VITEEE Exam 2014 - Previous Year Question paper with solutions
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VITEEE Exam 2014 Question paper with solution (Part 1)
1. A coil of resistance 10 Ω and an inductance 5 H is connected to a 100 V battery. The energy stored in the coil is
A) 325 erg
B) 125 J
C) 250 erg
D) 250 J
Answer: D) 250 J
Explanation: Current I = 100 / 10 = 10A E = 1 / 2 LI² = 1/2 * 5 * (10)² = 250 J
C) 250 erg
D) 250 J
Answer: D) 250 J
Explanation: Current I = 100 / 10 = 10A E = 1 / 2 LI² = 1/2 * 5 * (10)² = 250 J
2. A galvanometer has the current range of 15 mA and voltage range 750 mV. To convert this galvanometer into an ammeter of range 25 A, the required shunt is
A) 0.8 Ω
B) 0.93 Ω
C) 0.03 Ω
D) 2.0 Ω
Answer: C) 0.03 Ω
Explanation: V = 750 * 10-3 V Ig = 15 * 10-3 A and I = 25 A Using the relation a = V / Ig = 750 * 10-3 / 15 * 10-3 = 50 Ω Ig = S / S + a * I 15 * 10-3 = ( S / S + 50) * 25 S = 0.03 Ω
B) 0.93 Ω
C) 0.03 Ω
D) 2.0 Ω
Answer: C) 0.03 Ω
Explanation: V = 750 * 10-3 V Ig = 15 * 10-3 A and I = 25 A Using the relation a = V / Ig = 750 * 10-3 / 15 * 10-3 = 50 Ω Ig = S / S + a * I 15 * 10-3 = ( S / S + 50) * 25 S = 0.03 Ω
3. The Daniel cell is balanced on 125 cm length of a potentiometer. Now, the cell is short-circuited by a resistance of 2 Ω and the balance is obtained at 100 cm. The internal resistance of the Daniel cell is
A) 4/3 Ω
B) 1.5 Ω
C) 1.25 Ω
D) 0.5 Ω
Answer: D) 0.5 Ω
Explanation: r = R(I₁ / I₂ - 1) = 2(125 / 100 - 1) = 2((5 / 4) - 1) = 2 * 1 / 4 = 0.05 Ω
Answer: D) 7.5 * 10-2 A
Explanation: Here, the resistance for 10Ω, 60Ω and 100Ω are in series and they together are in parallel to 200Ω resistance when a potential difference of 15V is applied across 200Ω, then current through it is I = 15 / 200 = 7.5 * 10-2 A
5. A circuit has a self-inductance of 1 H and carries a current of 2A. To prevent sparking, when the circuit is switched off, a capacitor which can withstand 400 V is used. The least capacitance of the capacitor connected across the switch must be equal to
B) 1.5 Ω
C) 1.25 Ω
D) 0.5 Ω
Answer: D) 0.5 Ω
Explanation: r = R(I₁ / I₂ - 1) = 2(125 / 100 - 1) = 2((5 / 4) - 1) = 2 * 1 / 4 = 0.05 Ω
4. Four resistances of 10Ω, 60Ω, 100Ω and 200Ω respectively taken in order are used to form a Wheatstone's bridge. A 15V battery is connected to the ends of a 200 Ω resistance, the current through it will be
A) 7.5 * 10-5 A
B) 7.5 * 10-4 A
C) 7.5 * 10-3 A
D) 7.5 * 10-2 A
A) 7.5 * 10-5 A
B) 7.5 * 10-4 A
C) 7.5 * 10-3 A
D) 7.5 * 10-2 A
Answer: D) 7.5 * 10-2 A
Explanation: Here, the resistance for 10Ω, 60Ω and 100Ω are in series and they together are in parallel to 200Ω resistance when a potential difference of 15V is applied across 200Ω, then current through it is I = 15 / 200 = 7.5 * 10-2 A
5. A circuit has a self-inductance of 1 H and carries a current of 2A. To prevent sparking, when the circuit is switched off, a capacitor which can withstand 400 V is used. The least capacitance of the capacitor connected across the switch must be equal to
A) 50 μF
B) 25 μF
C) 100 μF
D) 12.5 μF
Answer: B) 25 μF
Explanation: Energy stored in capacitor is equal to energy stored in inductance 1/2 CV² = 1/2 LI² C = LI² / V² = 1 * (2)² / (400)² = 25μF
B) 25 μF
C) 100 μF
D) 12.5 μF
Answer: B) 25 μF
Explanation: Energy stored in capacitor is equal to energy stored in inductance 1/2 CV² = 1/2 LI² C = LI² / V² = 1 * (2)² / (400)² = 25μF
6. The output Y of the logic circuit shown in the figure is best represented as
A) A
B) B
C) C
D) D
