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 determine the dimensions of \( AB \) from the equation \( E = \frac{B - x^2}{At} \), we first establish the dimensions of the known variables:

1. Energy (\(E\)) has dimensions of \([M L^2 T^{-2}]\), where \(M\) denotes mass, \(L\) denotes length, and \(T\) denotes time.
2. Length (\(x\)) has dimensions of \([L]\).
3. Time (\(t\)) has dimensions of \([T]\).
4. The given equation is \( E = \frac{B - x^2}{At} \).

We now analyze the dimensions on both sides of the equation:

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

Right-hand side (RHS) dimensions: The RHS is \( \frac{B - x^2}{At} \). Since \( x^2 \) has dimensions of \([L^2]\), for consistency in the numerator, \(B\) must also have dimensions of \([L^2]\).

The dimensions of the RHS are thus:

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

By equating the dimensions of the LHS and RHS, we get:

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

Solving for the dimensions of \(A\):

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

Finally, we calculate the dimensions of \(AB\):

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

Simplifying, we obtain:

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

Therefore, the dimensions of \( AB \) are \([L^2 M^{-1} T^{-1}]\).

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