A lift is raised by light cables from rest to rest through a height 180 m. The greatest tension which the cables can safely bear is 9 times of lift's weight, if ascent is to made in shortest time then maximum speed of lift during its journey will be :- (g = 10 m/s2)

This problem can be solved using the concepts of physics, specifically kinematics and dynamics. Here are the steps:

1. First, we need to understand the problem. The lift is being raised through a height of 180 m. The maximum tension the cables can bear is 9 times the weight of the lift. We need to find the maximum speed of the lift during its journey.

2. The maximum tension that the cable can bear is 9 times the weight of the lift. This means that the maximum net force acting on the lift is 8 times its weight (since the weight itself is acting downwards).

3. According to Newton's second law, the net force acting on an object is equal to the mass of the object times its acceleration. So, the acceleration of the lift is 8g (where g is the acceleration due to gravity, given as 10 m/s²).

4. We know that the lift starts and ends at rest. So, the average speed of the lift is half of its maximum speed.

5. The average speed can also be calculated by dividing the total distance travelled by the total time taken. The total distance travelled by the lift is 180 m.

6. The time taken to travel this distance can be found using the equation of motion: s = ut + 0.5at². Here, s is the distance, u is the initial speed, a is the acceleration, and t is the time. Since the lift starts from rest, u is 0. So, the equation becomes 180 = 0.5 * 8g * t². Solving this equation gives t = sqrt(180 / (4g)).

7. Substituting this value of t into the equation for average speed gives the maximum speed of the lift as 2 * 180 / sqrt(180 / (4g)) = 4 * sqrt(180g) = 4 * sqrt(180 * 10) = 4 * 60 = 240 m/s.

So, the maximum speed of the lift during its journey is 240 m/s.