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Chapter 1: Problem 152

Pressure \(P=\left(a t^{2} / b x\right)\) where \(x=\) distance, \(t=\) time find the dimensional formula for \(\mathrm{a} / \mathrm{b}\) (a) \(\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{-4}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{-2}\) (d) \(\mathrm{M}^{+1} \mathrm{~L}^{0} \mathrm{~T}^{-2}\)

Short Answer

Expert verified
The short answer is: The dimensional formula for \(a/b\) is \(M^1L^0T^{-2}\).

Step by step solution

01

Identify the dimensions of known variables

Pressure (P) has dimensions of force per unit area, which is [ML^(-1)T^(-2)]. Time (t) has dimensions of [T], and distance (x) has dimensions of [L].
02

Write the given formula in terms of dimensions

We can replace each variable in the given equation with its dimensional formula \[P = \frac{at^2}{bx}\] as \[[ML^(-1)T^(-2)] = \frac{a[T]^2}{b[L]}\]
03

Equate dimensions on both sides of the equation

Now, equate the dimensions on the left side of the equation with the dimensions on the right side. \[ML^(-1)T^(-2) = \frac{a}{b} \cdot T^{2} L^{-1}\]
04

Solve for the dimensions of a/b

To find the dimensions of a/b, we need to find the exponents of M, L, and T in terms of a/b. From the previous equation, we can see: \[\frac{a}{b} = ML^0 T^{-2}\] The dimensional formula for a/b is therefore \[M^1L^0T^{-2}\] The correct answer is (d) M^1 L^0 T^{-2}.

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