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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 tapping key is at $$43$$ cm mark from end $$A$$. If the end correction for end $$A$$ is $$2$$ cm, then the determined value of $$R$$ will be ______ $$\Omega$$.
Correct Answer: 19
given:
balancing length from A = 43 cm
end correction at A = +2 cm
so corrected length near A:
l₁ = 43 + 2 = 45 cm
remaining length:
l₂ = 100 − 43 = 57 cm
meter bridge condition:
$$\frac{R}{15}=\frac{l_2}{l_1}$$
$$\frac{R}{15}=\frac{57}{45}$$
solve:
$$R=15\times\frac{57}{45}=19Ω$$
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