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Question 1Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

A body starts from rest and moves with uniform acceleration. The ratio of the distances covered in the 1st, 2nd and 3rd seconds of its motion is:

  1. 1 : 3 : 5 Correct answer.
  2. 1 : 4 : 9
  3. 1 : 2 : 3
  4. 1 : 5 : 9

Explanation

For uniform acceleration from rest, the distance in the nth second is proportional to (2n − 1). So the 1st, 2nd and 3rd seconds give 1 : 3 : 5, the classic odd-number ratio derived from s = u + a(2n − 1)/2.

Question 2Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

A projectile is launched at speed u at angle θ to the horizontal on level ground. Its horizontal range is maximum when θ equals:

  1. 30°
  2. 45° Correct answer.
  3. 90°
  4. 60°

Explanation

Range R = u² sin2θ / g is largest when sin2θ = 1, i.e. 2θ = 90°, giving θ = 45°. At this angle the sine factor attains its maximum value of one.

Question 3Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

An ideal gas undergoes an isothermal reversible expansion at temperature T. Which statement about the gas is correct?

  1. Its pressure remains constant throughout the process.
  2. No heat is exchanged with the surroundings.
  3. Its internal energy decreases because work is done by the gas.
  4. Heat absorbed equals the work done by the gas, and internal energy is unchanged. Correct answer.

Explanation

For an ideal gas, internal energy depends only on temperature, so an isothermal process has ΔU = 0. The first law then gives q = W, meaning the heat absorbed equals the work done by the gas.

Question 4Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

The IUPAC name of CH3–CH(CH3)–CH2–CH2–OH is:

  1. 3-methylbutan-1-ol Correct answer.
  2. 2-methylbutan-4-ol
  3. 2-methylbutan-1-ol
  4. 3-methylbutan-4-ol

Explanation

Numbering must start from the end nearer the –OH group so it gets the lowest locant, making the carbon bearing OH position 1. The methyl branch then falls on carbon 3 of a four-carbon chain, giving 3-methylbutan-1-ol.

Question 5Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

If f(x) = x³ − 3x² + 4, the value of f′(2) is:

  1. 8
  2. 4
  3. 0 Correct answer.
  4. −4

Explanation

Differentiating gives f′(x) = 3x² − 6x. Substituting x = 2 gives 12 − 12 = 0, so the tangent at x = 2 is horizontal, indicating a stationary point there.

Question 6Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

A cell of emf 2 V and internal resistance 0.5 Ω is connected to an external resistance of 1.5 Ω. The terminal potential difference across the cell is:

  1. 1.5 V Correct answer.
  2. 2.0 V
  3. 0.5 V
  4. 1.0 V

Explanation

Current I = ε/(R + r) = 2/2 = 1 A. The terminal voltage is V = ε − Ir = 2 − 1(0.5) = 1.5 V, equivalently IR = 1 × 1.5.

Question 7Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

Which of the following sets of quantum numbers is not permitted for an electron in an atom?

  1. n = 2, l = 1, m = 0
  2. n = 3, l = 2, m = −2
  3. n = 2, l = 2, m = 1 Correct answer.
  4. n = 4, l = 0, m = 0

Explanation

The azimuthal quantum number l can take values from 0 to n − 1 only. For n = 2 the allowed l values are 0 and 1, so l = 2 is impossible regardless of m.

Question 8Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

In a Mendelian monohybrid cross between two heterozygous tall pea plants (Tt × Tt), the expected phenotypic ratio of tall to dwarf offspring is:

  1. 3 : 1 Correct answer.
  2. 9 : 3 : 3 : 1
  3. 1 : 2 : 1
  4. 1 : 1

Explanation

A Tt × Tt cross gives genotypes TT : Tt : tt in 1 : 2 : 1. Since T is dominant, both TT and Tt appear tall, so the phenotypic ratio is 3 tall : 1 dwarf.

Question 9Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

The value of the limit as x → 0 of (sin 5x)/x is:

  1. 5 Correct answer.
  2. 1
  3. Does not exist
  4. 0

Explanation

Write (sin 5x)/x as 5 · (sin 5x)/(5x). As x → 0 the factor (sin 5x)/(5x) → 1, so the limit is 5. This uses the standard result lim θ→0 (sin θ)/θ = 1.

Question 10Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

Which of the following aqueous solutions will have the highest boiling point, assuming complete dissociation and equal molality?

