BITSAT Practice Set 1
BITSAT · 2026 · with answers and explanations
- questions
- 30
- Total marks
- 89
- Duration
- 180 minutes
- Negative marking
- −1
Subjects: Physics · Chemistry · Mathematics
Question 1
Physics · 3 marks · −1A ball is thrown vertically upward with a velocity of 20 m/s from the top of a building 60 m high. The time taken by the ball to reach the ground is (g = 10 m/s²):
- A.2 s
- B.4 s
- C.6 s
- D.8 s
Correct answer: C
Explanation: Taking upward as positive: s = -60 m, u = 20 m/s, a = -10 m/s². Using s = ut + ½at²: -60 = 20t - 5t², so 5t² - 20t - 60 = 0, t² - 4t - 12 = 0, (t-6)(t+2) = 0. Since t must be positive, t = 6 s.
Question 2
Physics · 3 marks · −1A block of mass 5 kg is placed on a rough inclined plane making an angle of 30° with the horizontal. If the coefficient of friction is 1/√3, the acceleration of the block sliding down the plane is (g = 10 m/s²):
- A.5 m/s²
- B.2.5 m/s²
- C.0 m/s²
- D.g m/s²
Correct answer: C
Explanation: Along the incline: a = g(sinθ - μcosθ) = 10(sin30° - (1/√3)cos30°) = 10(0.5 - (1/√3)(√3/2)) = 10(0.5 - 0.5) = 0 m/s². The block is in equilibrium on the incline.
Question 3
Physics · 3 marks · −1A body of mass 4 kg moving with velocity 6 m/s collides with another body of mass 2 kg at rest. If the collision is perfectly inelastic, the kinetic energy lost in the collision is:
- A.12 J
- B.24 J
- C.36 J
- D.48 J
Correct answer: B
Explanation: Initial KE = ½(4)(6²) = 72 J. After perfectly inelastic collision, v = m₁u₁/(m₁+m₂) = 4×6/6 = 4 m/s. Final KE = ½(6)(4²) = 48 J. Loss = 72 - 48 = 24 J.
Question 4
Physics · 3 marks · −1A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of inclination θ. The acceleration of the centre of mass is:
- A.gsinθ
- B.gsinθ/2
- C.2gsinθ/3
- D.gsinθ/3
Correct answer: C
Explanation: For a solid cylinder rolling without slipping, a = gsinθ/(1 + I/(MR²)). For a solid cylinder, I = MR²/2, so a = gsinθ/(1 + 1/2) = 2gsinθ/3.
Question 5
Physics · 3 marks · −1The gravitational potential energy of a body of mass m at a height h above the surface of the Earth (radius R) is -mgR. The escape velocity from height h is (given h = R):
- A.vₑ/2
- B.vₑ/√2
- C.vₑ√2
- D.2vₑ
Correct answer: B
Explanation: At height h = R from the surface (distance 2R from centre), escape velocity = √(2GM/(2R)) = √(GM/R) = v_e/√2, where v_e = √(2GM/R) is the escape velocity from the surface.
Question 6
Physics · 3 marks · −1Three capacitors of 2 μF, 3 μF, and 6 μF are connected in series across a 12 V battery. The charge on each capacitor is:
- A.6 μC
- B.12 μC
- C.24 μC
- D.36 μC
Correct answer: B
Explanation: In series, 1/C_eq = 1/2 + 1/3 + 1/6 = (3+2+1)/6 = 1, so C_eq = 1 μF. Charge Q = C_eq × V = 1 × 12 = 12 μC. In series, each capacitor carries the same charge.
Question 7
Physics · 3 marks · −1A wire of resistance 12 Ω is bent into a circle. The effective resistance between two diametrically opposite points is:
- A.12 Ω
- B.6 Ω
- C.3 Ω
- D.1.5 Ω
Correct answer: C
Explanation: The wire becomes two semicircular arcs each of resistance 6 Ω, connected in parallel between the two diametrically opposite points. R_eff = 6×6/(6+6) = 3 Ω.
Question 8
Physics · 3 marks · −1A circular coil of 200 turns and radius 10 cm carries a current of 2 A. The magnetic field at the centre of the coil is:
- A.4π × 10⁻⁴ T
- B.8π × 10⁻⁴ T
- C.2π × 10⁻⁴ T
- D.16π × 10⁻⁴ T
Correct answer: B
Explanation: B = μ₀NI/(2R) = 4π×10⁻⁷ × 200 × 2/(2 × 0.1) = 4π×10⁻⁷ × 2000 = 8π×10⁻⁴ T ≈ 25.1×10⁻⁴ T.
