Answer Key-Deducing Relations Among Physical Quantities Using Dimensional Analysis:

 Here is the Answer Key for the NEP-based worksheet on Deducing Relations Among Physical Quantities Using Dimensional Analysis:

✅ Section A: Conceptual Understanding

1. Definitions:
   (a) Dimensional Formula: An expression showing which base quantities and their powers represent a physical quantity. E.g., volume = [M⁰ L³ T⁰].
   (b) Dimensional Equation: An equation equating the physical quantity with its dimensional formula. E.g., [F] = [M L T⁻²].
   (c) Dimensionless Quantity: A physical quantity with no dimensions. E.g., angle (radian), refractive index.

2. Limitations:

   • It cannot determine the dimensionless constants (like 2π, ½).

   • It cannot differentiate between two physical quantities with the same dimensions (e.g., torque and work).

3. An equation is dimensionally consistent if all the terms in it have the same dimensions.

4. Because dimensional analysis only considers powers of base units, and constants like $2\pi$ are pure numbers without dimensions.

✅ Section B: Application-Based Questions

5. Simple Pendulum

   $$T = k \cdot l^x \cdot g^y$$

   Dimensions:
   $[T] = [L]^x [L T^{-2}]^y = L^{x+y} T^{-2y}$
   Equating dimensions:

   $$x + y = 0,\quad -2y = 1 \Rightarrow y = -\tfrac{1}{2},\ x = \tfrac{1}{2}$$ $$\boxed{T = k \sqrt{\frac{l}{g}}}$$

6. Wave Speed

   $$v = k \cdot T^a \cdot \mu^b$$

   Dimensions:
   $[L T^{-1}] = [M L T^{-2}]^a [M L^{-1}]^b$
   $= M^{a+b} L^{a - b} T^{-2a}$
   Equating powers:

   $$a + b = 0,\quad a - b = 1,\quad -2a = -1 \Rightarrow a = \tfrac{1}{2},\ b = -\tfrac{1}{2}$$ $$\boxed{v = k \sqrt{\frac{T}{\mu}}}$$

7. Free Fall Time

   $$t = k \cdot h^x \cdot g^y$$

   Dimensions:
   $[T] = [L]^x [L T^{-2}]^y = L^{x+y} T^{-2y}$

   $$x + y = 0,\quad -2y = 1 \Rightarrow y = -\tfrac{1}{2},\ x = \tfrac{1}{2}$$ $$\boxed{t = k \sqrt{\frac{h}{g}}}$$

✅ Section C: Higher-Order Thinking

8. Angular Momentum $Q = m^x v^y r^z$
   $[Q] = [M L^2 T^{-1}]$,
   $[m]^x [L T^{-1}]^y [L]^z = M^x L^{y+z} T^{-y}$
   Matching exponents:

   $$x = 1,\quad y + z = 2,\quad -y = -1 \Rightarrow y = 1,\ z = 1$$ $$\boxed{Q = k \cdot m \cdot v \cdot r}$$

9. Potential Energy:
   The formula $U = -\frac{G M m}{r}$ contains a negative sign and constant $G$.
   Dimensional analysis can confirm form like:

   $$U \propto \frac{Mm}{r}$$

   But cannot determine the value or sign of the gravitational constant $G$.
   ✅ Yes, it can be verified dimensionally but not derived exactly.

10. Centripetal Force

$$F = k \cdot m^x v^y r^z \Rightarrow [M L T^{-2}] = M^x L^{y+z} T^{-y}$$

Matching:

$$x = 1,\quad y + z = 1,\quad -y = -2 \Rightarrow y = 2,\ z = -1$$ $$\boxed{F = k \cdot \frac{m v^2}{r}}$$

✅ Section D: Competency Activity

Given:

$$\Phi = k \cdot t^a \cdot q^b \cdot l^c$$

Assuming $\Phi$ behaves like electric potential (V).

$$[V] = \frac{W}{q} = \frac{M L^2 T^{-2}}{A T} = M L^2 T^{-3} A^{-1}$$

Dimensions:

$$[M L^2 T^{-3} A^{-1}] = [T]^a [A T]^b [L]^c = A^b T^{a+b} L^c$$

Matching exponents:

$$A^{-1} = A^b \Rightarrow b = -1$$ $$T^{-3} = T^{a + b} \Rightarrow a + (-1) = -3 \Rightarrow a = -2$$ $$L^2 = L^c \Rightarrow c = 2$$ $$\boxed{a = -2,\ b = -1,\ c = 2} \Rightarrow \Phi = k \cdot \frac{l^2}{q \cdot t^2}$$