(A) In a meter bridge,the balanced condition is given by $\frac{P}{Q} = \frac{l_1 + \alpha}{l_2 + \beta}$,where $P = 15\,\Omega$,$Q = R$,$l_1 = 43\,cm

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(A) In a meter bridge,the balanced condition is given by $\frac{P}{Q} = \frac{l_1 + \alpha}{l_2 + \beta}$,where $P = 15\,\Omega$,$Q = R$,$l_1 = 43\,cm$,$l_2 = (100 - 43) = 57\,cm$,$\alpha = 2\,cm$ (end correction at $A$),and $\beta = 1\,cm$ (end correction at $B$).Substituting the values:$\frac{15}{R} = \frac{43 + 2}{57 + 1}$$\frac{15}{R} = \frac{45}{58}$$R = \frac{15 \times 58}{45}$$R = \frac{58}{3} \approx 19.33\,\Omega$.Given the options,the closest integer value is $19\,\Omega$.

Class 12 Physics Current Electricity Meter Bridge

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$A$ meter bridge setup is shown in the figure. It is used to determine an unknown resistance $R$ using a given resistor of $15\,\Omega$. The galvanometer $(G)$ shows null deflection when the tapping key is at the $43\,cm$ mark from end $A$. If the end correction for end $A$ is $2\,cm$ and for end $B$ is $1\,cm$,then the determined value of $R$ will be . . . . . . $\Omega$.

JEE MAIN 2022 All JEE MAIN

A
$19$
B
$20$
C
$24$
D
$18$

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