Question

use dimensional analysis. consider the equation ( v = \frac{1}{5} z x t^2 ). the dimensions of the variables ( v ), ( x ), and ( t ) are ( l/t ), ( l ), and ( t ) respectively. the numerical factor 5 is dimensionless. what must be the dimensions of the variable ( z ), such that both sides of the equation have the same dimensions? (circ) t (circ) ( 1/t^3 ) (circ) ( 1/t^2 ) (circ) ( 1/t )

Explanation:

Step1: Write dimension equation

Given \( v=\frac{1}{5}zxt^{2} \), substitute dimensions: \( \frac{[L]}{[T]} = z \cdot [L] \cdot [T]^{2} \) (since \( \frac{1}{5} \) is dimensionless).

Step2: Solve for \( z \)'s dimension

Rearrange: \( z=\frac{[L/T]}{[L][T]^{2}} \). Simplify: \( z=\frac{1}{[T]^{3}} \)? Wait, no—wait, \( \frac{[L]}{[T]} \div ([L][T]^{2}) = \frac{1}{[T]^{3}} \)? Wait, no, let's recalculate. \( \frac{[L]}{[T]} = z \cdot [L] \cdot [T]^2 \). Divide both sides by \( [L][T]^2 \): \( z = \frac{[L/T]}{[L][T]^2} = \frac{1}{[T]^3} \)? Wait, no, the options have \( 1/[T]^3 \)? Wait the options: one is \( 1/[T]^3 \)? Wait the user's options: let's check again. Wait the user's options: "1/[T]^3", "1/[T]^2", "1/[T]". Wait maybe I made a mistake. Wait \( v \) is \( [L/T] \), \( x \) is \( [L] \), \( t^2 \) is \( [T]^2 \). So \( v = z x t^2 \) (dimensionally). So \( [L/T] = z [L] [T]^2 \). Then \( z = \frac{[L/T]}{[L][T]^2} = \frac{1}{[T]^3} \). Wait but the options include "1/[T]^3"? Wait the user's options: the third option is "1/[T]^2"? Wait no, the user's options:

• [T]
• 1/[T]^3
• 1/[T]^2
• 1/[T]

Wait maybe I miscalculated. Wait \( v \) is \( L/T \), \( x \) is \( L \), \( t^2 \) is \( T^2 \). So \( L/T = z \times L \times T^2 \). Divide both sides by \( L \times T^2 \): \( z = (L/T) / (L T^2) = 1 / T^3 \). So the dimension of \( z \) is \( 1/[T]^3 \).

Answer:

\( \frac{1}{[T]^3} \) (corresponding to the option "1/[T]^3")