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Question: In a meter bridge experiment resistances are connected as shown in figure. Initially resistance \(P ...

In a meter bridge experiment resistances are connected as shown in figure. Initially resistance P=4ΩP = 4\Omega and the neutral point NN is at 60cm60cm from AA. Now an unknown resistance RR is connected in series to PP and the new position of the neutral point is at 80cm80cm from AA. The value of unknown resistance RR is:

(A) 6Ω6\Omega
(B) 7Ω7\Omega
(C) 335Ω\dfrac{{33}}{5}\Omega
(D) 203Ω\dfrac{{20}}{3}\Omega

Explanation

Solution

Using the formula of a meter bridge, first calculate the value of the resistance QQ. Then with this value of QQ we can determine the value of the unknown resistance RR using the same formula.
Formula used
Xl=R100l\dfrac{X}{l} = \dfrac{R}{{100 - l}}, where XX is the known resistance and RRis the unknown resistance and ll is in cmcm.

Complete step by step solution
A meter bridge is an electrical instrument that works on the principle of a Wheatstone bridge. It is used in finding the unknown resistance of a conductor.
It consists of a long wire of 1m1m which is separated into two sections. In the left section we attach the known resistance and in the right section, we attach the unknown resistance. A jockey is present to detect the balance point. The galvanometer indicates the balance point. The balance point is the point on the wire where the galvanometer shows zero deflection.
Let XX be the known resistance and RR be the unknown resistance.
Then we can write,
Xl=R100l\dfrac{X}{l} = \dfrac{R}{{100 - l}}
Using this formula we can determine the value of the unknown resistance.
The distance between AA and BB is 100cm100cm
It is given that AN=60cmAN = 60cm
So, NBNB would be equal to (10060)cm=40cm\left( {100 - 60} \right)cm = 40cm
From this diagram we can see that,
PAN=QNB 460=Q40 Q=166 Q=83Ω  \dfrac{P}{{AN}} = \dfrac{Q}{{NB}} \\\ \Rightarrow \dfrac{4}{{60}} = \dfrac{Q}{{40}} \\\ \Rightarrow Q = \dfrac{{16}}{6} \\\ \Rightarrow Q = \dfrac{8}{3}\Omega \\\
Now when another resistance RR is connected in series with PP, ANAN becomes 80cm80cm
So, the value of the unknown resistance becomes,
P+R80=Q20 4+R4=83 12+3R=32 3R=20 R=203Ω  \dfrac{{P + R}}{{80}} = \dfrac{Q}{{20}} \\\ \Rightarrow \dfrac{{4 + R}}{4} = \dfrac{8}{3} \\\ \Rightarrow 12 + 3R = 32 \\\ \Rightarrow 3R = 20 \\\ \Rightarrow R = \dfrac{{20}}{3}\Omega \\\
So, the value of the unknown resistance RR is 203Ω\dfrac{{20}}{3}\Omega

Therefore, the correct option is D.

Note: The main function of a meter bridge is to find the value of an unknown resistance. Another one of its functions is to compare two different resistances.