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Answers
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