A lift of total mass M is raised by cables from rest through a height h . The greatest tension which the cables can safely bear is n Mg. The maximum speed of lift during journey if the ascent is to make shortest time is :

Options

A. `sqrt(2gh((n+1)/n))`
B. `sqrt(2ghn)`
C. `sqrt(2gh((n)/(n+1))`
D. `sqrt(2gh(n-1))`

Answer

To solve the problem step by step, we will analyze the forces acting on the lift and apply the equations of motion.

### Step 1: Identify the forces acting on the lift
The forces acting on the lift are:
- The weight of the lift, which is \( W = Mg \) (where \( g \) is the acceleration due to gravity).
- The tension in the cable, which we denote as \( T \).

### Step 2: Apply Newton's second law
When the lift is accelerating upwards with an acceleration \( a \), we can apply Newton's second law:
\[ T - Mg = Ma \]
This can be rearranged to express the tension in terms of the mass and acceleration:
\[ T = Mg + Ma = M(g + a) \]

### Step 3: Determine the maximum tension
According to the problem, the maximum tension \( T \) that the cables can safely bear is given as \( nMg \). Therefore, we can set up the equation:
\[ M(g + a) = nMg \]

### Step 4: Solve for acceleration \( a \)
Now, we can solve for the maximum acceleration \( a \):
\[ g + a = n g \]
\[ a = n g - g = (n - 1)g \]

### Step 5: Use the equations of motion to find maximum speed
We will use the equation of motion to find the final velocity \( V \) of the lift after it has traveled a height \( h \). The equation we will use is:
\[ V^2 = U^2 + 2aS \]
Since the lift starts from rest, \( U = 0 \) and \( S = h \):
\[ V^2 = 0 + 2a h = 2ah \]
Substituting \( a = (n - 1)g \) into the equation:
\[ V^2 = 2(n - 1)gh \]

### Step 6: Solve for \( V \)
Taking the square root of both sides, we find:
\[ V = \sqrt{2(n - 1)gh} \]

### Final Answer
The maximum speed of the lift during its journey, if the ascent is to make the shortest time, is:
\[ V = \sqrt{2(n - 1)gh} \]

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