Junior Science Olympiad Y7-8 · Free Diagnostic Mock

🎁 FREE Diagnostic

12 questions to find exactly where your science is strong — and where it's not

This is a free diagnostic for the Junior Science Olympiad Y7-8 course. 12 real past-paper-style questions sample Physics, Chemistry, Biology and Earth Science.

It's an exam: answer all the questions first (you can change any answer freely), then submit the whole exam to see your report. The worked solutions and hints unlock only after you submit. Use the jump menu at the top to move around.

At the end you get a knowledge-point report: a bar for every topic you attempted, your weak spots flagged, and the exact lessons to revise.

~15 minutes. No login needed. No marks deducted for a wrong answer — never leave one blank.
Self-paced, no time limit — but try to commit to an answer before you review, just like a real test.

Year 7–8 students preparing for the ASI Junior Science Olympiad.

⭐ Q1JSO Y7-8Difficulty ★★★★Physics › Physics

The Problem

The gravitational field strength on the surface of an object is given by: g = k × m/r², where k is a constant, m is the mass of the object, and r is the radius. Mars has a mass about 0.107 times that of Earth, and a radius about 0.53 times that of Earth.

How does the gravitational field strength on the surface of Mars compare to that on the surface of Earth?

Your answer

❓ Pick the best answer:

A About 0.11 times as strong
B About 0.20 times as strong
C About 0.38 times as strong
D About 0.57 times as strong

🎯 Knowledge Point — what this tests

Physics reasoning: read the data/diagram, apply the relationship, and evaluate each option. This is a Physics question.

💡 Why the other options fail

A — Only multiplies by mass ratio (0.107), ignores radius term
B — Divides by radius ratio once (0.107/0.53 ≈ 0.20), not squared
D — Adds instead of multiplying the two ratios

g_Mars/g_Earth = m_Mars/m_Earth × (r_Earth/r_Mars)² = 0.107 × (1/0.53)² = 0.107 × 3.565 = 0.381 ≈ 0.38.

✅ Answer: C — About 0.38 times as strong

Revise: Physics lessons (P-series).

⭐ Q2JSO Y7-8Difficulty ★★★★Physics › Physics

The Problem

The gravitational field strength on the surface of an object is given by: g = k × m/r². A white dwarf star has approximately the same mass as the Sun but is compressed to about 0.01 times the Sun's radius.

Compared to the Sun's surface gravity, the white dwarf's surface gravity is approximately:

Your answer

❓ Pick the best answer:

A 100 times stronger
B 1,000 times stronger
C 10,000 times stronger
D 1,000,000 times stronger

🎯 Knowledge Point — what this tests

Physics reasoning: read the data/diagram, apply the relationship, and evaluate each option. This is a Physics question.

💡 Why the other options fail

A — Only takes 1/0.01 = 100 without squaring
B — Squares incorrectly or uses wrong ratio
D — Uses 0.01⁶ or similar over-calculation

g ∝ m/r². Same mass, r = 0.01r_Sun → ratio = (r_Sun/r_WD)² = (1/0.01)² = 100² = 10,000 times stronger.

✅ Answer: C — 10,000 times stronger

Revise: Physics lessons (P-series).

⭐ Q3JSO Y7-8Difficulty ★★★★★Physics › Physics

The Problem

The gravitational field strength on the surface of an object is given by: g = k × m/r². Scientists want to find a rocky body with exactly twice Earth's surface gravity. This body has the same mass as Earth.

What must the radius of this body be, compared to Earth's radius?

Your answer

❓ Pick the best answer:

A About 0.5 times Earth's radius
B About 0.71 times Earth's radius
C About 1.41 times Earth's radius
D About 2 times Earth's radius

🎯 Knowledge Point — what this tests

Physics reasoning: read the data/diagram, apply the relationship, and evaluate each option. This is a Physics question.