Question 9
Physics · 3 marks · −1A concave mirror of focal length 20 cm produces a real image that is three times the size of the object. The object distance is:
- A.20 cm
- B.80/3 cm
- C.40 cm
- D.60 cm
Correct answer: B
Explanation: For a real image, magnification m = -3 (inverted). m = -v/u, so v = 3u. Using mirror formula 1/v + 1/u = 1/f with sign convention (u = -x, v = -3x, f = -20): -1/(3x) - 1/x = -1/20, so 4/(3x) = 1/20, x = 80/3 ≈ 26.67 cm.
Question 10
Physics · 3 marks · −1In a common-emitter transistor amplifier, the input resistance is 1 kΩ, the output resistance is 5 kΩ, and the current gain (β) is 100. The voltage gain of the amplifier is:
- A.100
- B.200
- C.500
- D.1000
Correct answer: C
Explanation: Voltage gain A_v = β × (R_out/R_in) = 100 × (5000/1000) = 500.
Question 11
Chemistry · 3 marks · −1How many moles of oxygen are present in 64 g of O₂?
- A.1
- B.2
- C.3
- D.4
Correct answer: B
Explanation: Molar mass of O₂ = 32 g/mol. Number of moles = 64/32 = 2 moles.
Question 12
Chemistry · 3 marks · −1The maximum number of electrons that can be accommodated in the 3d subshell is:
- A.6
- B.8
- C.10
- D.14
Correct answer: C
Explanation: For d subshell, l = 2 and m_l ranges from -2 to +2 giving 5 orbitals. Each orbital holds 2 electrons, so the maximum is 5 × 2 = 10 electrons.
Question 13
Chemistry · 3 marks · −1Among BF₃, NH₃, NF₃, and BCl₃, the molecules with zero dipole moment are:
- A.BF₃ and NH₃
- B.NH₃ and NF₃
- C.BF₃ and BCl₃
- D.BF₃ and NF₃
Correct answer: C
Explanation: BF₃ and BCl₃ are trigonal planar (sp² hybridised) with 120° bond angles, so their individual bond dipoles cancel out giving zero net dipole moment. NH₃ and NF₃ are pyramidal with lone pairs, giving non-zero dipole moments.
Question 14
Chemistry · 3 marks · −1The pH of a 0.001 M HCl solution is:
- A.1
- B.2
- C.3
- D.4
Correct answer: C
Explanation: HCl is a strong acid that completely dissociates. [H⁺] = 0.001 = 10⁻³ M. pH = -log[H⁺] = -log(10⁻³) = 3.
Question 15
Chemistry · 3 marks · −1The standard enthalpy of formation of CO₂(g) is -393.5 kJ/mol. The enthalpy change when 24 g of carbon (graphite) is completely burnt in excess oxygen is:
- A.-393.5 kJ
- B.-787.0 kJ
- C.-196.75 kJ
- D.-1574 kJ
Correct answer: B
Explanation: Moles of C = 24/12 = 2 mol. The reaction C(s) + O₂(g) → CO₂(g) releases 393.5 kJ/mol. For 2 mol: ΔH = 2 × (-393.5) = -787 kJ.
Question 16
Chemistry · 3 marks · −1In the electrochemical series, the metal that is the strongest reducing agent is:
- A.Sodium
- B.Potassium
- C.Lithium
- D.Caesium
Correct answer: C
Explanation: In the electrochemical series, lithium (Li) has the most negative standard reduction potential (E° = -3.04 V), making it the strongest reducing agent among common metals.
Question 17
Chemistry · 3 marks · −1The element that shows the maximum number of allotropes in the p-block is:
- A.Sulphur
- B.Carbon
- C.Phosphorus
- D.Tin
Correct answer: B
Explanation: Carbon shows the maximum number of allotropes among p-block elements, including diamond, graphite, fullerenes (C₆₀), carbon nanotubes, graphene, and amorphous carbon.
Question 18
Chemistry · 3 marks · −1The IUPAC name of the complex [Co(NH₃)₅Cl]Cl₂ is:
- A.Pentaamminechloridocobalt(III) chloride
- B.Chloridopentaamminecobalt(III) chloride
- C.Pentaamminechloridocobalt(II) chloride
- D.Pentachloridoamminecobalt(III) chloride
Correct answer: A
Explanation: The complex cation has 5 NH₃ (pentaammine) and 1 Cl⁻ (chlorido) ligands coordinated to Co³⁺. Following IUPAC nomenclature: anionic ligands first alphabetically, then neutral ligands. Name: pentaamminechloridocobalt(III) chloride.
