VITEEE Exam 2015 - Download Question Paper with Solutions (Part 1)

VITEEE Exam 2015 - Question Paper with Solutions

Download VITEEE Exam 2015 previous year Mathematics question paper with solutions. To get admission in VIT University, students need to pass the VITEEE Exam for BTECH programmes. VITEEE Exam 2015 previous year Mathematics question paper will help you to easily prepare for the exam.

To get admission in Vellore Institute of Technology(VIT) for all the campuses (Vellore, Chennai, Bhopal, Amaravati), students need to clear the VITEEE examination, that's one in each of the toughest exams to enter into one in each of the foremost effective engineering university in associate country, VITEEE test 2015 previous year question papers with solutions will facilitate students to rearrange for the examination.

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VITEEE Exam 2015 - Download Question paper with Solutions (Part 2)

VITEEE Exam 2015 - Download Question paper with Solutions (Part 3)

VITEEE Exam 2019 - Previous year question paper, sample paper, mock test, syllabus

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So in this post, I will give you the VITEEE EXAM 2015 question paper with solutions and this is part 1 blog and you will get 40 questions in this blog. Remaining 80 questions of VITEEE EXAM 2015 will be available in the other 2 parts. I'm also a student and I know how hard it is to get admission and pass the online entrance exam. So that's why I'm giving you the Previous year question paper with solutions to easily pass the VITEEE EXAM.

There will be 40 questions out of 120 questions. Correct answer and explanation for each question are available after each question. Download links for this question paper will be available at the end of the page. Prepare well and Don't leave any questions.

VITEEE Exam 2015 - Question Paper with Solutions (Part 1)

1. Equal charges q each are placed at the vertices of an equilateral triangle of side r. The magnitude of electric field intensity at any vertex is

A) 2q / 4πε0r²

B) q / 4πε0r²

C) √3 q / 4πε0r²

D) √2 q / 4πε0r²

Answer: C) √3 q / 4πε0r²

Explanation: 

2. Two points masses, m each carrying charges -q and +q are attached to the ends of a massless rigid non-conducting wire of length 'L'. When this arrangement is placed in a uniform electric field, then it deflects through an angle i. The minimum time needed by a rod to align itself along the field is

A) 2π√mL/qE

B) π/2√mL/2qE

C) π√2mL/qE

D) 2π√3mL/qE

Answer: B) π/2√mL/2qE

Explanation: 

3. A condenser of capacitance C is fully charged by a 200V supply. It is then discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat 250 J/kg-K and of mass 100 g. If the temperature of the block rises by 0.4 K, then the value of C is

A) 300μF

B) 200μF

C) 400μF

D) 500μF

Answer: D) 500μF

Explanation: Given that V = 200 V m = 100g = 0.1 kg s = 250 Jkg-1K-1 Δθ = 0.4K C = ? The energy stored in the capacitor is given by U = 1/2 CV² = 1/2 C * (200)² = 2C * 104 J This energy is used to heat up the block. Let Δθ be the rise in temperature, then heat energy is given by Q = msΔθ = 0.1 * 2.50 * 0.4 = 10 J Now, 2C * 104 = 10 ⇒ C = 10 / 2 * 104 = 5 * 10-4 ∴ C = 500μF

4. The capacitance of a parallel plate capacitor with air as a medium is 3μF. As a dielectric is introduced between the plates, the capacitance becomes 15μF. The permittivity of the medium in C²N-1m-2 is

A) 8.15 * 10-11

B) 0.44 * 10-10

C) 15.2 * 1012

D) 1.6 * 10-14

Answer: B) 0.44 * 10-10

Explanation: The capacitance of air capacitor is given by C0 = ε0A / d = 3μF ...(i) When a dielectric of permittivity εr and dielectric constant K is introduced between the plates, then C = Kε0A / d = 15μF ...(ii) Dividing Eq. (ii) by Eq.(i), we get C / C0 = Kε0A / d / ε0A / d = 15 / 3 ⇒ K = 5 So, the prermittivity of the medium εr = ε0K = 8.85 * 10-12 * 5 = 0.44 * 10-10

5. The masses of three copper wires are in the ratio 2:3:5 and their lengths are in the ratio 5:3:2. Then, the ratio of their electrical resistances is

