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 Ω resistance is connected in scries with the smaller of the two resistances, the null point shifts by 20 cm to the right end. The value of smaller resistance in Ω is

Question

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 Ω resistance is connected in scries with the smaller of the two resistances, the null point shifts by 20 cm to the right end. The value of smaller resistance in Ω is (A) 12 (B) 24 (C) 36 (D) 48

Tardigrade Admin · Accepted Answer

let the reistance at right be R and at left be r (r/R)=(1x/lR)=(40/60), r is smaller (r+30/R)=(60/40) From above equation we get r =24 Ω

Q. Two unknown resistances are connected in two gaps of a meter-bridge. The null point is obtained at 40cm from left end. A 30Ω resistance is connected in scries with the smaller of the two resistances, the null point shifts by 20cm to the right end. The value of smaller resistance in Ω is

12

24

36

48

let the reistance at right be R and at left be r Rr​=lR​1x​​=6040​, r is smaller Rr+30​=4060​ From above equation we get r=24Ω

let the reistance at right be $R$ and at left be $r$
$\frac{r}{R} = \frac{1 _{x}}{l _{R}} = \frac{40}{60}$ ,
$r$ is smaller
$\frac{r + 30}{R} = \frac{60}{40}$
From above equation we get $r = 24Ω$