  1. 0.1 m CaCl2 Correct answer.
  2. 0.1 m glucose
  3. 0.1 m NaCl
  4. 0.1 m urea

Explanation

Boiling point elevation depends on the total number of particles, given by the van't Hoff factor i. CaCl2 gives i ≈ 3, NaCl gives i ≈ 2, and glucose and urea are non-electrolytes with i = 1, so CaCl2 produces the largest elevation.

Question 11Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

A particle of mass m moves in a circle of radius r with constant speed v. Its angular momentum about the centre of the circle is:

  1. mv/r
  2. mvr Correct answer.
  3. mv²/r
  4. zero

Explanation

Angular momentum L = mvr for motion perpendicular to the radius, since the velocity is tangential and the angle between r and v is 90°. The centripetal acceleration changes direction but not the magnitude of v, so L stays constant.

Question 12Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

The number of structural isomers of the molecular formula C4H10 is:

  1. 1
  2. 2 Correct answer.
  3. 4
  4. 3

Explanation

C4H10 corresponds to butane, which exists as n-butane (a straight chain) and 2-methylpropane (isobutane, a branched chain). No other skeleton is possible with four carbon atoms, so there are exactly two isomers.

Question 13Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

If the roots of the quadratic equation x² − 5x + k = 0 are equal, then the value of k is:

  1. 5
  2. −25/4
  3. 0
  4. 25/4 Correct answer.

Explanation

Equal roots require the discriminant to vanish: b² − 4ac = 25 − 4k = 0, giving k = 25/4. At this value the equation becomes a perfect square, (x − 5/2)² = 0.

Question 14Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

Light of wavelength λ in vacuum passes into a medium of refractive index 1.5. Inside the medium, the frequency and wavelength of the light respectively become:

  1. both frequency and wavelength reduced by 1.5
  2. unchanged frequency, wavelength λ/1.5 Correct answer.
  3. unchanged frequency, wavelength 1.5λ
  4. frequency reduced by 1.5, wavelength λ

Explanation

Refraction changes the speed of light to v = c/n but leaves the frequency fixed, since the source determines it. The wavelength therefore becomes λ/n = λ/1.5.

Question 15Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

Which of the following is the strongest oxidising agent among the given species, based on standard electrode potentials?

  1. Cu²⁺ (E° = +0.34 V)
  2. Fe²⁺ (E° = −0.44 V)
  3. Ag⁺ (E° = +0.80 V) Correct answer.
  4. Zn²⁺ (E° = −0.76 V)

Explanation

A species with a more positive reduction potential is more readily reduced and hence a stronger oxidising agent. Among the listed values, Ag⁺ has the highest reduction potential (+0.80 V), so it is the strongest oxidiser.

Question 16Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

The number of ways in which 4 distinct books can be arranged on a shelf is:

  1. 256
  2. 24 Correct answer.
  3. 16
  4. 4

Explanation

Arranging n distinct objects in a row gives n! permutations. For 4 books this is 4! = 4 × 3 × 2 × 1 = 24.

Question 17Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

An object is placed at the centre of curvature of a concave mirror. The image formed is:

  1. real, inverted and of the same size Correct answer.
  2. virtual, erect and magnified
  3. real, inverted and diminished
  4. formed at infinity

Explanation

For a concave mirror, an object at the centre of curvature C (distance 2f) produces an image at C itself. The image is real, inverted and the same size as the object, with magnification −1.

Question 18Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

In the reaction of benzene with a mixture of concentrated nitric acid and concentrated sulphuric acid, the electrophile that attacks the aromatic ring is:

  1. NO2⁺ Correct answer.
  2. SO3H⁺
  3. NO3⁻
  4. NO2

Explanation

Sulphuric acid protonates nitric acid, which loses water to generate the nitronium ion NO2⁺. This strongly electrophilic species attacks the electron-rich benzene ring to give nitrobenzene.

Question 19Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

The sum of the first 20 terms of the arithmetic progression 3, 7, 11, 15, … is:

  1. 820 Correct answer.
  2. 840
  3. 800
  4. 860

Explanation

Here a = 3 and d = 4, so the 20th term is 3 + 19 × 4 = 79. The sum is (n/2)(a + l) = 10 × (3 + 79) = 820.

