The resistances 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 1.6l1.The resistance R is

In the first case, At the balance point5R=l1100-l                              . . . . . (i)In the second case At the balance point5(R/2)=1.6l1100-1.6l1            . . . . . (ii)Divide eqn. (i) by eqn. (ii), we get 12=100-1.6l11.6100-l1160-1.6l1=200-3.2l11.6l1=40   or   l1=401.6=25 cmSubstituting this value in eqn. (i), we get 5R=2575R=37525Ω=15Ω