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Force and momentum

Chapter: Mechanics

Question 1 of 2 NDA MCQ

Two identical spring balances S 1 and S 2 are connected one after the other and are held vertically as shown in the figure. A mass of 10 kg is hanging from S 2. If the readings on S1 and S2 are W1 and W2 respectively, then : 

Detailed Explanation

Correct Answer

✓ Option (d) is correct. Both spring balances display the same reading.

📖 Governing Law / Core Concept

The readings of identical spring balances in series depend on the mass they support. According to Newton's second law, the force on each spring balance is the same when the system is in equilibrium.

🧪 Step-by-Step Breakdown

Given a mass \( m = 10 \, \text{kg} \), the force due to gravity (weight) can be expressed as:

\[ W = mg \] \

Where \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity). Thus, the total weight supported is:

\[ W = 10 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 98.1 \, \text{N} \] \

Since both spring balances are identical and measure the same weight applied vertically, thus:

\[ W_1 = W_2 = 10 \, \text{kg} \] \

🔍 Option Analysis

The other options fail to account for the equal distribution of weight across the identical spring balances.

⚡ Mnemonic / Speed-Run

For spring balance systems, remember: "Identical springs, identical readings." This can help quickly recall the principle.

Visual Suggestion

Visual Suggestion

Depict two identical springs vertically aligned, with arrows indicating the force exerted by the mass, labeled as \( W_1 \) and \( W_2 \), both showing a reading of 10 kg. This will provide clarity on how forces are distributed in spring systems.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 5 (Laws of Motion).

Question 2 of 2 NDA MCQ

One block of 2.0 kg mass is placed on top of another block of 3.0 kg mass. The coefficient of static friction between the two blocks is 0.2. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m/s², the maximum value of the frictional force is :

Detailed Explanation

Solution

Option (c) is correct.

Explanation:

The maximum static friction force \( f_s \) is determined by the equation:
\[ f_s = \mu_s N \]
Where: - \( \mu_s = 0.2 \) (coefficient of static friction) - \( N \) (normal force) First, we calculate the normal force \( N \) acting on the upper block (mass \( m_1 = 2.0 \, \text{kg} \)): Using \( N = m_1 g \):
\[ N = 2.0 \, \text{kg} \times 10 \, \text{m/s}^2 = 20 \, \text{N} \]
Now, substituting \( N \) back into the friction equation:
\[ f_s = 0.2 \times 20 \, \text{N} = 4 \, \text{N} \]
Thus, the maximum value of the frictional force is \( 4 \, \text{N} \).
📌 Hints / Properties Used:
  • Static friction formula: \( f_s = \mu_s N \)
  • Normal force \( N \): \( N = m g \)
📊 Visual Diagram Suggestion:

A force diagram showing both blocks with arrows indicating the gravitational force acting downwards and frictional force acting upwards between them. Label the forces \( N \) (normal force) and \( f_s \) (static friction force).

📖 Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 4 (Motion in a Plane)

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