3 Marks Question

Question 13 Marks

Mercury has an angle of contact equal to $140^{\circ}$ with soda-lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? Surface tension of mercury at the temperature of the experiment is 0.465 N $m ^{-1}$. Density of mercury $=13.6 \times 10^3 kgm ^{-3}$. .

Answer

Angle of contact between mercury and soda lime glass, $\theta=140^{\circ}$
Radius of the narrow tube, $r =1 mm=1 \times 10^{-3} m$
Surface tension of mercury at the given temperature, $s =0.465 N m ^{-1}$
Density of mercury, $\rho=13.6 \times 10^3 \frac{kg}{ m ^2}$
Dip in the height of mercury $= h$
Acceleration due to gravity, $g =9.8 \frac{m}{ s ^2}$
Surface tension is related with the angle of contact and the dip in the height as:
$
\begin{aligned}
& s=\frac{h \rho g r}{2 \cos \theta} \\
& \therefore h=\frac{2 s \cos \theta}{\operatorname{rog}} \\
& =\frac{2 \times 0.465 \times \cos 140}{1 \times 10^{-3} \times 13.6 \times 10^3 \times 9.8} \\
& =-0.00534 m \\
& =-5.31 mm
\end{aligned}
$
Here, the negative sign shows the decreasing level of mercury. Hence, the mercury level dips by 5.34 mm .

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Question 23 Marks

How is centripetal force provided in case of the following?
i. Motion of planet around the sun,
ii. Motion of moon around the earth.
iii. Motion of an electron around the nucleus in an atom.

Answer

i. The earth revolves round the sun. The earth is also acted upon by the centripetal force which is provided by the gravitational force of attraction between the sun and the earth.
ii. The motion of moon around the earth is also in circular path. The necessary centripetal force is provided by the gravitational attraction of the earth on the moon.
iii. In an atom, electrons revolve around the nucleus in various circular orbits. The necessary centripetal force for circular motion, is exerted by the electrostatic force of attraction between the positively-charged nucleus and the negatively charged electrons.

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Question 33 Marks

Show that Newton's third law of motion is contained in the second law.

Answer

Let $F _{ ba }$ be the force (action) exerted by A on B and $\frac{d p _B}{d t}$ be the resulting change of the momentum of B.
Let $F _{ AB }$ be the force (reaction) exerted by B on A and $\frac{d p _A}{d t}$ be the resulting change of momentum of A.
According to Newton's second law, $F =\frac{d p }{d t}$
Then, $F _{B A}=\frac{d p _B}{d t}$ and $F _{A B}=\frac{d p _A}{d t}$
$
\therefore \quad F _{B A}+ F _{A B}=\frac{d p _B}{d t}+\frac{d p _A}{d t}=\frac{d}{d t}\left( p _B+ p _A\right)\ldots(i)
$
In the absence of any external force, the rate of change of momentum of the whole system zero.
i.e. $\frac{d}{d t}\left( p _B+ p _A\right)=0$
So, $F _{ BA }+ F _{ AB }=0$ or $F _{ BA }=- F _{ AB }$
or Action $=-$ reaction
and it is a Newton's third law of motion.
Hence, proved.

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Question 43 Marks

A cylindrical vessel filled with water upto a height of 2 m stands on a horizontal plane. The side wall of the vessel has a plugged circular hole touching the bottom. Find the minimum diameter of the hole so that the vessel begin to move on the floor, if the plug is removed. The coefficient of friction between the bottom of the vessel and the plane is 0.4 and total mass of water plus vessel is 100 kg.

Answer

Velocity of efflux through the hole, $v =\sqrt{2 g h}$
$\because$ Distance moved by water in one second $v =\sqrt{2 g h}$
$\because$ Rate of the momentum $=(\rho A \sqrt{2 g h})(\sqrt{2 g h})=2$ ghA $\rho$
According to Newton's second law of motion,
Force due to the velocity of efflux $=2 gh A \rho$
Now, according to Newton's third law of motion,
Force on the vessel = Rate of the momentum Force on the vessel $=2 ghA \rho$
The vessel will move, if force on the vessel = force of friction
or $2 gh A \rho=\mu Mg$
or $A =\frac{\mu M}{2 h \rho}=\frac{0.4 \times 100}{2 \times 2 \times 1000}=\frac{1}{100}$
Since, the hole is circular,
$
\begin{aligned}
& A=\pi r^2=\frac{\pi D^2}{4} \\
& D=\sqrt{\frac{4 A}{\pi}}=\sqrt{\frac{4 \times 1}{100 \times 314}}=0.113 m
\end{aligned}
$
So, the diameter of a hole $D =0.113 m$

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Question 53 Marks

What is venturi-meter? On which principle does it work? How is the principle of venturi-meter applied in automobiles?

