On balancing condition R1(100−l)=(1000−R1)l .....(i) On Interchanging resistance balance point shifts left by 10cm
On balancing condition (1000−R1)(110−l)=R1(l−10) or, R1(l−10)=(1000−R1)(110−l) .....(ii) Dividing eqn (i) by (ii)
100−l
l−10
=
l
110−l
⇒(100−l)(110−l)=l(l−10) ⇒11000−100l−110l+l2=l2−10l ⇒11000=200l or, l=55 Putting the value of ' l ' in eqn (i) R1(100−55)=(1000−R1)55 ⇒R1(45)=(1000−R1)55 ⇒R1(9)=(1000−R1)11 ⇒20R1=11000 ∴R1=550KΩ