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1.Units, Dimensions and Measurement
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પ્રેશર હેડ માટે પારીમાણીક સૂત્ર.
A$[ M ^0 L ^0 T ^0]$
B$[ML^{-1}T ^{-2}]$
C$[ M ^0 L ^1 T ^{-2}]$
D$[ M ^0 L ^1 T ^0]$
Solution
$[\text { Presure head }]=\frac{[\text { Pressure }]}{[\text { Density }] \times[ g ]}$
Pressure $=\frac{\text { F orce }}{\text { Area }}$
But,Force $=$ 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,
$[\text { Presure head }]=\frac{[\text { Pressure }]}{[\text { Density }] \times[ g ]}=\frac{\left[ ML ^{-1} T ^{-2}\right]}{\left[ ML ^{-3}\right]\left[ LT ^{-2}\right]}=\left[ M ^0 L ^1 T ^0\right]$
Pressure $=\frac{\text { F orce }}{\text { Area }}$
But,Force $=$ 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
