JEE PYQ: Units & Measurements - Question ID 7def8494f47d (JEE Main 2024)

Core Formula & Concept:

In dimensional analysis, every physical quantity can be expressed in terms of fundamental dimensions: mass ($M$), length ($L$), time ($T$), and sometimes electric current ($I$), temperature ($\Theta$), etc. The given problem involves expressing mass ($m$) in terms of three fundamental constants:

• Speed of light in vacuum: $c$ (dimensions $LT^{-1}$)
• Gravitational constant: $G$ (dimensions $M^{-1}L^3T^{-2}$)
• Planck’s constant: $h$ (dimensions $ML^2T^{-1}$)

The equation provided is: $m = k \, c^P \, G^{-1/2} \, h^{1/2}$ where $k$ is a dimensionless constant. Our goal is to find the value of $P$ such that the dimensions on both sides of the equation match.

Step-by-Step Derivation:

Step 1: Write down the dimensions of each quantity involved.

• Mass: $[m] = M$
• Speed of light: $[c] = LT^{-1}$
• Gravitational constant: $[G] = M^{-1}L^3T^{-2}$
• Planck’s constant: $[h] = ML^2T^{-1}$

Step 2: Substitute the dimensions into the given equation.

The equation is: $m = k \, c^P \, G^{-1/2} \, h^{1/2}$ Ignoring the dimensionless constant $k$, we write the dimensional equation: $[m] = [c]^P \cdot [G]^{-1/2} \cdot [h]^{1/2}$ Substitute the dimensions: $M = (LT^{-1})^P \cdot (M^{-1}L^3T^{-2})^{-1/2} \cdot (ML^2T^{-1})^{1/2}$

Step 3: Simplify the exponents for each fundamental dimension.

Let’s break it down dimension by dimension: For Mass ($M$): - From $G^{-1/2}$: $(M^{-1})^{-1/2} = M^{1/2}$ - From $h^{1/2}$: $(M)^{1/2} = M^{1/2}$ - Total exponent of $M$ on RHS: $1/2 + 1/2 = 1$ - On LHS, exponent of $M$ is $1$. - Thus, $M$ exponents match: $1 = 1$ (consistent). For Length ($L$): - From $c^P$: $(L)^P = L^P$ - From $G^{-1/2}$: $(L^3)^{-1/2} = L^{-3/2}$ - From $h^{1/2}$: $(L^2)^{1/2} = L^1$ - Total exponent of $L$ on RHS: $P - 3/2 + 1 = P - 1/2$ - On LHS, exponent of $L$ is $0$ (since mass has no length dimension). - Thus, we have: $P - \frac{1}{2} = 0 \implies P = \frac{1}{2}$ For Time ($T$): - From $c^P$: $(T^{-1})^P = T^{-P}$ - From $G^{-1/2}$: $(T^{-2})^{-1/2} = T^1$ - From $h^{1/2}$: $(T^{-1})^{1/2} = T^{-1/2}$ - Total exponent of $T$ on RHS: $-P + 1 - 1/2 = -P + 1/2$ - On LHS, exponent of $T$ is $0$ (since mass has no time dimension). - Thus, we have: $-P + \frac{1}{2} = 0 \implies P = \frac{1}{2}$

Step 4: Verify consistency across all dimensions.

Both length and time dimensions yield $P = 1/2$, and mass dimension is already consistent. Thus, the value of $P$ is uniquely determined as $1/2$. Common Traps & Exam Tip:

1. Incorrect dimensional substitution: Students often confuse the dimensions of $G$ and $h$. Remember:

• $G$ has dimensions $M^{-1}L^3T^{-2}$ (not $ML^3T^{-2}$).
• $h$ has dimensions $ML^2T^{-1}$ (not $MLT^{-1}$).

2. Sign errors in exponents: When raising a quantity to a negative power (e.g., $G^{-1/2}$), ensure the exponents are correctly distributed. A common mistake is to write $G^{-1/2}$ as $M^{1/2}L^{-3/2}T^1$ instead of $M^{1/2}L^{-3/2}T^1$ (which is correct here, but the sign of the exponent in the original expression matters).

3. Forgetting to equate exponents for all dimensions: Some students only check one dimension (e.g., mass) and assume the others will automatically match. Always verify consistency for $M$, $L$, and $T$ separately.

4. Misinterpreting the role of $k$: The constant $k$ is dimensionless, so it does not affect the dimensional analysis. Ignore it during the derivation.

Exam Tip: When solving dimensional analysis problems, always:

1. Write down the dimensions of all quantities clearly.
2. Substitute them into the equation and simplify.
3. Equate the exponents of $M$, $L$, and $T$ separately to form equations.
4. Solve the system of equations to find the unknown exponent(s).

The correct value of $P$ is $\boxed{\frac{1}{2}}$, which corresponds to option C.