A) 1:9:15

B) 2:3:5

C) 5:3:2

D) 125:30:8

Answer: D) 125:30:8

Explanation: As we know that R = p(1/A) So R₁ : R₂ : R₃ = 1₁ / A₁ : 1₂ / A₂ : 1₃ / A₃ = 1²₁ / V₁ : 1²₂ / V₂ : 1²₃ / V₃ = 1²₁ / (m₁d) : 1²₂ / (m₂d) : 1²₃ / (m₃d) = 1²₁ / m₁ : 1²₂ / m₂ : 1²₃ / m₃ = 5² / 2 : 3² / 3 : 2² / 5 = 25 / 2 : 3: 4 / 5 = 125 : 30 : 8

6. A 30V-90W lamp is operated on a 120 V DC line. A resistor is connected in series with the lamp in order to glow it properly. The value of resistance is

A) 10 Ω

B) 30 Ω

C) 20 Ω

D) 40 Ω

Answer: B) 30 Ω

Explanation: The resistance of lamp is given by R0 = V² / P = (30)² / 90 = 10 Ω The current in the lamp will be I = V/R0 = 30/10 = 3A As the lamp is operated on 120 V DC, then resistence becomes R' = V'/i = 120/3 = 40Ω For proper glow, a resistence R is put in series with the bulb so that R' = R + R0 ⇒ R = R' - R0 = 40 - 10 = 30Ω

7. In a potentiometer experiment, the balancing length of a cell is 560 cm. When an external resistance of 10 Ω is connected in parallel to the cell, the balancing length changes by 60 cm. The internal resistance of a cell is 

A) 1.4Ω

B) 1.6Ω

C) 0.12Ω

D) 1.2Ω

Answer: D) 1.2Ω

Explanation: Consider a cell of emf E and balancing length 1₁ E = k11 The potential difference is balanced by length 1₂. V = k1₂ The internal resistance of the cell is given by r = (E - V / V)R = (E / V - 1)R = (1₁/1₂ - 1)R = (560 / 560 - 60 - 1)10 = (56/50 - 1)10 = 6/5 = 1.2Ω

8. Two sources of equal emf are connected to a resistance R. The internal resistance of these sources are r₁ and r₂ (r₁ > r₂). If the potential difference across the source having internal resistance r₂ is zero, then

A) R = r₁r₂ / r₂ - r₁

B) X R = r₂(r₁ + r₂ / r₂ - r₁)

C) R = r₁r₂ / r₂ + r₁

D) R = r₂ - r₁

Answer: D) R = r₂ - r₁

Explanation: Let E be the emf of each source. When they are connected in series, then the current in the circuit is given by I = Etot / Rtot = E + E / r₁ + r₂ + R = 2E / r₁ + r₂ + R So, potential drop across the cell of internal resistance r₂ is (2E / r₁ + r₂ + R)r₂ Hence, E - 2E / ( r₁ + r₂ + R) r₂ = 0 r₁ + r₂ + R = 2r₂ So, R = r₂ - r₁

9. An electron of mass 9.0x 10-31kg under the action of a magnetic field moves in a circle of radius 2 cm at a speed of 3 x 106 m/s. If a proton of mass 1.8 x 1027kg was to move in a circle of the same radius in the same magnetic field, then its speed will become

A) 1.5 * 103 m/s

B) 3 * 106 m/s

C) 6 * 104 m/s

D) 2 * 106 m/s

Answer: A) 1.5 * 103 m/s

Explanation: Here, the magnetic force will provide the necessary centripetal force ∴ Bqv = mv² / r Bqr = mv For electron and proton, the magnetic field B, charge q and radius r, all same. So, mv = constant i.e. meve = mp vp vp = (me / mp)ve = (9 * 10-31 / 1.8 * 10-27)3 * 106 vp = 1.5 * 10³ m/s

10. A horizontal rod of mass 0.01kg and length 10 cm is placed on a frictionless plane inclined at an angle 60° with the horizontal and with the length of rod parallel to the edge of the inclined plane. A uniform magnetic field is applied vertically downwards. If the current through the rod is 1.73 A, then the value of magnetic field induction B for which the rod remains stationary on the inclined plane is

