Question

On interchanging the resistances, the balance point of a metre bridge shifts to the left by 10 cm. The resistance of their series combination is 1 k`Omega`. How much was the resistance on the left slot before interchanging the resistances?

Options

• 550 `Omega`

• 910 `Omega`

• 990 `Omega`

• 505 `Omega`

Answer 1

550 `Omega`

Explanation:

Let balancing length be l,

∴ `"R"_1/"R"_2 = l/100 - l` ....(i)

f R₁ and R₂ are interchanged balancing length becomes, `(l - 10)

∴ `"R"_2/"R"_1 = (l -10)/[[100 - (l - 10)]] =  (l - 10)/(110 - l)` ....(ii)

From equations (i) and (ii),

` l/(100 - l) = (110 - l)/(l -10)`

∴` l^2 - 10  l = (110 xx 100) + (l^2 - 210 )`

∴ `200  l  = 110 xx I00`

∴ `l` = 55 cm

Substituting in equation (i), we get,

`"R"_1/"R"_2 = 55/45 = 11/9`  ....(iii)

When R₁ and R₂ are connected in series, R₁ + R₂ = 1000 `Omega`...(iv)

On solving equations (iii) and (iv), we get,

R₁ = 550  `Omega` and R₂ = 450  `Omega`