If the pressure of a liquid column at its bottom is 1000 Pa and height of the liquid is 20 cm, the density of the liquid is (keeping in mind the gravitational pull of earth)

50 g cm^-3

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.5 g cm^-3

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100 g cm^-3

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0.2 g cm^-3

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Solution

The correct option is B
.5 g cm^-3

We know P = $ρ g h$

1000Pa = $ρ$ * 10ms^-2 * 20cm

$ρ$ = 500 $\frac{N s^{2}}{m^{4}}$

Using the formula F= ma, we can resolve the N into $K g m s^{- 2}$

We get $ρ$ = $0.5 g / c m^{3}$

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If the pressure of a liquid column at its bottom is 1000 Pa and height of the liquid is 20 cm, the density of the liquid is (keeping in mind the gravitational pull of earth)

The correct option is B .5 g cm-3 We know P = ρgh 1000Pa = ρ * 10ms-2 * 20cm ρ = 500 Ns2m4 Using the formula F= ma, we can resolve the N into Kgms−2 We get ρ = 0.5g/cm3

The correct option is B
.5 g cm^-3

We know P = $ρ g h$

1000Pa = $ρ$ * 10ms^-2 * 20cm

$ρ$ = 500 $\frac{N s^{2}}{m^{4}}$

Using the formula F= ma, we can resolve the N into $K g m s^{- 2}$

We get $ρ$ = $0.5 g / c m^{3}$