VITEEE Exam 2016 - Question Paper with Solutions
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VITEEE Exam 2016 - Question Paper with Solutions (Part 1)
1. The potential of the electric field produced by point charge at any point (x, y, and z) is given by V = 3x² + 5, where x, y are in metres and V is in volts. The intensity of the electric field at (-2,1, 0) is
A) +17 Vm-1
B) -17 Vm-1
C) +12 Vm-1
D) -12 Vm-1
A) +17 Vm-1
B) -17 Vm-1
C) +12 Vm-1
D) -12 Vm-1
Answer: D) -12 Vm-1
Explanation: Potential (V) = 3x² + 5 Intensity of the electric field = dV /dx = 6x ∴ E at x = -2 = 6(-2) = -12 V/m
2. The potential of a large liquid drop when eight liquid drops are combined is 20 V. Then the potential of every single drop was
A) 10 V
B) 7.5 V
C) 5 V
D) 2.5 V
2. The potential of a large liquid drop when eight liquid drops are combined is 20 V. Then the potential of every single drop was
A) 10 V
B) 7.5 V
C) 5 V
D) 2.5 V
Answer: C) 5 V
Explanation: Volume of 8 drops = Volume of big drop ∴ (4/3 Ï€r³) * 8 = 4 / 3 Ï€R³ ⇒ 2r = R ...(i) According to charge conservation 8q = Q ...(ii) Potential of one small drop (V') = q / 4πε0r Similarly, potential of big drop (V) = Q / 4πε0R Now, V' / V = q / Q * R / r ⇒ V' / 20 = q / 8q * 2r / r [from Eqs.(i) and (ii)] ∴ V' = 5V
3. A and B are two metals with threshold frequencies 1.8 * 1014 Hz and 2.2 * 104 Hz. Two identical photons of energy 0.825 eV each are incident on them. Then photoelectrons are emitted by (Take h = 6.6 * 10-34 J-s)
A) B alone
B) A alone
C) Neither A nor B
D) Both A and B
3. A and B are two metals with threshold frequencies 1.8 * 1014 Hz and 2.2 * 104 Hz. Two identical photons of energy 0.825 eV each are incident on them. Then photoelectrons are emitted by (Take h = 6.6 * 10-34 J-s)
A) B alone
B) A alone
C) Neither A nor B
D) Both A and B
Answer: B) A alone
Explanation: Threshold energy of A is EA = hvA = 6.6 * 10-34 * 1.8 * 1014 = 11.88 * 10-20 J = 11.88 * 10-20 / 1.6 * 10-19 eV = 0.74 eV Similarly, EB = 0.91 eV since, the incident photons have energy greater than EA but less than EB. So, photoelectrons will be emitted from metal A only.4. In the Wheatstone's network given, P = 10Ω, Q = 20Ω, R = 15Ω, S = 30Ω, the current passing through the battery (of negligible internal resistance) is
A) 0.36 A
B) Zero
C) 0.18 A
D) 0.72 A
B) Zero
C) 0.18 A
D) 0.72 A
Answer: A) 0.36 A
Explanation: The balanced condition for Wheatstone's bridge is P / Q = R / S as is obvious from the given values. No, current flows through galvanometer is zero. Now, P and R are in series, so Resistance R₁ = P + R = 10 + 15 = 25Ω Similarly, Q and S are in series, so Resistance R₂ = R + S = 20 + 30 = 50Ω Net resistance of the network as R₁ and R₂ are in parallel 1 / R = 1 / R₁ + 1 / R₂ ∴ R = 25 * 50 / 25 + 50 = 50 / 3 Ω Hence, I V / R = 6 / 50 / 3 = 0.36 A
5. Three resistors 1Ω, 2Ω and 3Ω are connected to form a triangle. Across 3Ω resistor a 3 V battery is connected. The current through the 3Ω resistor is
A) 0.75 A
B) 1 A
C) 2 A
D) 1.5 A
5. Three resistors 1Ω, 2Ω and 3Ω are connected to form a triangle. Across 3Ω resistor a 3 V battery is connected. The current through the 3Ω resistor is
A) 0.75 A
B) 1 A
C) 2 A
D) 1.5 A
Answer: B) 1 A
6. In a common emitter amplifier, the input signal is applied across
A) anywhere
B) emitter-collector
C) collector-base
D) base-emitter
A) anywhere
B) emitter-collector
C) collector-base
D) base-emitter
Answer: B) emitter-collector
Explanation: 
7. the kinetic energy of an electron get tripled then the de-Broglie wavelength associated with it changes by a factor
A) 1/3
B) √3
C) 1/√3
D) 3
A) 1/3
B) √3
C) 1/√3
D) 3
Answer: C) 1/√3
Explanation: de-Broglie wavelength of an electron is given by γ = h / mv = h / √2mK or γ ∞ 1 / √K ∴ γ' / γ = 1 / √3K = √K / 1 = 1 / √3 or γ' = γ / √3 Hence, de-Broglie wavelength will change by factor 1 / √3.
