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PHYSICS [5 Mark Question Answer]

Pressure in Fluids · ICSE STD 9 (ICSE - English Medium). 4 questions.

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[5 Mark Question Answer]

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Question 14 Marks
State briefly, how and why the atmospheric pressure of a place varies with the altitude. Draw an approximate graph to illustrate this variation.
Answer
1. We know atmospheric pressure = height of air column $x$ density of air $x$ acceleration due to gravity; $P = h \rho g$ So, as we go up i.e. at higher altitudes, height of air column and hence atmospheric pressure decreases.
2. Also with the increase in altitude, density of air decreases and hence atmospheric pressure decreases.
If we take average density of air as $1.29 kgm ^{-3}$ and the density of mercury as 13 $600 kgm ^{-3}$
we can find the height
Column which will exert as much pressure as*' is exerted by 1 cm or ( 0.01 m ) column of mercury as
Height of air column $x$ density of air = height of mercury column $x$ density of mercury height of air column $\times 1.29 kg / m ^3$
$
=0.01 m \times 13600 kg / m^3
$
$
\therefore \quad \text { Height of air column }=\frac{136}{1.29} m= 1 0 5 m \text { (approx). }
$
Thus, 105 m of air column, on the average, will exert as much pressure as 1 cm column of mercury. Further, 1 cm of mercury column exerts pressure $=105 m$ of air column 76 cm of mercury column exerts pressure $=105 \times 76 m=7980 m= 8$ km (approx).
Thus, $8 ~ k m$ of air column will exert as much pressure as 76 cm of mercury column. However, it does not mean that atmosphere extends to only 8 km . As it is pointed out earlier, the density of atmosphere also changes with height. Thus, a fall of one cm in pressure does not mean that we have covered a vertical height of 105 m . On higher altitudes the vertical height of air is far in excess of 105 m , because of low density of air. A graph showing fall in pressure with height is shown in the figure.
Image
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Question 24 Marks
A beaker contains a liquid of density ‘ρ’ upto height ‘h’such that ‘PA’ is atmospheric pressure and ‘g’ is acceleration due to gravity. Answer the following questions :
(a) What is the pressure on the free surface of liquid?
(b) What is the pressure on the base of beaker?
(c) What is the lateral pressure at the base on the inner walls of beaker?

Answer
(a) Pressure on the free surface of liquid is equal to the atmospheric pressure $\left(P_{ z }\right)$.
(b) Consider a liquid contained in a beaker, such that ' $p$ ' is the density of liquid.
Consider a point B at the base of liquid and the liquid column of area of cross-section ' $a$ ' around it, such that ' $h$ 'is the height of the liquid column as shown in the figure.
Image

$\therefore \quad$ Volume of imaginary column of liquid
$
=\text { area of cross-section } \times \text { length }=a h
$
$\therefore \quad$ Mass of liquid column $=$ Volume $\times$ density
$
=V \times \rho=\text { a.h. } \rho
$
$\therefore \quad$ Weight of liquid column $=$ mass $\times g=m g$
$
=\text { a.h.p.g. }
$
$\therefore \quad$ Thrust exerted by liquid column on the base of the beaker
$
=\text { a.h.p.g. }
$
$\therefore \quad$ Pressure due to liquid column
$
\begin{array}{l}
P=\frac{\text { Force }}{\text { Area }}=\frac{F}{a}=\frac{a h \rho g}{a} \\
P=\text { h. } \rho . g .
\end{array}
$
So, pressure on the base of beaker $=h \rho g$
$\therefore \quad$ Total pressure at base of beaker $=$ Atmospheric pressure $+h \rho g$
$
=P_{a}+h \rho g
$
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Question 34 Marks
(a) The area of cross-sections of the pump plunger and press plunger of a hydraulic press are $0.02 m^2$ and $8 m^2$ respectively. If the hydraulic press overcomes a load of 800 kgf , calculate the force acting on pump plunger.
(b) If the mechanical advantage of the handle of pump plunger is 8 , calculate the force applied at the end of the handle of pump plunger.
Answer
(a) Load on the press plunger $= L =800 kgf$ Let the effort acting on the pump plunger $= E$
Area of cross-section of pump plunger $= A _1=0.02 m^2$
Area of cross-section of press plunger $= A _2=8 m^2$
Area of cross-section of pump plunger $=A_1=0.02 m^2$
Area of cross-section of press plunger $=A_2=8 m^2$
Now, $\frac{L}{E}=\frac{A_2}{A_1}$
$
\begin{array}{l}
\frac{800}{E}=\frac{8}{0.02}=\frac{800}{2}=400 \\
E=\frac{800}{400}=2 kgf
\end{array}
$
$\Rightarrow \quad$ Force acting on the pump plunger $=2 kgf$
(b) Mechanical advantage $=8$
But mechanical advantage $=\frac{ L }{ E }$
Here, $L =2 kgf$
Force applied at the of handle of pump plunger $= E =$ ?
$\Rightarrow$ M.A. $\frac{L}{E}$
$
\begin{array}{l}
8=\frac{2}{E} \\
E=\frac{2}{8}=\frac{1}{4}=0.25 kgf
\end{array}
$
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Question 44 Marks
Calculate the equivalent height of mercury, which will exert as much pressure as 960 m of sea water of density $1040 kgm ^{-3}$. Density of mercury is $13600 kgm ^{-3}$.
Answer
Height of mercury column $=h_{ Hg }=76 cm=0.76 m$
Vertical height of air column $=h_{\text {air }}=$ ?
Density of mercury $=\rho_{ Hg _{ g }}=13600 kgm ^{-3}$
Density of air $=\rho_{\text {air }}=1.29 kgm ^{-3}$
According to question:
Pressure due to air column $=$ Pressure due to Hg column
$
\begin{array}{l}
\rho_{air} \times h_{air} \times g=\rho_{Hg} \times h_{Hg} \times g \\
h_{air}=\frac{\rho_{Hg \times h_{H}}}{\rho_{\text {air }}} \\
h_{\text {air }}=\frac{13600 \times 0.76}{1.29}=8012.4 m
\end{array}
$
Density of mercury $=\rho_{ H _8}=13600 kgm ^{-3}$
Density of sea water $=\rho_{\text {water }}=1040 kgm ^{-3}$
Height of sea water column $=h_{\text {water }}=960 m$
Height of mercury column $=h_{ Hg }=$ ?
According to question,
Pressure due to mercury column = Pressure due to sea water column
$
\begin{array}{l}
\rho_{Hg} \times h_{Hg} \times g=\rho_{\text {water }} \times h_{\text {water }} \\
h_{Hg}=\frac{\rho_{\text {water }} \times h_{\text {water }}}{\rho_{Hg}}=\frac{1040 \times 960}{13600}=73.41 m
\end{array}
$
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