💡 Why the other options fail

A — Divides radius by 2 instead of by √2; confuses linear with squared relationship
C — Multiplies by √2 (inverts the relationship)
D — Doubles the radius (completely inverts the g ∝ 1/r² relationship)

2g = k × m / r². Since same mass: 2g/g = (r_E/r_new)² → 2 = (r_E/r_new)² → r_new = r_E / √2 ≈ 0.707 × r_E ≈ 0.71 r_E.

✅ Answer: B — About 0.71 times Earth's radius

Revise: Physics lessons (P-series).

⭐ Q4JSO Y7-8Difficulty ★★★★Chemistry › Chemistry

The Problem

Tritium (hydrogen-3) is a radioactive isotope with a half-life of approximately 12 years. Despite this short half-life, small but roughly constant amounts of tritium are found in ocean water throughout Earth's history.

Which of the following is the most plausible reason for the roughly constant level of tritium in ocean water?

Your answer

❓ Pick the best answer:

A All tritium was formed when the oceans first appeared on Earth
B Tritium is produced at a roughly constant rate in the upper atmosphere by cosmic ray interactions with nitrogen and oxygen
C Living organisms in the ocean continuously produce tritium during metabolism
D Tritium is chemically stable in salt water and does not decay
E Tritium is continuously recycled between the atmosphere and ocean without being lost

🎯 Knowledge Point — what this tests

Chemistry reasoning: read the data/diagram, apply the relationship, and evaluate each option. This is a Chemistry question.

💡 Why the other options fail

A — Half-life 12 years; Earth ~4.5 billion years old → all original tritium gone billions of years ago
C — Organisms do not produce tritium; they incorporate it but don't create nuclear isotopes
D — Tritium absolutely decays — this contradicts the definition of a radioactive isotope
E — Recycling doesn't create new atoms; total amount would still decline

With a half-life of only 12 years, all original Earth tritium would be gone after ~120 years (10 half-lives). The steady state is maintained by continuous production from cosmic ray spallation of atmospheric gases — exactly the same mechanism as C-14.

✅ Answer: B — Tritium is produced at a roughly constant rate in the upper atmosphere by cosmic ray interactions with nitrogen and oxygen

Revise: Chemistry lessons (C-series).

⭐ Q5JSO Y7-8Difficulty ★★★★Chemistry › Chemistry

The Problem

Nitrogen-14 (the most common nitrogen isotope) is NOT radioactive and makes up about 78% of Earth's atmosphere. This proportion has been roughly stable throughout much of Earth's history. Unlike carbon-14, nitrogen-14 does not need to be constantly replenished.

Which of the following BEST explains why the proportion of nitrogen-14 in the atmosphere remains stable WITHOUT constant replenishment?

Your answer

❓ Pick the best answer:

A Nitrogen-14 is produced by cosmic ray interactions in the upper atmosphere
B Nitrogen-14 is stable — it does not undergo radioactive decay
C Nitrogen-14 is constantly recycled by plants and animals through the nitrogen cycle
D Nitrogen-14 is replenished from the Earth's mantle through volcanic activity
E The half-life of nitrogen-14 is so long that its decay is undetectable

🎯 Knowledge Point — what this tests

Chemistry reasoning: read the data/diagram, apply the relationship, and evaluate each option. This is a Chemistry question.

💡 Why the other options fail

A — Replenishment mechanism — not needed if stable; also this applies to C-14, not N-14
C — Nitrogen cycle moves N around but doesn't address why it doesn't decrease (conflates recycling with stability)
D — Volcanic N₂ does contribute some nitrogen but this isn't the primary reason for stability
E — The question states N-14 is NOT radioactive; a very long half-life would still imply decay

The question states nitrogen-14 is NOT radioactive — it is inherently stable and does not decay. No replenishment is needed because there is no loss. This contrasts with C-14 which is radioactive and requires constant replenishment.

✅ Answer: B — Nitrogen-14 is stable — it does not undergo radioactive decay

Revise: Chemistry lessons (C-series).

⭐ Q6JSO Y7-8Difficulty ★★Biology › Biology

The Problem

When mammals are exposed to hot temperatures, the homeostatic mechanisms in their bodies respond to maintain their internal temperature.