Answer: D) D
Explanation:
7. A resistor of 6 kΩ with tolerance 10% and another resistance of 4 kΩ with tolerance 10% are connected in series. The tolerance of the combination is about
B) B
C) C
D) D
Answer: D) D
Explanation:
A) 5%
B) 10%
C) 12%
D) 15%
Answer: B) 10%
Explanation: In series combination, R = R₁ + R₂ = 6 + 4 = 10 kΩ Error in combination, ΔR = ΔR₁ + ΔR₂ = 10 / 100 * 6 + 10 / 100 * 4 = 0.6 + 0.4 = 1 ΔR / R = 1 / 10 = 10%
B) 10%
C) 12%
D) 15%
Answer: B) 10%
Explanation: In series combination, R = R₁ + R₂ = 6 + 4 = 10 kΩ Error in combination, ΔR = ΔR₁ + ΔR₂ = 10 / 100 * 6 + 10 / 100 * 4 = 0.6 + 0.4 = 1 ΔR / R = 1 / 10 = 10%
8. If we add an impurity to a metal those atoms also deflect electrons. Therefore,
A) the electrical and thermal conductivities both increase
B) the electrical and thermal conductivities both decrease
C) the electrical conductivity increases but thermal conductivity decreases
D) the electrical conductivity decrease but thermal conductivity increases
Answer: B) the electrical and thermal conductivities both decrease
Explanation: If the number of electrons increases, their number of a collision, increasing the thermal and electrical resistance. Hence, electrical and thermal conductivities both decrease
B) the electrical and thermal conductivities both decrease
C) the electrical conductivity increases but thermal conductivity decreases
D) the electrical conductivity decrease but thermal conductivity increases
Answer: B) the electrical and thermal conductivities both decrease
Explanation: If the number of electrons increases, their number of a collision, increasing the thermal and electrical resistance. Hence, electrical and thermal conductivities both decrease
9. A proton and an α-particle, accelerated through the same potential difference, enter a region of uniform magnetic field normally. If the radius of the proton orbit is 10 cm, then the radius of α-particle is
A) 10 cm
B) 10√2 cm
C) 20 cm
D) 5√2 cm
Answer: B) 10√2 cm
Explanation: Radius of path ri = 1 / β √ 2mv / q rα / rp = √ mα / mp √ qp / qα rα / 10 = √4 / 2 rα = 10√2 cm
B) 10√2 cm
C) 20 cm
D) 5√2 cm
Answer: B) 10√2 cm
Explanation: Radius of path ri = 1 / β √ 2mv / q rα / rp = √ mα / mp √ qp / qα rα / 10 = √4 / 2 rα = 10√2 cm
10. An ammeter and a voltmeter of resistance R are connected in series to an electric cell of negligible internal resistance. Their readings are A and V respectively. If another resistance R is connected in parallel with the voltmeter, then
A) both A and V will increase
B) both A and V will decrease
C) A will decrease and V will increase
D) A will increase and V will decrease
Answer: D) A will increase and V will decrease
Explanation: When resistance R is connected in parallel with the voltmeter, the effective resistance will decrease. According to Ohm's law, V = IR... R = V/I. Hence, R decrease, so V should decrease and I should increase
B) both A and V will decrease
C) A will decrease and V will increase
D) A will increase and V will decrease
Answer: D) A will increase and V will decrease
Explanation: When resistance R is connected in parallel with the voltmeter, the effective resistance will decrease. According to Ohm's law, V = IR... R = V/I. Hence, R decrease, so V should decrease and I should increase
11. A neutron is moving with velocity u. It collides head-on and elastically with an atom of mass number A. If the initial kinetic energy of the neutron is E. then how much kinetic energy will be retained by the neutron after reflection?
