Class 12 Model Question Solution | Physics | 2078/2079 Model Question Solution

Model Question Solution

Group 'A'

Which of the following is a correct formula for calculating radius of gyration of a rotating object?
Solution: Option (a) k\(^2=\frac{I}{m}\)
Explanation \[I=MK^2\] \[\therefore K^2=\frac{I}{M}\]

A horizontal stream of air is blown under one of the pan balance as shown in the figure. What will be the effect on the pan?

Solution: Option (b) goes down
Explanation Bernoulli's theorem for the stream line flow of an ideal liquid (incompressible and non viscous) gives, \[\begin{align} \frac{P}{\rho} + gh + \frac{v^2}{2}&=\text{constant}\\ \frac{P}{\rho} + \frac{v^2}{2}& =\text{constant} \hspace{0.1cm} \because \text{two pans are at same height}\\ \end{align}\] When v goes up, P goes down. For the above expression to be constant. So, pressure above the pan is high and that below is low. Due to this difference in pressure, the pan goes down.

What will be the height of a capillary on the surface of the Moon if it is 'h' on Earth?
Solution: Option (c) \(6h\)
Explanation The height of the capillary tube is , \[h=\frac{2Tcos\theta}{r \rho g}\] Here, T = surface tension of liquid, \(\theta\) = angle of contact, r = radius of hemisphere, \(\rho\) = density of liquid and g = acceleration due to gravity.
When we measure the height of the capillary tube in moon, the only factor that changes is \[g'=\frac{g}{6}\]. So, \[\therefore h'=6h\]

What is the coefficient of performance of an ideal refrigerator working between ice point and room temperature?
Solution: Option (d) 10
Explanation COP measures the coefficient of refrigerator. It is defined as the heat removed from the cold reservoir \(Q_{cold}\) (i.e., inside a refrigerator ) divided by work done W done to remove the heat (i.e., work done by compressor ).
\[\begin{align} \beta&=\frac{Q_2}{W}\\ &=\frac{Q_2}{Q_1 - Q_2}\\ \beta&=\frac{\frac{Q_2}{Q_1}}{\frac{Q_1}{Q_1}-\frac{Q_2}{Q_1}}\\ \end{align}\] For an ideal refrigerator, carnot cycle gives, \[\frac{Q_2}{Q_1}=\frac{T_2}{T_1}\] So, \[\begin{align} \beta&=\frac{\frac{T_2}{T_1}}{1-\frac{T_2}{T_1}}\\ \therefore \beta&=\frac{T_2}{T_1 - T_2}\\ \end{align}\] Given,
\(T_2\) = 0\(^\circ\) C = 273 K
\(T_1\) = 27\(^\circ\) C = 300 K
This gives, \[\beta \approx 10\] Note! COP is nearly equal to 4.5 in reality, for best refrigerator.

A thermodynamic system is taken from A to B via C and then returned to A via D as shown in the P-V diagram. The area of which segment of the graph represents the total work done by the system?

Solution: Option (c) ACBDA
Explanation The work done by the system is given by area of the enclosed graph of P-V diagram.

Which one of the following directly affects the quality of sound?
Solution: Option (d) wave form
Quality or timbre of sound is measure of complexity of sound which enables us to distinguish between two sounds of the same pitch and loudness produced by two different sources. It depends only on the waveform.
Note! The three characteristics of sound are pitch, loudness and quality.

A diffraction pattern is obtained using a beam of red light. What will be the effect on the diffraction pattern if the red light is replaced with white light?
Solution: Option (d) All bright fringes, except the central one, become colorful
Explanation See the diffraction demo in video solution of my channel.

In which one of the following diagrams the currents are related by the equation \(I_1-I_2=I_3-I_4\)?

Solution: Option (b)
Explanation At a junction, \[\Sigma I = 0\] i.e., algebraic sum of the current entering the junction is equal to the current leaving the junction.

