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

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

Inverse-square scaling: g ∝ m/r² — substitute both ratios, and square the radius ratio before dividing.

💡 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 — Misreads 1/0.53 as 5.3, giving 0.107 × 5.3 ≈ 0.57

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: Lessons S4 (Scaling) and P2a (Forces).

⭐ Q2JSO Y7-8Difficulty ★★★★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

Inverse-square scaling: shrinking r multiplies g by the square of the factor — here (1/0.01)² = 10,000.

💡 Why the other options fail

A — Only takes 1/0.01 = 100 without squaring
B — A factor-of-10 slip when squaring: 100² = 10,000, not 1,000
D — Cubes instead of squaring: (1/0.01)³ = 1,000,000

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: Lessons S4 (Scaling) and P2a (Forces).

⭐ Q3JSO Y7-8Difficulty ★★★★★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 0.71 times Earth's radius
B About 0.5 times Earth's radius
C About 1.41 times Earth's radius
D About 2 times Earth's radius

🎯 Knowledge Point — what this tests

Rearranging g ∝ m/r²: to double g at fixed mass, r must shrink by √2, not by 2.

💡 Why the other options fail

B — 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: A — About 0.71 times Earth's radius

Revise: Lessons S4 (Scaling) and P2a (Forces).

⭐ Q4JSO Y7-8Difficulty ★★★★Chemistry

The Problem

Tritium (hydrogen-3) is radioactive, with a half-life of about 12 years. Natural processes continually supply small amounts to the water cycle. In a simplified steady-state model, its average input balances radioactive loss.

Your answer

❓ Pick the best answer:

A All of the tritium was formed when the oceans first appeared on Earth, early in its history
B Tritium is chemically stable in salt water and does not decay
C Living organisms in the ocean continuously produce tritium as a by-product of their metabolism
D Tritium is produced at a roughly constant rate by cosmic rays in the upper atmosphere
E Tritium is continuously recycled between the atmosphere and the ocean without ever being lost

🎯 Knowledge Point — what this tests

Radioactive decay vs steady state: a 12-year half-life means today's tritium must be continuously produced — cosmic rays are the source.

Use the stated assumptions and distinguish the mechanism from broader claims.

After ten half-lives, 1/1024 of the original amount remains, not exactly zero. Natural tritium is produced by cosmic-ray-induced nuclear interactions in the atmosphere and enters the water cycle. Ongoing production can balance decay: D. Natural levels vary, and human nuclear activities can also add tritium.

✅ Answer: D — Tritium is produced at a roughly constant rate by cosmic rays in the upper atmosphere

Revise: Lesson C3 (Atoms & Ions).

⭐ Q5JSO Y7-8Difficulty ★★★★Chemistry

The Problem

Carbon-14 is radioactive and its natural atmospheric supply is replenished by cosmic-ray-induced reactions. Nitrogen-14 is a stable isotope. Why does nitrogen-14 not need replenishment specifically to compensate for radioactive decay?

Your answer

❓ Pick the best answer:

A Nitrogen-14 is stable — it does not undergo radioactive decay
B Nitrogen-14 is produced by cosmic ray interactions in the upper atmosphere
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

Nuclear stability: a stable isotope does not decay, so its amount needs no replenishment — the opposite of C-14.

Use the stated assumptions and distinguish the mechanism from broader claims.

A is correct: N-14 is stable and does not undergo radioactive decay. This says nothing about whether nitrogen moves between environmental reservoirs; it continuously cycles. Atmospheric N₂ is about 78% of present dry air, but that is not a claim that the composition has been constant throughout Earth history.

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

Revise: Lesson C3 (Atoms & Ions).

⭐ Q6JSO Y7-8Difficulty ★★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 Constrict, trapping heat in the body's core
C Dilate, increasing blood flow near the skin surface to release heat
D Dilate, reducing blood flow to prevent overheating
E Remain unchanged; skin blood vessels are not involved in thermoregulation

🎯 Knowledge Point — what this tests

Negative feedback in thermoregulation: overheating triggers vasodilation so warm blood reaches the skin and sheds heat.

Use the stated assumptions and distinguish the mechanism from broader claims.

During exercise, increased skin blood flow through vasodilation helps transfer body heat toward the surface: C. Heat loss also depends on the environment and sweating; blood-flow changes alone do not guarantee cooling in all hot conditions.

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

Revise: Lesson B6 (Genetics & Homeostasis).