A) (A / A + 1)² E
B) A / (A + 1)² E
C) (A - 1 / A + 1)² E
D) (A - 1) / (A + 1)² E
Answer: C) (A - 1 / A + 1)² E
Explanation:
B) A / (A + 1)² E
C) (A - 1 / A + 1)² E
D) (A - 1) / (A + 1)² E
Answer: C) (A - 1 / A + 1)² E
Explanation:
12. If a magnet is suspended at angle 30° to the magnet meridian, the dip of the needle makes an angle of 45° with the horizontal, the real dip is
A) tan-1 (√3 / 2)
B) tan-1 (√3)
C) tan-1 (√3/2)
D) tan-1 (2 / √3)
Answer: D) tan-1 (2 / √3)
Explanation: tan δ = tan δ / cos θ = tan 45° / cos 30° tan δ = 1 / √3 / 2 = 2 / √3 δ = tan-1 (2 / √3)
B) tan-1 (√3)
C) tan-1 (√3/2)
D) tan-1 (2 / √3)
Answer: D) tan-1 (2 / √3)
Explanation: tan δ = tan δ / cos θ = tan 45° / cos 30° tan δ = 1 / √3 / 2 = 2 / √3 δ = tan-1 (2 / √3)
13. Which has more luminous efficiency?
A) A 40 W bulb
B) A 40 W fluorescent tube
C) Both have the same
D) Cannot say
Answer: B) A 40 W fluorescent tube
Explanation: A 40 W fluorescent tube has greater luminous efficiency because, for the same power supply, it gives more light
B) A 40 W fluorescent tube
C) Both have the same
D) Cannot say
Answer: B) A 40 W fluorescent tube
Explanation: A 40 W fluorescent tube has greater luminous efficiency because, for the same power supply, it gives more light
14. The resistance of a germanium junction diode whose V - I is shown in the figure is (Vk =0.3 V)
A) 5 k Ω
B) 0.2 Ω
C) 2.3 Ω
D) (10/2.3)K Ω
Answer: B) 0.2 Ω
Explanation: R = ΔV / ΔI = 2.3 - 0.3 / 10 * 10-3 R = 2 /10 * 10³ R = 0.2 * 10³ Ω = 0.2 kΩ
B) 0.2 Ω
C) 2.3 Ω
D) (10/2.3)K Ω
Answer: B) 0.2 Ω
Explanation: R = ΔV / ΔI = 2.3 - 0.3 / 10 * 10-3 R = 2 /10 * 10³ R = 0.2 * 10³ Ω = 0.2 kΩ
15. In hydrogen discharge tube, it is observed that through a given cross-section 3.31 * 1015 electrons are moving from right to left and 3.12 * 105 protons are moving from left to right. The current in the discharge tube and its direction will be
A) 2 mA, towards left
B) 2 mA, towards right
C) 1 mA, towards the right
D) 2 mA, towards left
Answer: C) 1 mA, towards right
Explanation: Number of electrons ne = 3.13 * 1015 Number of protons np = 3.12 * 1015 Now, the current is given by the relation I = ne qe + np qp = 3.13 * 1015 * 1.6 * 10-17 + 3.12 * 1015 * 1.6 * 10-19 = 1 * 10-3 = 1 mA Now, due to excess the charge on electrons, the direction of the current will be towards right
A) 0 eV
B) 1 eV
C) 10 eV
D) 50 eV
Answer: B) 1 eV
Explanation: In conductance, the separation between conduction and volume bands is zero and in insulated, it is greater than 1 eV. Hence, in semiconductor, the separation between conduction and volume band is 1 eV
B) 2 mA, towards right
C) 1 mA, towards the right
D) 2 mA, towards left
Answer: C) 1 mA, towards right
Explanation: Number of electrons ne = 3.13 * 1015 Number of protons np = 3.12 * 1015 Now, the current is given by the relation I = ne qe + np qp = 3.13 * 1015 * 1.6 * 10-17 + 3.12 * 1015 * 1.6 * 10-19 = 1 * 10-3 = 1 mA Now, due to excess the charge on electrons, the direction of the current will be towards right
