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Daily Practice Sheet — 50 Questions
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Daily MCQ Paper — 14 April 2026
50 questions across all sections. Use the practice interface to attempt; review answers and explanations after submission.
- Q1. A spring of force constant k is compressed by x. The elastic PE stored is
- kx
- (1/2)kx
- (1/2)kx²
- kx²
- Q2. For a satellite in circular orbit of radius r around Earth (mass M), orbital velocity is
- sqrt(GM/r)
- sqrt(2GM/r)
- GM/r
- sqrt(GMr)
- Q3. Time constant of an RC charging circuit is
- R/C
- RC
- R+C
- 1/(RC)
- Q4. The work-energy theorem states that the net work done on a particle equals
- Change in its momentum
- Change in its kinetic energy
- Change in its potential energy
- Change in its total energy
- Q5. Escape velocity from Earth's surface (R, g) is
- sqrt(gR)
- sqrt(2gR)
- sqrt(gR/2)
- 2sqrt(gR)
- Q6. Kepler's third law states T² is proportional to
- a
- a²
- a³
- 1/a
- Q7. Power delivered by a constant force F to a body moving with velocity v is
- Fv
- Fv²
- F·v (dot product)
- F/v
- Q8. Total mechanical energy of a satellite in circular orbit of radius r is
- -GMm/r
- -GMm/(2r)
- +GMm/(2r)
- -GMm/(4r)
- Q9. Kirchhoff's junction rule is a consequence of
- Conservation of energy
- Conservation of charge
- Conservation of momentum
- Ohm's law
- Q10. In a balanced Wheatstone bridge (P,Q,R,S), the condition is
- P/Q = R/S
- P/R = Q/S
- P·Q = R·S
- PQRS = 1
- Q11. The efficiency of a Carnot engine working between T1 (hot) and T2 (cold) is
- 1 − T2/T1
- T1/T2 − 1
- T2/T1
- (T1−T2)/T2
- Q12. On a p-V diagram, the area under an isothermal expansion equals
- Internal energy change
- Work done by the gas
- Heat absorbed only
- Both work and heat (since ΔU=0)
- Q13. For an ideal gas in an isothermal process
- dU = 0
- dQ = 0
- dW = 0
- dS = 0
- Q14. Magnetic dipole moment of a current loop of N turns, area A and current I is
- NIA
- I/A
- NA/I
- N/IA
- Q15. The self-inductance of a long solenoid (n turns/m, area A, length l) is
- mu0*n*A*l
- mu0*n^2*A*l
- mu0*n^2*A/l
- mu0*n*A^2*l
- Q16. Visible light has wavelength range approximately
- 100-400 nm
- 400-700 nm
- 700-1000 nm
- 1-100 nm
- Q17. The rms speed of gas molecules of molar mass M at temperature T is
- sqrt(RT/M)
- sqrt(2RT/M)
- sqrt(3RT/M)
- sqrt(8RT/πM)
- Q18. The diagonal relationship in periodic table is between
- Li-Be
- Li-Mg
- Be-B
- Na-Mg
- Q19. LiCl is more covalent than NaCl primarily due to
- Smaller size of Li+
- Higher electronegativity of Li
- Polarising power of Li+ (Fajans rules)
- Li being lighter
- Q20. Diborane (B₂H₆) has bonding involving
- Only 2c-2e bonds
- Two 3c-2e (banana) bonds + four 2c-2e bonds
- Only 3c-2e bonds
- Hydrogen bonds only
- Q21. Borax bead test is based on formation of
- Sodium meta-borate
- Coloured metaborates of transition metals
- Sodium tetraborate
- Boric acid
- Q22. The most stable allotrope of carbon at standard conditions is
- Diamond
- Graphite
- Fullerene
- Amorphous carbon
- Q23. Zeolites belong to the class of
- Borosilicates
- Aluminosilicates
- Carbonates
- Phosphates
