EAMCET Question Paper (Physics)
EAMCET Question Paper (Physics)
Nuclear Physics
The following MCQ which appeared in EAMCET 2005 question paper highlights the law of conservation of momentum as applied to nuclear processes:
1.A nucleus of mass 218 amu in free state decays to emit an α-particle. Kinetic energy of the α-particle emitted is 6.7 MeV. The recoil energy (in MeV) of the daughter nucleus is
(a) 1
(b) 0.5
(c) 0.25
(d) 0.125
The kinetic energy (K) is given by
K = p2/2m where p is the momentum and m is the mass. The recoil momentum of the daughter nucleus is equal and opposite to the momentum of the α-particle.
We have p2/(2×4 amu) = 6.7 MeV, since the mass of the α-particle is 4 amu.
[You need not convert amu into kilogram and energy into joule and waste your time. You may imagine that p has proper unit to get the energy in MeV]
In the case of the daughter nucleus (of mass 218 – 4 = 214 amu), we have
p2/(2×214 amu) = K, whetre K is the recoil energy of the daughter (in MeV).
From the above two equations we obtain K = 0.125 MeV.
The following MCQ appeared in EAMCET 2000 question paper:
2.In a nuclear reactor using U235 as fuel the power output is 4.8 mega watts. The number of fissions per second is (energy released per fission of U235 = 200 MeV).
(a) 1.5×1017
(b) 3×1019
(c) 1.5×1025
(d) 3×1025
This is a very simple question. The energy produced per second is 4.8×106 joule. Since 1 electron volt is 1.6×10–19 joule, 200 MeV = 200×106×1.6×10–19 joule.
Therefore, the number of fissions per second = (4.8×106)/(200×106×1.6×10–19) = 1.5×1017.
Multiple choice questions on nuclear fission at the level expected from you will be simple. Here are three such questions:
(1)The energy released per fission of a U235 nucleus is around
(a) 0.02 eV
(b) 2 eV
(c) 2 MeV
(d) 20 MeV
(e) 200 MeV
The mass of the products of fission will be less than the mass of the U235 nucleus. The mass difference gets converted into energy. Even though different fragments can be produced, the energy released per fission of a U235 nucleus is around 200 MeV.
(2) From the following, pick out the most suitable energy of neutrons which will produce nuclear fission in a reactor
(a) 0.04 eV
(b) 40 eV
(c) 400 eV
(d) 2 MeV
(e) 20 MeV
Nuclear fission is induced most effectively by neutrons of thermal energy and hence the correct option is 0.04 eV
(3) The function of the moderator in a nuclear reactor is
(a) to absorb fast neutrons
(b) to adjust the power output to moderate levels
(c) to slow down fast neutrons
(d) to absorb slow neutrons
(e) to cool the reactor core
This is a simple question which repeatedly appears in entrance test papers with minor changes in the wrong options. The moderator is used to slow down fast neutrons to thermal energies [Option (c)].
Atomic Physics
(1) When the electron in a hydrogen atom of mass M undergoes transition from an orbit of higher quantum number n2 to an orbit of lower quantum number n1, the recoil velocity acquired by the atom is (Rydberg’s constant = R, Planck’s constant = h)
(a) (R/hM) (1/n12 – 1/n22)
(b) (hR/M) (n2 – n1)
(c) 1/hRM (1/n12 – 1/n22)
(d) h/RM) (1/n12 – 1/n22)
(e) (hR/M) (1/n12 – 1/n22)
The wave number of the photon emitted because of the electron transition is
ν' = 1/λ = R(1/n12 – 1/n22) where λ is the wave length of the photon and R is Rydberg’s constant.
The momentum of the photon is p = h/λ = hR(1/n12 – 1/n22) where h is Planck’s constant.
When the photon is emitted with this momentum, the atom recoils (like a gun firing a bullet) with an equal and opposite momentum. Therefore, the recoil velocity of the atom is given by
v = p/M = (hR/M)(1/n12 – 1/n22).
(2) If the radius of the innermost electron orbit in a hydrogen atom is R1, the de Broglie wave length of the electron in the second excited state is
(a) πR1
(b) 3πR1
(c) 4πR1
(d) 6πR1
(e) 9πR1
The wave length of the electron in the nth orbit is given by
λ = 2πRn/n where Rn is the radius of the nth orbit.4
[This follows because the angular momentum of the electron in the nth orbit is
mvRn = nh/2π.
