Practice Section

Timed practice sets: Easy → NEET → JEE Main → JEE Advanced

Practice Timer

00:00:00

How to Use Practice Sets:

Set timer based on difficulty level
Solve without looking at solutions first
Mark uncertain answers for review
Check solutions only after attempting all
Note your weak areas for focused revision

Easy Practice Set (CBSE Level)

Time: 15 minutes | Questions: 5 | Target: 4/5 correct

Q1. Nuclear Radius

The radius of a nucleus with mass number 64 is 4.8 × 10⁻¹⁵ m. Calculate the radius of a nucleus with mass number 125.

Hint: Use R = R₀A^(1/3). Find R₀ from first nucleus, then calculate for second.

Q2. Half-Life

A radioactive sample has 10⁶ atoms initially. After 30 minutes, 1.25 × 10⁵ atoms remain. Find the half-life in minutes.

Hint: N₀/N = 8 = 2³, so 3 half-lives have passed.

Q3. Binding Energy

Calculate the binding energy of an α-particle (He-4) if its mass is 4.0026 u. Given: mₚ = 1.0073 u, mₙ = 1.0087 u.

Hint: He-4 has 2 protons + 2 neutrons. Calculate mass defect, then BE = Δm × 931.5 MeV.

Q4. Decay Type

²³⁸U₉₂ undergoes α-decay. Write the decay equation and identify the daughter nucleus.

Hint: α-decay: A→A-4, Z→Z-2. New element has Z = 90.

Q5. Activity

A sample contains 10⁹ atoms with decay constant λ = 2 × 10⁻⁵ s⁻¹. Calculate the initial activity in Becquerel.

Hint: A = λN. Simple multiplication.

NEET Level Practice Set

Time: 30 minutes | Questions: 8 MCQs | Target: 6/8 correct

Q1. Nuclear Density

Which statement about nuclear density is correct?

(a) Increases with mass number
(b) Decreases with mass number
(c) Constant for all nuclei
(d) Zero for stable nuclei

Answer: (c) Constant for all nuclei

Nuclear density ρ ≈ 2.3 × 10¹⁷ kg/m³ is same for all nuclei because Volume ∝ A and Mass ∝ A.

Q2. Binding Energy Curve

Which nucleus has the maximum binding energy per nucleon?

(a) He-4
(b) Fe-56
(c) U-238
(d) C-12

Answer: (b) Fe-56

Fe-56 has maximum BE/A ≈ 8.8 MeV, making it the most stable nucleus.

Q3. Beta Decay

In β⁻ decay, which quantity remains unchanged?

(a) Mass number only
(b) Atomic number only
(c) Both A and Z
(d) Neither A nor Z

Answer: (a) Mass number only

In β⁻ decay, A remains same but Z increases by 1 (n → p + e⁻).

Q4. Half-Life Calculation

If 75% of a radioactive sample decays in 60 minutes, what is the half-life?

(a) 15 min
(b) 20 min
(c) 30 min
(d) 40 min

Answer: (c) 30 min

75% decayed means 25% remaining. N/N₀ = 1/4 = 1/(2²) → 2 half-lives. T½ = 60/2 = 30 min.

Q5. Q-Value

If Q-value of a nuclear reaction is negative, the reaction is:

(a) Exothermic
(b) Endothermic
(c) Spontaneous
(d) Radioactive

Answer: (b) Endothermic

Q < 0 means energy must be supplied (absorbed) for reaction to occur.

Q6. Mean Life vs Half-Life

The ratio of mean life (τ) to half-life (T½) is approximately:

(a) 0.693
(b) 1.00
(c) 1.44
(d) 2.00

Answer: (c) 1.44

τ = 1/λ and T½ = 0.693/λ. Therefore τ/T½ = (1/λ)/(0.693/λ) = 1/0.693 ≈ 1.44.

Q7. Fusion vs Fission

Which process powers the Sun?

(a) Nuclear fission
(b) Nuclear fusion
(c) Radioactive decay
(d) Chemical combustion

Answer: (b) Nuclear fusion

Sun converts H into He through fusion, releasing enormous energy (E = mc²).

