JEE Physics Dual Nature Of Radiation And Matter MCQs Set 24

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Multiple Choice Questions

Question 1: When a big drop of water is formed from \( n \) small drops of water, the energy loss is \( 3E \), where \( E \) is the energy of the bigger drop. If \( R \) is the radius of the bigger drop and \( r \) is the radius of the smaller drops, then the number of smaller drops \( n \) is
(a) \( \dfrac{4R}{r^2} \)
(b) \( \dfrac{4R}{r} \)
(c) \( \dfrac{2R^2}{r} \)
(d) \( \dfrac{4R^2}{r^2} \)
Show Answer & Explanation

Answer: (d) \( \dfrac{4R^2}{r^2} \)

Explanation:
1. Surface energy of the big drop \( E=4\pi R^2T \); energy lost \( =4\pi T(nr^2-R^2)=3E \).
2. \( nr^2-R^2=3R^2\Rightarrow n=\dfrac{4R^2}{r^2} \).

Teacher's Note:
1. Energy is released when surface area decreases.

Question 2: A tiny spherical oil drop carrying a net charge \( q \) is balanced in still air with a vertical uniform electric field of strength \( \dfrac{81\pi}7\times10^5 \) V m\(^{-1}\). When the field is switched off, the drop is observed to fall with terminal velocity \( 2\times10^{-3} \) m s\(^{-1}\). Given \( g=9.8 \) m s\(^{-2}\), viscosity of the air \( =1.8\times10^{-5} \) N s m\(^{-2}\) and the density of oil \( =900 \) kg m\(^{-3}\), the magnitude of \( q \) is
(a) \( 1.6\times10^{-19} \) C
(b) \( 3.2\times10^{-19} \) C
(c) \( 4.8\times10^{-19} \) C
(d) \( 8.0\times10^{-19} \) C
Show Answer & Explanation

Answer: (d) \( 8.0\times10^{-19} \) C

Explanation:
1. Terminal velocity: \( r=\sqrt{\dfrac{9\eta v}{2\rho g}}=4.29\times10^{-6} \) m, so \( m=\tfrac43\pi r^3\rho=2.97\times10^{-13} \) kg.
2. Balance: \( qE=mg\Rightarrow q=\dfrac{2.91\times10^{-12}}{3.64\times10^6}=8.0\times10^{-19} \) C.

Teacher's Note:
1. Use Stokes' law for the falling drop and \( qE=mg \) for balance.

Question 3:
Assertion (A): If the accelerating potential of an electron is doubled, then its velocity becomes 1.4 times.
Reason (R): It will move on a circular path with the same velocity.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (c) A is true, but R is false.

Explanation:
1. \( \tfrac12mv^2=eV\Rightarrow v\propto\sqrt V \), so doubling \( V \) multiplies \( v \) by \( \sqrt2\approx1.4 \): A is true.
2. The reason is not correct: the speed changed and there is no force providing circular motion in this situation, so R is false.

Teacher's Note:
1. The kinetic energy equals \( eV \).
2. A magnetic force alone would keep the speed constant, but the speed here has changed.

Question 4: Cathode rays enter a magnetic field making an oblique angle with the lines of magnetic induction. The nature of the path followed will be
(a) parabola
(b) helix
(c) circle
(d) straight line
Show Answer & Explanation

Answer: (b) helix

Explanation:
1. The velocity component perpendicular to \( B \) gives circular motion while the parallel component is unchanged, so the path is a helix.

Teacher's Note:
1. Perpendicular entry gives a circle; parallel entry gives a straight line.

Question 5: The linear momentum of an electron, initially at rest, accelerated through a potential difference of 100 V is
(a) \( 9.1\times10^{-24} \) kg m/s
(b) \( 6.5\times10^{-24} \) kg m/s
(c) \( 5.4\times10^{-24} \) kg m/s
(d) \( 1.6\times10^{-24} \) kg m/s
Show Answer & Explanation

Answer: (c) \( 5.4\times10^{-24} \) kg m/s

Explanation:
1. \( p=\sqrt{2meV}=\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times100}=5.4\times10^{-24} \) kg m/s.