Question 20Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

A body of mass 2 kg moving at 3 m/s collides head-on with a stationary body of mass 1 kg and they stick together. The common velocity after collision is:

  1. 1 m/s
  2. 2 m/s Correct answer.
  3. 1.5 m/s
  4. 3 m/s

Explanation

In a perfectly inelastic collision, momentum is conserved: (2 × 3) = (2 + 1)v, so v = 6/3 = 2 m/s. Kinetic energy is lost, but momentum of the system is conserved.

Question 21Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

Which of the following molecules has a non-zero net dipole moment?

  1. BF3
  2. CO2
  3. CCl4
  4. H2O Correct answer.

Explanation

H2O has a bent shape, so the two O–H bond dipoles do not cancel and the molecule is polar. CO2, CCl4 and BF3 are all highly symmetric (linear, tetrahedral and trigonal planar), so their bond dipoles cancel to give zero net moment.

Question 22Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

For a first-order reaction, the time required for the concentration to fall to half its initial value is:

  1. directly proportional to the initial concentration
  2. independent of the initial concentration Correct answer.
  3. inversely proportional to the square of the initial concentration
  4. directly proportional to the square of the rate constant

Explanation

For a first-order reaction, t½ = ln2 / k = 0.693/k, which contains no concentration term. The half-life is therefore constant regardless of starting concentration.

Question 23Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

A uniform rod of length L and mass M is rotated about an axis through one end perpendicular to its length. Its moment of inertia is:

  1. ML²
  2. ML²/12
  3. ML²/2
  4. ML²/3 Correct answer.

Explanation

For a uniform rod about its centre the moment is ML²/12; the parallel-axis theorem adds M(L/2)² = ML²/4. The total is ML²/12 + ML²/4 = ML²/3 about the end.

Question 24Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

The integral of x·e^x dx with respect to x equals:

  1. e^x(x − 1) + C Correct answer.
  2. x·e^x + C
  3. x²e^x/2 + C
  4. e^x(x + 1) + C

Explanation

Integrating by parts with u = x and dv = e^x dx gives x·e^x − ∫e^x dx = x·e^x − e^x + C = e^x(x − 1) + C. Differentiating this result returns x·e^x, confirming it.

Question 25Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

In a transformer, the primary has 500 turns and the secondary 50 turns. If the primary is supplied with 220 V AC, the secondary voltage is approximately:

  1. 44 V
  2. 22 V Correct answer.
  3. 2200 V
  4. 110 V

Explanation

For an ideal transformer, Vs/Vp = Ns/Np. Here Vs = 220 × (50/500) = 22 V, a step-down result because the secondary has fewer turns.

Question 26Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

A body starts from rest and moves with uniform acceleration 2 m/s². The distance covered by it in the 4th second of its motion is:

  1. 7 m Correct answer.
  2. 8 m
  3. 16 m
  4. 9 m

Explanation

The distance covered in the nth second is s_n = u + (a/2)(2n − 1). With u = 0, a = 2 m/s² and n = 4, s_4 = (2/2)(7) = 7 m. This formula subtracts the total displacement in (n−1) seconds from that in n seconds.

Question 27Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

A projectile is launched at 20 m/s at 30° to the horizontal (g = 10 m/s²). Its maximum height is:

  1. 10 m
  2. 5 m Correct answer.
  3. 20 m
  4. 2.5 m

Explanation

Maximum height H = u²sin²θ/(2g). Here sin 30° = 0.5, so H = (400 × 0.25)/(2 × 10) = 5 m. Only the vertical component of velocity determines the height reached.

Question 28Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

Two resistors of 6 Ω and 3 Ω are connected in parallel across a 6 V ideal battery. The total current drawn from the battery is:

  1. 3 A Correct answer.
  2. 2 A
  3. 0.67 A
  4. 9 A

Explanation

The parallel equivalent resistance is (6 × 3)/(6 + 3) = 2 Ω. The total current is I = V/R = 6/2 = 3 A. Alternatively, branch currents of 1 A and 2 A add to 3 A.

Question 29Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

A charged particle enters a uniform magnetic field with its velocity perpendicular to the field. Its path becomes a circle because the magnetic force:

  1. acts along the magnetic field direction
  2. increases the kinetic energy of the particle
  3. is always perpendicular to the velocity and does no work Correct answer.
  4. is parallel to the velocity and changes the speed

Explanation

The magnetic force F = qv × B is perpendicular to both v and B, so it acts as a centripetal force and changes only the direction of motion. Since it is perpendicular to velocity, it does zero work and the speed stays constant.

