Question

Match the column:

Column 1	Column 2
($h$(Planck’s constant)	$\left[ML{T}^{-1}\right]$
)$E$(Kinetic energy)	$\left[M{L}^2{T}^{-1}\right]$
$V$(Electric potential)	$\left[M{L}^2{T}^{-2}\right]$
$p$ (Linear momentum)	$\left[M{L}^2{I}^{-1}{T}^{-3}\right]$

Answer 1

Part A:

1. Planck’s constant is defined as the energy of a photon divided by its frequency.
2. It is expressed as,
   $h=\frac{E}{f}$ ($E=$energy of a photon, $f=$frequency of photon)
3. It is a fundamental dimensional constant.
4. $S.Iunit$of Planck's constant is$Js(Joule-second)$.
5. Its dimension is,
   $\begin{array}{rcl}& & \left[energy\right]\times \left[time\right]\\ & =& \left[M{L}^2{T}^{-2}\right]\times \left[T\right]\\ & =& \left[M{L}^2{T}^{-1}\right]\end{array}$

Hence,(A)-(II)

Part B:

1. Kinetic energy is defined as the energy of an object that possesses by the virtue of its motion.
2. It is expressed as,
   $E=\frac{1}{2}m{v}^2$ ($m=$mass of an object, $v=$velocity of the object)
3. $S.Iunit$ of it is$J\left(Joule\right)$
4. It is a dimensional variable, its dimension is,
   $\begin{array}{rcl}& & \left[mass\right]\times {\left[velocity\right]}^2\\ & =& \left[M\right]\times {\left[L{T}^{-1}\right]}^2\\ & =& \left[M{L}^2{T}^{-2}\right]\end{array}$

Hence,(B)-(III)

Part C:

1. Electric potential at a point in an electric field is defined as the work done in bringing a unit positive charge from infinity to that specific point in the electric field.
2. It is expressed as,
   $V=\frac{W}{q}$ ($W=$Work done, $q=$Positive test charge)
3. It is a dimensional variable, having$S.IunitJ{C}^{-1}(Joule/Coulomb)$.
4. Its dimension is,
   $$\begin{gathered} \frac{\left[Workdone\right]}{\left[charge\right]} \\ =\frac{\left[M{L}^2{T}^{-2}\right]}{\left[current\right]\left[time\right]} \\ =\frac{\left[M{L}^2{T}^{-2}\right]}{\left[IT\right]} \\ =\left[M{L}^2{T}^{-3}{I}^{-1}\right] \end{gathered}$$

Hence,(C)-(IV)

Part D:

1. Linear momentum is defined as the product of the mass and velocity of any moving object.
2. It is expressed as,
   $p=mv$
3. It is a dimensional variable, having$S.Iunitkgm{s}^{-1}$.
4. Its dimension is,
   $\begin{array}{rcl}& & \left[mass\right]\times \left[velocity\right]\\ & =& \left[M\right]\times \left[L{T}^{-1}\right]\\ & =& \left[ML{T}^{-1}\right]\end{array}$

Thus,(D)-(I)

Hence, the correct answer is (A) -(II), (B) (III), (C)- (IV), (D) → (I).