Question 1:
State three factors that affect the pressure in a liquid.
Answer:
Factors that affect the pressure in a liquid are
– depth (h),
– density of liquid (ρ),
– gravitational acceleration (g)
State three factors that affect the pressure in a liquid.
Answer:
Factors that affect the pressure in a liquid are
– depth (h),
– density of liquid (ρ),
– gravitational acceleration (g)
Question 2:
What is the water pressure at a depth of 24 m in a lake?
[Density of water, ρ = 1 000 kg m–3 and gravitational acceleration, g = 9.81 m s–2]
Answer:
$$ \begin{aligned} & P=? \\ & h=24 \mathrm{~m} \\ & \rho=1000 \mathrm{~kg} \mathrm{~m}^{-3} \\ & g=9.81 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned} $$
$$ \begin{aligned} &\text { Water pressure }\\ &\begin{aligned} & =h \rho g \\ & =(24) \times(1000) \times(9.81) \\ & =2.35 \times 10^5 \mathrm{~Pa} \end{aligned} \end{aligned} $$
What is the water pressure at a depth of 24 m in a lake?
[Density of water, ρ = 1 000 kg m–3 and gravitational acceleration, g = 9.81 m s–2]
Answer:
$$ \begin{aligned} & P=? \\ & h=24 \mathrm{~m} \\ & \rho=1000 \mathrm{~kg} \mathrm{~m}^{-3} \\ & g=9.81 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned} $$
$$ \begin{aligned} &\text { Water pressure }\\ &\begin{aligned} & =h \rho g \\ & =(24) \times(1000) \times(9.81) \\ & =2.35 \times 10^5 \mathrm{~Pa} \end{aligned} \end{aligned} $$
Question 3:
A diver dives to a depth of 35 m in the sea. What is the actual pressure acting on his body?
[Density of sea water, ρ = 1 060 kg m–3, gravitational acceleration, g = 9.81 m s–2 and atmospheric pressure = 100 kPa]
Answer:
$$ \begin{aligned} & h=35 \mathrm{~m} \\ & P=? \\ & \rho=1060 \mathrm{~kg} \mathrm{~m}^{-3} \\ & g=9.81 \mathrm{~m} \mathrm{~s}^{-2} \\ & P_{\mathrm{atm}}=100 \mathrm{kPa} \end{aligned} $$
$$ \begin{aligned} &\text { Water pressure, } P_{\text {water }}\\ &\begin{aligned} & =h \rho g \\ & =(35) \times(1060) \times(9.81) \\ & =363951 \mathrm{~Pa} \\ & \approx 364 \mathrm{kPa} \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Actual pressure }\\ &\begin{aligned} & =P_{\text {water }}+P_{\text {atm }} \\ & =364+100 \\ & =464 \mathrm{kPa} \end{aligned} \end{aligned} $$
A diver dives to a depth of 35 m in the sea. What is the actual pressure acting on his body?
[Density of sea water, ρ = 1 060 kg m–3, gravitational acceleration, g = 9.81 m s–2 and atmospheric pressure = 100 kPa]
Answer:
$$ \begin{aligned} & h=35 \mathrm{~m} \\ & P=? \\ & \rho=1060 \mathrm{~kg} \mathrm{~m}^{-3} \\ & g=9.81 \mathrm{~m} \mathrm{~s}^{-2} \\ & P_{\mathrm{atm}}=100 \mathrm{kPa} \end{aligned} $$
$$ \begin{aligned} &\text { Water pressure, } P_{\text {water }}\\ &\begin{aligned} & =h \rho g \\ & =(35) \times(1060) \times(9.81) \\ & =363951 \mathrm{~Pa} \\ & \approx 364 \mathrm{kPa} \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Actual pressure }\\ &\begin{aligned} & =P_{\text {water }}+P_{\text {atm }} \\ & =364+100 \\ & =464 \mathrm{kPa} \end{aligned} \end{aligned} $$