8. A radioactive substance contains 10000 nuclei and its half-life period is 20 days. The number of nuclei present at the end of 10 days is
A) 7070
B) 9000
C) 8000
D) 7500
8. A radioactive substance contains 10000 nuclei and its half-life period is 20 days. The number of nuclei present at the end of 10 days is
A) 7070
B) 9000
C) 8000
D) 7500
Answer: A) 7070
Explanation: We know, N / N0 = (1/2) t/T ⇒ N / 10000 = (1/2)10/20 ⇒ N = 10000 / √2 = 10000 / 1.414 = 7070
9. A direct X-ray photograph of the intestines is not generally taken by radiologists because
A) intestines would burst an exposure to X-rays
B) the X-rays would be not pass through the intestines
C) the X-rays will pass through the intestines without causing a good shadow for any useful diagnosis
D) a very small exposure of X-rays causes cancer in the intestines
9. A direct X-ray photograph of the intestines is not generally taken by radiologists because
A) intestines would burst an exposure to X-rays
B) the X-rays would be not pass through the intestines
C) the X-rays will pass through the intestines without causing a good shadow for any useful diagnosis
D) a very small exposure of X-rays causes cancer in the intestines
Answer: C) the X-rays will pass through the intestines without causing a good shadow for any useful diagnosis
10. Charge passing through a conductor of cross-section area A = 0.3 m² is given by q = 3t² + 5t + 2 in coulomb, where t is in second. What is the value of drift velocity at t = 2s? (Given, n = 2 * 1025 / m³)
A) 0.77 * 10-5 m/s
B) 1.77 * 10-5 m/s
C) 2.08 * 10-5 m/s
D) 0.57 * 10-5 m/s
Answer: B) 1.77 * 10-5 m/s
Explanation: A = 0.3 m² n = 2 * 1025 / m³ q = 3t² + 5t + 2 i = dq / dt = 6t + 5 = 17 i = neAvd Drift velocity, vd = i / neA = 17 / 2 * 1025 * 1.6 * 10-19 * 0.3 = 17 / 0.96 * 106 = 1.77 * 10-5 m/s
11. Two capacitors of capacities 1 μF and C μF are connected in series and the combination is charged to a potential difference of 120 V. If the charge on the combination is 80 μC, the energy stored in the capacitor of capacity C in μJ is
A) 1800
B) 1600
C) 14400
D) 7200
11. Two capacitors of capacities 1 μF and C μF are connected in series and the combination is charged to a potential difference of 120 V. If the charge on the combination is 80 μC, the energy stored in the capacitor of capacity C in μJ is
A) 1800
B) 1600
C) 14400
D) 7200
Answer: B) 1600
Explanation: Capacitances 1 μF and C μF are connected in series, then Ceq = C / 1+C Given, V = 120 V and q = 80 μC ∴ q = CeqV 80 = C / C + 1 * 20 or C = 2μF The energy stored in the capacitor of capacity C U = 1/2 q² / C = 1/2 * (80 * 10-6)² / 2 * 10-6; = 1/2 * 80 * 10-6 * 80 * 10-6 / 2 * 10-6 U = 1600 μJ
12. A hollow conducting sphere is placed in an electric field produced by a point charge placed at p as shown in the figure. Let VA, VB, VC be the potentials at points A, B and C respectively. Then
12. A hollow conducting sphere is placed in an electric field produced by a point charge placed at p as shown in the figure. Let VA, VB, VC be the potentials at points A, B and C respectively. Then
A) VC > VB
B) VB > VC
C) VA > VB
D) VA = VC
B) VB > VC
C) VA > VB
D) VA = VC
Answer: D) VA = VC
Explanation: Conducting surface behaves as an equipotential surface.
