A meter bridge setup is shown in the figure. It is used to determine an unknown resistance $R$ using a given resistor of $15Ω$ . The galvanometer $(G)$ shows null defection when tapping key is at $43 c m$ mark from end $A$ . If the end correction for end $A$ is $2 c m$ , then the determined value of $R$ will be $Ω$ .

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TestHub · Accepted Answer

Using the conditions of a balanced wheatstone bridge and adding the end correction.

$\frac{15}{R} = \frac{l + e}{( L + e ) - ( l + e )} \Rightarrow \frac{15}{( 43 + 2 )} = \frac{R}{( 102 - 45 )} \Rightarrow R = \frac{57}{45} \times 15$

$\Rightarrow R = 19Ω$

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