Question 30Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

In a Young's double-slit experiment the fringe width is β. If the separation between the slits is doubled while everything else is unchanged, the new fringe width is:

  1. 4β
  2. β/4
  3. 2β
  4. β/2 Correct answer.

Explanation

Fringe width is β = λD/d, so it is inversely proportional to slit separation d. Doubling d halves the fringe width, giving β/2.

Question 31Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

The work function of a metal is 2 eV. Light of photon energy 3.5 eV falls on it. The maximum kinetic energy of the emitted photoelectrons is:

  1. 5.5 eV
  2. 3.5 eV
  3. 0.5 eV
  4. 1.5 eV Correct answer.

Explanation

Einstein's photoelectric equation gives K_max = hν − φ. Substituting gives 3.5 − 2 = 1.5 eV. Emission occurs only because the photon energy exceeds the work function.

Question 32Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

Which of the following species is isoelectronic with the neon atom (10 electrons)?

  1. F
  2. Na⁺ Correct answer.
  3. Cl⁻
  4. Mg²⁺

Explanation

A sodium atom has 11 electrons, so losing one electron to form Na⁺ leaves 10 electrons, the same as neon. Isoelectronic species have equal total electron counts.

Question 33Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

The hybridisation of the carbon atom in the methyl carbocation (CH₃⁺) is:

  1. sp² Correct answer.
  2. sp
  3. sp³d
  4. sp³

Explanation

The carbocation carbon has three sigma bonds and an empty p orbital, giving a trigonal planar geometry with sp² hybridisation. The vacant p orbital makes the centre planar and highly electrophilic.

Question 34Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

For the equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the reaction is exothermic. According to Le Chatelier's principle, increasing the temperature will:

  1. have no effect on the equilibrium position
  2. shift the equilibrium towards the reactants Correct answer.
  3. shift the equilibrium towards the products
  4. increase the value of the equilibrium constant

Explanation

Heating an exothermic reaction supplies energy to the endothermic reverse direction, so the equilibrium shifts backwards towards N₂ and H₂. The forward reaction is favoured by removing heat, not by adding it.

Question 35Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

The pH of a 0.01 M aqueous solution of a strong monobasic acid is:

  1. 12
  2. 7
  3. 2 Correct answer.
  4. 1

Explanation

A strong acid dissociates completely, so [H⁺] = 0.01 = 10⁻² M. Then pH = −log(10⁻²) = 2. The concentration's exponent directly gives the pH for a decade value.

Question 36Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

In the complex [Co(NH₃)₆]Cl₃, the oxidation state of cobalt and the number of chloride ions ionisable in solution are respectively:

  1. +6 and 3
  2. +2 and 3
  3. +3 and 1
  4. +3 and 3 Correct answer.

Explanation

Six neutral NH₃ ligands contribute zero charge, so the +3 charge on the complex ion requires Co(III). The three Cl⁻ ions lie outside the coordination sphere and are all ionisable, precipitating as AgCl.

Question 37Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

Which of the following will undergo an SN2 reaction fastest with a given nucleophile?

  1. 2-Bromo-2-methylpropane
  2. Bromomethane Correct answer.
  3. 2-Bromobutane
  4. 2-Bromopropane

Explanation

SN2 proceeds through a backside attack, so steric hindrance at the carbon slows it. Bromomethane is primary with no bulky substituents, making it the fastest substrate in this series.

Question 38Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

The IUPAC name of CH₃CH(OH)CH₂CH₃ is:

  1. 2-Methylpropan-1-ol
  2. Butan-2-ol Correct answer.
  3. Butan-3-ol
  4. Butan-1-ol

Explanation

The four-carbon chain is numbered to give the hydroxyl group the lowest locant, which is 2 from either end. Hence the name is butan-2-ol.

Question 39Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

If the matrix A = [[2, 1], [1, 2]], then the eigenvalues of A are:

  1. 0 and 4
  2. 1 and 3 Correct answer.
  3. 2 and 2
  4. −1 and 5

Explanation

The characteristic equation is det(A − λI) = (2 − λ)² − 1 = 0, giving (2 − λ) = ±1. Thus λ = 1 and λ = 3, whose sum equals the trace 4 and product equals the determinant 3.

Question 40Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

The value of ∫₀^(π/2) sin²x dx is:

  1. π/4 Correct answer.
  2. π
  3. π/2
  4. 1

Explanation

Using sin²x = (1 − cos 2x)/2, the integral becomes [x/2 − sin 2x/4] from 0 to π/2, which evaluates to π/4. The symmetry of sin² and cos² over a quarter period also gives this value.