A coil having N turns and cross-section A carries current I. Which physical quantity does the product \(NIA\) represent?
Solution: Option (c) magnetic moment of the coil

What happens to the neutral temperature if the cold junction of a thermocouple is decreased?
Solution: Option (c) remains the same

What is the point where the seismic waves start?
Solution: Option (b) hypocentre

Group 'B'

Define surface tension.
Solution:
Surface tension is a property of liquid displayed by its acting as if it were a stretched elastic membrane. The SI unit of surface tension is Newton per metre.

Establish a relation between surface tension and surface energy of a liquid.
Solution:

Extra energy that a surface layer has is called the surface energy. Consider a rectangular frame with a sliding wire on its arm dipped on a soap solution and taken out.

Place it in a horizontal position. Net force of pull F on the wire due to the two surfaces of soap film is, \[F=2Tl\] Suppose the wire is slowly pulled out by the external force through a distance x so that area of two frame increases by lx and thus new surface of 2lx is created. Work done by the external force in this displacement is, \[W=Fx=2Tlx=T(2lx)\] This work is stored as the potential energy of the new surface and increase in surface energy (S) is, \[\begin{align} S&=W=T(2lx)\\ \implies T&=\frac{S}{2lx}=\frac{S}{A}\\ \end{align}\] Thus, surface tension is equal to the increase in surface energy per unit area.

Two spherical rain drops of equal size are falling vertically through air with a certain terminal velocity. If these two drops were to coalesce to form a single drop and fall with a new terminal velocity, explain how the terminal velocity of the new drop compares to the original terminal velocity.
Solution:
Since the drops are moving through the air, the upthrust due to the air being very small can be neglected.
We have, \[\begin{align} \text{volume of a larger drop} & = 2 \times \text{volume of small drop}\\ \frac{4}{3} \pi R^3& = 2 \times \frac{4}{3} \pi r^3\\ \therefore R = 2^\frac{1}{3} r \hspace{0.1cm} ... (i)\\ \end{align}\] From figure, \[\begin{align} 6\pi \eta r v&=\frac{4}{3} \pi r^3 \rho g \hspace{0.1cm} ... (ii)\\ 6 \pi \eta R v'&= \frac{4}{3} \pi R^3 \rho g \hspace{0.1cm} ... (iii)\\ \end{align}\] Dividing (iii) by (ii), we get,
\[\begin{align} \frac{Rv'}{rv}&=\left(\frac{R}{r}\right)^3\\ \frac{v'}{v}&=\left(\frac{R}{r}\right)^2\\ \frac{v'}{v}&=2^\frac{2}{3}\\ \therefore v'&=2^\frac{2}{3} \times v\\ \end{align}\]

Angular speed of a rotating body is inversely proportional to its moment of inertia.

1. Define 'moment of inertia'
Solution:
Moment of inertia is the quantitative measure of the rotational inertia of a body i.e., it is the property of a body to resist changes in its rotational state of motion.


2. Explain why angular velocity of the Earth increases when it comes closer to the Sun in its oribit.
Solution:
Since \(\omega \propto \frac{1}{I}\); the system consisting of sun and earth can be considered with sun being at the axis of rotation. As earth comes closer to the sun, \(I=Mr^2\) decreases and thus the angular velocity of the earth around the sun increases.

3. If the Earth were to shrink suddenly, what would happen to the length of the day? Give reason.
Solution:
Let \(R_e\) be the radius of earth before it shrinks and \(r\) be its radius after shrinking. Then the respective moment of inertia at this two cases are,
\[\begin{align} I&=MR_e^2 \hspace{0.1cm} ... (i)\\ I'&=Mr^2 \hspace{0.1cm} ... (ii)\\ \end{align}\] Since \(r \lt R_e\); \(I' \lt I\) and thus \(\omega' \gt \omega\). As the angular velocity of the earth increases after shrinking, this decreases the length of the day. Remember! \(\omega \propto \frac{1}{T}\).

