A man is standing on a weighing machine placed in a lift, when stationary, his weight is recorded as 40 kg. If the lift is accelerated upwards with an acceleration of 2m / s² , then the weight recorded in the machine will be (g=10 m/s² )

The correct option is 48 kg.

1. Understanding the Concept of Apparent Weight:
When a person stands on a weighing machine inside a lift (elevator), the reading of the weighing machine measures the normal reaction force ($N$) exerted by the machine on the person. This is known as the apparent weight of the person.

2. Case 1: The Lift is Stationary:
When the lift is stationary, the acceleration is zero. The forces acting on the man of mass $m$ are:
- The downward gravitational force (actual weight), given by $W=mg$.
- The upward normal reaction force, $N$.
Since there is no acceleration, the forces are balanced:

$N=mg$

We are given that the weight recorded when stationary is 40 kg. In physics terms, a weighing machine calibrated in kilograms measures the mass, but the force it experiences is equivalent to a mass of 40 kg under standard gravity $g$. Thus, we can write:
$m=40\text{kg}$

3. Case 2: The Lift is Accelerating Upwards:
When the lift accelerates upwards with an acceleration $a=2{\text{m/s}}^2$, the man also accelerates upwards at the same rate.
According to Newton's second law of motion, the net upward force is equal to the product of mass and acceleration:

$F_{\text{net}}=ma$

The net force is the difference between the upward normal reaction ($N^{\prime }$) and the downward gravitational force ($mg$):

$N^{\prime }-mg=ma$

Solving for the new normal reaction $N^{\prime }$ (the new apparent weight in Newtons):

$N^{\prime }=mg+ma=m(g+a)$

4. Calculating the Recorded Weight:
Substitute the given values into the equation:
- Mass, $m=40\text{kg}$
- Acceleration due to gravity, $g=10{\text{m/s}}^2$
- Upward acceleration, $a=2{\text{m/s}}^2$

$N^{\prime }=40\times (10+2)=40\times 12=480\text{N}$

To find the reading recorded by the machine in kg, we divide the normal force by $g=10{\text{m/s}}^2$:

$\text{Recorded Weight}=\frac{N^{\prime }}{g}=\frac{480}{10}=48\text{kg}$

Thus, the weight recorded by the machine when the lift is accelerating upwards is 48 kg.