A wire of length ' l ' and resistance 100 Ω is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are conne - Physics MCQ NEET (UG) 2024 2026 | ExamTest.live

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A wire of length ' l' and resistance 100 Ω is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

Solution & Step-by-step Explanation

Total resistance R = 100 Ω . Divided into 10 parts, so each part has resistance: r = \frac{100}{10} = 10 Ω First 5 parts in series: R_{s} = 5 r = 5 \times 10 = 50 Ω Next 5 parts in parallel: R_{p} = \frac{r}{5} = \frac{10}{5} = 2 Ω Final combination in series: R_{final} = R_{s} + R_{p} = 50 + 2 = 52 Ω .

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A wire of length ' l' and resistance 100 Ω is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

26 Ω

52 Ω

55 Ω

60 Ω

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