State Bernoulli's principle.
Solution:

It states that for the streamline flow of an ideal liquid, the total energy (i.e., the sum of the pressure energy, potential energy and kinetic energy) per unit mass remains constant at every cross-section throughout the flow. i.e., \[\frac{P}{\rho}+gh+\frac{v^2}{2}=\text{constant}\]

Derive Bernoulli's equation.
Solution:


Let us take the entire volume of the ideal fluid as in figure. From the work-energy theorem,\[W=(K.E.)_f-(K.E.)_i ... (i)\] The change in kinetic energy results from the change in speed between the ends of the tube and is, \[(K.E.)_f-(K.E.)_i=\frac{1}{2}m v_2^2-\frac{1}{2}m v_1^2=\frac{1}{2}\rho V(v_2^2-v_1^2) ... (ii)\] Where \(m=\rho V\) is the mass of fluid transferred (i.e. mass of the fluid that enters at the input end and leaves at the output end) during a small time interval \(\Delta t\).

During the vertical lift of the fluid of mass m , the work done by the gravitational force is , \[\begin{align*} W_g&=-mg(h_2-h_1)\\ &=-\rho g V (h_2-h_1) \end{align*} ... (iii)\] This work is negative because the upward displacement and the downward gravitational force have opposite directions.

Work must be done on the system (at the input end) to push the entering fluid into the tube and by the system (at the output end) to push forward the fluid that is located ahead of the emerging fluid.
In general, the work done by the force of magnitude F , acting on a fluid sample contained in a tube of area A to move the liquid through a distance \(x\) is, \[Fx=(PA)(x)=P(Ax)=PV\] So, the work done on the system is \(P_1V\) and the work done by the system is -\(P_2V\). Their sum is, \[W_p=-P_2 V + P_1 V=-(P_2-P_1)V ... (iv)\] Then from (i), (ii), (iii) and (iv), we get, \[\begin{align*} W_g+W_p&=\frac{1}{2}\rho V(v_2^2-v_1^2)\\ -\rho g V (h_2-h_1)-(P_2-P_1)V&=\frac{1}{2}\rho V(v_2^2-v_1^2)\\ P_1+\frac{1}{2} \rho v_1^2+\rho g h_1&=P_2+\frac{1}{2}\rho v_2^2+\rho g h_2\\ P+\frac{1}{2}\rho v^2+\rho g h&=\text{constant} ... (v) \end{align*}\] This proves the Bernoulli's theorem .

You can squirt water from a garden hose a considerably greater distance by partially covering the opening with your thumb. Explain how this works.
Solution:

From the principle of continuity, \[av=\text{constant}\] where a is tge cross-section of hose and v is the velocity of fluid.
When partially covered with the thumb, a decreases, so the speed of water increases and thus the water can squirt at a considerably greater distance.



Define 'harmonics' in music.
Solution:
Harmonics are the waves whose frequencies are integral multiple of frequency of fundamental wave. i.e., \[f_n=nf_1\] where, \(f_n\) is the n\(^{th}\) harmonic and \(f_1\) is the fundamental frequency of vibration or wave.

Calculate the frequency of a monotonous sound produced by a 30 cm long flute open at both ends and being played in the first harmonic.
Solution:
length l = 30 cm = 0.30 m
frequency f = ? (in first harmonic)
Now,
\[\begin{align} f&=\frac{v}{2l}\\ &= \frac{330}{2\times 0.30}\\ \therefore f&=550 \hspace{0.1cm} \text{Hz}\\ \end{align}\]

The flute mentioned in above question was being played by a passenger on a stationary bus. The bus then moves uniformly. Explain what change in the pitch of the sound, if any, a person sitting on a bench at the bus park will feel when the bus starts moving.
Solution:
The general formula of apparent shift in frequency is, \[f'=\frac{v \pm u_0}{v \pm u_s}\times f\] Since the observer is stationary, \(u_0=0\) and as the source is moving away from the observer, there is apparent decrease in frequency, Note that, if there is decrease in apparent frequency, use + sign in denominator and if there is increase in apparent frequency, use - sign in denominator.
So, \[\begin{align} f'&=\frac{v}{v+u_s} \times f\\ \therefore f' \lt f\\ \end{align}\] So, the person sitting on a bench at the bus park will feel the decrease in the pitch of the flute sound when the bus starts moving.

State the second law of thermodynamics.
Solution:
Second law of thermodynamics can be stated as:
Kelvin's Statement : It is impossible to get a continuous supply of work from a body by cooling it to a temperature lower than that of its surroundings.

