Chapter : 2. Force & Laws Of Motion
Examples on Laws of pressure
Example : 12
What will be the pressure in N/m2 at a depth of 1.5 m in brine of density 120 kg/cm3 ?
Solution. P = hdg
= 15 × 120 × 10 = 1800 N/m2
Example : 13
Calculate the density of a liquid if the pressure at a point 30 m below its surface is 32 × 104 N/m2.
Solution. P = hdg
= 1066.6 kg/m3
Example : 14
A force of 150 N is applied on an area of 1.5 m2. Calculate the pressure exerted.
Solution. Force, F = 150 N; area, A = 1.5 m2
Now, Pressure = Force / Area
Pressure,
Example : 15
A force of 500 dynes is applied on an area of 20 cm2. Calculate the pressure exerted.
Solution. Force, F = 500 dynes = 500 × 10–5 newton
Area, A = 20 cm2 = 20 × 10–4 m2
Pressure,
What will be the pressure in N/m2 at a depth of 1.5 m in brine of density 120 kg/cm3 ?
Solution. P = hdg
= 15 × 120 × 10 = 1800 N/m2
Example : 13
Calculate the density of a liquid if the pressure at a point 30 m below its surface is 32 × 104 N/m2.
Solution. P = hdg
= 1066.6 kg/m3Example : 14
A force of 150 N is applied on an area of 1.5 m2. Calculate the pressure exerted.
Solution. Force, F = 150 N; area, A = 1.5 m2
Now, Pressure = Force / Area
Pressure,
Example : 15
A force of 500 dynes is applied on an area of 20 cm2. Calculate the pressure exerted.
Solution. Force, F = 500 dynes = 500 × 10–5 newton
Area, A = 20 cm2 = 20 × 10–4 m2
Pressure,
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