[2 Mark Question Answer]

Question 12 Marks

Explain, why a gas bubble released at the bottom of a lake grows in size as it rises to the surface of the lake.

Answer

Bubble released at the bottom of a lake grows in size as it rises to the surface of the lake because the pressure exerted on it by water of the lake DECREASES hence by BOYLE'S LAW PV = constant the VOLUME of bubble INCREASES and the bubble grows in size.

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Question 22 Marks

State two advantages of aneroid barometer.

Answer

Advantages of aneroid barometer:
1. It is compact, portable and hence can be carried anywhere.
2. It does not contain any liquid and there is no chance of spilling over of liquid as in mercury barometer.

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Question 32 Marks

Calculate the hydrostatic pressure exerted by water at the bottom of a beaker. Take the depth of water as 40 cm , the density of water $1000 kgm ^{-3}$ and $g =9.8 ms^{-2}$.

Answer

$
\begin{array}{l}
\text { Pressure at the bottom of beaker }=P=\text { ? } \\
\text { Height (or depth) of water in beaker }=h=40 cm \\
h=0.4 m \\
\text { Density of water }=\rho \\
=1000 kgm^{-3} \text { Acceleration due to gravity }=g=9.8 ms^{-2} \\
P=h \rho g \\
P=0.4 \times 1000 \times 9.8 \\
=3920 Pa
\end{array}
$

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Question 42 Marks

The normal pressure of air is 76 cm of mercury. Calculate the pressure in SI units. [Density of mercury $=13600 kg / m ^3$ and $g =10 m / s ^2$ ]

Answer

Height of mercury column $= h =76 cm$
$
h=0.76 m
$
Density of mercury $=\rho=13600 kg / m ^3$
Acceleration due to gravity $= g =10 m / s ^2$
Pressure $= P =$ ?
$
\begin{array}{l}
P=h \rho g \\
P=0.76 \times 13600 \times 10=103360 N / m^2
\end{array}
$

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Question 52 Marks

Pressure at bottom of sea at some particular place is 8968960 Pa . If density of sea water is $1040 kgm ^3$ calculate the depth of sea. Take $g =9.8 ms^{-2}$. Neglect the pressure of the atmosphere.

Answer

Pressure at the bottom of the sea $= P =8968960 Pa$
$
\begin{array}{l}
P=h \rho g \\
h=P / \rho g \\
h=\frac{8968960}{1040 \times 9.8} \\
h=880 m
\end{array}
$

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Question 62 Marks

What vertical height of water will exert pressure of 333200 Pa ? Density of water is 1000 $kgnr ^3$ and $g =9.8 ms^{-2}$.

Answer

Vertical height of water $= h =$ ?
Pressure due to water column $= P =333200 Pa$ Acceleration due to gravity $= g =9.8 ms^{-2}$
$
\begin{array}{l}
P=h \rho g \\
h=P / \rho g \\
h=\frac{333200}{1000 \times 9.8} \\
h=34 m
\end{array}
$

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Question 72 Marks

66640 Pa pressure is exerted by 0.50 m vertical column of a liquid. If $g =9.8 Nkg ^{-1}$, calculate density of the liquid.

Answer

Pressure + P = 66640 Pa .
Vertical length of liquid column $= h =0.50 m$
Acceleration due to gravity $- g =9.8 ms^{-2}$
$
\begin{array}{l}
\rho=\frac{66640}{0.50 \times 9.8} \\
\rho=13600 kg m^{-3}
\end{array}
$

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Question 82 Marks

What is the pressure exerted by 75 cm vertical column of mercury of density 13600 $kgm ^{-3}$ in SI units.[Take $g =9.8 ms^{-2}$ ].

Answer

Vertical length of mercury column $= h =75 cm=0.75 m$
Density of mercury $=p=13600 kg m ^{-3}$
Acceleration due to gravity $= g =9.8 ms^{-2}$
Pressure = P = ?
$
\begin{array}{l}
p=h p g \\
P=0.75 \times 13600 \times 9.8 \\
P=99960 Pa
\end{array}
$

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Question 92 Marks

Calculate pressure exerted by 0.8 m vertical length of alcohol of density $0.80 gcnr ^5$ in SI units. [Take $g =10 ms^{-2}$ ].

Answer

Vertical length of alcohol column $= h =0.8 m$
Density of alcohol $=\rho=0.80 g cm ^{-3}$
$
\begin{array}{l}
\rho=\frac{0.80 \times 10^6}{10^3} kgm^{-3} \\
\rho=0.80 \times 1000 kgm^{-3} . \\
\rho=800 kg m^{-3} \\
\text { Pressure }=P=? \\
p=hpg \\
P=0.8 \times 800 \times 10 \\
=6400 Pa
\end{array}
$

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