Dimensional Analysis Marathon (JEE/NEET PYQs)

Dimensional Analysis Marathon (JEE/NEET PYQs)

Master the dimensions of key physical quantities with these three challenging questions.

Question 1: Dipole Moment Ratio

In an electromagnetic system, a quantity defined as the ratio of electric dipole moment and magnetic dipole moment has dimension of $[M^P L^Q T^R A^S]$. The value of P and Q are:

A) -1, 0 B) 0, -1 C) -1, 1 D) 1, -1

Solution 1

1. **Electric Dipole Moment ($\vec{p}$):** $p = \text{charge} \times \text{distance}$. Dimensions: $[A T] \times [L] = [M^0 L^1 T^1 A^1]$.
2. **Magnetic Dipole Moment ($\vec{m}$):** $m = \text{Current} \times \text{Area}$. Dimensions: $[A] \times [L^2] = [M^0 L^2 T^0 A^1]$.
3. **Ratio ($\text{Ratio} = p/m$):** $$[\text{Ratio}] = \frac{[M^0 L^1 T^1 A^1]}{[M^0 L^2 T^0 A^1]} = [M^0 L^{-1} T^1 A^0]$$

The ratio is $[M^P L^Q T^R A^S] = [M^0 L^{-1} T^1 A^0]$.

Therefore, $P=0$ and $Q=-1$. **Correct Option: B**.

Question 2: Dimensional Matching

Match List - I with List - II:

List - I (Quantity)	List - II (Dimension)
(A) Mass density	(I) $[M L^2 T^{-3}]$
(B) Impulse	(II) $[M L T^{-1}]$
(C) Power	(III) $[M L^2 T^{-2}]$
(D) Moment of inertia	(IV) $[M L^{-3} T^{0}]$

A) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) B) (A)-(II), (B)-(III), (C)-(IV), (D)-(I) C) (A)-(IV), (B)-(II), (C)-(I), (D)-(III) D) (A)-(IV), (B)-(II), (C)-(I), (D)-(III)

Solution 2

• **(A) Mass density ($\rho$):** Mass/Volume = $[M]/[L^3] = [M L^{-3} T^{0}]$ **(IV)**
• **(B) Impulse ($J$):** Force $\times$ Time = $[M L T^{-2}] \times [T] = [M L T^{-1}]$ **(II)** (Same as Momentum)
• **(C) Power ($P$):** Work/Time = $[M L^2 T^{-2}]/[T] = [M L^2 T^{-3}]$ **(I)**
• **(D) Moment of inertia ($I$):** Mass $\times$ Distance$^2$ = $[M] \times [L^2] = [M L^2 T^{0}]$ **(III)**

**Correct Match:** (A)-(IV), (B)-(II), (C)-(I), (D)-(III). **Correct Option: C** (Note: Option D is a typo in the image, C is the correct match).

Question 3: Dimension of $\sqrt{\mu_0/\epsilon_0}$

The dimension of $\sqrt{\frac{\mu_0}{\epsilon_0}}$ is equal to that of:

($\mu_0 = \text{Vacuum permeability}$ and $\epsilon_0 = \text{Vacuum permittivity}$)

A) Voltage B) Inductance C) Resistance D) Capacitance

Solution 3

1. **Speed of Light ($c$):** The speed of light in vacuum is related to $\mu_0$ and $\epsilon_0$ by the equation: $$c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$$
2. **Impedance of Free Space ($Z_0$):** The characteristic impedance of free space is defined as: $$Z_0 = \sqrt{\frac{\mu_0}{\epsilon_0}}$$
3. **Dimensionality:** Since impedance ($Z_0$) represents the opposition to the flow of electromagnetic energy, it has the same dimension as **Resistance** (which is measured in Ohms, $\Omega$).
4. **Dimensional Verification (Optional):** The dimension of Resistance is $[M L^2 T^{-3} A^{-2}]$. The dimension of $\sqrt{\mu_0/\epsilon_0}$ is also $[M L^2 T^{-3} A^{-2}]$.

The dimension of $\sqrt{\mu_0/\epsilon_0}$ is equal to that of **Resistance**. **Correct Option: C**.

Key Takeaways

• **Dimensional Homogeneity:** Only quantities with the same dimensions can be added or subtracted.
• **Ratio of $\mu_0$ and $\epsilon_0$ ($\sqrt{\mu_0/\epsilon_0}$):** This is the **Impedance of Free Space**, dimensionally equal to Resistance.
• **Impulse vs. Momentum:** They share the same dimension: $[M L T^{-1}]$.