With two resistance `R_1 and R_2 (>R_1)` in the two gaps of a metre bridge the balance was found to be 1/3 m from the zero end. When a `6 Omega `resistance is connected in series with the smaller of the two resistance, the point is shifted to 2/3 m from the same end, then `R_1 and R_2` are

Answer

` 2 Omega, 4 Omega `

Explanation

For two resistor P and Q
` (P)/(Q) = (1)/(100 - I) = ((1)/(3))/(1 - (1)/(3)) = (1)/(2) " "`...(i)
` Q = 2P, P lt Q `
Now, ` P. = P +6, I. = 2//3 `
` (P.)/(Q) = (I . )/((100 - I .)) = ((2)/(3))/((1)/(3)) = (2)/(1) `
` (P +6 )/( Q) = (2)/(1)" " `...(ii)
On solving (i) & (ii) , P ` = 2 Omega, Q = 4 Omega `