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 l1. The resistance ‘R’ is

Initial condition 5l1=R100−l1−−−−(1) Final condition 51.6l1=R2100−1.6l1 R1.6100−l1=R2100−1.6l1−−−−(2)  from equation (1)  1.6100−l1=2100−1.6l1 ⇒160−1.6l1=200−3.2l1 ⇒1.6l1=40 ⇒l1 = 40016 = 25 From equation (1) 525 = R100−25 ⇒ R = 15 Ω