A) 1 T

B)3 T

C) 2.5 T

D) 4 T

Answer: A) 1 T

Explanation: 

11. A current of 2 A is flowing in the sides of an equilateral triangle of side 9 cm. The magnetic field at the centroid of the triangle is

A) 1.66 * 10-5 T

B) 1.22 * 10-4 T

C) 1.33 * 10-5 T

D) 1.44 * 10-4 T

Answer: C) 1.33 * 10-5 T

Explanation: 

12. The direction of magnetic field dB due to current element dl at a distance r is the direction of

A) r * dl

B) dl * r

C) (rdl)r^

D) dl

Answer: B) dl * r

Explanation: The direction of dB is the direction of vector dl * r. From right-hand screw rule, if we place a right-handed screw at the point where the magnetic field is needed to be determined and turn its handle from dl to r, then the direction in which the screw advances gives the direction of the field dB.

13. A galvanometer with a scale divided into 100 equal divisions has a current sensitivity of 10 divisions per milliampere and voltage sensitivity of 2 divisions per millivolt. The galvanometer resistance will be

A) 4 Ω

B) 5 Ω

C) 3 Ω

D) 7 Ω

Answer: B) 5 Ω

Explanation: Here, the current sensitivity is 10 div/mA and there are 100 division on the scale. So, the current required for full scale deflection is Ig = 1/10 * 100 mA = 10 mA = 0.01A Also the voltage sensitivity is 2 div/mV so the voltage required for full scale deflection is Vg = 1/2 * 100 mV = 0.05V Galvanometer resistance is given by G = Vg / Ig = 0.05/0.01 = 5Ω

14. The earth is considered as a short magnet with its centre coinciding with the geometric centre of the earth. The angle of dip ∅ related to the magnetic latitude λ as

A) tan ∅ = 1 / 2 tan α

B) tan λ = 2 tan ∅

C) tan λ = 2 tan ∅

D) tan ∅ = 2 tan λ

Answer: D) tan ∅ = 2 tan λ

Explanation: 

15. Which of the following statement related to the hysteresis loop is incorrect?

A) The curve of B against H for a ferromagnetic material is called hysteresis loop

B) The area of B-H curve is a measure of power dissipated per cycle per unit area of the specimen

C) Coercitivity is a measure of the magnetic field required to destroy the residual magnetism of ferromagnetic material

D) The retentivity of a specimen is the measure of magnetic field remaining in the specimen when the magnetising field is removed

Answer: B) The area of B-H curve is a measure of power dissipated per cycle per unit area of the specimen

Explanation: The hysteresis loop i.e. area of B-H curve is a measure of energy dissipated per cycle per unit volume of the specimen. It depends on the nature of the magnetic material.

16. A magnetic needle lying parallel to the magnetic field requires W units of work to turn it through an angle of 45°. The torque required to maintain the needle in this position will be

A) √2 W

B) 1 / √3 W

C) (√2 -1)W

D) W / (√2 -1)

Answer: D) W / (√2 -1)

Explanation: Work done by a magnet to turn from angle θ₁ to θ₂ is W = MB(cosθ₁ - cosθ₂) = MB(cos 0° - cos 45°) = MB(1 - 1/√2) = (√2 - 1 / √2)MB Also torque acting on the magnet is given by τ = MB sin 45° = MB / √2 So W = (√2 - 1) ⋅ τ ∴ τ = W / (√2 - 1)

17. An induced emf has

A) a direction same as field direction

B) a direction opposite to the field direction

C) no direction of its own

D) None of the above

Answer: C) no direction of its own

Explanation: As we know from Lenz's law, the direction of induced emf in a circuit is such that it opposes the magnetic flux that produces it. So, if the magnetic flux linked with a closed circuit increases the induced current flows in a direction so as to develop a magnetic flux in the opposite direction of original flux. If the magnetic flux linked with a closed circuit decreases, then the induced current flows in a direction to develop a magnetic flux in the same direction of original flux. So, the induced emf has no direction of its own.

