Two resistors 2 Ω and 3 Ω are connected in the gaps of bridge as shown in figure. The null point is obtained with the contact of jockey at some point on wire XY. When an unknown resistor is connected in parallel with 3 Ω resistor, the null point is shifted by 22.5 cm toward Y. The resistance of unknown resistor i $s____Ω.$

Question

G1 Physics · Accepted Answer

Initially, let the balance point be x cm from X. For a meter bridge, 2/3 = x/(100-x). Therefore, 2(100-x) = 3x, so x = 40 cm. After connecting an unknown resistance R in parallel with 3 Ω, the new right-gap resistance is R(right) = 3R/(3+R). The balance point shifts by 22.5 cm toward Y. Hence the new balance length from X is 40+22.5 = 62.5 cm. The other wire length is 100-62.5 = 37.5 cm. Therefore, 2/[3R/(3+R)] = 62.5/37.5 = 5/3. Solving, 2(3+R)/(3R) = 5/3. Thus, 6(3+R) = 15R. Hence, 18+6R = 15…