Th resistance of the wire is 121 ohm. It is divided into 'n' equal parts and they are connected in parallel, then effective resistance is 1 ohm. the value of 'n' is

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Solution

The correct option is C 11
$\frac{1}{R_{e q}} = \frac{n}{R} + \frac{n}{R} + - - - - - - - n t e r m s$
$= n . \frac{n}{R}$ from date
$= \frac{n^{2}}{R}$ $R_{e} q .1$ $R = 121$
$R_{e q} = \frac{R}{n^{2}}$
$= \frac{121}{n^{2}}$
$n^{2} = 121$
$n = \sqrt{121} = 11$
$∴ n = 11$

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Th resistance of the wire is 121 ohm. It is divided into 'n' equal parts and they are connected in parallel, then effective resistance is 1 ohm. the value of 'n' is

The correct option is C 111Req=nR+nR+−−−−−−−n terms=n.nR from date=n2R Req.1 R=121 Req=Rn2=121n2n2=121n=√121=11∴n=11

The correct option is C 11
$\frac{1}{R_{e q}} = \frac{n}{R} + \frac{n}{R} + - - - - - - - n t e r m s$
$= n . \frac{n}{R}$ from date
$= \frac{n^{2}}{R}$ $R_{e} q .1$ $R = 121$
$R_{e q} = \frac{R}{n^{2}}$
$= \frac{121}{n^{2}}$
$n^{2} = 121$
$n = \sqrt{121} = 11$
$∴ n = 11$