Teacher's Note:
1. Use \( \tfrac{p^2}{2m}=eV \).

Question 6: When a cathode ray tube is operated at 2912 V, the velocity of electrons is \( 3.2\times10^7 \) m/s. The velocity of the cathode rays if the tube is operated at 5824 V is
(a) \( 2.4\times10^7 \) m/s
(b) \( 5.2\times10^7 \) m/s
(c) \( 4.525\times10^7 \) m/s
(d) \( 2.4\times10^6 \) m/s
Show Answer & Explanation

Answer: (c) \( 4.525\times10^7 \) m/s

Explanation:
1. \( v\propto\sqrt V \): \( v=3.2\times10^7\times\sqrt2=4.525\times10^7 \) m/s.

Teacher's Note:
1. Doubling the voltage raises the speed by \( \sqrt2 \).

Question 7: An \( \alpha \)-particle of mass \( 6.4\times10^{-27} \) kg and charge \( 3.2\times10^{-19} \) C is situated in a uniform electric field of \( 1.6\times10^5 \) V m\(^{-1}\). The velocity of the particle at the end of \( 2\times10^{-2} \) m path when it starts from rest is
(a) \( 2\sqrt3\times10^5 \) m s\(^{-1}\)
(b) \( 8\times10^5 \) m s\(^{-1}\)
(c) \( 16\times10^5 \) m s\(^{-1}\)
(d) \( 4\sqrt2\times10^5 \) m s\(^{-1}\)
Show Answer & Explanation

Answer: (d) \( 4\sqrt2\times10^5 \) m s\(^{-1}\)

Explanation:
1. \( \tfrac12mv^2=qEd\Rightarrow v^2=\dfrac{2\times3.2\times10^{-19}\times1.6\times10^5\times2\times10^{-2}}{6.4\times10^{-27}}=3.2\times10^{11} \).
2. \( v=5.66\times10^5=4\sqrt2\times10^5 \) m s\(^{-1}\).

Teacher's Note:
1. Work done by the field equals the kinetic energy gained.

Question 8: Millikan's oil drop experiment established that
(a) electric charge depends on velocity
(b) the specific charge of an electron is \( 1.76\times10^{11} \) C kg\(^{-1}\)
(c) electron has wave nature
(d) electric charge is quantised
Show Answer & Explanation

Answer: (d) electric charge is quantised

Explanation:
1. Millikan found all drop charges to be integral multiples of the electronic charge \( e \), proving charge quantisation.

Teacher's Note:
1. Thomson, not Millikan, measured \( e/m \).

Question 9: A strong argument for the particle nature of cathode rays is that they
(a) produce fluorescence
(b) travel through vacuum
(c) get deflected by electric and magnetic fields
(d) cast shadow
Show Answer & Explanation

Answer: (c) get deflected by electric and magnetic fields

Explanation:
1. Only charged particles are deflected by electric and magnetic fields, so deflection shows a particle nature.

Teacher's Note:
1. Shadows and fluorescence can also be shown by waves.

Question 10: An electron of mass \( m \) and charge \( e \) in an external field \( E \) experiences an acceleration
(a) \( \dfrac{e}{mE} \), in the opposite direction to the field
(b) \( \dfrac{eE}m \), in the direction of the field
(c) \( \dfrac{em}E \), in the direction of the field
(d) \( \dfrac{eE}m \), in the opposite direction to the field
Show Answer & Explanation

Answer: (d) \( \dfrac{eE}m \), in the opposite direction to the field

Explanation:
1. \( F=-eE \), so \( a=\dfrac{eE}m \) opposite to the field (the electron is negative).

Teacher's Note:
1. Negative charges move against the field.

Question 11: The specific charge of a proton is \( 9.6\times10^7 \) C kg\(^{-1}\). The specific charge of an \( \alpha \)-particle will be
(a) \( 9.6\times10^7 \) C kg\(^{-1}\)
(b) \( 19.2\times10^7 \) C kg\(^{-1}\)
(c) \( 4.8\times10^7 \) C kg\(^{-1}\)
(d) \( 2.4\times10^7 \) C kg\(^{-1}\)
Show Answer & Explanation

Answer: (c) \( 4.8\times10^7 \) C kg\(^{-1}\)

Explanation:
1. An \( \alpha \)-particle has charge \( 2e \) and mass \( \approx4m_p \), so \( \dfrac{q}{m}=\dfrac{2}{4}\times9.6\times10^7=4.8\times10^7 \).