Answer

Venturi-meter is a device used to measure the flow speed of a liquid. It is basically based on Bernoulli's principle and works on the principle that when a liquid flows in the tube from wide neck to a narrow constriction, the speed of flow increases and the pressure falls.
Bernoulli's principle states that with the increase in the velocity of the fluid its pressure decreases (or) there is a decrease in the fluid pressure energy. This decrease in the fluid pressure in the areas where the flow velocity is increased is called the Bernoulli effect.
It is utilised in the carburettor of automobiles. The carburettor has a venturi channel (fine nozzle) through which air flows with a large speed. The pressure is then lowered at the narrow neck as a result, the valve of petrol chambers opens and the petrol is sucked up in the chamber to provide the correct mixture of air and petrol necessary for combustion.

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Question 63 Marks

A certain substance has a mass of $50 g /$ mole. When 300 J of heat is added to 25 g of sample of this material, its temperature rises from $25^{\circ} C$ to $45^{\circ} C$. Calculate
i. the thermal capacity,
ii. specific heat capacity, and
iii. molar heat capacity of the sample.

Answer

i. Total heat supplied to sample $\Delta Q =300 J$ and rise in temperature $\Delta T = T _2- T _1=45-25=20^{\circ} C$
$\therefore$ Thermal capacity of substance is given as $=\frac{\Delta Q}{\Delta T}=\frac{300}{20}=15 J^0 C ^{-1}$
ii. mass, $m =25 g=0.025 kg$
$\therefore$ Specific heat capacity, $c =\frac{1}{m} \cdot \frac{\Delta Q}{\Delta T}=\frac{1}{0.025} \times 15=600 Jkg ^{-1}{ }^{ o } C ^{-1}$
iii. Number of moles in 25 g sample of $50 g / mol$ is
$\mu=\frac{25}{50}=0.5$ moles
$\therefore$ Molar heat capacity $C =\frac{1}{\mu} \frac{\Delta Q}{\Delta T}=\frac{1}{0.5} \times 15=30 Jmol ^{-10} C ^{-1}$.

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Question 73 Marks

Draw the following graphs for an object projected upward with a velocity $v _0$, which comes back to the same point after some time:
i. Acceleration versus time graph,
ii. Speed versus time graph,
iii. Velocity versus time graph.

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Question 83 Marks

If, at $50^{\circ} C$ and 75 cm of mercury pressure, a definite mass of a gas is compressed
i. slowly
ii. suddenly, then what will be the final pressure and temperature of the gas in each case if the final volume is one-fourth of the initial volume? $(\gamma=1.5)$

Answer

Here $V_2=\frac{1}{4} V_1, P_1=75 cm$ of Hg
$
T_1=50+273=323 K
$
i. When the gas is compressed slowly, the process is isothermal.
$
\therefore P_1 V_1=P_2 V_2 \text { or } 75 \times V_1=P_2 \times \frac{1}{4} V_1
$
or $P _2=75 \times 4=300 cm$ of Hg
As the process is isothermal, so $T _2=50^{\circ} C$
ii. When the gas is compressed suddenly, the process is adiabatic.
$
\begin{aligned}
& \therefore P_1 V_1^\gamma=P_2 V_2^\gamma \\
& \text { or } P_2=P_1\left(\frac{V_1}{V_2}\right)^\gamma=75\left(\frac{V_1}{\frac{V_1}{4}}\right)^{1.5}=75 \times 4^{1.5} \\
& =75 \times 4 \times 4^{\frac{1}{2}}=75 \times 4 \times 2=600 cm \text { of } Hg \\
& \text { Also, } T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1} \\
& \text { or } T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} \\
& =323 \times(4)^{0.5}=323 \times 2 \\
& =646 K=373^{\circ} C
\end{aligned}
$

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