[3 Mark Question Answer]

Question 13 Marks

(a) Define SI unit of pressure.
(b) The atmospheric pressure at a place is 650 mm of Hg . Calculate this pressure in Pascals $( Pa )$.

Answer

(a) SI unit of pressure is pascal $( Pa )$ or $Nm ^{-2}$
One Pascal: When a force of one newton acts normally on an area of one square metre $\left(1 m^2\right)$ then pressure acting on the surface acting on the surface is called one Pascal.
(b) Height of mercury column $= h =650 mm$
$
h=65 cm=0.65 m
$
Density of mercury $=\rho=13600 kg / m ^3$
Acceleration due to gravity $= g =10 ms^{-2}$
Pressure $(P)=650 mm$ of $Hg = h \rho g$
$=0.65 \times 13600=10$
$=88400 Pa$

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Question 23 Marks

The pressure in water pipe on the ground floor of a building is 40000 pascals, whereas on the first floor it's 10000 pascals. Find the height of first floor. (Acceleration due to gravity $g =10 ms^{-2}$ )

Answer

Pressure on the ground floor of a building $= P _1=40000 Pa$
Pressure on the first floor of a building $= P _2=10000 Pa$
$\rho=1000 kg / m ^3=$ density of water
Acceleration due to gravity $= g =10 m / s ^2$
Difference in pressure $= P ,- P _2=40000-10000=30000 Pa$
Let $h =$ height of first floor
Pressure of water due to height $( h )= h \rho g$
$
30000=h \times 1000 \times 10
$
$
h=\frac{30000}{10000}=3 m
$

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Question 33 Marks

The base of cylindrical vessel measures $300 cm^2$. Water is poured into it upto a depth of 6 cm . Calculate the pressure of water on the base in vessel.

Answer

Area of base of cylinder $= A =300 cm^2$
$
\begin{array}{l}
A=300 \times 10 m^2 \\
A=3 \times 10^2 m^2
\end{array}
$
Height (or depth) of water column $= h =6 cm$
$
h=\frac{6}{100} m
$
Density of water $=\rho=1000 kg / m ^3$
Acceleration due to gravity $= g =10 m / s ^2$
Pressure at the base in vessel $= P = h \rho g$
$
\begin{array}{l}
\rho=\frac{6}{100} \times 1000 \times 10 \\
P=600 Pa
\end{array}
$

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Question 43 Marks

At a given place, a barometer records 70 cm of Hg . If the mercury in barometer is replaced by water, what would be resulting reading? (Density of $Hg =13600 kg / m ^2$; Density of water $=1000 kg / m ^3$ )

Answer

Height of mercury column $= h =70 cm=0.70 m$
Density of mercury $=\rho=13600 kg / m ^3$
Acceleration due to gravity $= g =10 m / s ^2$
Pressure = hpg
For water:
Height of water column = $h ^{\prime}=$ ?
Pressure due to water column $= P ^{\prime}$
Density of water $=\rho_w=1000 kg / m ^3$
$
P^{\prime}=h^{\prime} \rho_w g
$
But, $P ^{\prime}= P$
$h^{\prime} \rho_w g=h \rho g$
$h^{\prime} \rho_w=h \rho$
$
h^{\prime}=\frac{h \rho}{\rho_w}=\frac{0.7 \times 13600}{1000}=9.52 m
$

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Question 53 Marks

State three factors on which the pressure at a point in a liquid depends.

Answer

Factors on which the pressure at a point in a liquid depends are:
1. Pressure in a liquid is directly proportional to its height or depth.
2. Pressure in a liquid is directly proportional to its density.
3. Pressure in a liquid is directly proportional to the acceleration due to gravity.
4. Pressure in a liquid is independent of the area of crosssection.

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Question 63 Marks

The radii of the press plunger and pump plunger are in ratio of $50: 4$. If an effort of 20 kgf acts on the pump plunger, calculate the maximum effort which the press plunger can over come.

Answer

Effort acting on the pump plunger $- E =20 kgf$
Load acting on the press plunger $= L =$ ?
So let radius of pump plunger $=50 x=R$ and radius of press plunger $=A x=r$
Now, $\frac{ L }{ E }=\frac{\pi R ^2}{\pi r^2}$
$
\begin{array}{l}
\frac{L}{20}=\frac{(50 x)^2}{(4 x)^2}=\frac{2500}{16}=\frac{625}{4} \\
L=\frac{625}{4} \times 20 \\
L=625 \times 5=3125 kgf
\end{array}
$

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Question 73 Marks

The pressure of water on ground floor is 160000 Pa . Calculate the pressure at the fifth floor, at a height of 15 m .

Answer

Pressure of water at ground floor $= P ,=160000 Pa$.
Pressure of water at fifth floor $= P _2=$ ?
Height of fifth floor $= h =15 m$
Density of water $=\rho=1000 kgm ^{-3}$
Difference in pressure of water at ground and fifth floor
$
=P_1-P_2
$
Pressure of water due to height $(h)=h \rho g$
$
\begin{array}{l}
=P_1-P_2=h \rho g \\
160000-P_2=15 \times 1000 \times 10 \\
P_2=160000-150000 \\
P_2=10000 Pa
\end{array}
$

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Question 83 Marks

The pressure of water on the ground floor, in a water pipe is 150000 Pa, whereas pressure on the fourth floor is 30000 Pa . Calculate height of fourth floor. Take $g =10$ $ms ^{-2}$.

Answer

Pressure of water at fourth floor $= P _2=3000$ Pa Let h be the height of fourth floor Difference in pressure of water at ground floor and fourth floor
$
\begin{array}{l}
=P_1-P_2=150000-30000 \\
=120000 Pa
\end{array}
$
Pressure of water due to height $(h)=h \rho g$
$
\begin{array}{l}
\Rightarrow \begin{aligned}
\Rightarrow & P_1-P_2=h p \\
\Rightarrow & P_1-P_2=h \rho g \\
& 120000=h \times 1000 \times 10 \\
& \quad\left[\rho=1000 kgm^{-3}=\text { Density of water }\right] \\
& h=\frac{120000}{10000}=12 m
\end{aligned}
\end{array}
$

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Question 93 Marks

Atmospheric pressure at sea level is 76 cm of mercury. Calculate the vertical height of air column exerting the above pressure. Assume the density of air $1.29 kgm ^{-3}$ and that of mercury is $13600 kgm ^{-3}$. Why the height calculated by you is far less than actual height of atmosphere?

Answer

Height of mercury column $=h_{ Hg }=76 cm=0.76 m$
Vertical height of air column $=h_{\text {air }}=$ ?
Density of mercury $=\rho_{ H _{ g }}=13600 kgm ^{-3}$
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_{\text {air }} \times h_{\text {air }} \times g=\rho_{Hg_8} \times h_{Hg} \times g \\
h_{\text {air }}=\frac{\rho_{Hg \times h_{Hg}}}{\rho_{\text {air }}} \\
h_{\text {air }}=\frac{13600 \times 0.76}{1.29}=8012.4 m
\end{array}
$

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