Teacher's Note:
1. Charge doubles; mass quadruples.

Question 12: A uniform electric field and a uniform magnetic field exist in a region in the same direction. An electron is projected with a velocity pointed in the same direction. Then the electron will
(a) be deflected to the left without increase in speed
(b) be deflected to the right without increase in speed
(c) not be deflected but its speed will decrease
(d) not be deflected but its speed will increase
Show Answer & Explanation

Answer: (c) not be deflected but its speed will decrease

Explanation:
1. The magnetic force is zero (velocity parallel to \( B \)). The electric force on the electron is opposite to \( E \), i.e. opposite to its velocity, so it slows down.

Teacher's Note:
1. A force parallel to the velocity changes only the speed.

Question 13: A charged oil drop is suspended in a uniform field of \( 3\times10^4 \) V m\(^{-1}\) so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge \( =9.9\times10^{-15} \) kg and \( g=10 \) m s\(^{-2}\))
(a) \( 3.3\times10^{-18} \) C
(b) \( 3.2\times10^{-18} \) C
(c) \( 1.6\times10^{-18} \) C
(d) \( 4.8\times10^{-18} \) C
Show Answer & Explanation

Answer: (a) \( 3.3\times10^{-18} \) C

Explanation:
1. \( qE=mg\Rightarrow q=\dfrac{9.9\times10^{-15}\times10}{3\times10^4}=3.3\times10^{-18} \) C.

Teacher's Note:
1. Electric force balances weight.

Question 14: Cathode rays of velocity \( 10^5 \) m s\(^{-1}\) describe an approximate circular path of radius 1 m in an electric field of 300 V cm\(^{-1}\). If the velocity of the cathode rays is doubled, the value of the electric field so that the rays describe the same circular path will be
(a) 2400 V cm\(^{-1}\)
(b) 600 V cm\(^{-1}\)
(c) 1200 V cm\(^{-1}\)
(d) 12000 V cm\(^{-1}\)
Show Answer & Explanation

Answer: (c) 1200 V cm\(^{-1}\)

Explanation:
1. \( eE=\dfrac{mv^2}{r} \), so \( E\propto v^2 \) for the same radius: \( E=4\times300=1200 \) V cm\(^{-1}\).

Teacher's Note:
1. Doubling \( v \) multiplies the required \( E \) by 4.

Question 15: When a high energy UV photon beam enters an electric field, it will be
(a) accelerated
(b) retarded
(c) undeflected
(d) none of these
Show Answer & Explanation

Answer: (c) undeflected

Explanation:
1. Photons carry no charge, so an electric field does not deflect them.

Teacher's Note:
1. Photons are neutral.

Question 16: An \( \alpha \)-particle of energy 5 MeV is scattered through \( 180^\circ \) by a fixed uranium nucleus. The distance of the closest approach is of the order of
(a) 1 Å
(b) \( 10^{-10} \) cm
(c) \( 10^{-12} \) cm
(d) \( 10^{-15} \) cm
Show Answer & Explanation

Answer: (c) \( 10^{-12} \) cm

Explanation:
1. \( r=\dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(92e)}{K}=\dfrac{9\times10^9\times184\times(1.6\times10^{-19})^2}{5\times1.6\times10^{-13}} \).
2. \( r\approx5\times10^{-14} \) m \( =5\times10^{-12} \) cm.

Teacher's Note:
1. Initial kinetic energy equals electrostatic potential energy at closest approach.

Question 17: In a region, steady and uniform electric and magnetic fields are present. These two fields are parallel to each other. A charged particle is released from rest in this region. The path of the particle will be a
(a) helix
(b) straight line
(c) ellipse
(d) circle
Show Answer & Explanation

Answer: (b) straight line

Explanation:
1. Starting from rest, the particle has no magnetic force and the electric force is along the fields, so it moves in a straight line.

Teacher's Note:
1. Magnetic force needs a velocity component perpendicular to \( B \).