Question 41Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

The derivative of sin⁻¹x with respect to x, for −1 < x < 1, is:

  1. −1/√(1 − x²)
  2. 1/(1 + x²)
  3. 1/√(1 − x²) Correct answer.
  4. √(1 − x²)

Explanation

The standard result is d(sin⁻¹x)/dx = 1/√(1 − x²) on the open interval where the function is differentiable. This follows from differentiating x = sin y implicitly.

Question 42Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

The eccentricity of the ellipse x²/25 + y²/9 = 1 is:

  1. 5/4
  2. 16/25
  3. 3/5
  4. 4/5 Correct answer.

Explanation

Here a² = 25 and b² = 9, so a = 5 and b = 3. Eccentricity e = √(a² − b²)/a = √16/5 = 4/5, which lies between 0 and 1 as required for an ellipse.

Question 43Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5 without repetition, with the condition that the number is even?

  1. 24
  2. 60
  3. 120
  4. 48 Correct answer.

Explanation

An even number must end in 2 or 4, giving 2 choices for the units digit. The remaining three places are filled from the other four digits in 4 × 3 × 2 = 24 ways, so the total is 2 × 24 = 48.

Question 44Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

In a binomial distribution with n = 10 and p = 0.5, the mean and variance are respectively:

  1. 5 and 5
  2. 10 and 5
  3. 5 and 2.5 Correct answer.
  4. 2.5 and 5

Explanation

For a binomial distribution, mean = np = 10 × 0.5 = 5 and variance = np(1 − p) = 10 × 0.5 × 0.5 = 2.5. The variance is smaller than the mean unless p = 0.5 gives equality, which is not the case here.

Question 45Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

A bag contains 3 red and 2 blue balls. Two balls are drawn at random without replacement. The probability that both are red is:

  1. 1/2
  2. 3/10 Correct answer.
  3. 9/25
  4. 2/5

Explanation

The probability is (3/5) × (2/4) = 6/20 = 3/10, or equivalently C(3,2)/C(5,2) = 3/10. Without replacement, the second draw uses the reduced total.

Question 46Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

The general solution of the differential equation dy/dx = y is:

  1. y = C + x
  2. y = Cx
  3. y = Ce^x Correct answer.
  4. y = C ln x

Explanation

Separating variables gives dy/y = dx, and integrating gives ln y = x + C′, so y = Ce^x. The constant C is determined by an initial condition.

Question 47Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

The domain of the function f(x) = √(x − 3) is:

  1. (−∞, ∞)
  2. [3, ∞) Correct answer.
  3. (3, ∞)
  4. (−∞, 3]

Explanation

The expression under a real square root must be non-negative, so x − 3 ≥ 0, giving x ≥ 3. The endpoint 3 is included because √0 = 0 is defined.

Question 48Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

If the vectors a = 2i + 3j − k and b = i − j + 2k, then a · b equals:

  1. −3 Correct answer.
  2. −7
  3. 3
  4. 7

Explanation

The dot product is (2)(1) + (3)(−1) + (−1)(2) = 2 − 3 − 2 = −3. Multiply corresponding components and add the results.

Question 49Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

A survey of 100 students found that 60 study Mathematics, 50 study Physics, and 30 study both. The number of students who study at least one of the two subjects is:

  1. 110
  2. 90
  3. 80 Correct answer.
  4. 70

Explanation

By the principle of inclusion–exclusion, n(M ∪ P) = n(M) + n(P) − n(M ∩ P) = 60 + 50 − 30 = 80. Subtracting the intersection avoids double counting.

Question 50Integrated First Degree (B.E., M.Sc. & B.Pharm.) admission test

A uniform solid cylinder of mass M and radius R rolls without slipping down an incline. Its acceleration compared with that of a frictionless sliding block on the same incline is:

  1. smaller, because rotational kinetic energy takes part of the gravitational energy Correct answer.
  2. larger, because rolling adds rotational kinetic energy to the motion
  3. zero, because static friction prevents motion
  4. equal, since both have the same mass and gravity acts equally

Explanation

For a rolling cylinder, a = g sinθ/(1 + I/MR²) = (2/3)g sinθ, whereas a sliding block has a = g sinθ. Part of the potential energy becomes rotational kinetic energy, leaving less for translation.