Question

A lift raises 50 passengers each having average weight \(600\ \text{N}\) to a height of \(100\ \text{m}\) at a constant speed in time \(T\). If the average power of \(15\ \text{kW}\) is required by the lift, then the value of \(T\) in seconds is

• A. 150
• B. 100
• C. 300
• D. 200

Hint:

Average power is calculated using \[ P=\frac{W}{T}. \] For lifting bodies vertically, work done is equal to total weight multiplied by height.

Answer 1

The Correct Option is D

Step 1: Compute the work done by the lift.
Total weight \( = 50 \times 600 = 30000\,\text{N} \). Work done \( W = F \times h = 30000 \times 100 = 3\times10^6\,\text{J} \).

Step 2: Find time using Power = Work / Time.
\[ T = \frac{W}{P} = \frac{3\times10^6}{15\times10^3} = 200\,\text{s} \] \[ \boxed{200\text{ seconds}} \]