Question

Sometimes it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity. In one such system, dimensions of different quantities are given in terms of a quantity X as follows: [position] = [$X^{\alpha }$] ; [speed] = [$X^{\beta }$] ; [acceleration] =[$X^p$] ; [linear momentum] = [$X^q$] ; [force] = [$X^r$]. Then ,

• A. $\alpha +p=2\beta$
• B. $p+q-r=\beta$
• C. $p-q+r=\alpha$
• D. $p+q+r=\beta$

Answer 1

Answer: (A), (B)

$p+q-r=\beta$

Given L = x${}^{\alpha }$ ……(1)
$L{T}^{-1}=X^{\beta }$……(2)
$L{T}^{-2}=X^p$ ……(3)
$ML{T}^{-1}=X^q$ ……(4)
$ML{T}^{-2}=X^r$ ……(5)

Dividing equation (1) and (2) :

$\frac{\left(1\right)}{\left(2\right)}\Rightarrow T=X^{\alpha -\beta }$

From equation (3)

$\frac{X^{\alpha }}{X^{2(\alpha -\beta )}}=X^p$

$\alpha +p=2\beta$
Oprion a is correct.

From equation (5) :

$X^q=x^r{X}^{\alpha -\beta }$

$\Rightarrow \alpha +r-q=\beta$……(6)

Replacing value of '$\alpha$' in equation (6) from option (a) :
$2\beta -p+r-q=\beta$

$\Rightarrow p+q-r=\beta$
Option b is correct.

Replacing value of '$\beta$' in equation (6) from option (a) :

$2\alpha +2r-2q=\alpha +p$

$\alpha =p+2q-2r$

option c,d are incorrect