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Key Facts: BCECE Exam

Cutoff

Qualifying Mark

Varies by year/category

300

Total Questions

100 per subject paper

90m

Time per Paper

1.5 hours per subject

₹1,000

Application Fee

For General/OBC

+4 / -1

Marking Scheme

With negative marking

The BCECE is a state-level offline entrance exam conducted by BCECEB for admission into professional Pharmacy, Agriculture, and Health Science courses in Bihar. Candidates are tested in 3 papers (PCM or PCB), each containing 100 multiple-choice questions with a +4/-1 marking scheme and a 90-minute time limit per paper. Standard application fee is ₹1,000 for general/OBC candidates.

Sample BCECE Practice Questions

Try these sample questions to test your BCECE exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Which of the following physical quantities has the same dimensional formula as that of Planck's constant (h)?
A.Linear momentum
B.Angular momentum
C.Torque
D.Force
Explanation: Planck's constant has the dimensions of [M L^2 T^-1]. Angular momentum is defined as L = r x p, which also has the dimensions [L] * [M L T^-1] = [M L^2 T^-1]. Therefore, both Planck's constant and angular momentum share the same dimensional formula.
2A projectile is thrown with an initial velocity of 20 m/s at an angle of 30 degrees with the horizontal. Taking g = 10 m/s^2, what is the maximum height attained by the projectile?
A.5.0 meters
B.10.0 meters
C.15.0 meters
D.20.0 meters
Explanation: The maximum height H is given by the formula H = (u^2 * sin^2(theta)) / (2g). Plugging in the given values: H = (20^2 * sin^2(30)) / (2 * 10) = (400 * 0.25) / 20 = 100 / 20 = 5.0 meters.
3A force F = (3i + 4j) N acts on a particle and displaces it from position r1 = (i + j) m to position r2 = (3i + 5j) m. What is the work done by the force on the particle?
A.10 Joules
B.14 Joules
C.22 Joules
D.25 Joules
Explanation: The displacement vector dr is r2 - r1 = (3i + 5j) - (i + j) = (2i + 4j) m. Work done is the dot product of force and displacement: W = F . dr = (3i + 4j) . (2i + 4j) = (3 * 2) + (4 * 4) = 6 + 16 = 22 Joules.
4A bullet of mass 10 g is fired horizontally from a gun of mass 4 kg with a muzzle velocity of 400 m/s. What is the recoil velocity of the gun?
A.-1.0 m/s
B.-2.0 m/s
C.-4.0 m/s
D.-10.0 m/s
Explanation: By conservation of linear momentum, the total momentum before firing is zero, so m1*v1 + m2*v2 = 0. Here, m1 = 0.01 kg (bullet mass), v1 = 400 m/s, and m2 = 4 kg. Thus, (0.01 * 400) + 4 * v2 = 0, which gives 4 + 4*v2 = 0, leading to v2 = -1.0 m/s (opposite direction to the bullet).
5A block of mass 5 kg is placed on a rough inclined plane making an angle of 30 degrees with the horizontal. If the coefficient of static friction is 0.6, what is the frictional force acting on the block? (Take g = 10 m/s^2)
A.25 N
B.26 N
C.30 N
D.43 N
Explanation: First, find the force pulling the block down the incline: F_down = m * g * sin(theta) = 5 * 10 * sin(30) = 25 N. The maximum limit of static friction is f_max = mu_s * m * g * cos(theta) = 0.6 * 5 * 10 * cos(30) = 30 * 0.866 = 25.98 N. Since F_down (25 N) is less than f_max (25.98 N), the block remains at rest, and the static frictional force exactly balances the downward force. Thus, the static friction force is 25 N.
6A solid cylinder and a hollow sphere, having the same mass and radius, are rotating about their respective standard axes of symmetry. What is the ratio of their moments of inertia (I_cylinder / I_sphere)?
A.3:4
B.4:5
C.5:4
D.3:5
Explanation: The moment of inertia of a solid cylinder of mass M and radius R about its longitudinal axis is I_cylinder = (1/2) * M * R^2. The moment of inertia of a hollow sphere about its diameter is I_sphere = (2/3) * M * R^2. The ratio I_cylinder / I_sphere is (1/2) / (2/3) = 3/4.
7If the radius of the Earth shrinks by 1% while its mass remains constant, how will the acceleration due to gravity (g) at the surface of the Earth change?
A.Decrease by 1%
B.Increase by 1%
C.Decrease by 2%
D.Increase by 2%
Explanation: The acceleration due to gravity is g = G * M / R^2. Differentiating this relation gives the fractional change: dg/g = -2 * dR/R. Since the radius shrinks, dR/R = -1% (-0.01). Therefore, dg/g = -2 * (-1%) = +2%. The value of g increases by 2%.
8According to Kepler's Third Law of planetary motion, the square of the period of revolution (T) of a planet is directly proportional to the cube of the semi-major axis of its orbit (R). If the distance of a planet from the Sun becomes 4 times its present value, what will its new orbital period become?
A.2 times the original period
B.4 times the original period
C.8 times the original period
D.16 times the original period
Explanation: Kepler's law states that T^2 is proportional to R^3, which means T is proportional to R^(3/2). If the distance R becomes 4R, the new period T' becomes proportional to (4R)^(3/2) = 4^(3/2) * R^(3/2) = 8 * T. Thus, the orbital period becomes 8 times its original value.
9A solid sphere of mass M and radius R rolls down an inclined plane of inclination theta without slipping. What is the linear acceleration of the sphere down the incline?
A.g * sin(theta)
B.(5/7) * g * sin(theta)
C.(2/3) * g * sin(theta)
D.(3/5) * g * sin(theta)
Explanation: For a rolling body down an incline, the linear acceleration is given by a = g * sin(theta) / (1 + I / (M * R^2)). For a solid sphere, I = (2/5) * M * R^2. Substituting this in the formula: a = g * sin(theta) / (1 + 2/5) = g * sin(theta) / (7/5) = (5/7) * g * sin(theta).
10A particle of mass m moving with speed v collides head-on elastically with another particle of mass 2m at rest. What is the fraction of kinetic energy lost by the colliding particle of mass m?
A.1/9
B.4/9
C.8/9
D.1/3
Explanation: Using elastic collision equations, the final velocity v1' of the first particle is v1' = (m1 - m2)/(m1 + m2) * v1. Here, m1 = m, m2 = 2m, and v1 = v. So, v1' = (m - 2m)/(m + 2m) * v = -v/3. The final kinetic energy is KE_f = (1/2) * m * (-v/3)^2 = (1/9) * ((1/2) * m * v^2) = (1/9) * KE_i. The kinetic energy lost by the first particle is KE_i - KE_f = (8/9) * KE_i. Thus, the fraction of kinetic energy lost is 8/9.

