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NEET UG PHYSICS MCQ: UNITS & MEASUREMENTS

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Application +4 | -1
In an experiment, the following observation’s were recorded: L=2.820  mL = 2.820\;m, M=3.00 kg,l=0.087 cm, diameter D=0.041 cm. Taking g=9.81 m/s2m/{s^2} using the formula,
Y=4MgLπD2lY = \frac{{4MgL}}{{\pi {D^2}l}}, the maximum permissible error in Y is
View answer and explanation
Correct option: C
0.065
Option C (0.065) is correct because the maximum permissible error in Y is calculated by summing the percentage errors of each component, weighted by their exponents in the formula.

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