JEE Physics 2026 Question with Solution

Given: $k_1 = (10 \pm 0.2)\, \text{N/m}$ and $k_2 = (20 \pm 0.3)\, \text{N/m}$.

Find: Percentage error in the equivalent spring constant when the springs are connected in parallel.

For springs in parallel, the equivalent spring constant is the sum of the individual spring constants.

$$k_{eq} = k_1 + k_2$$

Substituting the measured values:

$$k_{eq} = 10 + 20 = 30 \, \text{N/m}$$

For addition, absolute errors add.

$$\Delta k_{eq} = \Delta k_1 + \Delta k_2 = 0.2 + 0.3 = 0.5 \, \text{N/m}$$

So, the equivalent spring constant is $k_{eq} = (30 \pm 0.5)\, \text{N/m}$.

Now percentage error is

$$\%\,\text{error} = \frac{\Delta k_{eq}}{k_{eq}} \times 100\%$$ $$\%\,\text{error} = \frac{0.5}{30} \times 100\% = \frac{5}{3}\% \approx 1.67\%$$

Therefore, the percentage error in the equivalent spring constant is $1.67\%$. The correct option is B.

Given: Two measured spring constants with absolute errors.

Find: How the error combines in a parallel combination.

The key idea is to distinguish between rules for addition and multiplication.

• For addition or subtraction, absolute errors are added.
• For multiplication or division, relative errors are added.

Here the parallel combination uses addition:

$$k_{eq} = k_1 + k_2$$

Hence the absolute error becomes

$$\Delta k_{eq} = \Delta k_1 + \Delta k_2$$

Using the given values:

$$\Delta k_{eq} = 0.2 + 0.3 = 0.5 \, \text{N/m}$$

and

$$k_{eq} = 10 + 20 = 30 \, \text{N/m}$$

Now convert absolute error into percentage error:

$$\%\,\text{error} = \frac{0.5}{30} \times 100\%$$ $$= 1.666\ldots\% \approx 1.67\%$$

Therefore, the correct option is B.