A rope of negligible mass can support a load of M kg. What will be the mass of the greatest load raised by the rope? Where g is the acceleration due to gravity and h is the height through which the said load rises from rest with uniform acceleration in time t.

Options

A. `M/(1+ (2h)/("gt"^(2))) kg`
B. `M/((2h)/("gt"^(2)) kg`
C. `("Mgt"^(2))/(2h-"gt"^(2))kg`
D. `(Mh^(2))/(2h + "gt"^(2))kg`

Explanation

To solve the problem, we need to find the maximum mass \( m \) that can be raised by a rope supporting a load \( M \) kg, given that the load rises to a height \( h \) with uniform acceleration in time \( t \).

### Step-by-Step Solution:

1. **Understand the Forces Acting on the Mass**:
- The mass \( m \) is being lifted by the tension \( T' \) in the rope.
- The weight of the mass \( m \) acting downwards is \( mg \).
- The tension \( T' \) must overcome this weight and provide the necessary upward acceleration.

2. **Apply Newton's Second Law**:
- The net force acting on the mass \( m \) can be expressed as:
\[ T' - mg = ma \]
- Rearranging gives us:
\[ T' = mg + ma \]

3. **Express the Tension in Terms of Maximum Load**:
- The maximum tension \( T \) that the rope can support is equal to the weight of the load \( M \):
\[ T = Mg \]
- Therefore, we can set \( T' \) equal to \( T \) for the maximum load condition:
\[ Mg = mg + ma \]

4. **Solve for the Maximum Mass \( m \)**:
- Rearranging the equation gives:
\[ Mg = m(g + a) \]
- Thus, the maximum mass \( m \) can be expressed as:
\[ m = \frac{Mg}{g + a} \]

5. **Determine the Acceleration \( a \)**:
- To find \( a \), we can use the second equation of motion:
\[ h = ut + \frac{1}{2} a t^2 \]
- Since the initial velocity \( u = 0 \):
\[ h = \frac{1}{2} a t^2 \]
- Rearranging gives:
\[ a = \frac{2h}{t^2} \]

6. **Substitute \( a \) Back into the Equation for \( m \)**:
- Substitute \( a \) into the expression for \( m \):
\[ m = \frac{Mg}{g + \frac{2h}{t^2}} \]
- This can be simplified to:
\[ m = \frac{Mg}{g + \frac{2h}{t^2}} = \frac{Mg t^2}{gt^2 + 2h} \]

### Final Expression for Maximum Mass \( m \):
Thus, the maximum mass \( m \) that can be raised by the rope is:
\[ m = \frac{Mg t^2}{gt^2 + 2h} \]