Question

30. ( frac{R_{1}+10}{R_{2}}=frac{50}{50}=1 )
( therefore quad R_{1}+10=R_{2} )
( begin{array}{ll}text { Again, } & frac{R_{1}}{R_{2}}=frac{40}{60}=frac{2}{3} therefore & 3 R_{1}=2 R_{2}end{array} )
Substituting the value of ( R_{2} ) from ( mathrm{Eq} . ) (i), we get ( 3 R_{1}=2left(R_{1}+10right) )
( therefore quad R_{1}=20 Omega )
Ans.

# 30. An unknown resistance R is connected in series with a resistance of 10 2. This combination is connected to one gap of a meter bridge, while other gap is connected to another resistance R,. The balance point is at 50 cm. Now, when the 10 22 resistance is removed, the balance point shifts to 40 cm. Then, the value of R, is (a) 60 22 (b) 40 12 (c) 20 2 (d) 10 22

Solution