- Q24. First transition series elements typically show
- Only +2 oxidation state
- Variable oxidation states due to participation of (n-1)d electrons
- Only +3
- +1 only
- Q25. The intense colour of transition-metal complexes is due to
- s-s transitions
- p-p transitions
- d-d transitions
- f-f transitions
- Q26. The catalytic activity of d-block metals is attributed to
- Inert pair effect
- Variable oxidation states + ability to provide adsorption sites
- Diamagnetism
- Lanthanoid contraction
- Q27. The element with strongest reducing nature in the first transition series in aqueous solution (E°) is
- Sc
- Cr
- Mn
- Cu
- Q28. Tyndall effect is shown by
- True solutions
- Colloidal solutions
- Suspensions
- Both colloidal and suspensions
- Q29. The bond enthalpy of H-H is approximately (kJ/mol)
- 436
- 347
- 414
- 498
- Q30. Faraday's first law of electrolysis states
- m ∝ Q
- m ∝ V
- m ∝ I²
- m ∝ 1/Q
- Q31. Van't Hoff factor i for NaCl in water (assuming complete dissociation) is
- 1
- 2
- 3
- 4
- Q32. The osmotic pressure equation is
- pi = nRT
- pi*V = nRT
- pi = nRT/V
- pi*V = R/T
- Q33. The Henderson-Hasselbalch equation for a buffer is
- pH = pKa + log([salt]/[acid])
- pH = pKa − log([salt]/[acid])
- pH = pKa + log([acid]/[salt])
- pH = pKw − pKa
- Q34. For a spontaneous process at constant T and P
- ΔG > 0
- ΔG = 0
- ΔG < 0
- ΔS = 0
- Q35. The cross product a×b yields a vector
- Parallel to a
- Parallel to b
- Perpendicular to both a and b
- In the plane of a and b
- Q36. The dot product a·b of two vectors a and b is
- |a||b|sinθ
- |a||b|cosθ
- |a×b|
- |a|+|b|
- Q37. The equation of a line through point A(a) parallel to vector b in vector form is
- r = a
- r = a + λb
- r·b = 0
- r×b = a
- Q38. The shortest distance between two skew lines r = a₁+λb₁ and r = a₂+μb₂ is
- |(a₂-a₁)·(b₁×b₂)| / |b₁×b₂|
- |a₂-a₁|
- |b₁×b₂|
- |(a₂-a₁)×(b₁×b₂)|
- Q39. If A is a 3×3 matrix with |A| ≠ 0, then A⁻¹ equals
- adj(A)/|A|
- |A|·adj(A)
- adj(A)·|A|²
- (adj A)⁻¹
- Q40. The scalar triple product [a b c] gives
- Area of parallelogram
- Volume of parallelepiped
- A vector
- Zero always
- Q41. Bayes theorem expresses P(A|B) as
- P(A)P(B)/P(A∩B)
- P(B|A)P(A)/P(B)
- P(A∪B)/P(B)
- P(A)+P(B)-P(A∩B)
- Q42. For a binomial distribution B(n,p), mean and variance are
- np, np
- np, npq
- n²p, npq
- np, p²q
- Q43. The system AX = B has a unique solution if
- |A| = 0
- |A| ≠ 0
- rank(A) < n
- A is singular
- Q44. If two rows (or columns) of a determinant are interchanged, the determinant
- Remains same
- Becomes zero
- Changes sign
- Doubles
- Q45. If y = f(g(x)), then dy/dx equals
- f'(g(x))
- g'(x)
- f'(g(x))·g'(x)
- f'(x)·g'(x)
- Q46. The locus of a point equidistant from a fixed point and a fixed line is a
- Circle
- Parabola
- Ellipse
- Hyperbola
- Q47. The equation of a plane in normal form is
- x + y + z = 0
- lx + my + nz = p
- ax + by + c = 0
- x = 0
- Q48. If A is a 3×3 matrix with |A| = 5, then |adj A| =
- 5
- 25
- 125
- 1/5
- Q49. Equation of parabola with focus (a,0) and directrix x = -a is
- y² = 4ax
- x² = 4ay
- x² = -4ay
- y² = -4ax
- Q50. Sum of squares of first n natural numbers is
- n(n+1)/2
- n(n+1)(2n+1)/6
- [n(n+1)/2]²
- n²