Question

B Two resistances are connected in the two gaps of a meter bridge. The balance point is 25 ctn from the
zero end. A resistance of 1022 is connected in series with the smaller of the two resistances when the
null point shifts to 50cm. The smaller of the two resistances has the value
​

Answer 1

Answer:

Value of resistance = 5 Ω

Explanation:

Given:

• When two resistances are connected in the two gaps of a metre bridge, the balance point is 25 cm from the zero end.
• When a resistance of 10 Ω is connected in series, the smaller of the two resistances the null point shifts to 50 cm.

To Find:

• The value of the smaller resistance

Solution:

Metre bridge works on the principle of Wheatstone bridge.

It is a bridge of network of four resistors. When the galvanometer shows zero deflection, that is when no current flows through the galvanometer, the bridge is said to be balanced.

In a metre bridge,

$\sf \dfrac{P}{Q} =\dfrac{l}{100-l}$

where P is the unknown resistance and l is the balance point.

According to the given question,

$\sf \dfrac{P}{Q} =\dfrac{25}{100-25}$

$\sf \dfrac{P}{Q} =\dfrac{25}{75}=\dfrac{1}{3}$

Q = 3P -----(1)

According to the second case given,

When a resistance of 10 Ω is added to the smaller of the two resistances, the null points shifts to 50 cm.

Let us assume that P is the resistance with the lower value,

Hence,

$\sf \dfrac{P+10}{Q} =\dfrac{50}{100-50}$

$\sf \dfrac{P+10}{Q} =\dfrac{50}{50}=1$

Q = P + 10 ----(2)

From equations 1 and 2,

3P = P + 10

2P = 10

P = 5 Ω

Hence the value of the smaller of the two resistances is 5 Ω.

Answer 2

Given :-

Two  resistances are connected in the two gaps of a meter bridge. The balance point is 25 cm from the  zero end. A resistance of 1022 is connected in series with the smaller of the two resistances when the  null point shifts to 50cm.

To Find :-

Small  of the two resistances has the value

​Solution :-

Let the resistances be R1 and R2

So,

The equation formed

$\sf \dfrac{R_1}{R_2} = \dfrac{100 - l}{l}$

$\sf \dfrac{R_1}{R_2} = \dfrac{25}{100-25}$

$\sf \dfrac{R_1}{R_2} = \dfrac{25}{75}$

$\sf \dfrac{R_1}{R_2} = \dfrac{1}{3}$

Now

By cross multiplication

$\sf 3(R_1) = 1(R_2)$

$\sf 3R_1 = R_2(1)$

For 2nd case

$\sf \dfrac{R_1 + 10}{R_2} = \dfrac{50}{100}$

$\sf \dfrac{R_1+10}{R_2} = \dfrac{50}{50}$

$\sf \dfrac{R_1 + 10}{R_2} = 1$

By cross multiplication

$\sf 1(R_2) = R_{1}+10$

$\sf R_{2} = R_{1} + 10(2)$

Now

By using 1 and 2

$\sf R_{2} = R_{2}$

$\sf 3R_{1} = R_{1} + 10$

$\sf 3R_{1} - R_{1} = 10 \star$

$\sf 2R_{1} = 10$

$\sf R_1 = \dfrac{10}{2}$

$\sf R_{1} = 5$