A) \[\simeq 1.47V\]
B) \[\simeq 1.57V\]
C) \[\simeq 1.37V\]
D) \[\simeq 1.67V\]
Correct Answer: D
Solution :
: As \[\frac{{{\varepsilon }_{1}}}{{{\varepsilon }_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\] where \[{{\varepsilon }_{1}}\] and \[{{\varepsilon }_{2}}\] are the emfs of two cells and \[{{l}_{1}}\] and \[{{l}_{2}}\] are the corresponding balancing lengths of the potentiometer wire. \[\therefore \] \[{{\varepsilon }_{2}}=\frac{{{l}_{2}}}{{{l}_{1}}}{{\varepsilon }_{1}}\] Here, \[{{\varepsilon }_{1}}=1.25V,\,{{l}_{1}}=30cm,\,{{l}_{2}}=40cm\] \[\therefore \] \[{{\varepsilon }_{2}}=\frac{40}{30}\times 1.25=1.67V\]
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