13. In a hydrogen discharge tube, it is observed that through a given cross-section 3.13 * 1015 electrons are moving from right to left and 3.12 * 1015 protons are moving from left to right. What is the electric current in the discharge tube and what is its direction?
A) 1 mA towards right
B) 1 mA towards left
C) 2 mA towards left
D) 2 mA towards right
13. In a hydrogen discharge tube, it is observed that through a given cross-section 3.13 * 1015 electrons are moving from right to left and 3.12 * 1015 protons are moving from left to right. What is the electric current in the discharge tube and what is its direction?
A) 1 mA towards right
B) 1 mA towards left
C) 2 mA towards left
D) 2 mA towards right
Answer: A) 1 mA towards right
Explanation: I = neqe + nPqE = 1 mA (towards right)
14. In CuSO₄ solution when electric current equal to 2.5 faradays is passed, the gm equivalent deposited on the cathode is
A) 1
B) 1.5
C) 2
D) 2.5
14. In CuSO₄ solution when electric current equal to 2.5 faradays is passed, the gm equivalent deposited on the cathode is
A) 1
B) 1.5
C) 2
D) 2.5
Answer: A) 1
Explanation: 1 faraday deposited 1 g equivalent
15. In hydrogen an atom, an electron is revolving in the orbit of radius 0.53 â„« with 6.6 * l015 radiations/s. The magnetic field produced at the centre of the orbit is
A) 0.125 Wb/m²
B) 1.25 Wb/m²
C) 12.5 Wb/m²
D) 125 Wb/m²
15. In hydrogen an atom, an electron is revolving in the orbit of radius 0.53 â„« with 6.6 * l015 radiations/s. The magnetic field produced at the centre of the orbit is
A) 0.125 Wb/m²
B) 1.25 Wb/m²
C) 12.5 Wb/m²
D) 125 Wb/m²
Answer: C) 12.5 Wb/m²
Explanation: The magnetic field B = μ0 / 4Ï€ ⋅ 2Ï€ (qv) / r = 10-7 * 2 * 3.14 * (1.6 * 10-19 * 1.6 * 1015) / 0.53 * 10-10 = 12.5 Wb / m³
16. The dipole moment of a short bar magnet is 12.5 A-m². The magnetic field on its axis at a distance of 0.5 m from the centre of the magnet is
A) 1.0 * 10-4 N/A-m
B) 4 * 10-2 N/A-m
C) 2 * 10-6 N/A-m
D) 6.64 * 10-8 N/A-m
16. The dipole moment of a short bar magnet is 12.5 A-m². The magnetic field on its axis at a distance of 0.5 m from the centre of the magnet is
A) 1.0 * 10-4 N/A-m
B) 4 * 10-2 N/A-m
C) 2 * 10-6 N/A-m
D) 6.64 * 10-8 N/A-m
Answer: C) 2 * 10-6 N/A-m
Explanation: The magnetic field B = μ0 / 4Ï€ ⋅ 2N/d³ = 10-7 * 2 * 1.25 / (0.5)³ = 2 * 10-6 N / A-m
17. The turn ratio of a transformer is given as 2:3. If the current through the primary coil is 3 A, thus calculate the current through the load resistance
A) 1 A
B) 4.5 A
C) 2 A
D) 1.5 A
17. The turn ratio of a transformer is given as 2:3. If the current through the primary coil is 3 A, thus calculate the current through the load resistance
A) 1 A
B) 4.5 A
C) 2 A
D) 1.5 A
Answer: C) 2 A
Explanation: IP / IS = nS / nS i.e., 3 / IS = 3 / 2 IS = 2 and
18. In an AC circuit, the potential across an inductance and resistance joined in series are respectively 16 V and 20 V. The total potential difference across the circuit is
A) 20.0 V
B) 25.6 V
C) 31.9 V
D) 33.6 V
18. In an AC circuit, the potential across an inductance and resistance joined in series are respectively 16 V and 20 V. The total potential difference across the circuit is
A) 20.0 V
B) 25.6 V
C) 31.9 V
D) 33.6 V
Answer: B) 25.6 V
Explanation: Voltage V = √V² R + V² C = √ (20)² + (16)² = √656 = 25.6 V
19. If a hydrogen atom is its ground state absorbs 10.2 eV of energy. The orbital angular momentum is increased by
A) 1.05 * 10-34 J/s
B) 3.16 * 10-34 J/s
C) 2.11 * 10-34 J/s
D) 4.22 * 10-34 J/s
19. If a hydrogen atom is its ground state absorbs 10.2 eV of energy. The orbital angular momentum is increased by
A) 1.05 * 10-34 J/s
B) 3.16 * 10-34 J/s
C) 2.11 * 10-34 J/s
D) 4.22 * 10-34 J/s
Answer: A) 1.05 * 10-34 J/s
Explanation: Electron after absorbing 10.2 eV energy goes to its first excited state (n = 2) from ground state (n = 1) ∴ Increase in momentum = h / 2Ï€ = 6.6 * 10-34 / 6.28 = 1.05 * 10-34 J-s
20. Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of the nucleus to that of the Helium nucleus is (14)1/3. The atomic number of the nucleus will be
A) 25
B) 26
C) 56
D) 30
20. Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of the nucleus to that of the Helium nucleus is (14)1/3. The atomic number of the nucleus will be
A) 25
B) 26
C) 56
D) 30
Answer: B) 26
Explanation: By using R = R0 A1/3 R₁ / R₂ = (A₁ / A₂)1/3 R / RHe = (A / 4)1/3 (14)1/3 = (A/4)1/3 A = 56 So, Z = 56 - 30 = 26
21. Each resistance shown in the figure is 2Ω. The equivalent resistance between A and B is
21. Each resistance shown in the figure is 2Ω. The equivalent resistance between A and B is
A) 2 Ω
B) 4 Ω
C) 8 Ω
D) 1 Ω
B) 4 Ω
C) 8 Ω
D) 1 Ω
Answer: A) 2 Ω
Explanation:
22. If in a triode value amplification factor is 20 and plate resistance is 10 kΩ, then its mutual conductance is
A) 2 milli mho
B) 20 milli mho
C) (1/2) milli mho
D) 200 milli mho
A) 2 milli mho
B) 20 milli mho
C) (1/2) milli mho
D) 200 milli mho
Answer: A) 2 milli mho
Explanation: Amplification factor μ = 20 Plate resistance RP = 10 kΩ = 10 * 10³ Ω ∴ Mutual conductance gm = μ / RP = 20 / 10 * 10³ = 2 * 10-3 mho = 2 milli mho
23. The output waveform of full-wave rectifier is
23. The output waveform of full-wave rectifier is
A) A
B) B
C) C
D) D
B) B
C) C
D) D
Answer: C) C
24. Calculate the energy released when three α-particles combined to form a 12C nucleus, the mass defect is (Atomic mass of ₂; He4 is 4.002603 u)
A) 0.007809 u
B) 0.002603 u
C) 4.002603 u
D) 0.5 u
Answer: A) 0.007809 u
Explanation: Mass defect Δm = Total mass of α - particles - mass of 12C nucleus = 3 * 4.002603 - 12 = 12.007809 - 12 = 0.007809 unit
25. In the figure shown, the magnetic field induction at point O will be
25. In the figure shown, the magnetic field induction at point O will be
A) μ0 i / 2πr
B) (μ0 / 4π) (i / r) (π + 2)
C) (μ0 / 4π)(i / r)(π + 1)
D) μ0 / 4π i / r (π - 2)
B) (μ0 / 4π) (i / r) (π + 2)
C) (μ0 / 4π)(i / r)(π + 1)
D) μ0 / 4π i / r (π - 2)
Answer: B) (μ0 / 4π) (i / r) (π + 2)
Explanation:
26. In the photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is
A) 1.3 V
B) 0.5 V
C) 2.3 V
D) 1.8 V
A) 1.3 V
B) 0.5 V
C) 2.3 V
D) 1.8 V
Answer: B) 0.5 V
Explanation: Stopping potential = Maximum KE eV = KEmax so, option (b) is correct
27. A current of 2 A flows through a 2Ω resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9Ω resistor. The internal resistance of the battery is
A) 1/3 Ω
B) 1/4 Ω
C) 1 Ω
D) 0.5 Ω
27. A current of 2 A flows through a 2Ω resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9Ω resistor. The internal resistance of the battery is
A) 1/3 Ω
B) 1/4 Ω
C) 1 Ω
D) 0.5 Ω
Answer: A) 1/3 Ω
Explanation: Current i = E / R + r 2 = E / 2 + r ... (i) 0.5 = E / 9 + r ... (ii) From Eqs. (i) and (ii), we have 2 / 0.5 = 9 + r / 2 + r 4 = 9 + r / 2 + r 3r = 1 r = 1/3 Ω
28. The current i is a coil varies with time as shown in the figure. The variation of induced emf with time would be
28. The current i is a coil varies with time as shown in the figure. The variation of induced emf with time would be
A) A
B) B
C) C
D) D
B) B
C) C
D) D
Answer: D) D
Explanation: We know e = -L di/dt During 0 to T / 4 , di / dt = constant So, e = -ve For T/4 to T/2 , di/dt = 0 e = 0 For T/2 to 3T/4, di/dt = constant e = +ve
29. A transistor is operated in common emitter configuration at VC = 2 V such that a change in the base current from 100 μA to 300 μA produces a change in the collector current from 10 mA to 20 mA. The current gain is
A) 75
B) 100
C) 25
D) 50
29. A transistor is operated in common emitter configuration at VC = 2 V such that a change in the base current from 100 μA to 300 μA produces a change in the collector current from 10 mA to 20 mA. The current gain is
