In a metre bridge experiment,the ratio of the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) In a metre bridge,the balance condition is given by the formula $\frac{P}{Q} = \frac{l}{100-l}$,where $P$ is the resistance in the left gap,$Q$ is

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) In a metre bridge,the balance condition is given by the formula $\frac{P}{Q} = \frac{l}{100-l}$,where $P$ is the resistance in the left gap,$Q$ is the resistance in the right gap,and $l$ is the balance point distance from the left end in $cm$.Given the ratio $\frac{P}{Q} = \frac{2}{3}$.Substituting the values into the formula:$\frac{2}{3} = \frac{l}{100-l}$Cross-multiplying gives:$2(100 - l) = 3l$$200 - 2l = 3l$$200 = 5l$$l = \frac{200}{5} = 40 ~cm$.Therefore,the balance point from the left is $40 ~cm$.

In a metre bridge experiment,the ratio of the left gap resistance to the right gap resistance is $2:3$. The balance point from the left is: (in $~cm$)

Similar Questions

Which of the following is a Lewis acid?

What does a mature sperm contain?

In which of the following pairs do both species have $sp^3$ hybridization?

The ratio of the magnetic field at the centre of a current-carrying circular wire and the magnetic field at the centre of a square coil made from the same length of wire will be

If $|x| < 1$ and $y = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$,then $x$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module