Question 18: The material used for making a thermionic cathode must have
(a) low work function and low melting point
(b) low work function and high melting point
(c) high work function and high melting point
(d) high work function and low melting point
Show Answer & Explanation

Answer: (b) low work function and high melting point

Explanation:
1. A low work function gives easy emission and a high melting point lets it be heated strongly without melting.

Teacher's Note:
1. Tungsten with thorium coating is used.

Question 19: An oil drop having a mass \( 4.8\times10^{-10} \) g and charge \( 2.4\times10^{-18} \) C stands still between two charged horizontal plates separated by a distance of 1 cm. If now the polarity of the plates is changed, the instantaneous acceleration of the drop is (take \( g=10 \) m s\(^{-2}\))
(a) 5 m s\(^{-2}\)
(b) 10 m s\(^{-2}\)
(c) 15 m s\(^{-2}\)
(d) 20 m s\(^{-2}\)
Show Answer & Explanation

Answer: (d) 20 m s\(^{-2}\)

Explanation:
1. Initially \( qE=mg \). After reversal the electric force also points down: \( a=\dfrac{mg+qE}m=2g=20 \) m s\(^{-2}\).

Teacher's Note:
1. Both forces then act downward.

Question 20: In Millikan's oil drop experiment, an oil drop of mass \( 16\times10^{-6} \) kg is balanced by an electric field of \( 10^6 \) V m\(^{-1}\). The charge in coulomb on the drop is (assume \( g=10 \) m s\(^{-2}\))
(a) \( 6.2\times10^{-11} \)
(b) \( 16\times10^{-9} \)
(c) \( 16\times10^{-11} \)
(d) \( 16\times10^{-13} \)
Show Answer & Explanation

Answer: (c) \( 16\times10^{-11} \)

Explanation:
1. \( q=\dfrac{mg}E=\dfrac{16\times10^{-6}\times10}{10^6}=16\times10^{-11} \) C.

Teacher's Note:
1. Balance: \( qE=mg \).

Question 21: A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected along the direction of the fields with a certain velocity, then
(a) its velocity will decrease
(b) its velocity will increase
(c) it will turn towards the right of the direction of motion
(d) it will turn towards the left of the direction of motion
Show Answer & Explanation

Answer: (a) its velocity will decrease

Explanation:
1. The magnetic force is zero; the electric force on the electron is opposite to the field, so the electron decelerates.

Teacher's Note:
1. Negative charge accelerates against \( \vec E \).

Question 22: If an electron moves from rest from a point at which potential is 50 V to another point at which potential is 70 V, then its kinetic energy in the final state will be
(a) \( 3.2\times10^{-20} \) J
(b) \( 3.2\times10^{-18} \) J
(c) \( 3.2\times10^{-19} \) J
(d) zero
Show Answer & Explanation

Answer: (b) \( 3.2\times10^{-18} \) J

Explanation:
1. \( K=e\Delta V=1.6\times10^{-19}\times20=3.2\times10^{-18} \) J.

Teacher's Note:
1. Electrons move toward higher potential and gain energy.

Question 23: The resistance of a discharge tube is
(a) zero
(b) ohmic
(c) non-ohmic
(d) infinity
Show Answer & Explanation

Answer: (c) non-ohmic

Explanation:
1. The current through a discharge tube is not proportional to the voltage, so its resistance is non-ohmic.

Teacher's Note:
1. Ionisation changes with the voltage.

Question 24: If the distance of a 100 W lamp is increased from a photocell, the saturation current \( i \) in the photocell varies with the distance \( d \) as
(a) \( i\propto d^2 \)
(b) \( i\propto d \)
(c) \( i\propto\dfrac1d \)
(d) \( i\propto\dfrac1{d^2} \)
Show Answer & Explanation

Answer: (d) \( i\propto\dfrac1{d^2} \)

Explanation:
1. Intensity falls as \( 1/d^2 \) and the saturation current is proportional to intensity.

Teacher's Note:
1. Inverse square law.