16. In a semiconductor, the separation between conduction and valence band is of the order of
A) 0 eV
B) 1 eV
C) 10 eV
D) 50 eV
Answer: B) 1 eV
Explanation: In conductance, the separation between conduction and volume bands is zero and in insulated, it is greater than 1 eV. Hence, in semiconductor, the separation between conduction and volume band is 1 eV
17. If 1000 droplets each of potential 1V and radius r are mixed to form a big drop. Then, the potential of the drop as compared to small droplets will be
A) 1000 V
B) 800 V
C) 100 V
D) 20 V
Answer: C) 100 V
Explanation: The potential of the big drop = q * n2/3 = 100 V
B) 800 V
C) 100 V
D) 20 V
Answer: C) 100 V
Explanation: The potential of the big drop = q * n2/3 = 100 V
18. A Zener diode, having breakdown voltage equal to 15 V is used in a voltage regulator circuit shown in the figure. The current through the diode is
A) 10 mA
B) 15 mA
C) 20 mA
D) 5 mA
Answer: D) 5 mA
Explanation:
B) 15 mA
C) 20 mA
D) 5 mA
Answer: D) 5 mA
Explanation:
19. The activity of a radioactive sample is measured as N0 counts per minute at t = 0 and N0 / C counts per minute at t = 5 min. The time, (in a minute) at which the activity reduces to half its value, is
A) loge 2/5
B) 5 / loge 2
C) 5 log10 2
D) 5 loge 2
Answer: D) 5 loge 2
Explanation: Fraction remains after n half-lives N / N0 = (1/2)n = (1/2)t/7 N = N0 / e = eN0 / eN0 = (1/2)5/7 1 / e = (1/2)5/7 Taking log on both sides, we get log 1 - log e = 5/7 log 1/2 -1 = 5/7(-log 2) T = 5 loge 2 Now, let t' be the time after which activity reduces to half (1/2) = (1/2)t'/5loge2 t' = 5 loge 2
B) 5 / loge 2
C) 5 log10 2
D) 5 loge 2
Answer: D) 5 loge 2
Explanation: Fraction remains after n half-lives N / N0 = (1/2)n = (1/2)t/7 N = N0 / e = eN0 / eN0 = (1/2)5/7 1 / e = (1/2)5/7 Taking log on both sides, we get log 1 - log e = 5/7 log 1/2 -1 = 5/7(-log 2) T = 5 loge 2 Now, let t' be the time after which activity reduces to half (1/2) = (1/2)t'/5loge2 t' = 5 loge 2
20. If the electron in the hydrogen atom jumps from the third orbit to the second orbit, the wavelength of the emitted radiation in term of Rydberg constant is
A) 6 / 5R
B) 36 / 5R
C) 64 / 7R
D) None of these
Answer: B) 36 / 5R
Explanation: 1/λ = R(1/2² - 1/3²) = R(1/4 - 1/9) 1/λ = R(9-4 / 36) = 5R / 36 λ = 36 / 5R
B) 36 / 5R
C) 64 / 7R
D) None of these
Answer: B) 36 / 5R
Explanation: 1/λ = R(1/2² - 1/3²) = R(1/4 - 1/9) 1/λ = R(9-4 / 36) = 5R / 36 λ = 36 / 5R
21. Silver has a work function of 4.7 eV. When ultraviolet light of wavelength 100 nm is incident on it a potential of 7.7 V is required to stop the photoelectrons from reaching the collector plate. How much potential will be required to stop photoelectrons, when the light of wavelength 200 nm is incident on it?
