Question

Two persons are pulling a rope towards themselves with a force of 200 N each. If the Young's modulus is 2 x 10¹¹ N/m² and area of cross-section is 2 cm² for the rope, the elongation in the rope is ____.
(distance between the persons holding the ropes is 2 m.)

• A. 10 μm
• B. 20 μm
• C. 5 μm
• D. 40 μm

Answer 1

The Correct Option is A

To solve this problem, we will use the formula for elongation derived from Hooke's Law and the definition of Young's modulus. The elongation or extension (\Delta L) in the rope can be calculated using the formula:

\Delta L = \frac{F L}{A Y}

where:

• F is the force applied,
• L is the original length of the rope,
• A is the cross-sectional area of the rope, and
• Y is Young's modulus.

Given values:

• F = 200\, \text{N} (force applied by each person, but since they pull against each other, the tension in the rope remains 200 N),
• L = 2\, \text{m} (original length),
• A = 2\, \text{cm}^2 = 2 \times 10^{-4}\, \text{m}^2 (conversion from cm² to m²),
• Y = 2 \times 10^{11}\, \text{N/m}^2.

Now, substitute these values into the formula:

\Delta L = \frac{200 \times 2}{2 \times 10^{-4} \times 2 \times 10^{11}}

Simplifying the calculation:

\Delta L = \frac{400}{4 \times 10^7}

\Delta L = \frac{400}{4 \times 10^7} = 10^{-5}\, \text{m}

Converting the result to micrometers (since 1 m = 10⁶ μm):

\Delta L = 10\ \mu m

Therefore, the elongation in the rope is 10 μm.

Correct Answer: 10 μm