Question

A meterbridge setup (as shown) is used to determine end correction at $\text{A}$ and $\text{B}$. When a resistance of $15\Omega$ is used in left gap and of $20\Omega$ in right gap then null point is $42\text{cm}$ from $\text{A}$, now if these resistances are intercharged null point is $57\text{cm}$ from $\text{A}$ then end correction at $\text{A}$ & $\text{B}$ are

• A. $3\text{cm},2\text{cm}$
• B. $6\text{cm},1\text{cm}$
• C. $1.5\text{cm},1.5\text{cm}$
• D. $1\text{cm},2\text{cm}$

Answer 1

The correct option is A $3\text{cm},2\text{cm}$

Since, a null point is achieved by balancing potential drops in a circuit. The potential difference between the points is zero, i.e. points are at the same potential.

Let $x,y$ be end correction at $\text{A, B}$ and $l_1$ be null point from $\text{A}$

we know that,

$\frac{R}{l_1+x}=\frac{S}{100-l_1+y}$

$\frac{15}{42+x}=\frac{20}{58+y}$

$4x-3y=6..............\left(1\right)$

When resistances are interchanged,

$\Rightarrow \frac{20}{57+x}=\frac{15}{43+y}$

$4(43+y)=3(57+x)$

$172+4y=171+3x$

$3x-4y=1...............\left(2\right)$

On solving $\left(1\right)$ & $\left(2\right)$ we get,

$x=3\text{cm},y=2\text{cm}$