18. A coil of area 5 cm² having 20 turns is placed in a uniform magnetic field of 10³ gauss. The normal to the plane of coil makes an angle 30° with the magnetic field. The flux through the coil is

A) 6.67 * 10-4 wb

B) 3.2 * 10-5 wb

C) 5.9 * 10-4 wb

D) 8.65 * 10-4 wb

Answer: D) 8.65 * 10-4 wb

Explanation: Given that N = 20 B = 10³ gauss = 10³ * 10-4 T = 0.1T A = 5cm² = 5 * 10-4 m² θ = 80° ∴ The flux through the coil is given by ∅ = NBA cosθ = 20 * 0.1 * 5* 10-4 * cos 30° = 10 * 10-4 * √3 / 2 = 5√3 * 10 -4 = 8.65 * 10-4 wb

19. The current graph for resonance in the LC circuit is

A) A

B) B

C) C

D) D

Answer: C) C

Explanation: In LC circuit , if XL = XC then ω = 1 / √LC I0 = ∞, so Z = E0 / I0 = 0. As 1/√LC is the natural frequency of LC circuit, therefore for an LC circuit if the frequency of applied AC becomes equal to the natural frequency of an AC circuit then the amplitude of current becomes infinite due to zero impedance. Hence, the correct diagram is an option (c).

20. The value of inductance L for which the current is maximum in series LCR circuit with C = 10 μF and ω = 1000 rad/s

A) 10 mH

B) 50 mH

C) 200 mH

D) 100 mH

Answer: D) 100 mH

Explanation: In resonance condition, maximum current flows in the circuit. Current in the LCR circuit is given by i = V / √ R² + (XL - XC)² For current to be maximum denominator should be minimum ( XL - XC)² = 0 ⇒ XL = XC ⇒ ωL = 1 / ωC ∴ L = 1 / ω²C = 1 / (1000)² * 10 * 10-6 = 1 /10 H ∴ L = 0.1 H = 100 mH

21. A ray of light is incident on a plane mirror at an angle of 30°. At what angle with the horizontal must a plane mirror be placed so that the reflected ray becomes vertically upwards?

A) 40°

B) 20°

C) 30°

D) 60°

Answer: C) 30°

Explanation: 

22. A compound microscope having magnifying power 35 with its eye-piece of focal length 10 cm. Assume that the final image is at least distance of distinct vision then the magnification produced by the objective is 

A) -4

B) 5

C) 10

D) -10

Answer: D) -10

Explanation: For a compound microscope, the magnifying power is given by MP = me * m0 When the final image is at least distance of distance vision then Me = 1 + D/fe So, MP = m0[1 + D/fe] ⇒ -35 = mo[1 + 25/10] ⇒ -35 = m0 * 3.5 ⇒ m0 = -10 The ngative sign indicates that the image formed by objective is inverted.

23. The refractive index for a prism is given as μ = cot A / 2. Then, the angle of minimum deviation in terms of the angle of the prism is

A) 90° - A

B) 2A

C) 180° -A

D) 180° -2A

Answer: D) 180° -2A

Explanation: The refractive index of a prism is given by μ = sin(A + δm)/2 / sin(A/2) where, A = angle of prism δm = angle of minimum deviation Given, μ = cot(A/2) = cos(A/2) / sin(A/2) So, from Eq.(i), cos(A/2) / sin(A/2) = sin(A + δm / 2) / sin(A/2) ⇒ sin(π/2 - A/2) = sin(A/2 + δm / 2) ⇒ δm = π - 2A δm = 180° - 2A

24. Two convex lenses of power 2D and 5D are separated by a distance (1/3)m. The power of the optical system formed is

A) + 2 D

B) - 2 D

C) - 3 D

D) + 3 D

Answer: D) + 3 D

Explanation: Given, P₁ = 2D P₂ = 3D d = 1/3 m We know that 1/F = 1/f₁ + 1/f₂ - d/f₁ . f₂ ∴ P = P₁ + P₂ - dP₁ . P₂ = 2 + 3 - 1/3 * 2 * 3 ⇒ P = 3D

25. Two light rays having the same wavelength in the vacuum are in phase initially. Then, the first ray travels a path L₁, through a medium of refractive index μ₁ while the second ray travels a path L₂ through a medium of refractive index μ₂. The two waves are then combined to observe interference. The phase difference between the two waves is

