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# Which of the following has the same dimensions?$A)$ impulse and momentum.$B)$ specific heat and latent heat.$C)$ moment of inertia and force.$D)$ thrust and surface tension.  Hint: We will have to find the dimensional formula for every pair of quantities and compare them. We need to find a pair where the formulae match. We need to know the basic definition given for every quantity and have to use dimensional formulae of basic quantities that are used to derive the given quantities.

Formula used:
\begin{align} & \text{Impulse = force }\!\!\times\!\!\text{ time} \\ & \text{momentum = mass }\!\!\times\!\!\text{ velocity} \\ & \text{specific heat}=\dfrac{Q}{m\times \Delta T} \\ & \text{latent heat}=\dfrac{Q}{m} \\ & \text{moment of inertia}=M{{R}^{2}} \\ & \text{force}=m\times a \\ & \text{thrust = force} \\ & \text{surface tension = }\dfrac{\text{force}}{\text{length}} \\ \end{align}

We will determine the dimensional formula of every given quantities option wise.
$A)$Here we have impulse and momentum.
Formula of impulse is given by,
$\text{Impulse = force }\!\!\times\!\!\text{ time}$
Dimensional formula of force is $\left[ ML{{T}^{-2}} \right]$ and time is $\left[ T \right]$. Now, dimensional formula of impulse will be,
$\text{Impulse = }\left[ ML{{T}^{-2}} \right]\left[ T \right]=\left[ ML{{T}^{-1}} \right]$
Now, for momentum we need dimensional formulae of mass and velocity, which are $\left[ M \right]$and$\left[ L{{T}^{-1}} \right]$ respectively.
So, dimensional formula of momentum is, $\text{momentum =}\left[ M \right]\left[ L{{T}^{-1}} \right]=\left[ ML{{T}^{-1}} \right]$
Therefore, impulse and momentum have the same dimensional formula.

$B)$ Here we have specific heat and latent heat. $Q$ is heat energy, $m$ is mass and $\Delta T$is change in temperature. The dimensional formula for these quantities are,
\begin{align} & Q=\left[ M{{L}^{2}}{{T}^{-2}} \right] \\ & m=\left[ M \right] \\ & \Delta T=\left[ K \right] \\ \end{align}
Now, the dimensional formula of specific heat will be,
$\text{specific heat}=\dfrac{\left[ M{{L}^{2}}{{T}^{-2}} \right]}{\left[ M \right]\left[ K \right]}=\left[ {{M}^{0}}{{L}^{2}}{{T}^{-2}}{{K}^{-1}} \right]$
Dimensional formula of latent heat is,
$\text{latent heat}=\dfrac{\left[ M{{L}^{2}}{{T}^{-2}} \right]}{\left[ M \right]}=\left[ {{M}^{0}}{{L}^{2}}{{T}^{-2}} \right]$
Therefore, specific heat and latent heat have different dimensional formulas.

$C)$Here, we have moments of inertia and force. The dimensional formula of moment of inertia is,
$\text{moment of inertia}=\left[ M \right]\left[ {{L}^{2}} \right]=\left[ M{{L}^{2}} \right]$
We already have the dimensional formula for force as $\left[ ML{{T}^{-2}} \right]$.
Therefore, moments of inertia and force have different dimensions.

$D)$Here we have thrust and surface tension. Thrust is a force. So it will have the same dimensional formula as $\left[ ML{{T}^{-2}} \right]$.
Dimensional formula of surface tension is, $\text{surface tension = }\dfrac{\left[ ML{{T}^{-2}} \right]}{\left[ L \right]}=\left[ M{{L}^{0}}{{T}^{-2}} \right]$
Therefore, thrust and surface tension have different dimensional formulas.

So, we can conclude that in the given options, only option a have quantities with same dimensional formulae.

Note:
We can make this question easier to solve by remembering the dimensional formulas of each quantity. If we know the dimensional formulae directly, then we can just compare them and find which are having the same dimensional formula. But if it is not possible, we can always derive the dimensional formula for different quantities using basic definition.
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