Question

If force (F), velocity (V) and time (T) are considered as fundamental physical quantity, then dimensional formula of density will be :

• A. \(F^2V^{–2}T^6\)
• B. \(FV^4T^{–6}\)
• C. \(FV^{–2}T^2\)
• D. \(FV^–4T^{–2}\)

Answer 1

The Correct Option is D

Given:

Force (\( F \)), velocity (\( V \)), and time (\( T \)) are fundamental quantities.

Step 1: Dimensional Formula for Density

Let the dimensional formula for density (\( \rho \)) be:

\[ [\rho] = F^x V^y T^z, \]

where \( x, y, \) and \( z \) are constants to be determined.

Step 2: Dimensions of Fundamental Quantities

• \( [F] = [MLT^{-2}] \) (force)
• \( [V] = [LT^{-1}] \) (velocity)
• \( [T] = [T] \) (time)
• \( [\rho] = [ML^{-3}] \) (density)

Step 3: Substitute Dimensions and Equate Powers

Substituting the dimensions into \( [\rho] \):

\[ [ML^{-3}] = [MLT^{-2}]^x [LT^{-1}]^y [T]^z. \]

Expanding the dimensions on the right-hand side:

\[ [ML^{-3}] = M^x L^{x+y} T^{-2x-y+z}. \]

Equating the powers of \( M, L, \) and \( T \):

• \( M: x = 1 \)
• \( L: x + y = -3 \)
• \( T: -2x - y + z = 0 \)

Step 4: Solve the System of Equations

• From \( x = 1 \), substitute into \( x + y = -3 \):
• Substitute \( x = 1 \) and \( y = -4 \) into \( -2x - y + z = 0 \):

Step 5: Write the Dimensional Formula for Density

Substituting \( x = 1 \), \( y = -4 \), and \( z = -2 \):

\[ [\rho] = F^1 V^{-4} T^{-2}. \]

Final Answer:

The dimensional formula for density is \( F V^{-4} T^{-2} \).