During intense exercise on a hot day, what happens to the blood vessels in a person's skin?

Your answer

❓ Pick the best answer:

A Constrict, reducing blood flow near the skin surface
B Dilate, increasing blood flow near the skin surface to release heat
C Constrict, trapping heat in the body's core
D Dilate, reducing blood flow to prevent overheating
E Remain unchanged; skin blood vessels are not involved in thermoregulation

🎯 Knowledge Point — what this tests

Biology reasoning: read the data/diagram, apply the relationship, and evaluate each option. This is a Biology question.

💡 Why the other options fail

A — Vasoconstriction is the cold response (opposite of heat response); common misconception of reversing the direction
C — Constriction in heat would cause dangerous overheating (hyperthermia)
D — 'Dilate' is correct but 'reducing blood flow' contradicts dilation definition
E — Skin blood vessels are THE primary site of thermoregulatory blood flow changes

In heat: vasodilation brings warm blood close to cool skin surface, allowing heat to dissipate to the environment — a negative feedback mechanism reducing core temperature.

✅ Answer: B — Dilate, increasing blood flow near the skin surface to release heat

Revise: Biology lessons (B-series).

⭐ Q7JSO Y7-8Difficulty ★★Biology › Biology

The Problem

Blood glucose concentration is regulated by homeostatic mechanisms in the body. When blood glucose rises above the normal range (after a meal, for example), the pancreas responds to bring levels back to normal.

Which correctly describes the homeostatic response to high blood glucose?

Your answer

❓ Pick the best answer:

A The pancreas releases glucagon, which causes cells to absorb more glucose
B The pancreas releases insulin, which stimulates cells to take up glucose from the blood
C The pancreas releases insulin, which causes the liver to release more glucose into the blood
D The pancreas releases glucagon, which stimulates the liver to break down glycogen into glucose
E The liver spontaneously converts excess glucose to fat without any hormonal signal

🎯 Knowledge Point — what this tests

Biology reasoning: read the data/diagram, apply the relationship, and evaluate each option. This is a Biology question.

💡 Why the other options fail

A — Glucagon has the OPPOSITE effect (raises blood glucose); this confuses insulin and glucagon
C — Insulin causes cells to absorb glucose, not release it; C reverses the direction
D — Glucagon breaks down glycogen → raises blood glucose; this is the response to LOW glucose
E — Hormone signals are required; spontaneous conversion is incorrect

High blood glucose → insulin released → cells (especially liver and muscle) take up glucose → blood glucose returns to normal. Classic negative feedback.

✅ Answer: B — The pancreas releases insulin, which stimulates cells to take up glucose from the blood

Revise: Biology lessons (B-series).

⭐ Q8JSO Y7-8Difficulty ★★★Biology › Biology

The Problem

A student hypothesises: 'When exposed to cold water, the human body dilates blood vessels in the skin so that warm blood reaches the skin quickly, heating it up and protecting the person from hypothermia.'

Which of the following BEST identifies the flaw in this hypothesis?

Your answer

❓ Pick the best answer:

A The hypothesis correctly describes vasoconstriction, not vasodilation
B The hypothesis incorrectly states the response — vasoconstriction (not vasodilation) occurs, which REDUCES heat loss from the blood to the cold water
C The hypothesis is correct; warm blood at the skin surface does protect against hypothermia by preventing heat loss
D The hypothesis is flawed because blood temperature is always the same regardless of vessel width
E The hypothesis is flawed because only the heart, not blood vessels, responds to cold temperatures

🎯 Knowledge Point — what this tests

Biology reasoning: read the data/diagram, apply the relationship, and evaluate each option. This is a Biology question.

