Physics Q48 – JEE-MAIN 2025

Home Physics Mechanical Properties of SolidsQuestion #292

The length of a light string is

1.4 m

when the tension on it is

5 N

. If the tension increases to

7 N

, the length of the string is

1.56 m

. The original length of the string is _______ m.

Enter Calculated Numerical Value:

Clue 1 of 3: Elastic Behavior

Understand that the string behaves elastically, meaning its extension is proportional to the applied tension.

🧠 Full Solution Path

Step 1: Formulate equations based on Hooke's Law✦ Active

According to Hooke's Law, the tension ( $F$ ) in an elastic string is proportional to its extension ( $Δ L = L - L_{0}$ ), where $L$ is the stretched length and $L_{0}$ is the original length. Let $k$ be the spring constant.

F = k (L - L_{0})

For the first condition:

5 N = k (1.4 m - L_{0})

(Equation 1) For the second condition:

7 N = k (1.56 m - L_{0})

(Equation 2)

Step 2: Solve the system of equations for

L_{0}

○ Expand

Divide Equation 1 by Equation 2 to eliminate $k$ :

\frac{5}{7} = \frac{k (1.4 - L_{0})}{k (1.56 - L_{0})}

\frac{5}{7} = \frac{1.4 - L_{0}}{1.56 - L_{0}}

5 (1.56 - L_{0}) = 7 (1.4 - L_{0})

7.8 - 5 L_{0} = 9.8 - 7 L_{0}

7 L_{0} - 5 L_{0} = 9.8 - 7.8

2 L_{0} = 2

L_{0} = 1.0 m

💡 Teacher's Secret Hint

Ensure careful algebraic manipulation to avoid calculation errors.

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Physics Question 48 – JEE-MAIN 2025

Clue 1 of 3: Elastic Behavior

🧠 Full Solution Path