Clausius Statement : It is impossible for a self acting machine unaided by any external agency to convey heat from one body to another at a higher temperature i.e., we cannot make heat flow from a colder body to a warmer body without doing external work.

A refrigerator transfers heat from a cold body to hot body. Does this not violate the second law of thermodynamics? Give reason.
Solution:
It does not violate the second law of thermodynamics. It is in accordance with the 'Clausius Statement' of second law of thermodynamics as the refrigerator transfers heat from a cold body to a hot body with the aid of an external agency (i.e., compressor).

In the given figure, a heat engine absorbs \(Q_1\) amount of heat from a source at temperature \(T_1\) and rejects \(Q_2\) amount of heat to a sink at temperature \(T_2\) doing external work W.


1. Obtain an expression for the efficiency of this heat engine.
Solution:
\[\begin{align} \eta& = \frac{\text{external work obtained}}{\text{heat energy absorbed from the source}}\\ &=\frac{W}{Q_1}\\ &=\frac{Q_1-Q_2}{Q_1}\\ &=1-\frac{Q_2}{Q_1}\\ \end{align}\] This gives the expression for the efficiency of this heat engine.


2. Under what condition does the efficiency of such engine become zero percentage, if at all?
Solution:
From the above expression, we can deduce that \(\eta\) will be zero when the heat from the source \(Q_1\) is totally rejected to the sink \(Q_2\).


A students wants to measure the magnetic flux density between the poles of two weak bar magnets mounted on a steel yoke as shown in the figure. The magnitude of the flux density is between 0.02 T and 0.04 T.

1. Define magnetic flux density.
Solution:
Magnetic flux density is defined as the number of magnetic lines of force passing through a unit area of a surface of a material. It is denoted by B and has unit as Tesla (T).


2. One way of measuring the magnetic flux density could be the use of a Hall probe. Suggest one reason why Hall probe is not a suitable instrument to measure the magnetic flux density for the arrangement shown in the above figure.
Solution:
The hall voltage is given by, \[V_H=\frac{BI}{net}\] where B is the magnetic flux density, I is the current flowing through the Hall probe, n is the charge carrier concentration, e is the charge of electron and t is the thickness of hall element in hall probe. Magnetic flux density of 0.02 - 0.04 T will produce a very small hall voltage which would be too small to meaure with a standard voltmeter. So, Hall probe is not suitable instrument to measure the magnetic flux density for the arrangement shown in figure (in question).


3. Another method of measuring the magnetic flux density for the arrangement shown in the above figure is to insert a current carrying wire between poles of the magnet. Explain how the magnetic flux density can be determined using this method. You are allowed to use any additional apparatus.
Solution:
The magnetic flux density of the magnet can be measured by inserting the wire between the poles of the magnet as shown in the figure (for figure, click on the video) with the arrangement explained below:

1. Held the wire on a wooden platform.
2. Place the arrangement on the top pan balance with an aid of wooden support.
When the current is switched on, the wire will experience a force on a downward direction (Fleming's left hand rule) so that the mass difference is seen in the pan balance. We can write, \[F=\Delta m \times g \hspace{0.1cm} ... (i)\] where \(g=9.8 \hspace{0.1cm} m/s^2\). Putting the values of mass difference (\(\Delta m\)), we can find F . Also, we know, the force experienced by the straight current carrying wire is \(F=BIl\). Here, B is the magnetic flux density, I is the current flowing through the wire and l is the length of the wire. After obtaining the value of F from (i), substituting this in \[B=\frac{F}{Il}\] , we can obtain B .
In this way, we can find the value of magnetic flux density ( B ).




Law of electromagnetic induction can be expressed mathematically as \(\epsilon = - N \frac{d\phi}{dt}\).
1. State what the symbols \(\epsilon\) and \(\frac{d\phi}{dt}\) represent in the equation.
Here, \(\epsilon\) is the e.m.f induced in the loop, N is the number of turns in the coil, and \(\frac{d\phi}{dt}\) is the change in magnetic flux through the loop.