A) 75
B) 100
C) 25
D) 50
Answer: D) 50
Explanation: β = ΔIC / ΔIB = (20 - 10) mA / (300 - 100) mA = 10 * 10-3 / 200 * 10-6 = 50
30. A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron
A) speed will decrease
B) speed will increase
C) will turn towards the left of the direction of motion
D) will turn towards the right of direction a motion
30. A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron
A) speed will decrease
B) speed will increase
C) will turn towards the left of the direction of motion
D) will turn towards the right of direction a motion
Answer: A) speed will decrease
Explanation: Field B has not applied only force. Field E will apply a force opposite to velocity of the electron hence, speed will decreases.
31. Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency Æ’ Hz. The magnitude of magnetic induction at the centre of the ring is
A) μ0q ƒ / 2R
B) μ0q / 2ƒ R
C) μ0q / 2πƒ R
D) μ0q ƒ / 2π R
31. Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency Æ’ Hz. The magnitude of magnetic induction at the centre of the ring is
A) μ0q ƒ / 2R
B) μ0q / 2ƒ R
C) μ0q / 2πƒ R
D) μ0q ƒ / 2π R
Answer: A) μ0q ƒ / 2R
Explanation: We know that β = μ0i / 2R q = it ⇒ i = q / t = q Æ’ β = μ0q Æ’ / 2R
32. A galvanometer of resistance, G is shunted by a resistance S ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is
A) S² / (S + G)
B) SG / (S + G)
C) G² / (S + G)
D) G / (S + G)
32. A galvanometer of resistance, G is shunted by a resistance S ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is
A) S² / (S + G)
B) SG / (S + G)
C) G² / (S + G)
D) G / (S + G)
Answer: C) G² / (S + G)
Explanation:
33. Three charges, each + q, are placed at the corners of an isosceles triangle ABC of sides BC and AC, 2a. D and E are the mid-points of BC and CA. The work done in taking a charge Q from D to E is
A) eqQ / 8πε0 a
B) qQ / 4πε0a
C) Zero
D) 3qQ / 4πε0a
B) qQ / 4πε0a
C) Zero
D) 3qQ / 4πε0a
Answer: C) Zero
Explanation:
34. A square loop, carrying a steady current I, is placed in a horizontal plane near a long straight conductor carrying a steady current I₁ at a distance d from the conductor as shown in the figure. The loop will experience
A) a net repulsive force away from the conductor
B) a net torque acting upward perpendicular to the horizontal plane
C) a net torque acting downward normal to the horizontal plane
D) a net attractive force towards the conductor
B) a net torque acting upward perpendicular to the horizontal plane
C) a net torque acting downward normal to the horizontal plane
D) a net attractive force towards the conductor
Answer: D) a net attractive force towards the conductor
Explanation:
35. The threshold frequency for a photo-sensitive metal is 3.3 * l014 Hz. If the light of frequency 8.2 * 1014 Hz is incident on this metal, the cut-off voltage for the photo-electric emission is nearly
A) 2 V
B) 3 V
C) 5 V
D) 1 V
A) 2 V
B) 3 V
C) 5 V
D) 1 V
Answer: A) 2 V
Explanation: V0 = E - V / e = h (v - v0) / e = 6.62 * 10-34 (8.2 * 1014 - 3.3 * 1014) / 1.6 * 10-19 = 6.62 * 10-34 / 1.6 * 4.9 * 1033 = 6.62 * 4.9 * 10-1 / 1.6 V0 = 2 volt
36. For the given circuit of p-n junction diode, which of the following statement is correct
36. For the given circuit of p-n junction diode, which of the following statement is correct
A) In forward biasing the voltage across R is V
B) In forward biasing the voltage across R is 2 V
C) In reverse biasing the voltage across R is V
D) In reverse biasing the voltage across R is 2 V
B) In forward biasing the voltage across R is 2 V
C) In reverse biasing the voltage across R is V
D) In reverse biasing the voltage across R is 2 V
Answer: A) In forward biasing the voltage across R is V
Explanation: In forward biasing, the resistance of p-n junction diode is zero, so the whole voltage appears across the resistance.
