Question

velocity is equal root pressure upon X ,then write the dimensions of x

Answer 1

Ans: M^1/2L^-3/2T^0

Explanation: V=√p/x

x=√p/v

= √ML^-1T^-2 / LT^-1

= M^1/2L^-3/2T^0

Answer 2

The correct answer is $M^{\frac{1}{2} }L^{\frac{-3}{2} }$.

Given: $v=\frac{\sqrt{P} }{x}$

To Find: The dimensions of x.

Solution:

$v=\frac{\sqrt{P} }{x}$

V is velocity here.

$v=\frac{m}{sec} =L^{1} T^{-1}$

P is pressure here.

$P =\frac{N}{m^{2} } = M^{1}L^{-1}T^{-2}$

$\sqrt{P} = \frac{\sqrt{N} }{m} =M^{\frac{1}{2} }L^{\frac{-1}{2} }T^{-1}$

$x=\frac{\sqrt{P} }{v}$

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

x = $M^{\frac{1}{2} }L^{\frac{-3}{2} }$

Hence, the dimensions of x are $M^{\frac{1}{2} }L^{\frac{-3}{2} }$.

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