A) 2π / λ (L₁ / μ₁ - L₂ / μ₂)

B) 2π / λ (L₂ - L₁)

C) 2π / λ (μ₂L₁ - μ₁L₂)

D) 2π / λ (μ₁L₁ - μ₂L₂)

Answer: D) 2π / λ (μ₁L₁ - μ₂L₂)

Explanation: Optical path for first ray = μ₁L₁ Optical path for second ray = μ₂L₂ So, phase difference is given by Δ∅ = 2π / λ * path difference = 2π / λ * Δx ⇒ Δ∅ = 2π / λ(μ₁L₁ - μ₂L₂)

26. Two polaroids are kept crossed to each other. If one of them is rotated an angle of 60°, the percentage of incident light now transmitted through the system is

A) 10%

B) 20%

C) 25%

D) 12.5%

Answer: D) 12.5%

Explanation: Let I0 be the intensity of unpolarized light, so the intensity of the first polaroid is I0 / 2. On rotating through 60° , the intensity of light from second polaroid is I = (I0 / 2)(cos60)² = I0 / 2 1 / 4 = I0 / 8 = 0.125I0 So, percentage of incident light transmitted through the system is 12.5%.

27. An electromagnetic wave propagating along north lies its electric field vertically upward. The magnetic field vector points towards

A) downward

B) east

C) north

D) south

Answer: B) east

Explanation: As the electromagnetic wave is the crossed field of electric and magnetic waves, So, the direction of propagation of EM wave is the direction of vector E * B. Here E is upward and (E * B) is towards the north. So, from right-hand thumb rule B will be along the east.

28. Pick out the wrong statement.

A) A

B) B

C) C

D) D

Answer: C) C

Explanation: An electromagnetic wave is that type of wave which is radiated by an accelerated charge and which propagates through space as coupled electric and magnetic field. These fields are oscillating perpendicular to each other. Hence, the option (c) is wrong.

29. When sunlight is scattered by atmospheric atoms and molecules the amount of scattering of light of wavelength 880nm is A. Then, the amount of scattering of light of wavelength 330 nm is approximately

A) 10 A

B) 20 A

C) 40 A

D) 50.5 A

Answer: D) 50.5 A

Explanation: We know from Rayleigh's law of scattering I ∝ 1 / λ4 So, I₁ / I₂ = (λ₂ / λ₁)4 ⇒ A / I₂ = (330 / 880)4 = (3/8)4 = 81 / 4096 I₂ = 4096 / 81 A = (50.557)A

30. The ratio of the volume occupied by an atom to the volume of the nucleus is

A) 105 :1

B) 1020:1

C) 1015:1

D) 1:1015

Answer: C) 1015:1

Explanation: Radius of an atom = IA ≈ 10-10 m Radius of nucleus = IN ≈ 10-15 m So, ratio of their volumes is VA / VN = 4/3πrA³ / 4/3πrN³ = (rA / rN)³ = (10-10 / 10-15)³ ⇒ VA : VN = 1015 : 1

31. When a hydrogen atom is raised from ground energy level to excited energy level, then

A) potential energy increases and kinetic energy decreases

B) kinetic energy increases and potential energy decreases

C) Both KE and PE increase

D) Both KE and PE decrease

Answer: A) potential energy increases and kinetic energy decreases

Explanation: As r increases, the potential energy increases. Thus, it decreases kinetic energy of the hydrogen atom. So, when an atom jumps from one energy level to the higher level, its potential energy increases and kinetic energy decreases.

32. The half-life for α-decay of uranium 92U228 is 4.47 * 108 yr. If a rock contains 60% of original 92U228 atoms, then its age is [take log 6 = 0.778, log 2 = 0.3]

A) 1.2 * 107 yr

B) 3.3 * 108 yr

C)  4.2 * 109 yr

D) 6.5 * 109 yr

Answer: B) 3.3 * 108 yr

Explanation: Given T1/2 = 4.47 * 108 yr N/N0 = 60 / 100 = (1/2)n ⇒ 2n = 10/6 Apply logarithm on both sides nlog2 = log10 - log6 ⇒ n * 0.3 = 1 - 0.778 = 0.22 n = 0.222 / 0.3 = 0.74 So t = nT1/2 = 0.74 * 4.47 * 108 t = 33 * 108 yr