2. Explain the significance of the negative sign.
Negative sign indicates that the e.m.f is induced such that it opposes the change in the magnetic flux.


3. Two identical copper balls are dropped from the same height as shown in the figure. Ball P passes through a region of uniform horizontal magnetic field of flux density B. Explain why ball P takes longer than ball Q to reach the ground.

In the video, I have discussed the solution to this by discussing the diamagnetic property of copper. Here, I will completely discuss this in terms of Lenz's law .
As ball P falls through the uniform magnetic field, the emf is induced in the ball and hence the current is induced in it. According to Lenz's law, 'the direction of an induced current is always such as to oppose the change in the magnetic field that produces it.'. This will produce the magnetic field in the ball P so as to oppose the uniform magnetic field. Due to this reason, the ball P will fall with the effective acceleration less than the acceleration due to gravity and hence is not in the free fall. However, ball Q remains unaffected by the magnetic field and continues to be in a free fall. Thus, ball P takes longer time than ball Q to reach the ground.

Ultraviolet radiation of frequency 1.5\(\times\)10\(^{15}\) Hz is incident on the surface of an aluminium plate whose work function is 6.6\(\times\)10\(^{-19}\) J.

1. Show that the maximum speed of the electrons emitted from the surface of the aluminium is 8.6\(\times\)10\(^5\) m/s.
Given,
frequency, f = 1.5\(\times\)10\(^{15}\) Hz
work function, \(\phi\) = 6.6\(\times\)10\(^{-19}\) J
maximum speed, \(v_{max}\) = ?
From Einstein's photoelectric equation, \[\begin{align*} E&=\phi + \frac{1}{2} m v_{max}^2\\ hf&= \phi + \frac{1}{2} m v_{max}^2\\ 6.62\times 10^{-34} \times 1.5 \times 10^{15}& = 6.6\times 10^{-19} + \frac{1}{2} \times 9.1 \times 10^{-31} \times v_{max}^2\\ \therefore v_{max}&= 8.6\times 10^5 \hspace{0.1cm} \text{m/s}\\ \end{align*}\]


2. State and explain what change, if any, occurs to the maximum speed of the emitted electrons when the intensity of the ultraviolet radiation is increased.
When the intensity of the UV radiation is increased, the number of photoelectrons emitting from the plate surface increases. This has no relation with the kinetic energy of the electrons. Hence, there is no change in the maximum speed of the emitted electrons when the intensity of the UV radiation is increased.


Next Question
1. State Bohr's postulates of atomic model.
I'll leave this to you.


2. The figure shows Lymen series of energy transmission in hydrogen atom. Calculate the frequency of a photon emitted by an electron jumping from the second excited state to the ground.

When electron jumps from second excited state (i.e., n=3) to the ground state (i.e., n=1), it emits a photon with frequency f and thus energy hf ,
\[\begin{align*} E_3-E_1&=hf\\ (-1.51-(-13.6)) \times 1.6\times 10^{-19} & = 6.62 \times 10^{-34} \times f\\ \therefore f&=2.92\times 10^{15} \hspace{0.1cm} \text{Hz}\\ \end{align*}\]


Next Question (Please follow the video for the complete discussion of the following question).
1. Sketch the symbol of a p-n junction diode and indicate the polarity of its ends.
2. Copy the outline of a diode bridge rectifier and complete it by adding diodes in the gaps.
3. Explain what will happen if one of the four diodes is damaged so that it stops conducting totally in any direction. Sketch a graph to show the p.d. across the load \(R_L\) would vary with time in this situation.


Earthquake sets rocks and buildings in motion. When a rock is subjected to compression, a restoring force develops inside it. This restoring force is given by an equation \(F=-Ax\) where x is displacement and A is a constant.

1. Prove that this force will make the rock vibrate with simple harmonic motion.
From Newton's second law of motion,
\[\begin{align*} ma&=-Ax\\ a&=-\frac{A}{m}x\\ \end{align*}\] Since, A and m are constants, \[a\propto x\] As acceleration is directly proportional to the displacement from the mean position, the rock will vibrate with the simple harmonic motion.


