Problem 74 - A Complete Resource Book in Physics for JEE Main

Question

Two bodies of masses \(10 \mathrm{~kg}\) and \(20 \mathrm{~kg}\), respectively, kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force \(F=600 \mathrm{~N}\) is applied to body of mass \(10 \mathrm{~kg}\). What is the tension in the string in each case? (A) \(200 \mathrm{~N}\) (B) \(100 \mathrm{~N}\) (C) \(400 \mathrm{~N}\) (D) \(600 \mathrm{~N}\)

Step-by-Step Solution

Answer

The tension in the string is \(400 \mathrm{~N}\).

Step 1: 1. Determine the total mass of the system

Add both masses together to get the total mass of the system: \[M_{total} = 10 kg + 20 kg = 30 kg\]

Step 2: 2. Calculate the acceleration of the system

Use Newton's second law to find the acceleration of the system: \[F = M_{total} \times a\] Solve for a, the acceleration: \[a = \frac{F}{M_{total}} = \frac{600 N}{30 kg} = 20 \frac{m}{s^2}\]

Step 3: 3. Calculate the force exerted on the 20 kg mass (tension)

The tension in the string is the force experienced by the 20 kg mass due to the acceleration of the system. Use Newton's second law again: \[T = m \times a\] Where T is the tension, m is the mass of the 20 kg body, and a is the acceleration of the system: \[T = 20 kg \times 20 \frac{m}{s^2} = 400 N\]

Step 4: 4. Determine the correct option

Comparing the calculated tension to the given options, we find that the correct answer corresponds to: (C) \(400 \mathrm{~N}\)