1.Units, Dimensions and Measurement
normal

પ્રેશર હેડ માટે પારીમાણીક સૂત્ર.

A

$\left[ M ^0 L ^0 T ^0\right]$

B

$\left[ ML ^{-1} T ^{-2}\right]$

C

$\left[M^0 L^1 T^{-2}\right]$

D

$\left[M^0 L^1 T^0\right]$

Solution

(d)

$[\text { Presure head }]=\frac{[\text { Pressure }]}{[\text { Density }] \times[ g ]}$

Pressure $=\frac{\text { Force }}{\text { Area }}$

But, F orce $=$ Mass $\times$ Accelaration $=[ M ]\left[ LT ^{-2}\right]=\left[ MLT ^{-2}\right]$

Then,

$[\text { Pressure }]=\frac{\left[ MLT ^{-2}\right]}{\left[ L ^2\right]}=\left[ ML ^{-1} T ^{-2}\right]$

$[$ Density $]=\frac{[ Mass ]}{[\text { V olume }]}=\frac{[ M ]}{\left[ L ^3\right]}= ML ^{-3}$

Now,

$[\text { Presure head }]=\frac{[\text { Pressure }]}{[\text { Density }] \times[ g ]}=\frac{\left[ ML ^{-1} T ^{-2}\right]}{\left[ ML ^{-3}\right]\left[ LT ^{-2}\right]}=\left[ M ^0 L ^1 T ^0\right]$

Standard 11
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.