JEE Main · 2004 · 24 Apr 2004 · Offline · Current Electricity

In a metre bridge experiment null point is obtained at 20 cm from one end of the wire when resistance

X

is balanced against another resistance

Y

. If

X < Y

, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X against Y?

Answer: (A)

Solution

In a metre bridge the balance condition is

\frac{R _{1}}{R _{2}} = \frac{ℓ _{1}}{100 - ℓ _{1}}

. The first balance

\frac{X}{Y} = \frac{20}{80} = \frac{1}{4}

tells us

Y = 4 X

. Balancing

4 X

against

Y

now makes the ratio

\frac{4 X}{4 X} = 1 = \frac{ℓ}{100 - ℓ}

, so

ℓ = 100 - ℓ

and the null point shifts to

50

cm.

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option