3. The base of a rectangular vessel measures
10 cm x 18 cm. Water is poured into it up to a
depth of 4 cm. What is the pressure on the base?
What is the thrust on the base? Assume density of
water = 1000 kg m and g = 10 ms?.
[Ans. 400 Nm", 7.2 N]
. An open U-tube pressure gauge containing water
Answers
Answer:
pressure
=40kg/m^2
Thrust
=0.72kg
Explanation:
pressure=density*depth
=(1000kg/m cube)*(4/100)m
=40 kg/m square
Thrust or force=pressure*area
=(40kg/m square)*(10/100)*18/100)
=(400*18)/(10000)
=72/100=0.72 kg
Given :
The base of a rectangular vessel measures = 10 cm x 18 cm.
Height of water in the vessel = 4 cm
To Find :
(i) What is the pressure on the base?
(ii) What is the thrust on the base?
Solution :
(i) We know, Pressure = hρg
Where, h = Height of the water level; ρ = density of water; g = acceleration due to gravity
Pressure on the base = hρg
= 4×10⁻² × 1000 × 10
= 400 N/m²
Therefore, the Pressure on the base is 400 Nm⁻² or 400 N/m²
(ii) Thrust = Pressure × Area
= 400 × 10 × 10⁻² × 18 × 10⁻² (As 1 cm = 10⁻² m)
= 7.2 N
Therefore, Thrust on the base is 7.2 N