JEE Physics 2025 Question with Solution

Given: When tension is $T_1 = 5 \, \text{N}$, the length is $L_1 = 1.40 \, \text{m}$. When tension is $T_2 = 7 \, \text{N}$, the length is $L_2 = 1.56 \, \text{m}$.

Find: The original unstretched length $L_0$ of the string.

Concept Used: For an elastic string, extension is proportional to tension:

$$T = kx = k(L - L_0)$$

where $k$ is the force constant.

Step 1: Write equations for the two cases:

$$T_1 = k(L_1 - L_0), \quad T_2 = k(L_2 - L_0)$$

Step 2: Eliminate $k$:

$$\frac{T_1}{L_1 - L_0} = \frac{T_2}{L_2 - L_0}$$

Step 3: Substitute values:

$$\frac{5}{1.40 - L_0} = \frac{7}{1.56 - L_0}$$

Step 4: Cross multiply and solve:

$$5(1.56 - L_0) = 7(1.40 - L_0)$$ $$7L_0 - 5L_0 = 7(1.40) - 5(1.56)$$ $$2L_0 = 9.8 - 7.8 = 2.0$$ $$L_0 = 1.0 \, \text{m}$$

Therefore, the original length of the string is $1.0 \, \text{m}$.

Given: The string length changes from $1.40 \, \text{m}$ at $5 \, \text{N}$ to $1.56 \, \text{m}$ at $7 \, \text{N}$.

Find: The natural length of the string.

Since

$$T = k(L - L_0)$$

we can write length as a linear function of tension:

$$L = L_0 + \frac{T}{k}$$

Using the two data points:

$$1.40 = L_0 + \frac{5}{k}$$ $$1.56 = L_0 + \frac{7}{k}$$

Subtracting the first equation from the second:

$$1.56 - 1.40 = \frac{7}{k} - \frac{5}{k}$$ $$0.16 = \frac{2}{k}$$ $$k = 12.5 \, \text{N/m}$$

Now substitute into

$$1.40 = L_0 + \frac{5}{12.5}$$ $$1.40 = L_0 + 0.40$$ $$L_0 = 1.00 \, \text{m}$$

Hence, the original length is $1$.