💡 Why the other options fail

A — Partially right (mentions vasoconstriction) but doesn't explain WHY vasodilation is wrong
C — Accepts the student's flawed hypothesis; vasodilation in cold water would greatly INCREASE heat loss
D — Blood temperature does vary; arterial blood is ~37°C; this claim is incorrect
E — Blood vessels do respond to cold via autonomic nervous system

The student has the direction of the response wrong. In cold water, VASOCONSTRICTION occurs (not vasodilation). Vasoconstriction reduces blood flow to the skin surface, minimising the temperature difference between blood and cold water, thereby REDUCING heat loss — protecting against hypothermia.

✅ Answer: B — The hypothesis incorrectly states the response — vasoconstriction (not vasodilation) occurs, which REDUCES heat loss from the blood to the cold water

Revise: Biology lessons (B-series).

⭐ Q9JSO Y7-8Difficulty ★★★★Chemistry › Chemistry

The Problem

A chemist prepares solutions by dissolving potassium nitrate in water. The table shows the maximum mass of potassium nitrate (g) that can dissolve in 100 mL of water at different temperatures.

Based on the trend in the table, approximately how many grams of potassium nitrate could dissolve in 100 mL of water at 60°C?

Your answer

❓ Pick the best answer:

A 100 g
B 110 g
C 140 g
D 170 g

🎯 Knowledge Point — what this tests

Chemistry reasoning: read the data/diagram, apply the relationship, and evaluate each option. This is a Chemistry question.

💡 Why the other options fail

A_100 — Rounds down too early or uses linear extrapolation with too small a step
C_140 — Assumes the solubility doubles each step
D_170 — Extrapolates too aggressively (uses linear trend from 20-50 only)

Differences per 10°C: 32-21=11, 46-32=14, 64-46=18, 86-64=22. Differences increase by approximately 3-4 per step. Next step (50→60°C): 86 + (22+4) ≈ 86+26 = 112 ≈ 110g.

✅ Answer: B — 110 g

Revise: Chemistry lessons (C-series).

⭐ Q10JSO Y7-8Difficulty ★★★★Earth Science › EarthScience

The Problem

Like oxygen, hydrogen has two common stable isotopes: ¹H (light hydrogen) and ²H (deuterium, heavy hydrogen). Water molecules containing ²H require more energy to evaporate than those containing ¹H. During condensation (rainfall), water molecules with ²H tend to condense first. The same fractionation principles that apply to oxygen isotopes also apply to hydrogen isotopes in the water cycle.

Based on this information, which of the following statements are TRUE? I. Water vapour in the atmosphere has a LOWER proportion of ²H than the ocean water it evaporated from. II. Rainwater in tropical regions has a HIGHER proportion of ²H than rainfall in polar regions. III. During an ice age (when more water is locked in ice caps), ocean water would have a LOWER proportion of ²H than during a warm period.

Your answer

❓ Pick the best answer:

A I only
B II only
C III only
D I and II
E I and III
? (F) II and III
? (G) All three

🎯 Knowledge Point — what this tests

Earth Science reasoning: read the data/diagram, apply the relationship, and evaluate each option. This is a Earth Science question.

💡 Why the other options fail

G_all — III is the reverse: ice ages → more ¹H locked in ice → ocean RICHER in ²H, not poorer
E_I_III — Includes incorrect statement III

I ✓: lighter ¹H evaporates more easily → vapour enriched in ¹H → lower ²H than ocean. II ✓: ²H condenses first near equator; by time air reaches poles, ²H depleted → tropical rain has more ²H. III ✗: ice is enriched in ¹H (forms preferentially from lighter water vapour) → locking up ¹H in ice INCREASES ocean ²H proportion during ice ages (opposite of the statement).

✅ Answer: D — I and II

Revise: Earth Science lessons (E-series).

⭐ Q11JSO Y7-8Difficulty ★★★★Earth Science › EarthScience

The Problem

Carbon has two stable isotopes: ¹²C (lighter) and ¹³C (heavier). During photosynthesis, plants preferentially absorb ¹²CO₂ over ¹³CO₂ because reactions involving lighter molecules proceed slightly faster. As a result, plant material (wood, leaves) is slightly depleted in ¹³C compared to the atmosphere. When plants decompose or are burned, the ¹³C-depleted carbon is released back to the atmosphere.