Question 25: What is the de-Broglie wavelength of the electron accelerated through a potential difference of 100 V?
(a) 12.27 Å
(b) 1.227 Å
(c) 0.1227 Å
(d) 0.001227 Å
Show Answer & Explanation

Answer: (b) 1.227 Å

Explanation:
1. \( \lambda=\dfrac{12.27}{\sqrt V}\text{ Å}=\dfrac{12.27}{10}=1.227 \) Å.

Teacher's Note:
1. Use \( \lambda=12.27/\sqrt V \) Å for electrons.

Question 26: The energy of a photon is equal to the kinetic energy of a proton. If \( \lambda_1 \) is the de-Broglie wavelength of the proton, \( \lambda_2 \) the wavelength associated with the photon and if the energy of the photon is \( E \), then \( \lambda_1/\lambda_2 \) is proportional to
(a) \( E^4 \)
(b) \( E^{1/2} \)
(c) \( E^2 \)
(d) \( E \)
Show Answer & Explanation

Answer: (b) \( E^{1/2} \)

Explanation:
1. \( \lambda_1=\dfrac h{\sqrt{2mE}} \) and \( \lambda_2=\dfrac{hc}E \), so \( \dfrac{\lambda_1}{\lambda_2}=\dfrac{E}{c\sqrt{2mE}}\propto E^{1/2} \).

Teacher's Note:
1. Express both wavelengths in terms of \( E \).

Question 27: In a hydrogen-like atom, an electron makes a transition from an energy level with quantum number \( n \) to another with quantum number \( (n-1) \). If \( n\gg1 \), the frequency of radiation emitted is proportional to
(a) \( \dfrac1n \)
(b) \( \dfrac1{n^2} \)
(c) \( \dfrac1{n^4} \)
(d) \( \dfrac1{n^3} \)
Show Answer & Explanation

Answer: (d) \( \dfrac1{n^3} \)

Explanation:
1. \( \nu\propto\dfrac1{(n-1)^2}-\dfrac1{n^2}\approx\dfrac{2}{n^3} \) for large \( n \).

Teacher's Note:
1. Expand \( (n-1)^{-2} \).

Question 28: The work functions of silver and sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the stopping potential versus frequency plot for silver to that of sodium is
(a) 4
(b) 3
(c) 1
(d) 7
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. \( V_0=\dfrac he\nu-\dfrac{\phi}e \): the slope \( h/e \) is the same for every metal, so the ratio is 1.

Teacher's Note:
1. Slope is a universal constant.

Question 29: The potential energy of a particle of mass \( m \) varies as \( U(x)=E_0 \) for \( 0\le x\le1 \) and \( U(x)=0 \) for \( x>1 \). The de-Broglie wavelength of the particle in the range \( 0\le x\le1 \) is \( \lambda_1 \) and that in the range \( x>1 \) is \( \lambda_2 \). If the total energy of the particle is \( 2E_0 \), find \( \lambda_1/\lambda_2 \).
(a) \( \sqrt2 \)
(b) \( \sqrt3 \)
(c) \( \sqrt{1/2} \)
(d) \( \sqrt{2/3} \)
Show Answer & Explanation

Answer: (a) \( \sqrt2 \)

Explanation:
1. Kinetic energies: \( E_0 \) in the first region and \( 2E_0 \) in the second. \( \lambda\propto\dfrac1{\sqrt K} \), so \( \dfrac{\lambda_1}{\lambda_2}=\sqrt{\dfrac{2E_0}{E_0}}=\sqrt2 \).

Teacher's Note:
1. Lower kinetic energy means a longer wavelength.

Question 30: The de-Broglie wavelength for a 100 g bullet moving at 900 m s\(^{-1}\) is (\( h=6.63\times10^{-34} \) J s)
(a) \( 3.7\times10^{-35} \) m
(b) \( 7.4\times10^{-36} \) m
(c) \( 7.8\times10^{-37} \) m
(d) \( 8.2\times10^{-39} \) m
Show Answer & Explanation

Answer: (b) \( 7.4\times10^{-36} \) m

Explanation:
1. \( \lambda=\dfrac h{mv}=\dfrac{6.63\times10^{-34}}{0.1\times900}=7.4\times10^{-36} \) m.

Teacher's Note:
1. Macroscopic wavelengths are immeasurably small.

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