |
So, voltmeter reads \[\frac{500E}{504}\]instead of E. |
Hence, per cent error in reading is |
Per cent error \[=\frac{\left( E-\frac{500}{504}E \right)}{\left( \frac{500}{504}E \right)}\times 100\approx 0.85\] |
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