Therefore, de Broglie wave length, λ = h/mv = 2πRn/n ]
The second excited state has quantum number n = 3 (Third orbit). The radius of the 3rd orbit in terms of the radius R1 of the first orbit is given by (remembering Rn = n2 R1)
R3 = 9R1
Therefore, λ = 2πRn/n = 2π×9R1/3 = 6πR1
[It will be convenient to remember that the de Broglie wave length of the electron in the nth orbit is n times the the wave length in the innermost orbit].
Electric Power
The following questions are simple and you are expected to solve them in a couple of minutes:
(1) A current I ampere gets divided and flows into a parallel network of resistors as shown in the adjoining figure. If the power dissipated in the 1 Ω resistor is P watt, what is the total power (in watt) dissipated in the branch containing the two 3 Ω resistors?
(a) P
(b) 6P
(c) 3P
(d) P/2
(e) 3P/2
If the current through the 1 Ω resistor is I1, the current through the branch containing the two 3 Ω resistors is I1/2.
[Note that the P.D across the three branches is the same and hence the current through the 6 Ω branch must be half the current through the 3 Ω branch].
The power dissipated in the 1 Ω resistor is
I12 ×1= P
Therefore, the power dissipated in the branch containing the two 3 Ω resistors is
(I1/2) 2×6 = 6P/4 = 3P/2 watt.
(2) A battery of internal resistance R ohm and output V volt is connected across a variable resistor. The heat generated in the variable resistor is maximum when the current in it is
(a) V/2R
(b) V/4R
(c) 4V/R
(d) V/R
(e) None of the above
You may be remembering the maximum power transfer theorem which states that a direct current source will transfer maximum power to a load when the resistance of the load is equal to the internal resistance of the source.
In the above problem, since the internal resistance of the battery is R, the power dissipated in the variable resistor is maximum when its resistance also is R. The total resistance in the circuit in this condition is R+R = 2R and the current is V/2R.
Courtesy:physicsplus.blogspot.com
EAMCET Question Paper (Physics)
Nuclear Physics
The following MCQ which appeared in EAMCET 2005 question paper highlights the law of conservation of momentum as applied to nuclear processes:
1.A nucleus of mass 218 amu in free state decays to emit an α-particle. Kinetic energy of the α-particle emitted is 6.7 MeV. The recoil energy (in MeV) of the daughter nucleus is
(a) 1
(b) 0.5
(c) 0.25
(d) 0.125
The kinetic energy (K) is given by
K = p2/2m where p is the momentum and m is the mass. The recoil momentum of the daughter nucleus is equal and opposite to the momentum of the α-particle.
We have p2/(2×4 amu) = 6.7 MeV, since the mass of the α-particle is 4 amu.
[You need not convert amu into kilogram and energy into joule and waste your time. You may imagine that p has proper unit to get the energy in MeV]
In the case of the daughter nucleus (of mass 218 – 4 = 214 amu), we have
p2/(2×214 amu) = K, whetre K is the recoil energy of the daughter (in MeV).
From the above two equations we obtain K = 0.125 MeV.
The following MCQ appeared in EAMCET 2000 question paper:
2.In a nuclear reactor using U235 as fuel the power output is 4.8 mega watts. The number of fissions per second is (energy released per fission of U235 = 200 MeV).
(a) 1.5×1017
(b) 3×1019
(c) 1.5×1025
(d) 3×1025
This is a very simple question. The energy produced per second is 4.8×106 joule. Since 1 electron volt is 1.6×10–19 joule, 200 MeV = 200×106×1.6×10–19 joule.
Therefore, the number of fissions per second = (4.8×106)/(200×106×1.6×10–19) = 1.5×1017.
Multiple choice questions on nuclear fission at the level expected from you will be simple. Here are three such questions:
(1)The energy released per fission of a U235 nucleus is around
(a) 0.02 eV
(b) 2 eV
(c) 2 MeV
(d) 20 MeV
(e) 200 MeV
The mass of the products of fission will be less than the mass of the U235 nucleus. The mass difference gets converted into energy. Even though different fragments can be produced, the energy released per fission of a U235 nucleus is around 200 MeV.
