Verified Solution • 2025–26 Session moderate

QNEET-2024-Ph-151: NEET 2024 Physics - Class 11 NEET-2024-Ph-151 Solution: In an electrical circuit the voltage is measured as \(V=(200…

In an electrical circuit the voltage is measured as V=(200±4) voltV=(200\pm 4)\ \text{volt} and the current is measured as I=(20±0.2) AI=(20\pm 0.2)\ \text{A}. The value of the resistance is: (10±4.2)Ω(10\pm 4.2)\Omega (10±0.3)Ω(10\pm 0.3)\Omega (10±0.1)Ω(10\pm 0.1)\Omega (10±0.8)Ω(10\pm 0.8)\Omega

Multiple Choice Options

  • A (10±4.2)Ω(10 \pm 4.2) \Omega ( 10 � 4.2 ) �
  • B (10±0.3)Ω(10 \pm 0.3) \Omega ( 10 � 0.3 ) �
  • C (10±0.1)Ω(10 \pm 0.1) \Omega ( 10 � 0.1 ) �
  • D (10±0.8)Ω(10 \pm 0.8) \Omega ( 10 � 0.8 ) �

Step-by-Step Verified Solution

To determine the resistance R using Ohm's law, we use the formula: R = V I Given: V = ( 200 � 4 ) volts I = ( 20 � 0.2 ) amperes First, calculate the nominal value of the resistance: R = 200 20 = 10 � Next, we need to calculate the uncertainty in the resistance. For division, the relative uncertainties add up. The relative uncertainty for voltage is: _ V V = 4 200 = 0.02 and for current: _ I I = 0.2 20 = 0.01 Adding these relative uncertainties gives us the relative uncertainty for the resistance: _ R R = _ V V + _ I I = 0.02 + 0.01 = 0.03 Now, we calculate the absolute uncertainty for the resistance: _ R = R _ 0.03 = 10 _ 0.03 = 0.3 � Therefore, the resistance with its uncertainty is: R = 10 � 0.3 � Thus, the correct answer is: Option B ( 10 � 0.3 ) �

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