JEE PYQ: Units & Measurements - Question ID d0d1ec9386a1 (JEE Main 2022)

In problems involving dimensional correctness, we rely on the principle of dimensional homogeneity. Every physically meaningful equation must have the same dimensions on both sides. The key formula we use is:

• The time period \( T \) has the dimension of time: \([T] = \mathrm{M}^0 \mathrm{L}^0 \mathrm{T}^1\).
• Surface tension \( S \) has dimensions of force per unit length: \([S] = \mathrm{M}^1 \mathrm{L}^0 \mathrm{T}^{-2}\).
• Density \( \rho \) has dimensions of mass per unit volume: \([\rho] = \mathrm{M}^1 \mathrm{L}^{-3} \mathrm{T}^0\).
• Radius \( r \) has the dimension of length: \([r] = \mathrm{M}^0 \mathrm{L}^1 \mathrm{T}^0\).

We will check whether the given expression for \( T \) has the same dimensions as time.

Step 1: Write down the given expression for \( T \).

The assertion claims: $T = K \sqrt{\frac{\rho r^3}{S^{3/2}}}$ where \( K \) is dimensionless.

Step 2: Compute the dimensions of the numerator \( \rho r^3 \).

\([\rho] = \mathrm{M}^1 \mathrm{L}^{-3} \mathrm{T}^0\) and \([r^3] = \mathrm{M}^0 \mathrm{L}^3 \mathrm{T}^0\). Thus, $[\rho r^3] = \mathrm{M}^1 \mathrm{L}^{-3} \cdot \mathrm{L}^3 = \mathrm{M}^1 \mathrm{L}^0 \mathrm{T}^0.$

Step 3: Compute the dimensions of the denominator \( S^{3/2} \).

\([S] = \mathrm{M}^1 \mathrm{L}^0 \mathrm{T}^{-2}\), so $[S^{3/2}] = \mathrm{M}^{3/2} \mathrm{L}^0 \mathrm{T}^{-3}.$

Step 4: Compute the dimensions of the fraction \( \frac{\rho r^3}{S^{3/2}} \).

$$\left[\frac{\rho r^3}{S^{3/2}}\right] = \frac{\mathrm{M}^1 \mathrm{L}^0 \mathrm{T}^0}{\mathrm{M}^{3/2} \mathrm{L}^0 \mathrm{T}^{-3}} = \mathrm{M}^{1 - 3/2} \mathrm{L}^0 \mathrm{T}^{0 - (-3)} = \mathrm{M}^{-1/2} \mathrm{L}^0 \mathrm{T}^3.$$

Step 5: Compute the dimensions of the square root of the fraction.

$\left[\sqrt{\frac{\rho r^3}{S^{3/2}}}\right] = \left(\mathrm{M}^{-1/2} \mathrm{L}^0 \mathrm{T}^3\right)^{1/2} = \mathrm{M}^{-1/4} \mathrm{L}^0 \mathrm{T}^{3/2}.$

Step 6: Compare with the dimension of time \( T \).

The dimension of time is \([T] = \mathrm{M}^0 \mathrm{L}^0 \mathrm{T}^1\). The expression inside the square root yields \(\mathrm{M}^{-1/4} \mathrm{L}^0 \mathrm{T}^{3/2}\), which does **not** match the dimension of time.

Step 7: Analyze the Reason (R).

The Reason (R) states: "Using dimensional analysis we get R.H.S. having different dimension than that of time period." This matches exactly with our calculation. The R.H.S. does **not** have the dimension of time, so the expression is dimensionally **incorrect**.

Step 8: Evaluate the Assertion (A) and Reason (R).

• The Assertion (A) claims the expression is dimensionally correct. This is **false**.
• The Reason (R) correctly states that the R.H.S. has a different dimension than time. This is **true**.

Thus, the correct choice is: D: (A) is false but (R) is true.

Students often make the following mistakes in such questions:

1. Ignoring fractional exponents in dimensional analysis: Many forget to raise dimensions to fractional powers (like \( S^{3/2} \)), leading to incorrect cancellations.
2. Miscounting dimensions of surface tension: Surface tension has dimensions of force per unit length (\( \mathrm{M} \mathrm{T}^{-2} \)), not force per unit area. Misremembering this leads to wrong results.
3. Assuming the formula is correct without checking: Since the question mentions \( K \) is dimensionless, students may assume the formula is correct without verifying dimensions.
4. Confusing Assertion and Reason logic: Students may think that if the Reason (R) is true, the Assertion (A) must also be true, which is not the case here.

Exam Tip: Always verify dimensions of every term in the equation, especially when fractional powers are involved. Never assume correctness based on the presence of a dimensionless constant.