$l_x=40 \ cm, l_R=60 \ cm$ $\frac{x}{R}=\frac{l_x}{l_R}=\frac{40}{60}=\frac{2}{3} \ldots \ldots . .(i)$ $\therefore \frac{x+30}{R}=\frac{60}{4

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

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Class 11 Chemistry Data Entry - Chemistry Level 0 of all question bank

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

MHT CET 2017 All MHT CET

A
$12$
B
$24$
C
$36$
D
$48$

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