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Current Electricity — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsCurrent Electricity1 Mark
MCQ
Two unknown resistances are connected in two gaps of a meter$-$bridge. The null point is obtained at $40 \ cm$ from left end. A $30 \Omega$ resistance is connected in series with the smaller of the two resistances, the null point shifts by $20 \ cm$ to the right end. The value of smaller resistance in $\Omega$ is
  • A
    $12$
  • ✓
    $24$
  • C
    $36$
  • D
    $48$
✓

Answer

Correct option: B.
$24$
$l_x=40 \ cm ,$
$l_R=60 \ cm$
$\frac{x}{R}=\frac{l_x}{l_R}=\frac{40}{60}=\frac{2}{3}. . . . .(i)$
$\therefore \frac{x+30}{R}=\frac{60}{40}=\frac{3}{2} ;$
$2(x+30)=3 R$
$R=\frac{2(x+30)}{3}......(ii)$
$($From $(i)$ and $(ii))$
$\frac{3 x}{2(x+30)}=\frac{2}{3}$
$\Rightarrow 9 x=4 x+120$
$ 5 x=120$
$\Rightarrow x=24 \Omega$

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