Question

Consider two physical quantities \( A \) and \( B \) related to each other as \( E = \frac{B - x^2}{At} \) where \( E \), \( x \), and \( t \) have dimensions of energy, length, and time, respectively. The dimension of \( AB \) is:

• A. \( L^{-2} M T^0 \)
• B. \( L^2 M^{-1} T^{-1} \)
• C. \( L^{-2} M^{-1} T^1 \)
• D. \( L^0 M^{-1} T^1 \)

Answer 1

The Correct Option is B

To solve this problem, we need to identify the dimension of \( AB \) from the given equation \( E = \frac{B - x^2}{At} \).

1. \(E\) represents energy, whose dimensions are \([M L^2 T^{-2}]\), where \(M\) is mass, \(L\) is length, and \(T\) is time.
2. \(x\) represents length, so its dimensions are \([L]\).
3. \(t\) represents time, so its dimensions are \([T]\).
4. The equation is given as \( E = \frac{B - x^2}{At} \).

Let's analyze the dimensions of both sides of the equation:

Left-hand side dimensions (LHS): The dimension of energy \( E \) is \([M L^2 T^{-2}]\).

Right-hand side dimensions (RHS): The RHS is \( \frac{B - x^2}{At} \). Since \( x^2 \) is \([L^2]\), the dimension of \( B \) must also be \([L^2]\) to make the dimensions inside the numerator consistent.

The RHS simplifies to:

\[\frac{[L^2]}{[A][T]}\]

The LHS-dimensional formula is:

\([M L^2 T^{-2}] = \frac{[L^2]}{[A][T]}\)

Equating the dimensions, we get:

Step 1: Since \([M L^2 T^{-2}]\) is on the left side and \([A L^2 T^{-1}]\) is on the right side after substituting the dimensions, we can equate these:

\([M L^2 T^{-2}] = \frac{[L^2]}{[A][T]}\)

Step 2: Solving for the dimension of \(A\):

\([A] = \frac{[L^2]}{[M L^2 T^{-2} T]} = [M^{-1} T^{-1}]\)

Step 3: The dimension of \(AB\) is:

\([A][B] = [M^{-1} T^{-1}][L^2]\)

Simplifying:

\(= [L^2 M^{-1} T^{-1}]\)

Thus, the dimension of \( AB \) is \(L^2 M^{-1} T^{-1}\), which matches with one of the provided options. Therefore, the correct answer is:

Correct Answer: \([L^2 M^{-1} T^{-1}]\)

Answer 2

Given:

\[ E = \frac{B - x^2}{At}. \]

The dimensions of \(E\), \(x\), and \(t\) are:

\[ [E] = ML^2T^{-2}, \quad [x] = L, \quad [t] = T. \]

The term \(B - x^2\) must have the same dimensions as \(E\), so:

\[ [B] = L^2. \]

Rearrange the equation to find the dimensions of \(A\):

\[ A = \frac{B - x^2}{E \cdot t} = \frac{L^2}{ML^2T^{-2} \cdot T} = M^{-1}T. \]

Therefore:

\[ [A] = M^{-1}T. \]

The dimensions of \(AB\) are:

\[ [AB] = [A][B] = (M^{-1}T)(L^2) = L^2M^{-1}T. \]

Thus, the answer is:

\[ L^2M^{-1}T. \]