(i) The null point distance for a cell of emf \[1.0\] volt is\[312.5\text{ }cm\]. |
(ii) The null point distance for a cell of emf \[1.5\] volt is\[512.5\text{ }cm\]. |
(iii) The null point distance for a cell of emf \[1.8\] volt is\[750\text{ }cm\]. |
(iv) The null point distance for a cell of emf \[1.6\] volt is\[500\text{ }cm\]. |
The correct statements are |
A) (i),(ii) and (iii)
B) (ii), (hi) and (iv)
C) (i),(ii) and (iv)
D) (i), (hi) and (iv)
Correct Answer: D
Solution :
[d] Resistances of wires in ratio \[=\frac{\rho \times 5}{\pi {{r}^{2}}}:\frac{\rho \times 5}{4\pi {{r}^{2}}}=4:1\] Potential difference across 1st wire \[=2\times \frac{4}{5}=1.6V\] Potential difference across 2nd wire \[=2\times \frac{1}{5}=0.4V\] Hence cell of emf 1 volt and cell \[1.5\]volt will be balanced in the first wire but the cell \[1.8V,2V\]will be balanced on second wire. Length required to balance first cell \[1=\frac{1.6}{5}I,\] \[I=\frac{5}{1.6}=3.125\,m=312.5\,cm\] Length required to balance second cell \[1.5=\frac{1.6}{5}I,\] \[I=\frac{1.5\times 5}{1.6}=468.7\,cm\] Length required to balance third cell (emf\[=1.8V\]) \[=5\times \frac{5}{0.4}\times 0.2=7.5=750\,cm\]You need to login to perform this action.
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