Question

Find dimension of a×b if p=b-x^2/at where x =distance t=time p=pressure​

Answer 1

Answer:

Hii Here is your ans

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Answer 2

$a\times b=M^{-1} L^{5}T$

Explanation:

$p=\frac{b-x^2}{at}$ where P is the pressure, t is time and x is distance.

We can write given equation as,

$p=\frac{b}{at}-\frac{x^2}{at}$

$\frac{b}{at}$  and  $\frac{x^2}{at}$  will have the same dimension as Pressure p.

Equate both the expression

$\frac{b}{at}=\frac{x^2}{at}$

$b=x^2$

write dimension of the given equation in MLT form,

$b=L^2$

Hence dimension of $b$ is $L^2$.

$p=\frac{b}{at}$

$M L^{-1}T^{-2}=\frac{L^2}{aT}$

$M L^{-3}T^{-1}=\frac{1}{a}$

$a=M^{-1} L^{3}T$

Required Dimension is,

$a\times b=M^{-1} L^{3}TL^2\\a\times b=M^{-1} L^{5}T$

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