Question 19
Chemistry · 3 marks · −1The number of isomers possible for the alkene C₄H₈ is:
- A.2
- B.3
- C.4
- D.5
Correct answer: B
Explanation: C₄H₈ (butene) has three structural isomers: but-1-ene (CH₂=CHCH₂CH₃), but-2-ene (CH₃CH=CHCH₃), and 2-methylpropene ((CH₃)₂C=CH₂). But-2-ene has cis and trans forms, so total = 4 stereoisomers, but 3 structural isomers.
Question 20
Chemistry · 3 marks · −1Which of the following amino acids is NOT essential for the human body?
- A.Valine
- B.Leucine
- C.Alanine
- D.Isoleucine
Correct answer: C
Explanation: Alanine is a non-essential amino acid as the human body can synthesise it. Valine, leucine, and isoleucine are all essential amino acids that must be obtained from the diet.
Question 21
Mathematics · 3 marks · −1If A = {1, 2, 3, 4} and B = {3, 4, 5, 6}, then A ∩ B is:
- A.{1, 2, 3, 4, 5, 6}
- B.{3, 4}
- C.{1, 2, 5, 6}
- D.{1, 2}
Correct answer: B
Explanation: The intersection A ∩ B contains elements common to both sets. Common elements are 3 and 4. So A ∩ B = {3, 4}.
Question 22
Mathematics · 3 marks · −1If one root of the equation x² - 5x + k = 0 is 2, then the value of k is:
- A.4
- B.5
- C.6
- D.7
Correct answer: C
Explanation: Substituting x = 2: (2)² - 5(2) + k = 0, so 4 - 10 + k = 0, giving k = 6.
Question 23
Mathematics · 3 marks · −1The coefficient of x³ in the expansion of (1 + x)¹ is:
- A.56
- B.84
- C.120
- D.126
Correct answer: B
Explanation: Using binomial theorem, the coefficient of x³ in (1+x)⁹ is C(9,3) = 9!/(3!6!) = (9×8×7)/(3×2×1) = 84.
Question 24
Mathematics · 2 marksThe sum of the first 20 terms of the arithmetic progression 3, 7, 11, 15, ... is:
- A.780
- B.820
- C.860
- D.900
Correct answer: B
Explanation: First term a = 3, common difference d = 4. S₂₀ = (20/2)[2(3) + (20−1)(4)] = 10[6 + 76] = 10 × 82 = 820.
Question 25
Mathematics · 3 marks · −1The value of sin²(π/6) + cos²(π/3) + tan²(π/4) is:
- A.1
- B.3/2
- C.2
- D.5/2
Correct answer: B
Explanation: sin(π/6) = 1/2, cos(π/3) = 1/2, tan(π/4) = 1. So sin²(π/6) + cos²(π/3) + tan²(π/4) = 1/4 + 1/4 + 1 = 3/2.
Question 26
Mathematics · 3 marks · −1The slope of the line passing through the points (2, 3) and (5, 9) is:
- A.1
- B.2
- C.3
- D.1/2
Correct answer: B
Explanation: Slope m = (y₂ - y₁)/(x₂ - x₁) = (9 - 3)/(5 - 2) = 6/3 = 2.
Question 27
Mathematics · 3 marks · −1The value of lim(x→0) (sin 5x)/(3x) is:
- A.1
- B.3/5
- C.5/3
- D.0
Correct answer: C
Explanation: lim(x→0) sin(5x)/(3x) = lim(x→0) [sin(5x)/(5x)] × (5x/3x) = 1 × 5/3 = 5/3.
Question 28
Mathematics · 3 marks · −1The value of ∫₀² (3x² + 2x) dx is:
- A.8
- B.10
- C.12
- D.14
Correct answer: C
Explanation: ∫(3x² + 2x)dx = x³ + x². Evaluating from 0 to 2: [2³ + 2²] - [0] = 8 + 4 = 12.
Question 29
Mathematics · 3 marks · −1A bag contains 5 red and 3 blue balls. If two balls are drawn at random without replacement, the probability that both are red is:
- A.5/14
- B.5/8
- C.25/64
- D.10/28
Correct answer: A
Explanation: P(both red) = (5/8) × (4/7) = 20/56 = 5/14.
Question 30
Mathematics · 3 marks · −1If A is a square matrix of order 3 such that |A| = 5, then |2A| is:
- A.10
- B.20
- C.40
- D.80
Correct answer: C
Explanation: |kA| = kⁿ|A| for a matrix of order n. So |2A| = 2³ × |A| = 8 × 5 = 40.