A) 15.4 V
B) 2.35 V
C) 3.85 V
D) 1.5 V
Answer: D) 1.5 V
Explanation: ev0 + ∅ = hc / λ and ev'0 + ∅'0 = hc / λ' ev'0 + ∅ / ev0 + ∅ = λ / λ' = 100 / 200 = 1/2 2ev'0 + 2∅ = ev0 + ∅ ev'0 = ev0 - ∅ / 2 = 7.7 - 4.7 / 2 = 1.5 V
22. If the distance of 100 W lamp is increased from a photocell, the saturation current I in the photocell varies with the distance d as
B) 2.35 V
C) 3.85 V
D) 1.5 V
Answer: D) 1.5 V
Explanation: ev0 + ∅ = hc / λ and ev'0 + ∅'0 = hc / λ' ev'0 + ∅ / ev0 + ∅ = λ / λ' = 100 / 200 = 1/2 2ev'0 + 2∅ = ev0 + ∅ ev'0 = ev0 - ∅ / 2 = 7.7 - 4.7 / 2 = 1.5 V
22. If the distance of 100 W lamp is increased from a photocell, the saturation current I in the photocell varies with the distance d as
A) I ∞ d²
B) I ∞ d
C) I ∞ 1/d
D) I ∞ 1/d²
Answer: D) I ∞ 1/d²
Explanation: Photoelectric current depends on the intensity of incident radiation i.e., I ∞ I But, the intensity of radiation I ∞ 1 / d² I ∞ 1 / d²
B) I ∞ d
C) I ∞ 1/d
D) I ∞ 1/d²
Answer: D) I ∞ 1/d²
Explanation: Photoelectric current depends on the intensity of incident radiation i.e., I ∞ I But, the intensity of radiation I ∞ 1 / d² I ∞ 1 / d²
23. Following process is known as hv ↔ e+ + e-
A) pair production
B) photoelectric effect
C) Compton effect
D) Zeeman Effect
Answer: A) pair production
Explanation: Pair production refers to the creation of an elementary particle and its anti-particle usually from a photon (or another neutral boson). This is allowed, provided there is enough energy available to create the pair
B) photoelectric effect
C) Compton effect
D) Zeeman Effect
Answer: A) pair production
Explanation: Pair production refers to the creation of an elementary particle and its anti-particle usually from a photon (or another neutral boson). This is allowed, provided there is enough energy available to create the pair
24. During charging a capacitor, variations of potential V of the capacitor with time t is shown as
A) A
B) B
C) C
D) D
Answer: A) A
Explanation:
B) B
C) C
D) D
Answer: A) A
Explanation:
25. When a resistor of 11Ω is connected in series with an electric cell. The current flowing in it is 0.5 A. Instead when a resistor of 5Ω is connected to the same electric cell in series, the current increases by 0.4 A. The internal resistance of the cell is
A) 1.5 Ω
B) 2 Ω
C) 2.5 Ω
D) 3.5 Ω
Answer: C) 2.5 Ω
Explanation: I = E / R + r 0.5 = E / 11 + r E = 5.5 + 0.5r 0.9 = E / 5 + r E = 4.5 + 0.9r on solving these equation, r = 2.5 Ω
B) 2 Ω
C) 2.5 Ω
D) 3.5 Ω
Answer: C) 2.5 Ω
Explanation: I = E / R + r 0.5 = E / 11 + r E = 5.5 + 0.5r 0.9 = E / 5 + r E = 4.5 + 0.9r on solving these equation, r = 2.5 Ω
26. A battery is charged at a potential of 15 V in 8 h when the current, flowing is 10A. The battery on discharge supplies a current of 5A for 15 h. The mean terminal voltage during discharge is 14V. The watt-hour efficiency of the battery is
A) 80 %
B) 90 %
C) 87.5 %
D) 82.5 %
Answer: C) 87.5 %
Explanation: The power of the battery, when charged is given by P₁ = V₁I₁ The electrical energy dissipated is given by Et = P₁t₁ E₁ = V₁I₁t₁ = 15 * 10 * 8 = 1200 Wh Similarly, the electrical energy dissipated during the discharge of battery is given by E₂ = V₂I₂t₂ = 14 * 5 * 15 = 1050 Wh Hence, watt-hour efficiency of the battery is given by η = E₂ / E₁ * 100 = 0.875 * 100 = 87.5%
