In a meter bridge experiment resistances are connected as shown in the figure. Initially resistance P = 4 Ω and the neutral point N is at 60 cm from A. Now an unknown resistance R is connected in series to P and the new position of the neutral point is at 80 cm from A. The value of unknown resistance R is :

Question

In a meter bridge experiment resistances are connected as shown in the figure. Initially resistance P = 4 Ω and the neutral point N is at 60 cm from A. Now an unknown resistance R is connected in series to P and the new position of the neutral point is at 80 cm from A. The value of unknown resistance R is : (A) (33/5) Ω (B) 6 Ω (C) 7 Ω (D) (20/3) Ω

Tardigrade Admin · Accepted Answer

Initially (4/60)=(Q/40) Q=(16/6)=(8/3) Ω Finally (4+R/80)=(Q/20)=(8/60) 4+R=(64/6) R=(64/6)-4=(64-24/6)=(40/6) =(20/3) Ω

$\frac{33}{5} Ω$

$6Ω$

$7Ω$

$\frac{20}{3} Ω$

Initially $\frac{4}{60} = \frac{Q}{40} Q = \frac{16}{6} = \frac{8}{3} Ω$
Finally $\frac{4 + R}{80} = \frac{Q}{20} = \frac{8}{60}$
$4 + R = \frac{64}{6}$
$R = \frac{64}{6} - 4 = \frac{64 - 24}{6} = \frac{40}{6}$
$= \frac{20}{3} Ω$

533​Ω

6Ω

7Ω

320​Ω

Initially 604​=40Q​Q=616​=38​Ω Finally 804+R​=20Q​=608​ 4+R=664​ R=664​−4=664−24​=640​ =320​Ω

$\frac{33}{5} Ω$

$6Ω$

$7Ω$

$\frac{20}{3} Ω$

Initially $\frac{4}{60} = \frac{Q}{40} Q = \frac{16}{6} = \frac{8}{3} Ω$
Finally $\frac{4 + R}{80} = \frac{Q}{20} = \frac{8}{60}$
$4 + R = \frac{64}{6}$
$R = \frac{64}{6} - 4 = \frac{64 - 24}{6} = \frac{40}{6}$
$= \frac{20}{3} Ω$