Question

Resistances 4 Ω and 6 Ω are connected across the left gap and right gap respectively of a metre bridge. When a 2 Ω resistance is connected in series with the 4 Ω resistance in the left gap, the shift in the balance point is *
(a) 10 cm
(b) 15 cm
(c) 20 cm
(d) 25 cm
​

Answer 1

Explanation:

Case 1

⇒

4

R

=

1

1

R = 4

⇒

2

4

=

100−x

x

200 = 3x

x=

3

200

Shift =

3

200

−50=

3

50

=16.7 cm

Answer 2

Answer:

The shift in the balance point is 10cm.

Explanation:

Given the resistances 4 Ω and 6 Ω at the left gap and right gap respectively of a metre bridge.

2 Ω resistance is connected in series with 4 Ω resistance in the left gap.

Meter bridge is an instrument used to find the unknown resistance of a conductor, and works on the principle of a Wheatstone bridge.

The balance condition for resistances $R_{1}~ \&~ R_{2}$ is given by

$\frac{R_{1} }{R_{2}} =\frac{L}{100 - L}$  where L is the balancing length.

Thus the condition of balance when 4Ω and 6Ω are connected is

$\frac{4 }{6} =\frac{L}{100 - L}$

$\implies 6L =4(100 - L)$

$\implies 6L+4L =400 \implies L=40~cm$                          ...(1)

When a 2Ω resistance is connected in series with 4Ω resistance, the equivalent resistance in the leftt gap is

$R=R_{1} + R_{2}=2+4=6\Omega$

Thus the balancing length is

$\frac{6}{6} =\frac{L}{100 - L}$

$\implies L =100 - L\implies L =50~cm$                            ...(2)

The shift in the balance point is the difference in the balancing lengths,

i.e., $50-40 = 10~ cm$

Therefore, the shift in the balance point is 10cm.