B) 90 %
C) 87.5 %
D) 82.5 %
Answer: C) 87.5 %
Explanation: The power of the battery, when charged is given by P₁ = V₁I₁ The electrical energy dissipated is given by Et = P₁t₁ E₁ = V₁I₁t₁ = 15 * 10 * 8 = 1200 Wh Similarly, the electrical energy dissipated during the discharge of battery is given by E₂ = V₂I₂t₂ = 14 * 5 * 15 = 1050 Wh Hence, watt-hour efficiency of the battery is given by η = E₂ / E₁ * 100 = 0.875 * 100 = 87.5%
27. A circular current carrying coil has a radius R. The distance from the centre of the coil on the axis, where the magnetic induction will be 1/8th to its value at the centre of the coil is
A) R / √3
B) R √3
C) 2 √3 R
D) 2 / √3 R
Answer: B) R √3
Explanation: Bcentre / Baxis = ( 1 + x² / R²)3/2 Also, Baxis = 1 / 8 Bcentre 8 / 1 = ( 1 + x² / R²)3/2 4 = 1 + x² / R² 3 = x² / R² x² = 3R² x = √3 R
B) R √3
C) 2 √3 R
D) 2 / √3 R
Answer: B) R √3
Explanation: Bcentre / Baxis = ( 1 + x² / R²)3/2 Also, Baxis = 1 / 8 Bcentre 8 / 1 = ( 1 + x² / R²)3/2 4 = 1 + x² / R² 3 = x² / R² x² = 3R² x = √3 R
28. The incorrect statement regarding the lines of force of the magnetic field B is
A) magnetic intensity is a measure of lines of force passing through a unit area held normal to it
B) magnetic lines of force from a closed curve
C) inside a magnet, its magnetic lines of force move from the north pole of a magnetic towards its south pole
D) due to magnetic lines of force never cut each other
Answer: C) inside a magnet, its magnetic lines of force move from the north pole of a magnetic towards its south pole
Explanation: Inside a magnet, magnetic lines of force move from south pole to north pole
B) magnetic lines of force from a closed curve
C) inside a magnet, its magnetic lines of force move from the north pole of a magnetic towards its south pole
D) due to magnetic lines of force never cut each other
Answer: C) inside a magnet, its magnetic lines of force move from the north pole of a magnetic towards its south pole
Explanation: Inside a magnet, magnetic lines of force move from south pole to north pole
29. Two coils have a mutual inductance 0.55 H. The current changes in the first coil according to the equation I = I0 sin ωt. Where, I0 = 10A and ω = 100π rad/s. The maximum value of emf in the second coil is
A) 2 π
B) 5 π
C) π
D) 4 π
Answer: B) 5 π
Explanation: e = M di/dt = 0.0005 * d / dt (i0 sin ω t) = 0.005 * i0 cos ω t emax = 0.005 * 10 * 100π = 5π
B) 5 π
C) π
D) 4 π
Answer: B) 5 π
Explanation: e = M di/dt = 0.0005 * d / dt (i0 sin ω t) = 0.005 * i0 cos ω t emax = 0.005 * 10 * 100π = 5π
30. An L-C-R circuit contains R = 50Ω, L = 1 mH and C = 0.1μF. The impedance of the circuit will be minimum for a frequency of
A) 105 / 2 π Hz
B) 106 / 2 π Hz
C) 2 π * 105 Hz
D) 2 π * 106 Hz
Answer: A) 105 / 2 π Hz
Explanation: Impedance of L-C-R circuit will be minimum for a resonant frequency so, v0 = 1 / 2Ï€√LC = 1 / 2Ï€ √ 1 * 10-3 * 0.1 * 10-6 = 105 / 2Ï€ Hz
B) 106 / 2 π Hz
C) 2 π * 105 Hz
D) 2 π * 106 Hz
Answer: A) 105 / 2 π Hz
Explanation: Impedance of L-C-R circuit will be minimum for a resonant frequency so, v0 = 1 / 2Ï€√LC = 1 / 2Ï€ √ 1 * 10-3 * 0.1 * 10-6 = 105 / 2Ï€ Hz
31. An eye can detect 5 * 104 photons per square metre per sec of green light (λ = 5000 â„«) while the ear can detect 10-13 W/m². The factor by which the eye is more sensitive as a power detector then the ear is close to
