# NEET Questions Solved

PMT - 1995

Two wires of same metal have the same length but their cross-sections are in the ratio 3 : 1. They are joined in series. The resistance of the thicker wire is 10 Ω. The total resistance of the combination will be

(1) 40 Ω

(2) $\frac{40}{3}\Omega$

(3) $\frac{5}{2}\Omega$

(4) 100 Ω

(1) For same material and same length

$\frac{{R}_{2}}{{R}_{1}}=\frac{{A}_{1}}{{A}_{2}}=\frac{3}{2}$${R}_{2}=3{R}_{1}$

Resistance of thick wire ${R}_{1}=10\Omega$

∴ Resistance of thin wire ${R}_{2}=30\Omega$

Total resistance in series = 10 + 30 = 40 Ω

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