The resistance in the two arms of the meter bridge are $5 Ω$ and $R Ω$ respectively. When the resistance ‘R’ is shunted with an equal resistance , the new balance point is at $1.6 l_{1}$ . The resistance ‘R’ is

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$10 Ω$

$25 Ω$

$15 Ω$

$20 Ω$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$I n i t i a l c o n d i t i o n : \frac{5}{l_{1}} = \frac{R}{100 - l_{1}} - - - - (1)$

$F i n a l c o n d i t i o n : \frac{5}{1.6 l_{1}} = \frac{(\frac{R}{2})}{100 - 1.6 l_{1}} - - - - (2)$

$\Rightarrow \frac{R}{1.6 (100 - l_{1})} = \frac{R}{2 (100 - 1.6 l_{1})} [f r o m e q u a t i o n (1) a n d (2)]$

$\Rightarrow 1.6 (100 - l_{1}) = 2 (100 - 1.6 l_{1})$

$\Rightarrow 160 - 1.6 l_{1} = 200 - 3.2 l_{1}$

$\Rightarrow 1.6 l_{1} = 40$

$\Rightarrow l_{1} = \frac{400}{16} = 25$

$F r o m e q u a t i o n (1), \frac{5}{25} = \frac{R}{100 - 25} \Rightarrow R = 15 Ω$

Watch 3-min video & get full concept clarity