A) 5
B) 10
C) 106
D) 15
Answer: A) 5
Explanation: E = 12375 / 5000 = 2.475 eV = 4 * 10-19 J So, the minimum intensity to which the eye can respond. Ieye = (photon flux) * energy of a photon Ieye = (5 * 104) * 4 * 10-19 = 2 * 10-14 (W/m²) Now, lesser the intensity required by a detector for detection more sensitive it will be = Iear / Ieye = 10-13 / 2 * 10-14 = 5
B) 10
C) 106
D) 15
Answer: A) 5
Explanation: E = 12375 / 5000 = 2.475 eV = 4 * 10-19 J So, the minimum intensity to which the eye can respond. Ieye = (photon flux) * energy of a photon Ieye = (5 * 104) * 4 * 10-19 = 2 * 10-14 (W/m²) Now, lesser the intensity required by a detector for detection more sensitive it will be = Iear / Ieye = 10-13 / 2 * 10-14 = 5
32. The amplification factor of a triode is 50. If the grid potential is decreased by 0.20 V. What increase, in plate potential will keep the plate current unchanged?
A) 5 V
B) 10 V
C) 0.2 V
D) 50 V
Answer: B) 10 V
Explanation: μ = ΔVp / ΔVg ⇒ ΔVp = -μ * ΔVs = -50(-20) = 10 V
B) 10 V
C) 0.2 V
D) 50 V
Answer: B) 10 V
Explanation: μ = ΔVp / ΔVg ⇒ ΔVp = -μ * ΔVs = -50(-20) = 10 V
33. If in nuclear fission, a piece of uranium of mass 5.0 g is lost, the energy obtained in kWh is
A) 1.25 * 107
B) 2.25 * 107
C) 3.25 * 107
D) 0.25 * 107
Answer: A) 1.25 * 107
Explanation: E = Δmc² = 0.5 * 10-3 * (3 * 108)² = 4.5 * 1013 E = 4.5 * 1013 / 3.6 * 106 = 1.25 * 107 kWh
B) 2.25 * 107
C) 3.25 * 107
D) 0.25 * 107
Answer: A) 1.25 * 107
Explanation: E = Δmc² = 0.5 * 10-3 * (3 * 108)² = 4.5 * 1013 E = 4.5 * 1013 / 3.6 * 106 = 1.25 * 107 kWh
34. Current in the circuit will be
A) 5 / 40 A
B) 5 / 50 A
C) 5 / 10 A
D) 5 / 20 A
Answer: B) 5 / 50 A
Explanation: The diode in lower branch is forward and diode in upper branch is reversed biased, so I = 5 / 20 + 30 = 5 / 50 A
B) 5 / 50 A
C) 5 / 10 A
D) 5 / 20 A
Answer: B) 5 / 50 A
Explanation: The diode in lower branch is forward and diode in upper branch is reversed biased, so I = 5 / 20 + 30 = 5 / 50 A
35. An installation consisting of an electric motor driving a water pump left 75 L of water per second to a height of 4.7 m. If the motor consumes a power of 5 kW, then the efficiency of the installation is
A) 39%
B) 69%
C) 93%
D) 96%
Answer: B) 69%
Explanation: Power consumed by the motor = 5 kW = 5 * 10³ W = 5000 W Power used in lifting the water = mgh / t = 7.5 * 9.8 * 4.7 = 3454.5kW Efficiency = Power used / power consumed * 100% = 3454.5 / 5000 * 100 = 69%
B) 69%
C) 93%
D) 96%
Answer: B) 69%
Explanation: Power consumed by the motor = 5 kW = 5 * 10³ W = 5000 W Power used in lifting the water = mgh / t = 7.5 * 9.8 * 4.7 = 3454.5kW Efficiency = Power used / power consumed * 100% = 3454.5 / 5000 * 100 = 69%
36. The potential difference across the terminals of a battery is 50 V when 11 A current is drawn and 60 V, when 1 A current is drawn. The emf and the internal resistance of the battery are
A) 62 V, 2 Ω
B) 63 V, 1 Ω
C) 61 V, 1 Ω
D) 64 V, 2 Ω
Answer: C) 61 V, 1 Ω
Explanation:
B) 63 V, 1 Ω
C) 61 V, 1 Ω
D) 64 V, 2 Ω
Answer: C) 61 V, 1 Ω
Explanation:
37. Beyond which frequency, the ionosphere bands any incident electromagnetic radiation but do not reflect it back towards the earth?