2. Show that the speed of an object undergoing simple harmonic motion is given by the expression \(v=\pm \omega \sqrt{(A^2-x^2)}\) where the symbols carry standard meaning.
The equation of a simple harmonic motion is,
\[x=A cos\omega t\] \[\begin{align*} v&=\frac{dx}{dt}\\ &=\frac{d(A cos\omega t)}{dt}\\ &=-A \omega sin\omega t\\ &=-A \omega \sqrt{1-cos^2 \omega t}\\ &=-A \omega \sqrt{1-\frac{x^2}{A^2}}\\ \therefore v&=\pm \omega \sqrt{A^2-x^2}\\ \end{align*}\]


3. Calculate the maximum speed of a building shaken by S-waves of 21 Hz and amplitude 0.05 m.
\[\begin{align*} v_{max}&=A\omega\\ &=0.05 \times 2\pi f\\ &=0.05 \times 2 \pi \times 21\\ \therefore v_{max}&=6.60\hspace{0.1cm} \text{m/s}\\ \end{align*}\]


4. Explain why tall buildings are more susceptible to damage by S-waves which generally have low frequency.
Tall buildings are characterized by their low frequencies. Resonance occurs when the frequency of excitation (or forced vibration) is close to ta system's natural frequency. Therefore, tall buildings are more susceptible to damage.
Note! Higher the length, lower is the natural frequency and higher is the chance to collapse with low frequency S-waves. Short buildings have shorter length and hence higher natural frequency, and thus lower risk to S-waves.


The figure below shows the variation of emf and current with typical LRC circuit.


1. Explain whether the phase constant is positive or negative.
In general, current is taken as the reference phasor. Since, emf is leading the current in this case (see figure), the phase constant is positive.


2. Sketch a phasor diagram for the given case.




3. Is the circuit more inductive or capacitive? Explain.
The only difference between this two conditions relies on the phase difference between the emf and the current. Since, emf is leading the current in this case, the circuit is inductive in nature. If it was the other way, the circuit would be capacitive instead.


4. To increase the rate at which energy is transferred to the load, should the inductance be increased or decreased? Justify your answer.
For LRC circuit, the power factor is, \[cos\phi = \frac{R}{\sqrt{R^2+\left(L\omega - \frac{1}{C\omega}\right)^2}}\] When the inductance L is increased, the power factor decreases and hence the energy transfer decreases to the load. When L is decreased, the power factor decreases and hence the energy transfer increases to the load.
Or Question A student sets up a circuit as shown in the figure given below to measure the emf of a test cell.

1. Explain why he is unable to find a balance point and state the change he must make in order to achieve the balance.
He is unable to find the balance point due to the connection error he has made while arranging the potentiometer circuit. The polarity of the 'driver cell' and the 'test cell' should be same in order to get the experimentation correct and find the balance point in the experiment. So, he must change the polarity of 'test cell' so as to match the polarity of 'driver cell' in order to achieve the balance.


2. State how he would recognize the balance point.
He would recognize the balance point with the help of a galvanometer. When the galvanometer shows no deflection while the jockey touches the particular point in the potentiometer wire, that would be recognized as the balance point.


3. He obtained the balance point for distance 37.5 cm using standard cell of emf 1.50 V. And for the test cell, the balance distance AB was 25.0 cm. Calculate the emf of the test cell.
Solution:
For the standard cell,
balancing length, \(l_1\) = 37.5 cm
emf, \(E_1\) = 1.50 V
For the test cell,
balancing length, \(l_2\) = 25.0 cm
emf, \(E_2\)=?

From the principle of potentiometer, \[E \propto l\] \[\begin{align*} \frac{E_1}{E_2}&=\frac{l_1}{l_2}\\ \frac{1.50}{E_2}&=\frac{37.5}{25.0}\\ \therefore E_2&=1 \hspace{0.1cm} \text{V}\\ \end{align*}\] So, the emf of the test cell is 1.00 V.