Q8. Carbon Dating Limitation

C-14 dating cannot be used for:

(a) Ancient wood
(b) Fossil bones
(c) Rocks and minerals
(d) Old manuscripts

Answer: (c) Rocks and minerals

C-14 method only works for organic materials (once living). Rocks are inorganic—never had C-14.

JEE Main Level Practice Set

Time: 36 minutes (3 min/problem) | Questions: 12 | Target: 9/12 correct

JEE Main Strategy: Aim for 3 minutes per problem. If stuck after 1 minute, mark for review and move on. Negative marking exists!

Q1. Numerical: Binding Energy Per Nucleon

A nucleus ¹⁶O has atomic mass 15.994 u. Calculate its BE/A. (mₚ = 1.0073 u, mₙ = 1.0087 u)

Step 1: Identify Z = 8, N = 8, A = 16

Step 2: Expected mass = 8(1.0073) + 8(1.0087) = 16.128 u

Step 3: Δm = 16.128 - 15.994 = 0.134 u

Step 4: BE = 0.134 × 931.5 = 124.82 MeV

Step 5: BE/A = 124.82/16 = 7.80 MeV

Q2. Decay Constant from Activity

A sample of 10²⁰ atoms has activity 5 × 10¹⁵ Bq. Calculate (a) decay constant λ, (b) half-life.

(a) λ: A = λN → λ = A/N = (5 × 10¹⁵)/(10²⁰) = 5 × 10⁻⁵ s⁻¹

(b) T½: T½ = 0.693/λ = 0.693/(5 × 10⁻⁵) = 13,860 s ≈ 3.85 hours

Q3. Graph Interpretation

A graph of log₁₀(A) vs t for a radioactive sample has slope -0.0301 day⁻¹. Find half-life in days.

log₁₀(A) = log₁₀(A₀) - (λ/2.303)t

Slope = -λ/2.303 = -0.0301

λ = 0.0301 × 2.303 = 0.0693 day⁻¹

T½ = 0.693/λ = 0.693/0.0693 = 10 days

Note: Full JEE Main set would have 12 problems. For space efficiency, 3 representative problems shown. The actual test should include problems on: Q-value, successive decay, nuclear radius ratio, BE curve applications, and more decay problems.

JEE Advanced Level Practice Set

Time: 45 minutes | Questions: 6 | Target: 4/6 correct

JEE Advanced Reality: Questions are multi-concept, require deep thinking. Don't panic if you can't solve all—partial marks exist for correct approach.

Q1. Threshold Energy (Conceptual + Numerical)

Consider the reaction ¹⁶O + ¹H → ¹³N + ⁴He. Given: mass(¹⁶O) = 15.995 u, mass(¹H) = 1.008 u, mass(¹³N) = 13.006 u, mass(⁴He) = 4.003 u. (a) Calculate Q-value. (b) Is this exothermic or endothermic? (c) Qualitatively explain if threshold energy is needed.

(a) Q-value:

Reactants: 15.995 + 1.008 = 17.003 u

Products: 13.006 + 4.003 = 17.009 u

Δm = 17.003 - 17.009 = -0.006 u

Q = -0.006 × 931.5 = -5.59 MeV

(b) Q < 0 → Endothermic (energy absorbed)

(c) Yes, threshold energy needed because:

Reaction is endothermic (Q < 0)
Minimum KE ≥ |Q| must be supplied
Conservation of momentum requires threshold > |Q|

Q2. Decay Chain Problem

Nucleus X (T½ = 10 days) decays to Y (T½ = 5 days) which decays to stable Z. Initially only X present (N₀ atoms). Derive an expression for number of Y atoms after time t. (Assume t >> T½(Y))

Rate equation for Y:

dN_Y/dt = λ_X N_X - λ_Y N_Y (production - decay)

For t >> T½(Y), equilibrium reached:

Production rate ≈ Decay rate

λ_X N_X ≈ λ_Y N_Y

N_X = N₀ e^(-λ_X t)

Therefore: N_Y = (λ_X/λ_Y) N_X = (T½_Y/T½_X) N₀ e^(-λ_X t)

N_Y = (5/10) N₀ e^(-0.693t/10) = 0.5 N₀ e^(-0.0693t)

Note: Full JEE Advanced set would have 6 complex problems covering multi-step reasoning, graph analysis, conceptual depth, and application to unfamiliar scenarios.

Exam Strategy → Quick Revision