A) 50 MHz
B) 40 MHz
C) 30 MHz
D) 20 MHz
Answer: B) 40 MHz
Explanation: The ionosphere can reflect electromagnetic waves of frequency less than 40 MHz but do not reflect electromagnetic wave frequency more than 40 MHz
B) 40 MHz
C) 30 MHz
D) 20 MHz
Answer: B) 40 MHz
Explanation: The ionosphere can reflect electromagnetic waves of frequency less than 40 MHz but do not reflect electromagnetic wave frequency more than 40 MHz
38. A metallic surface ejects electrons. When exposed to green light of intensity I but no photoelectrons are emitted, when exposed to the yellow light of intensity I. It is possible to eject an electron from the same surface by
A) yellow light of the same intensity which is more than
B) green light of any intensity
C) red light of any intensity
D) None of the above
Answer: A) yellow light of the same intensity which is more than
Explanation: λred > λviolet VIBGYOR pattern shows that VIBG all have λ less than that of yellow colour and hence can initiate photoelectric effect irrespective of intensity
A) 2.9 cm
B) 3.9 cm
C) 2.35 cm
D) 2 cm
Answer: C) 2.35 cm
Explanation: The formula for radius of circular path is r = mV / B = V / (e/m)B r = 6 * 107 / 1.7 * 1011 * 1.5 * 10-2 = 2.35 * 10-2m r = 2.35m
A) 1.5 H
B) 2.5 H
C) 3.5 H
D) 0.5 H
Answer: A) 1.5 H
Explanation: Self-inductance of the solenoid L = μr ⋅ μ0 N² A / I = 600 * 4Ï€ * 10-7 * (2000)² * (1.5) * 10-4 / 0.3 = 1.5H
B) green light of any intensity
C) red light of any intensity
D) None of the above
Answer: A) yellow light of the same intensity which is more than
Explanation: λred > λviolet VIBGYOR pattern shows that VIBG all have λ less than that of yellow colour and hence can initiate photoelectric effect irrespective of intensity
39. An electron moves at a right angle to a magnetic field of 5 * 10-2 T with a speed of 6 * 107 m/s. If the specific charge of the electron is 1.7 * 1011 C/kg. The radius of the circular path will be
B) 3.9 cm
C) 2.35 cm
D) 2 cm
Answer: C) 2.35 cm
Explanation: The formula for radius of circular path is r = mV / B = V / (e/m)B r = 6 * 107 / 1.7 * 1011 * 1.5 * 10-2 = 2.35 * 10-2m r = 2.35m
40. A solenoid 30 cm long is made by winding 2000 loops of wire on an iron rod whose cross-section is 1.5 cm². If the relative permeability of the iron is 6000. What is the self-inductance of the solenoid?
B) 2.5 H
C) 3.5 H
D) 0.5 H
Answer: A) 1.5 H
Explanation: Self-inductance of the solenoid L = μr ⋅ μ0 N² A / I = 600 * 4Ï€ * 10-7 * (2000)² * (1.5) * 10-4 / 0.3 = 1.5H
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