4. He could have used an ordinary voltmeter to measure the emf of the test cell directly. The student, however, argues that the above instrument is more precise than an ordinary voltmeter. Justify the logic.
Ordinary voltmeter, in practice, has the internal resistance due to which the potential drops across it and the reading is not accurate. However, the potentiometer circuit draws no current from the circuit due to which it gives the correct reading of emf of the test cell.


Next Question:
1. Explain what is meant by quantization of charge.
Quantization of charge explains the discretization of charge. It signifies that the charge of any object is equal to the integral multiple of the fundamental charge or the elementary charge. The value of elementary charge is equal to the magnitude of the charge of the electron or a proton. Thus, if a body has charge q , the quantization of charge can be expressed as, \[q=ne\] where, n is the integer, and e is the elementary charge.


2. In a Millikan's oil drop experiment, an oil drop of weight 1.5\(\times\)10\(^{-14}\) N is held stationary between plates 10 mm apart by applying a p.d. of 470 V between the plates.
1. State the condition necessary for the drop to remain stationary. Also, sketch the forces acting on the oil drop.
2. Calculate the charge on the oil drop.
3. Explain what would happen if the above oil drop is suddenly struck by a stray alpha particle.
1.Solution: Neglecting the viscous force and upthrust due to air, we can see(in figure) that the two forces: electrostatic force and the weight acting on the body, balances each other when the drop is stationary.


2.Solution: \[\begin{align*} F_e&=W\\ qE&=W\\ q&=\frac{W}{E}\\ &=\frac{W}{\frac{V}{d}}\\ &=\frac{Wd}{V}\\ &=\frac{1.5 \times 10^{-14} \times 10 \times 10^{-3}}{470}\\ &=3.19 \times 10^{-19} \hspace{0.1cm} \text{C}\\ \therefore q&=2e\\ \end{align*}\] So, the charge on oil drop is 2 times the elementary charge. It is negative in nature.
3.Solution: Stray alpha particle are the low-energetic, unwantedly wandering alpha partticles. Alpha particles carries the positive charge of 2e. When the stray alpha particle strikes the negatively charged oil drop (q=-2e), it combines to form the elemental helium i.e, He.


Or Question
1. Derive an expression \(N=N_0 e^{-\lambda t}\) for a radioactive process where the symbols carry their standard meanings.
Solution:
Let \(N_0\) be the initial number of atoms present in the radioactive sample at time, t=0 and \(N\) be the number of atoms left after time t.
From the decay law, \[\frac{dN}{dt} \propto N\] \[\frac{dN}{dt}=-\lambda N\] where \(\lambda\) is the decay constant. This can be expressed as, \[\frac{dN}{N}=-\lambda dt\] Now, taking integration on both sides, \[\begin{align*} \int_{N_0}^{N} \frac{dN}{N}&=-\lambda \int_{0}^{t} dt\\ [ln N]_{N_0}^{N}&=-\lambda[t]_{0}^{t}\\ lnN-lnN_0&=-\lambda t\\ ln{\frac{N}{N_0}}=-\lambda t\\ \therefore N&=N_0 e^{-\lambda t}\\ \end{align*}\]


2. A student measured the activity of sample of radioactive rock. Her results are presented in the graph.

1. Explain why the data are scattered.
Radioactive processes are the purely stochastic/statistical process. Statistical process are random in nature. When performing the large number of observations, the curve of radioactive process though being random follows a exponential decay. The data are scattered due to the randomness of radioactive process and not due to the measurement error which many of you may have thought of Guys!!Believe me, I have not seen any of the the pure statistical and random process as radioactive process in nature.

2. Determine the half-life of this sample.
Solution:
Half-life is the time taken by the radioactive sample to decay half of its original number of atoms.
At t = 40 s, N = 30 cps
At t = 90 s, N = 15 cps
Since, the number of radioactive sample reduced to half in 50 seconds, half-life of the given sample is 50 seconds.

3. How will the shape of this curve will change if she repeats the experiment with a sample with a larger decay constant. Give reason to your answer.
From the relation, \[T_{\frac{1}{2}}=\frac{0.693}{\lambda}\] it can be inferred that the sample having larger decay constant will have small half-life. It means that the sample will radiate faster than the earlier case.

Last Updated on July 8, 2022.