33. A nuclear transformation is given by Y (n,α) → 3Li7. The nucleus of element Y is

A) 5Be11

B) 5B10

C) 5B9

D) 6C12

Answer: B) 5B10

Explanation: Y(n,α) means the nucleus splits into α-particle and neutrons i.e. zYA + 0n¹ → ₃Li7 + ₂He4 So A + 1 = 7 + 4 ⇒ A = 10 and Z + 0 = 3 + 2 Z = 5 Hence, the nucleus of element Y is boron i.e. 5Y10 = 5B10

34. The angular momentum of an electron in Bohr's hydrogen atom whose energy is -3.4eV, is 

A) 5h / 2π

B) h / 2π

C) h / π

D) 2h / 3π

Answer: C) h / π

Explanation: Energy of electron in nth orbit of hydrogen atom is given by En = -(13.6 / n²)eV ⇒ 3.4 = -(13.6 / n²) ⇒ n² = 4 ⇒ n = 2 The angular momentum of electron is given by L = nh / 2π = 2h / 2π = h / π

35. When the momentum of a photon is changed by an amount p' then the corresponding change in the de-Broglie wavelength is found to be 0.20%. Then, the original momentum of the photon was

A) 300 p'

B) 500 p'

C) 400 p'

D) 100 p'

Answer: B) 500 p'

Explanation: λ = h / p ∴ λ ∝ 1 / p ⇒ Δp / p = -Δλ / λ ∴ |Δp / p| = |Δλ / λ| ⇒ p' / p = 0.20 / 100 = 1 / 500 ⇒ p = 500p'

36. Suppose a beam of electrons with each electron having energy 0 incident on a metal surface kept in an evacuated chamber. Then,

A) electrons can be emitted with any energy with a maximum of E0

B) no electrons will be emitted as only photons can emit electrons

C) electrons can be emitted but all with an energy E0

D) electrons can be emitted with any energy with a maximum of E0 - ∅ , where ∅ being a work function

Answer: A) electrons can be emitted with any energy with a maximum of E0

Explanation: The emitted electrons may lie near the surface and so can have a maximum amount of energy E0. If they are from deep inside, then energy is less than E0.

37. An n-type semiconductor is

A) neutral

B) positively charged

C) negatively charged

D) negatively or positively charged depending on the amount of impurity added

Answer: A) neutral

Explanation: The n-type semiconductor has an excess of free electrons for conduction. These electrons are unbound. Since the total number of electrons in an atom is equal to the total number of protons in the nucleus. So, an n-type semiconductor is neutral.

38. In the half-wave rectifier circuit operating with 50 Hz mains frequency. The fundamental frequency in the ripple will be

A) 100 Hz

B) 20 Hz

C) 50 Hz

D) 25 Hz

Answer: C) 50 Hz

Explanation: In half wave rectifier, the output is obtained for half cycle only. Therefore, the frequency of the ripple is the same as that of the input i.e. 50 Hz.

39. The input resistance of a common emitter amplifier is 330 Ω and the load resistance is 5 kΩ. A change of base current is 15 μA results in the change of collector current by 1 mA. The voltage gain of the amplifier is 

A) 1000

B) 10001

C) 1010

D) 1100

Answer: C) 1010

Explanation: Given ΔIC = 1mA = 10-3 A ΔIb = 15μA = 15 * 10-6 A RL = 5 k Ω = 5 * 10³ Ω R₁ = 330 Ω The voltage gain of an amplifier is given by Ar = ΔIC * RL / ΔIb * Ri = 10-3 * 5 * 10³ / 15 * 10-6 * 330 = 100000 / 99 ≈ 1010

40. To get an output y = 0 from the circuit shown in the figure, the input C must be

A) 0

B) 1

C) either 0 or 1

D) None of these

Answer: A) 0

Explanation: The output of OR gate is Y = A + B The output of AND gate is Y' = Y ⋅ C ⇒ Y' = (A + B) ⋅ C If C = 0 irrespective of A and B, then output Y must be zero.

Download link: VITEEE EXAM 2015 Question paper with Solutions

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