37. If the binding energy per nuclear in Li7 and He4 nuclei are respectively 5.60 MeV and 7.06 MeV, then the energy of reactor Li7 + P → 2 ₂He4 is
A) 19.6 MeV
B) 2.4 MeV
C) 8.4 MeV
D) 17.3 MeV
37. If the binding energy per nuclear in Li7 and He4 nuclei are respectively 5.60 MeV and 7.06 MeV, then the energy of reactor Li7 + P → 2 ₂He4 is
A) 19.6 MeV
B) 2.4 MeV
C) 8.4 MeV
D) 17.3 MeV
Answer: D) 17.3 MeV
Explanation: BE of Li7 = 39.20 MeV and He4 = 28.24 MeV Hence binding energy of ₂He4 = 56.84 MeV Energy of reaction = 56.84 - 39.20 = 17.28 MeV
38. The graph between the square root of the frequency of a specific line of characteristic spectrum of X-ray and the atomic number of the target will be
38. The graph between the square root of the frequency of a specific line of characteristic spectrum of X-ray and the atomic number of the target will be
A) A
B) B
C) C
D) D
B) B
C) C
D) D
Answer: B) B
Explanation: √V ∞ (Z - b)
39. A resistor R, an inductor L and capacitor C are connected in series to an oscillar of frequency n. If the resonant frequency is nr, then the current lags behind voltage, when
A) n = 0
B) n < nr
C) n = nr
D) n > nr
39. A resistor R, an inductor L and capacitor C are connected in series to an oscillar of frequency n. If the resonant frequency is nr, then the current lags behind voltage, when
A) n = 0
B) n < nr
C) n = nr
D) n > nr
Answer: D) n > nr
Explanation: The current will lag behind the voltage when reactance of inductance is more than the reactance of condenser. Thus ωL > 1 / ωc or ω > 1 / √LC or n > 1 / 2Ï€√LC or n > nr where nr = resonant frequency
40. A parallel plate capacitor has capacitance C. If it is equally filled with parallel layers of materials of dielectric constant K₁ and K₂ its capacity becomes C₁ The ratio of C₁ and C is
A) K₁ + K₂
B) K₁K₂ / K₁ + K₂
C) K₁ + K₂ / K₁K₂
D) 2 K₁K₂ / K₁ + K₂
40. A parallel plate capacitor has capacitance C. If it is equally filled with parallel layers of materials of dielectric constant K₁ and K₂ its capacity becomes C₁ The ratio of C₁ and C is
A) K₁ + K₂
B) K₁K₂ / K₁ + K₂
C) K₁ + K₂ / K₁K₂
D) 2 K₁K₂ / K₁ + K₂
Answer: D) 2 K₁K₂ / K₁ + K₂
Explanation: CA = K₁ε0A / d/2, CB = K₂ε0A / d/2 Ceq = C₁ / C₂ = 2K₁K₂ / K₁ + K₂ = CA CB / CA + CB = (2K₁K₂ / K₁ + K₂) ε0 A / d (∴ e = ε0 A / d)
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