QJEEM24Phc468: JEE Main 2024 Physics - Class 11 JEEM24Phc468 Solution: To find the spring constant of a spring experimentally, a…

To determine the spring constant $k$ of a spring experimentally, we can use the formula derived from Hooke's Law and the period of oscillation for a mass-spring system:

$T = 2 \pi \sqrt{\frac{m}{k}}$

Here, $T$ is the period of oscillation, $m$ is the mass, and $k$ is the spring constant. Rearranging the formula to solve for $k$ , we get:

$k = \frac{4 \pi^2 m}{T^2}$

To find the error in $k$ , we have to consider the errors in both the measurements of $T$ and $m$ . Let's denote the percentage errors as follows:

$\Delta T / T \cdot 100\% = 2\%$ (positive error)

$\Delta m / m \cdot 100\% = -1\%$ (negative error)

According to the rules of error propagation, the relative error in $k$ can be found by adding the relative errors in the measurements, each multiplied by the respective powers to which they affect $k$ . Since $T$ is squared in the denominator and $m$ is linear in the numerator, the calculation is as follows:

$\frac{\Delta k}{k} = \left| -2 \cdot \frac{\Delta T}{T} \right| + \left| 1 \cdot \frac{\Delta m}{m} \right|$

Substituting the percentage errors:

$\frac{\Delta k}{k} = \left| -2 \cdot 0.02 \right| + \left| 1 \cdot (-0.01) \right|$

$\frac{\Delta k}{k} = 0.04 + 0.01$

$\frac{\Delta k}{k} = 0.05$

Thus, the percentage error in determining the value of $k$ is:

$\frac{\Delta k}{k} \cdot 100\% = 5\%$

The correct answer is Option A: 5%