(2) From the following, pick out the most suitable energy of neutrons which will produce nuclear fission in a reactor
(a) 0.04 eV
(b) 40 eV
(c) 400 eV
(d) 2 MeV
(e) 20 MeV
Nuclear fission is induced most effectively by neutrons of thermal energy and hence the correct option is 0.04 eV
(3) The function of the moderator in a nuclear reactor is
(a) to absorb fast neutrons
(b) to adjust the power output to moderate levels
(c) to slow down fast neutrons
(d) to absorb slow neutrons
(e) to cool the reactor core
This is a simple question which repeatedly appears in entrance test papers with minor changes in the wrong options. The moderator is used to slow down fast neutrons to thermal energies [Option (c)].
Atomic Physics
(1) When the electron in a hydrogen atom of mass M undergoes transition from an orbit of higher quantum number n2 to an orbit of lower quantum number n1, the recoil velocity acquired by the atom is (Rydberg’s constant = R, Planck’s constant = h)
(a) (R/hM) (1/n12 – 1/n22)
(b) (hR/M) (n2 – n1)
(c) 1/hRM (1/n12 – 1/n22)
(d) h/RM) (1/n12 – 1/n22)
(e) (hR/M) (1/n12 – 1/n22)
The wave number of the photon emitted because of the electron transition is
ν' = 1/λ = R(1/n12 – 1/n22) where λ is the wave length of the photon and R is Rydberg’s constant.
The momentum of the photon is p = h/λ = hR(1/n12 – 1/n22) where h is Planck’s constant.
When the photon is emitted with this momentum, the atom recoils (like a gun firing a bullet) with an equal and opposite momentum. Therefore, the recoil velocity of the atom is given by
v = p/M = (hR/M)(1/n12 – 1/n22).
(2) If the radius of the innermost electron orbit in a hydrogen atom is R1, the de Broglie wave length of the electron in the second excited state is
(a) πR1
(b) 3πR1
(c) 4πR1
(d) 6πR1
(e) 9πR1
The wave length of the electron in the nth orbit is given by
λ = 2πRn/n where Rn is the radius of the nth orbit.4
[This follows because the angular momentum of the electron in the nth orbit is
mvRn = nh/2π.
Therefore, de Broglie wave length, λ = h/mv = 2πRn/n ]
The second excited state has quantum number n = 3 (Third orbit). The radius of the 3rd orbit in terms of the radius R1 of the first orbit is given by (remembering Rn = n2 R1)
R3 = 9R1
Therefore, λ = 2πRn/n = 2π×9R1/3 = 6πR1
[It will be convenient to remember that the de Broglie wave length of the electron in the nth orbit is n times the the wave length in the innermost orbit].
Electric Power
The following questions are simple and you are expected to solve them in a couple of minutes:
(1) A current I ampere gets divided and flows into a parallel network of resistors as shown in the adjoining figure. If the power dissipated in the 1 Ω resistor is P watt, what is the total power (in watt) dissipated in the branch containing the two 3 Ω resistors?
(a) P
(b) 6P
(c) 3P
(d) P/2
(e) 3P/2
If the current through the 1 Ω resistor is I1, the current through the branch containing the two 3 Ω resistors is I1/2.
[Note that the P.D across the three branches is the same and hence the current through the 6 Ω branch must be half the current through the 3 Ω branch].
The power dissipated in the 1 Ω resistor is
I12 ×1= P
Therefore, the power dissipated in the branch containing the two 3 Ω resistors is
(I1/2) 2×6 = 6P/4 = 3P/2 watt.
(2) A battery of internal resistance R ohm and output V volt is connected across a variable resistor. The heat generated in the variable resistor is maximum when the current in it is
(a) V/2R
(b) V/4R
(c) 4V/R
(d) V/R
(e) None of the above
You may be remembering the maximum power transfer theorem which states that a direct current source will transfer maximum power to a load when the resistance of the load is equal to the internal resistance of the source.
In the above problem, since the internal resistance of the battery is R, the power dissipated in the variable resistor is maximum when its resistance also is R. The total resistance in the circuit in this condition is R+R = 2R and the current is V/2R.