Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

The determination of the value of acceleration due to gravity ( $g$ ) by simple pendulum method employs the formula, $g = 4\pi^2 \frac{L}{T^2}$ . The expression for the relative error in the value of $g$ is:

$\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\left(\frac{\Delta T}{T}\right)$

$\frac{\Delta g}{g} = 4\pi^2 \left[\frac{\Delta L}{L} - 2\frac{\Delta T}{T}\right]$

$\frac{\Delta g}{g} = 4\pi^2 \left[\frac{\Delta L}{L} + 2\frac{\Delta T}{T}\right]$

$\frac{\Delta g}{g} = \frac{\Delta L}{L} - 2\left(\frac{\Delta T}{T}\right)$

Step-by-Step Solution

Given formula for acceleration due to gravity is: $g = 4\pi^2 \frac{L}{T^2}$ In measuring physical quantities, to find the maximum possible relative error, the relative errors of the individual quantities are added, multiplied by their respective powers. Since $4\pi^2$ is a constant, its error is zero. The relative error in $g$ is given by: $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}$

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