Question

If the dimensional formula of (Energy $\times$ speed) is $[M^a L^b T^c]$ then $a, b$ and $c$ are

• A. $(1, 3, -3)$
• B. $(1, 2, 2)$
• C. $(1, 2, 3)$
• D. $(1, 3, -2)$

Hint:

To find the dimensional formula of a product, multiply the dimensional formulas of each term by adding the exponents of corresponding dimensions (M, L, T).

Answer 1

The Correct Option is A

To find the dimensional formula of (Energy $\times$ speed), we first determine the dimensional formulas of energy and speed.
- Energy: Energy can be represented by kinetic energy, $\frac{1}{2}mv^2$, where $m$ is mass and $v$ is velocity. The dimensional formula of mass $m$ is $[M]$. Velocity $v$ is distance per time, so its dimensional formula is $[L T^{-1}]$. Thus, the dimensional formula of energy is:
\[ [m][v]^2 = [M][L T^{-1}]^2 = [M L^2 T^{-2}] \] - Speed: Speed is distance per time, so its dimensional formula is: \[ [L T^{-1}] \] Now, calculate the dimensional formula of (Energy $\times$ speed): \[ [\text{Energy} \times \text{speed}] = [M L^2 T^{-2}] \times [L T^{-1}] \] Combine the dimensions: - Mass: $[M^1] \times [M^0] = [M^1]$
- Length: $[L^2] \times [L^1] = [L^{2+1}] = [L^3]$
- Time: $[T^{-2}] \times [T^{-1}] = [T^{-2-1}] = [T^{-3}]$
Thus, the dimensional formula of (Energy $\times$ speed) is: \[ [M^1 L^3 T^{-3}] \] This corresponds to $[M^a L^b T^c]$, where $a = 1$, $b = 3$, and $c = -3$. Comparing with the options, this matches option (1): $(1, 3, -3)$.
Thus, the correct answer is (1).