Match List-I with List-II : List-I:
`{:(,"List-I", ,"List-II"),((a),"h(Plank's constant)",(i),[M LT^(-1)]),((b),"E(Kinetic energy)",(ii),[M L^(2)T^(-1)]),((c),"V(elecic potential)",(iii),[M L^(2)T^(-2)]),((d),"P(linear momentum)",(iv),[ML^(2)I^(-1)T^(-3)]):}`
Choose the correct answer from the options given below :

Options

A. `(a) to (iii), (b) to (iv), (c) to (ii), (d) to (i)`
B. `(a) to (ii), (b) to (iii), (c) to (iv), (d) to (i)`
C. `(a) to (i), (b) to (ii), (c) to (iv), (d) to (iii)`
D. `(a) to (iii), (b) to (ii), (c) to (iv), (d) to (iii)`

Answer

To solve the problem of matching List-I with List-II based on the dimensional formulas, we will analyze each item in List-I and determine its dimensional formula. Then, we will match it with the corresponding item in List-II.

### Step 1: Determine the Dimensional Formula of Planck's Constant (h)

Planck's constant (h) relates energy (E) and frequency (ν) through the equation:
\[ E = hν \]
We can also express momentum (p) in terms of Planck's constant:
\[ p = \frac{h}{\lambda} \]
Where \( \lambda \) is the wavelength.

From the momentum formula:
\[ h = p \cdot \lambda \]
The dimensional formula for momentum (p) is:
\[ [p] = [M][L][T^{-1}] \]
And the dimensional formula for wavelength (\( \lambda \)) is:
\[ [\lambda] = [L] \]

Thus, we have:
\[ [h] = [p] \cdot [\lambda] = [M][L][T^{-1}] \cdot [L] = [M][L^2][T^{-1}] \]
So, the dimensional formula for Planck's constant is:
\[ [h] = [M][L^2][T^{-1}] \]

### Step 2: Determine the Dimensional Formula of Kinetic Energy (E)

The formula for kinetic energy (E) is:
\[ E = \frac{1}{2} mv^2 \]
Where \( m \) is mass and \( v \) is velocity.

The dimensional formula for mass (m) is:
\[ [m] = [M] \]
And the dimensional formula for velocity (v) is:
\[ [v] = \frac{[L]}{[T]} \]

Thus, the dimensional formula for kinetic energy is:
\[ [E] = [M] \cdot \left(\frac{[L]}{[T]}\right)^2 = [M][L^2][T^{-2}] \]

### Step 3: Determine the Dimensional Formula of Electric Potential (V)

The formula for electric potential (V) is given by:
\[ V = \frac{W}{Q} \]
Where \( W \) is work done and \( Q \) is charge.

Work done (W) can be expressed as:
\[ W = F \cdot d \]
Where \( F \) is force and \( d \) is displacement.

The dimensional formula for force (F) is:
\[ [F] = [M][L][T^{-2}] \]
And for displacement (d):
\[ [d] = [L] \]

Thus, the dimensional formula for work done is:
\[ [W] = [F][d] = [M][L][T^{-2}] \cdot [L] = [M][L^2][T^{-2}] \]

Now, the dimensional formula for charge (Q) is:
\[ [Q] = [I][T] \]
Where \( I \) is current.

Thus, the dimensional formula for electric potential is:
\[ [V] = \frac{[W]}{[Q]} = \frac{[M][L^2][T^{-2}]}{[I][T]} = [M][L^2][T^{-3}][I^{-1}] \]

### Step 4: Determine the Dimensional Formula of Linear Momentum (P)

The formula for linear momentum (P) is:
\[ P = mv \]
Where \( m \) is mass and \( v \) is velocity.

Thus, the dimensional formula for linear momentum is:
\[ [P] = [M] \cdot \left(\frac{[L]}{[T]}\right) = [M][L][T^{-1}] \]

### Step 5: Match List-I with List-II

Now we can match the dimensional formulas obtained with those in List-II:

- **(a) Planck's constant (h)**: \( [M][L^2][T^{-1}] \) → Matches with (ii) \( [M][L^2][T^{-1}] \)
- **(b) Kinetic energy (E)**: \( [M][L^2][T^{-2}] \) → Matches with (iii) \( [M][L^2][T^{-2}] \)
- **(c) Electric potential (V)**: \( [M][L^2][T^{-3}][I^{-1}] \) → Matches with (iv) \( [M][L^2][I^{-1}][T^{-3}] \)
- **(d) Linear momentum (P)**: \( [M][L][T^{-1}] \) → Matches with (i) \( [M][L][T^{-1}] \)

### Final Matching:
- (a) → (ii)
- (b) → (iii)
- (c) → (iv)
- (d) → (i)

### Conclusion:
The correct answer is option 2.