Determine the dimensions of the Universal Gravitational Constant G and Pressure.

The formula for the gravitational force ($F$) between two masses ($m_1$, $m_2$) separated by a distance ($r$) is:

Where $G$ is the universal gravitational constant.

• Dimensions of Force ($F$): $\text{M L T}^{-2}$
• Dimensions of Distance ($r$): $\text{L}$
• Dimensions of Mass ($m_1$, $m_2$): $\text{M}$

The correct combination of dimensions for the Universal Gravitational Constant G is $\boxed{\text{a) M}^{-1}\text{L}^3\text{T}^{-2}}$.

Pressure ($P$) is defined as force ($F$) per unit area ($A$):

• Dimensions of Force ($F$): $\text{M L T}^{-2}$
• Dimensions of Area ($A$): $\text{L}^2$

The correct combination of dimensions for Pressure is $\boxed{\text{a) ML}^{-1}\text{T}^{-2}}$.

The period ($T$) of a simple pendulum is given by:

Where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.

Let $T_1$ be the initial period and $L_1$ be the original length.
Let $T_2$ be the new period and $L_2$ be the new length.
Given:

• $T_1 = 7.0 \text{ s}$
• The length is shortened by $1.5 \text{ m}$, so $L_2 = L_1 - 1.5 \text{ m}$.
• The new period is $T_2 = 8.5 \text{ s}$.

The equations are:

Divide equation (1) by equation (2):

Square both sides:

Substitute $T_1 = 7.0 \text{ s}$, $T_2 = 8.5 \text{ s}$, and $L_2 = L_1 - 1.5 \text{ m}$:

Cross-multiply:

Rearrange to solve for $L_1$:

*Note: The calculated original length is negative, which is physically impossible for a pendulum. This indicates an inconsistency in the problem statement, as shortening the length of a pendulum should decrease its period, but here the period increased