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
____________________

PLEASE ANSWER THIS QUESTION IT'S URGENT​

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,

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

Q

P

=

100−l

l

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

According to the given question,

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

Q

P

=

100−25

25

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

Q

P

=

75

25

=

3

1

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,

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

Q

P+10

=

100−50

50

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

Q

P+10

=

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

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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

$\huge\bf\red{ \mid{ \overline{ \underline{ANSWER}}} \mid}$

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.