Based on this information, which of the following statements are TRUE? I. Living plant tissue has a LOWER proportion of ¹³C than atmospheric CO₂. II. Atmospheric CO₂ from burning fossil fuels (ancient plant matter) would have a LOWER proportion of ¹³C than CO₂ from volcanic eruptions. III. Ocean water has a HIGHER proportion of ¹³C than the atmosphere, because oceans preferentially absorb ¹²CO₂.

Your answer

❓ Pick the best answer:

A I only
B II only
C I and II
D I and III
E II and III
? (F) All three

🎯 Knowledge Point — what this tests

Earth Science reasoning: read the data/diagram, apply the relationship, and evaluate each option. This is a Earth Science question.

💡 Why the other options fail

F_all — III is not supported; nothing in the stimulus indicates ocean preferentially absorbs ¹²CO₂
A_only_I — Misses the logical extension to fossil fuels (same mechanism as II)

I ✓: photosynthesis preferentially uses ¹²CO₂ → plant material lower in ¹³C than atmosphere. II ✓: fossil fuels = ancient plant matter = ¹³C-depleted; volcanic CO₂ comes from geological sources with average ¹³C. III ✗: oceans don't preferentially absorb ¹²CO₂ via a biological mechanism; this claim is not supported by the stimulus.

✅ Answer: C — I and II

Revise: Earth Science lessons (E-series).

⭐ Q12JSO Y7-8Difficulty ★★★★★Earth Science › EarthScience

The Problem

Scientists drill ice cores from Antarctica and measure the ratio of O-18 to O-16 in the ice (expressed as δ¹⁸O, where more negative values = less O-18). During ice ages, Antarctic ice has δ¹⁸O values around −55 to −60 per mille (‰). During warm interglacial periods, values are around −40 to −45 ‰. Meanwhile, ocean sediment records show the OPPOSITE trend: ocean water becomes MORE enriched in O-18 during ice ages.

A scientist finds an ice core sample with δ¹⁸O = −58 ‰. Based on this information, which conclusion is MOST SUPPORTED?

Your answer

❓ Pick the best answer:

A The ice formed during a warm interglacial period, because very negative δ¹⁸O indicates high O-18 content
B The ice formed during an ice age, because very negative δ¹⁸O indicates low O-18 content consistent with cold conditions
C The ice formed during a warm period, because ocean water δ¹⁸O would be negative during warm periods
D The ice formed during an ice age because ocean records confirm ice ages have more O-18 than warm periods

🎯 Knowledge Point — what this tests

Earth Science reasoning: read the data/diagram, apply the relationship, and evaluate each option. This is a Earth Science question.

💡 Why the other options fail

A — Confuses 'more negative δ¹⁸O' with 'more O-18'; more negative = LESS O-18
C — Confuses ice δ¹⁸O with ocean δ¹⁸O trends; they move in opposite directions
D — Ocean records show more O-18 during ice ages (correct) but this applies to OCEAN not ice; conflates the two records

δ¹⁸O = −58‰ is within the ice age range (−55 to −60‰). Very negative δ¹⁸O in ice means depleted in O-18 (more O-16). Cold air masses lose O-18 preferentially through precipitation before reaching Antarctica, leaving the remaining vapour and ice very ¹⁸O-depleted.

✅ Answer: B — The ice formed during an ice age, because very negative δ¹⁸O indicates low O-18 content consistent with cold conditions

Revise: Earth Science lessons (E-series).

🏁 Finish

That's the end of the exam

Go back and change any answers you like. When you're ready, submit the whole exam to lock it in and see your knowledge report — then you can revisit every question to read the full worked solution.

The report stays locked until you submit — you can't open it early.

📊 Diagnostic Report

Your knowledge-point report

Here is how you performed on every knowledge area you attempted. Bars show your accuracy per topic — green is strong, amber is shaky, red needs work. Only questions you answered are scored.

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