JEE Physics 2026 Question with Solution

Given: $T=2\pi\sqrt{\frac{m}{k}}$, error in mass measurement is $10\%$, total time for $50$ oscillations is $60\,\text{s}$, and watch resolution is $2\,\text{s}$.

Find: Percentage error in determination of spring constant $k$.

From the relation,

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

So, the fractional error is

$$\frac{\Delta k}{k}=\frac{\Delta m}{m}+2\frac{\Delta T}{T}$$

Given,

$$\frac{\Delta m}{m}=10\%=0.10$$

For the measured total time $t=60\,\text{s}$ and resolution $\Delta t=2\,\text{s}$,

$$\frac{\Delta t}{t}=\frac{2}{60}=\frac{1}{30}$$

Now,

$$T=\frac{t}{50}=\frac{60}{50}=1.2\,\text{s}$$

and

$$\Delta T=\frac{\Delta t}{50}=\frac{2}{50}=0.04\,\text{s}$$

Therefore,

$$\frac{\Delta T}{T}=\frac{0.04}{1.2}=\frac{1}{30}$$

Substituting,

$$\frac{\Delta k}{k}=0.10+2\left(\frac{1}{30}\right)=0.10+0.0667=0.1667$$

Hence, percentage error is

$$0.1667\times100\approx16.67\%$$

Therefore, the working in the solution gives $16.67\%$. However, this value is not present in the options. The solution labels option B as correct, so the defensible marked answer is B, though there is a clear discrepancy between the working and the listed options.

Given: $k=\frac{4\pi^2 m}{T^2}$ after rearranging the SHM formula.

Find: How the error in $m$ and $T$ contributes to the error in $k$.

Since $k$ is directly proportional to $m$ and inversely proportional to $T^2$, percentage errors add as

$$\frac{\Delta k}{k}=\frac{\Delta m}{m}+2\frac{\Delta T}{T}$$

The factor of $2$ appears because $T$ is squared.

The total measured time is for $50$ oscillations, so dividing both the measured time and its absolute error by $50$ leaves the same relative error:

$$T=\frac{t}{50}, \qquad \Delta T=\frac{\Delta t}{50}$$

Hence,

$$\frac{\Delta T}{T}=\frac{\Delta t/50}{t/50}=\frac{\Delta t}{t}$$

Now use

$$\frac{\Delta t}{t}=\frac{2}{60}=\frac{1}{30}=0.0333$$

So,

$$\frac{\Delta k}{k}=0.10+2(0.0333)=0.1667$$

Thus, the percentage error is $16.67\%$.

Therefore, the numerical result from the shown steps is $16.67\%$, but the solution's still marks B as the correct option.