In an experiment to find the $emf$ of a cell using a potentiometer,the length of the null point for a cell of $emf$ $1.5 \ V$ is found to be $60 \ cm$. If this cell is replaced by another cell of $emf$ $E$,the length of the null point increases by $40 \ cm$. The value of $E$ is $\frac{x}{10} \ V$. The value of $x$ is $............$
- A$24$
- B$25$
- C$23$
- D$22$
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$A$ cell in the secondary circuit gives a null deflection for $2.5 \,m$ length of a potentiometer wire having a total length of $10 \,m$. If the length of the potentiometer wire is increased by $1 \,m$ without changing the cell in the primary circuit, the new position of the null point is: (in $m$)
$A$ potentiometer wire is $4 \,m$ long and a potential difference of $3 \,V$ is maintained between its ends. The e.m.f. of the cell which balances against a length of $100 \,cm$ of the potentiometer wire is: (in $\,V$)
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View SolutionThe length of a potentiometer wire is $L$. $A$ cell of e.m.f. $E$ is balanced at a length $\frac{L}{3}$ from the positive end of the wire. If the length of the wire is increased by $\frac{L}{2}$,at what distance will the same cell give a balance point?
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