A) \[\frac{1}{8l}\]
B) \[\frac{1}{4l}\]
C) \[8l\]
D) \[16l\]
Correct Answer: C
Solution :
In a meter bridge the ratio of two resistances is \[\frac{R}{R}=\frac{l}{l}\] where \[l\] and \[l\] are balancing lengths Resistance \[R=\frac{\rho l}{A}=\frac{\rho l}{\pi {{r}^{2}}}\] If material remains same \[\rho =\rho \] Given \[l=2l\] \[r=\frac{r}{2}\] \[\therefore \] \[R=\frac{\rho l}{A}\] \[=\frac{\rho 2l}{\pi {{\left( \frac{r}{2} \right)}^{2}}}\] \[=\frac{8\rho l}{\pi {{r}^{2}}}\] \[R=8R\] Therefore, the new balancing point is expected to be \[8l\].
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