JEE Main | 2018Current Electricity and Electrical InstrumentEasy

Question

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

Options

A.

505 kΩ

B.

550 kΩ

C.

910 kΩ

D.

990 kΩ

Solution

R1 + R2 = 1000 ⇒ R2 = 1000 - R1


On balancing condition
R1(100 - l) = (1000 - R1)l     ...(1)
On Inter changing resistance


On balancing condition
(1000 - R1) (110 - l) = R1 (l - 10)
or R1(l - 10 ) = (1000 - R1)(110 - l)    ....(2)
(1) ÷ (2)
100-ll-10=l110-l
⇒ (100 - l)(110 - l)  = l(l - 10)
⇒ 11000 - 100l - 110l + l2 = l2 -10l
⇒ 11000 = 200l
l = 55
Put in eq(1)
R1(100 - 55) = (1000 - R1)55
R1(45) = (1000 - R1)55
R1(9) = (1000 - R1)11
20 R1 = 11000
R1 = 550

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