Two unknown resistances X and Y are connected to left and right gaps of a meter bridge and the balancing point is obtained 80cm from left when circuit is switched on. Now 10 ohm resistance is connected in parallel to X, it is observed that the balancing point is 50 cm from left when circuit is switched on. The values of X and Y respectively are

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$9 Ω, 40 Ω$

$3 Ω, 10 Ω$

$40 Ω, 9 Ω$

$30 Ω, 7.5 Ω$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The balancing length in first case is 80cm
Balancing length for meter bridge is $\frac{R}{S} = \frac{l}{100 - l}$
$\frac{X}{Y} = \frac{80}{20}$
$\Rightarrow X = 4 Y$
When $10 Ω$ resistance is connected in parallel to X .
The resistance of the left gap is $\frac{10 X}{10 + X}$
according to balancing condition of meter bridge $\frac{R}{S} = \frac{l}{100 - l}$
$\frac{10 X}{(10 + X) Y} = \frac{50}{50}$
$(\frac{R}{S} = \frac{l'}{100 - l'}), l' is given 50 cm$
On solving above two equations we get $X = 30 Ω$ and $Y = 7.5 Ω$
Therefore, the correct answer is (C).