QNEET-2024-Ph-83: NEET 2024 Physics - Class 12 NEET-2024-Ph-83 Solution: A wire of length 'l' and resistance \(100 \) is divided into 10 …

To solve this problem, we first need to understand how the resistance changes when we cut the wire into equal parts and how it behaves when connected in different configurations (series and parallel). Given: Original length of the wire, l $l$ l . Total resistance of the wire R = 100 � $R = 100 \Omega$ R = 100� . The wire is divided into 10 equal parts. Resistance of each part: Since the wire is divided into 10 equal parts, the length of each part is l 10 $\frac{l}{10}$ 10 l _ . Resistance is proportional to length (as long as the cross-sectional area and material of the wire remain constant). Therefore, the resistance of each part, denoted as r $r$ r , is 1 10 $\frac{1}{10}$ 10 1 _ th of the total resistance: r = R 10 = 100 � 10 = 10 � First 5 parts in series: When resistors are connected in series, the total resistance is the sum of the individual resistances: R series = 5 _ 10 � = 50 � Next 5 parts in parallel: When resistors are connected in parallel, the total resistance R parallel $R_{\text{parallel}}$ R parallel _ can be calculated using the reciprocal formula: 1 R parallel = 1 10 � + 1 10 � + 1 10 � + 1 10 � + 1 10 � = 5 _ 1 10 � = 5 10 � = 1 2 � Thus, R parallel = 2 � Final combination in series: The total resistance of the combination, where the series and parallel groups are again connected in series, will be: R total = R series + R parallel = 50 � + 2 � = 52 � Thus, the correct answer to the resistance of the final combination is: Option B 52 