About the BCECE Exam

State-level entrance examination in Bihar for admission to undergraduate courses in Pharmacy, Agriculture, Health Sciences, and allied professional fields.

Questions

300 scored questions

Time Limit

4.5 hours total (90 mins per paper)

Passing Score

Cutoff percentile (varies by course and category)

Exam Fee

₹1,000 (Bihar Combined Entrance Competitive Examination Board (BCECEB))

BCECE Exam Content Outline

33.3%

Physics

Mechanics, Thermodynamics, Waves & Oscillations, Electromagnetism, Optics, Modern Physics

33.3%

Chemistry

Physical Chemistry, Inorganic Chemistry, Organic Chemistry

33.3%

Mathematics / Biology

Algebra, Calculus, Coordinate Geometry, Probability, Vectors / Plant & Human Physiology, Cell Structure, Genetics, Ecology

How to Pass the BCECE Exam

What You Need to Know

  • Passing score: Cutoff percentile (varies by course and category)
  • Exam length: 300 questions
  • Time limit: 4.5 hours total (90 mins per paper)
  • Exam fee: ₹1,000

Keys to Passing

  • Complete 500+ practice questions
  • Score 80%+ consistently before scheduling
  • Focus on highest-weighted sections
  • Use our AI tutor for tough concepts

BCECE Study Tips from Top Performers

1Focus on the NCERT textbooks for Class 11 and 12, as the entire syllabus is closely aligned with them.
2Since there is a negative marking of -1 for every wrong answer, avoid blind guessing on questions where you are unsure.
3Practice solving numerical questions in Physics and Physical Chemistry to improve speed and calculation accuracy.
4Allocate equal revision time to all three chosen subjects, as each subject paper contains 100 questions and carries 400 marks.
5Take timed mock tests of 90 minutes per subject to mimic actual exam conditions and improve time management.

Frequently Asked Questions

What is the BCECE exam?

BCECE (Bihar Combined Entrance Competitive Examination) is a state-level entrance test conducted by the Bihar Combined Entrance Competitive Examination Board (BCECEB). It is used for admissions into professional degree courses in Pharmacy, Agriculture, Public Health (Nursing, Physiotherapy, Occupational Therapy), and other allied health science fields in government and private institutions in Bihar.

What is the exam pattern and marking scheme for BCECE?

The BCECE is conducted offline in pen-and-paper mode. Each subject paper (Physics, Chemistry, Mathematics, Biology, Agriculture) contains 100 multiple-choice questions. For each correct answer, 4 marks are awarded, and 1 mark is deducted for each incorrect answer (negative marking). Candidates have 90 minutes to complete each paper.

What is the application fee for the BCECE exam?

For standard 3-subject combinations (PCM or PCB), the application fee is ₹1,000 for General, BC, and EBC candidates, and ₹500 for SC, ST, and Disabled Quota (DQ) candidates. For 4-subject combinations (PCMB), the fee is ₹1,100 for General/OBC and ₹550 for Reserved categories.

Are candidates from other states eligible to write the BCECE?

Generally, only permanent residents of Bihar or candidates whose parents are employees of the Government of Bihar, PSUs in Bihar, or central government employees posted in Bihar are eligible to apply for seats under the state quota.

What is the syllabus of the BCECE exam?

The syllabus of the BCECE is based on the Class 11 and Class 12 curriculum of Physics, Chemistry, Mathematics, and Biology as prescribed by the NCERT/CBSE and Bihar State Board (BSEB).