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 at1.6l1. The resistance R is

 Known resistance, R = 5Ω

Initial unknown resistance = RΩ

New balancing point L = 1.6l1

We have,

R/L = S/100−L——— 1

Where,

R is the known resistance

S is the unknown resistance

Initially, the balancing point is at l1.

Substitute the R=5Ω and L=l1 in equation 1, we get,

5/l1 = S/100−l1 ———– 2

When the resistance R is shunted with an equal resistance R, the resistance in that arm becomes R2

Then,

5L = R/2/100−L———– 3

Substitute the value of L1 in the equation 3. We get,

5/1.6l1 = R/2/100−1.6l1——- 4

Divide equation 2 by 4, then,

1.6/2=(100−1.6l1)/100−l1

⇒ 160−1.6l1=200−3.2l1

⇒ 1.6l1=40

⇒ l1=401.6=25cm

Substitute l1=25cm in equation 1, we get,

5/25 = R/100−25

⇒ R = 15Ω