2019 A/L Physics Past Paper · Question 02 (English Medium)

Answer: (4) $L^3M^{-1}T^{-2}$

What the question asks

We want the dimensions of the gravitational constant $G$.

It is in Newton's law of gravitation.

Write them using $M$, $L$ and $T$.

Then find the matching option.

Relevant theory

Symbolic expressions which show the relation of a physical quantity to basic units is called its dimension.

If the equation is correct, the dimension of its both sides should be equal.

Start with the law.

$$F = \dfrac{Gm_1m_2}{r^2}$$

Make $G$ the subject.

$$G = \dfrac{Fr^2}{m_1m_2}$$

Both sides of an equation have the same dimensions.

Write the dimensions of each part.

$[F] = MLT^{-2}$

$[r^2] = L^2$

$[m_1m_2] = M^2$

$$[G] = \dfrac{MLT^{-2} \cdot L^2}{M^2} = L^3M^{-1}T^{-2}$$

Reading the options

Compare the powers of $L$, $M$ and $T$ in each option with $(3, -1, -2)$.

asked

L^3 M^{-1} T^{-2}

$G = F r^2 / (m_1 m_2)$

1

L^2 M^{-1} T^{-1}

$L^2 M^{-1} T^{-1}$

2

L^2 M^{-2}

$L^2 M^{-2}$

3

L^2 M^{-2} T^{-1}

$L^2 M^{-2} T^{-1}$

4

$L^3 M^{-1} T^{-2}$

5

L^3 M^{-2} T^{-2}

$L^3 M^{-2} T^{-2}$

Options (1), (2) and (3) have $L^2$.

We need $L^3$. So they are out.

Option (5) has $L^3$ and $T^{-2}$ right.

But it has $M^{-2}$. We need $M^{-1}$. So it is out too.

Only (4) is left.

Checks

Check the answer.

$$\dfrac{[G][m_1m_2]}{[r^2]} = \dfrac{L^3M^{-1}T^{-2} \times M^2}{L^2}$$

$$= MLT^{-2}$$

This is $[F]$. Both sides match.

Check the unit too.

The unit of $G$ is $\text{N m}^2\text{ kg}^{-2}$.

A newton is $\text{kg m s}^{-2}$.

$$\text{N m}^2\text{ kg}^{-2} = \text{kg}^{-1}\text{ m}^3\text{ s}^{-2}$$

This matches $L^3M^{-1}T^{-2}$.

Conclusion

$[G] = L^3M^{-1}T^{-2}$

$[G] = L^3M^{-1}T^{-2}$.

Only option (4) matches. The answer is (4).

To find the dimensions of a constant, write the law it is in.

Make the constant the subject.

Then write each quantity using $M$, $L$ and $T$.

Exam tip