Statement-I : Linear momentum and moment of force have same dimensions.

Statement-II : Planck’s constant and angular momentum have same dimension.

In the light of the above statements, choose the correct answer from the options given below :

The correct option/answer is: Statement-I is false while statement-II is correct.

Let's analyze both statements step-by-step using dimensional analysis.

Step 1: Analyze Statement-I
Statement-I states that linear momentum and moment of force have the same dimensions.

The formula for linear momentum ($p$) is:
$p=mv$
where $m$ is mass and $v$ is velocity.
The dimensions of mass are $\left[M\right]$ and the dimensions of velocity are $\left[L{T}^{-1}\right]$.
Therefore, the dimensional formula of linear momentum is:
$\left[p\right]=\left[ML{T}^{-1}\right]$

The moment of force (torque, $\tau$) is given by:
$\tau =r\times F$
where $r$ is distance and $F$ is force.
The dimensions of distance are $\left[L\right]$ and the dimensions of force are $\left[ML{T}^{-2}\right]$.
Therefore, the dimensional formula of moment of force is:
$\left[\tau \right]=\left[L\right]\left[ML{T}^{-2}\right]=\left[M{L}^2{T}^{-2}\right]$

Comparing the two dimensional formulas:
$\left[ML{T}^{-1}\right]\ne \left[M{L}^2{T}^{-2}\right]$
Thus, Statement-I is false.

Step 2: Analyze Statement-II
Statement-II states that Planck’s constant and angular momentum have the same dimensions.

According to Planck's equation, the energy ($E$) of a photon is given by:
$E=h\nu$
where $h$ is Planck's constant and $\nu$ is frequency.
So, $h=\frac{E}{\nu }$.
The dimensions of energy are $\left[M{L}^2{T}^{-2}\right]$ and the dimensions of frequency are $\left[T^{-1}\right]$.
Therefore, the dimensional formula of Planck's constant is:
$\left[h\right]=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[T^{-1}\right]}$

The formula for angular momentum ($L$) is:
$L=mvr$
where $m$ is mass, $v$ is velocity, and $r$ is radius.
Therefore, the dimensional formula of angular momentum is:
$\left[L\right]=\left[M\right]\left[L{T}^{-1}\right]\left[L\right]=\left[M{L}^2{T}^{-1}\right]$

Comparing the two dimensional formulas:
$\left[h\right]=\left[L\right]=\left[M{L}^2{T}^{-1}\right]$
Thus, Statement-II is correct.

Conclusion:
Statement-I is false while statement-II is correct.