Two unknown resistances X and  Y are connected in the left and right gaps of a meter bridge and the balancing point is obtained at 60 cm from the left. When a 20 Ω resistance is connected in parallel to X  the balance point is 50 cm. Calculate X and Y (in ohm)

# Two unknown resistances X and  Y are connected in the left and right gaps of a meter bridge and the balancing point is obtained at  from the left. When a $20\text{\hspace{0.17em}}\mathrm{\Omega }$ resistance is connected in parallel to X  the balance point is  Calculate X and Y (in ohm)

1. A

$10,\frac{20}{3}$

2. B

$\frac{10}{3},20$

3. C

4. D

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### Solution:

Resistance R = X Resistance  S = Y Balancing length $l=60\mathrm{cm}$ Balancing length $l=50\mathrm{cm}$ Balancing condition for meter bridge $\frac{\mathrm{R}}{\mathrm{S}}=\frac{\mathrm{l}}{100-\mathrm{l}}\frac{\mathrm{X}}{\mathrm{Y}}=\frac{60}{40}⇒\frac{3}{2}$ ...(1)

When a 20 $\Omega$ resistance is connected in parallel to X
$\frac{X}{Y}=\frac{{l}^{\mathrm{\prime }}}{100-{l}^{\mathrm{\prime }}}\frac{\frac{20X}{20+X}}{Y}=\frac{50}{50}⇒1\frac{20}{20+X}\left(\frac{3}{2}\right)=1⇒30=20+X\therefore X=10\mathrm{\Omega }\therefore Y=\frac{20}{3}\mathrm{\Omega }$  Register to Get Free Mock Test and Study Material

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