Solveeit Logo
Login

Question

Question: Given \(p = \dfrac{{a - {t^2}}}{{bx}}\) where p\( = \) pressure x\( = \) distance and t\( = \) times...

Given p=a−t2bxp = \dfrac{{a - {t^2}}}{{bx}} where p== pressure x== distance and t== times. Find the dimension of a/b.a/b.
A.  [M2LT−3] B. [MT−2] C.   [ML3T−1] D.   [LT−3]  A.\;\left[ {{M^2}L{T^{ - 3}}} \right] \\\ B.{\text{ [M}}{{\text{T}}^{ - 2}}] \\\ C.\;{\text{ [M}}{{\text{L}}^3}{T^{ - 1}}] \\\ D.\;{\text{ [L}}{{\text{T}}^{ - 3}}] \\\

Explanation

Solution

Here we will use the concept of the dimensional analysis. The dimensional analysis is the mathematical tool to calculate the physical parameters and its analysis involves the fundamental units of the dimensions M (mass), Length (L) and T (Time). It is used to determine the relationships between the numbers of variables. Also, use the basic rules of power and exponents to simplify.

Complete step by step answer:
Now, given that p is the pressure and it can be defined as the force per unit Area.
p=FAp = \dfrac{F}{A}
Write dimensions in the left hand side of the equation.
p=[M1L1T−2][L2]p = \dfrac{{[{M^1}{L^1}{T^{ - 2}}]}}{{[{L^2}]}}
When exponent with the same base and in division, then powers are subtracted.
p=[M1L−1T−2]p = [{M^1}{L^{ - 1}}{T^{ - 2}}]
X is the distance so its length.
Therefore dimension of x=[L]x = [L]
Dimension of t=[T]t = [T]
Given that: (a−t2)(a - {t^2}), but we can subtract two terms if they have the same dimensions. Since we know that the dimension of time “t”. Here the dimensions of “t” and “a” are same.
[a]=[t2]=[T2] ......(1)[a] = [{t^2}] = [{T^2}]{\text{ }}......{\text{(1)}}
⇒[p]=[a−t2][b][x]\Rightarrow[p] = \dfrac{{[a - {t^2}]}}{{[b][x]}}
Make the unknown [b] the subject –
[b]=[a−t2][p][x][b] = \dfrac{{[a - {t^2}]}}{{[p][x]}}
Place the dimensions of all the terms in the left hand side of the equation –
[b]=[T2][M1L−1T−2][L][b] = \dfrac{{[{T^2}]}}{{[{M^1}{L^{ - 1}}{T^{ - 2}}][L]}}
Simplify using the laws of powers and exponents
[b]=[M−1L0T4] .....(2)[b] = [{M^{ - 1}}{L^0}{T^4}]{\text{ }}.....{\text{(2)}}
By using the equations (1)  and (2)(1)\;{\text{and (2)}}
The required relation –
ab=[T2][M−1L0T4]\dfrac{a}{b} = \dfrac{{[{T^2}]}}{{[{M^{ - 1}}{L^0}{T^4}]}}
Simplify –
ab=[M1L0T−2]\dfrac{a}{b} = [{M^1}{L^0}{T^{ - 2}}]
Hence, from the given multiple choices- the option B is the correct answer.

Note: The mathematical expression which shows the powers to which the fundamental units are to be raised to get one unit of the derived quantity is called the dimensional formula of that quantity. If “x” is the unit of the derived quantity represented by x=MaLbTcx = {M^a}{L^b}{T^c} where MaLbTc{M^a}{L^b}{T^c}is called the dimensional formula and the exponents a, b and c are called the dimensions.