The acceleration due to gravity is found upto an accuracy of 4% on a planet. The energy supplied to a simple pendulum of known mass 'm' to undertake oscillations of time period T is being estimated. lf time period is measured to an accuracy of 3%, the accuracy to which E is known as ___________%.

Answer

To solve the problem, we need to understand how the accuracy in the measurements of gravity (g) and time period (T) affects the accuracy in the energy (E) supplied to a simple pendulum.

### Step-by-Step Solution:

1. **Understanding the Energy of a Simple Pendulum**:
The total mechanical energy (E) of a simple pendulum is given by the potential energy at its maximum height, which can be expressed as:
\[ E = mgh \]
where \( h \) is the height, \( m \) is the mass, and \( g \) is the acceleration due to gravity.

2. **Relating Height to Gravity**:
The height \( h \) can be related to the amplitude of the swing and the gravitational acceleration. However, for small angles, we can consider that the energy primarily depends on \( g \) and the maximum height reached.

3. **Error Propagation in Energy**:
The accuracy of the energy \( E \) can be derived from the accuracy of \( g \) and the mass \( m \). Since the mass is constant, we only need to consider the accuracy in \( g \):
\[ \frac{\Delta E}{E} = \frac{\Delta g}{g} \]

4. **Given Accuracy Values**:
- The accuracy in \( g \) is given as 4%, which means:
\[ \frac{\Delta g}{g} \times 100 = 4\% \]
- The accuracy in \( T \) is given as 3%, but since \( T \) does not directly affect the energy calculation, we will not consider it in our final energy accuracy.

5. **Final Calculation of Energy Accuracy**:
Since the energy \( E \) is primarily dependent on \( g \) and the mass \( m \) (which is constant), the accuracy in energy \( E \) will also be the same as the accuracy in \( g \):
\[ \text{Accuracy in } E = 4\% \]

Thus, the accuracy to which the energy \( E \) is known is **4%**.