AIIMS 2007 Physics Newton's Laws of Motion MCQ Question

To solve the problem, we need to analyze the scenario of a man weighing himself in a lift that is either ascending or descending at constant speeds.

1. Understanding Weight in Different Scenarios:

   • The weight of an object is defined as the force exerted on it due to gravity, given by the formula: $W = mg$ where $m$ is the mass of the object, and $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$).

2. Weight in an Ascending Lift:

   • When the lift is ascending with a uniform speed, the acceleration of the lift is $0 \, \text{m/s}^2$. The apparent weight (the reading on the weighing machine) of the man can be calculated as: $W_{\text{up}} = mg + ma$ Since the lift's speed is constant, $a = 0$: $W_{\text{up}} = mg + m(0) = mg$ Hence, the apparent weight remains: $W_{\text{up}} = 60 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 588 \, \text{N}$

3. Weight in a Descending Lift:

   • When the lift is descending with a uniform speed, again, the acceleration $a = 0 \, \text{m/s}^2$. The apparent weight can be calculated similarly: $W_{\text{down}} = mg - ma$ Again, since the lift's speed is constant, $a = 0$: $W_{\text{down}} = mg - m(0) = mg$ Thus, the apparent weight remains: $W_{\text{down}} = 60 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 588 \, \text{N}$

4. Calculating the Ratio:

   • The ratio of the weights recorded when the lift is ascending to when it is descending can be expressed as: $\text{Ratio} = \frac{W_{\text{up}}}{W_{\text{down}}} = \frac{588 \, \text{N}}{588 \, \text{N}} = 1$

5. Analyzing the Options:

   • The answer options provided were: A) 0.5 B) 1 C) 2 D) 1.5
   • The calculated ratio is $1$, which corresponds to option B.

6. Conclusion:

   • Therefore, the correct answer is B (1), as the weight of the man recorded in both cases remains the same due to the constant velocities (no net acceleration).

Why Other Options Are Incorrect:

• Option A (0.5): This would imply that the weight in one scenario is half of the other, which does not hold true as both weights are equal.
• Option C (2): This suggests that the weight in the ascending lift is double that in the descending lift, which is not accurate as both are the same.
• Option D (1.5): This would mean that one weight is 1.5 times the other, which is also incorrect since both weights are equal.

In conclusion, the uniform speeds in both scenarios mean that the net force acting on the man is the same, leading to an equal reading on the weighing machine.