Question

the resistances in the two arms of metre Bridge are 5Ω and R Ω respectively when the resistance R is shunted with an equal resistance the new balance point is at 1.6 l1 the resistance R, is​

Answer 1

=> R= 15 Ω

For the first case balance condition if meter bridge will be

5/l1 = R/ ( 100-l1 ) _[i]

Now, by shunting resistance R by an equal resistanceR, new resistance and that am become R/2 So, new balance condition will be

5/1.6 l1 = R/2 / ( 100 - 1.6 l1 ) _[ii]

From Eqs [i] and [ii],

1.6 /1 = [ ( 100 - 1.6 l1 ) / 100-l1 ] × 2

=> 160 - 1.6 l1 = 200 - 3.2 l1

1.6 l1 = 40

l1 = 40/1.6 = 25 m

=> from Eq [i], 5/25 = R/75

=> R = 15Ω

∴The resistance R is 15Ω !

Answer 2

Answer:

♣️The resistance R is 15 Ω♣️

♥️Hope it will be helpful ☺️ thank you!♥️