A) 2 V
B) 3 V
C) 1 V
D) 1.5 V
Correct Answer: C
Solution :
Total current drawn from the battery \[i=\frac{E}{R+r}=\frac{6}{100+0}=0.06A\] Resistance of 50 cm wire is \[R=\frac{\rho l}{A}=\left( \frac{\rho }{A} \right)l\] \[=\left( \frac{R}{l} \right)l\] \[\left( \because R=\frac{\rho l}{A} \right)\] \[=\frac{100}{300}\times 50\] So, \[R=\frac{50}{3}\Omega \] Hence, the potential difference between two points on the wire separated by a distance\[l\]is \[V=iR=0.06\times \frac{50}{3}=1\,V\]
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