Question

The apparent weight of a girl in a moving lift is 25% more than her true weight. If the lift starts from rest, the distance travelled by the girl in the first 2 s is (acceleration due to gravity = 10 ms\(^2\))

• A. 25 m
• B. 10 m
• C. 5 m
• D. 15 m

Hint:

In a moving lift, the apparent weight is used to determine the acceleration, and from there, we can calculate the distance travelled.

Answer 1

The Correct Option is C

The apparent weight is greater than the true weight by 25%. This indicates that the lift is accelerating upwards. Let the acceleration of the lift be \( a \).

Using the equation for apparent weight:

\[ \text{Apparent weight} = \text{True weight} \times \left(1 + \frac{a}{g} \right) \]

Since the apparent weight is 25% more:

\[ 1 + \frac{a}{g} = 1.25 \] \[ \frac{a}{g} = 0.25 \] \[ a = 0.25 \times 10 = 2.5 \, \text{m/s}^2 \]

Now, using the equation of motion to calculate the distance travelled:

\[ s = ut + \frac{1}{2} a t^2 \]

Since the lift starts from rest, \( u = 0 \), and \( t = 2 \) s:

\[ s = 0 + \frac{1}{2} \times 2.5 \times (2)^2 \] \[ s = 0 + \frac{1}{2} \times 2.5 \times 4 = 5 \, \text{m} \]

Final Answer:
5 m.