⭐ Q7JSO Y7-8Difficulty ★★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 liver spontaneously converts excess glucose to fat without any hormonal signal
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 pancreas releases insulin, which stimulates cells to take up glucose from the blood

🎯 Knowledge Point — what this tests

Blood glucose homeostasis: high glucose → pancreas releases insulin → cells absorb glucose and the liver stores glycogen.

Use the stated assumptions and distinguish the mechanism from broader claims.

High blood glucose stimulates pancreatic insulin release. Insulin promotes glucose uptake especially in muscle and adipose tissue and promotes storage while suppressing glucose production by the liver. These actions help lower blood glucose: E.

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

Revise: Lesson B6 (Genetics & Homeostasis).

⭐ Q8JSO Y7-8Difficulty ★★★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 names the wrong response — cold triggers vasoconstriction, not vasodilation
B Vasoconstriction widens skin vessels and increases heat transfer to the 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 does not depend on vessel width
E The hypothesis is flawed because the heart, rather than the blood vessels, responds to cold

🎯 Knowledge Point — what this tests

Evaluating a hypothesis: in cold water the body vasoconstricts to cut heat loss — the reverse of the proposed vasodilation.

Use the stated assumptions and distinguish the mechanism from broader claims.

Cold exposure normally causes skin vasoconstriction, reducing blood flow near the surface and limiting heat transfer from the core to the environment: A. The hypothesis incorrectly names vasodilation. This response helps conserve heat but does not guarantee protection from hypothermia.

✅ Answer: A — The hypothesis names the wrong response — cold triggers vasoconstriction, not vasodilation

Revise: Lesson B6 (Genetics & Homeostasis).

⭐ Q9JSO Y7-8Difficulty ★★★★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.

Temperature (°C)	10	20	30	40	50
Max. dissolved (g / 100 mL)	21	32	46	64	86

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

Extrapolating a data trend: the solubility steps grow (11, 14, 18, 22 g), so continue the growing pattern — not a flat or doubling one.

💡 Why the other options fail

A — Uses the average step (≈16 g per 10 °C), giving 86 + 16 ≈ 102 — but the steps are growing
C — Adds the whole 20–50 °C rise again (86 + 54 = 140) instead of continuing the growing steps
D — Doubles the 50 °C value (86 × 2 = 172) — no step in the table doubles

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: Lesson C1 (Matter & States).

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

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

Isotope fractionation in the water cycle: light ¹H evaporates preferentially and heavy ²H condenses first.

💡 Why the other options fail

I ✓ — lighter ¹H₂O evaporates more readily, so the vapour is depleted in ²H.
II ✓ — heavy ²H condenses and rains out first; air reaching the poles is depleted, so tropical rain is richer.
III ✗ — reversed: ice ages lock ¹H-rich ice away, leaving the ocean RICHER in ²H. Any option including III is wrong, and any option missing I or II is incomplete — the answer is D (I and II).

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: Lesson E4 (Fossils & Climate).

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

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. For this question, take the volcanic CO₂ source to have a higher C-13 proportion than the fossil-fuel CO₂ source.

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

Isotope fractionation in the carbon cycle: photosynthesis prefers ¹²C, so plant carbon — and fossil-fuel CO₂ — is ¹³C-depleted.

Use the stated assumptions and distinguish the mechanism from broader claims.

I follows from the stated preference for C-12. Under the explicit comparison supplied, fossil-fuel CO₂ has a lower C-13 proportion than volcanic CO₂, so II also follows. III gives the wrong causal direction: preferentially adding C-12 would lower, not raise, the C-13 proportion in that added carbon. Ocean isotope behaviour also involves several physical and biological processes. Answer: C, I and II.

✅ Answer: C — I and II

Revise: Lesson E4 (Fossils & Climate).

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

The Problem

In this hypothetical calibrated ice-core dataset: 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 sample is enriched in O-18 because ocean water is enriched in O-18 during ice ages.

🎯 Knowledge Point — what this tests

Ice-core δ¹⁸O as a climate proxy: colder periods give more strongly negative values in precipitation and ice.

Use the stated assumptions and distinguish the mechanism from broader claims.

In the hypothetical dataset supplied, −58‰ falls in the ice-age range: B. This indicates a lower O-18/O-16 ratio than the stated reference, consistent with colder conditions in this calibration. Preferential removal of heavier water during rainout leaves polar vapour depleted in heavy isotopes. Absolute ice-core ranges depend on location and conditions; these numbers are not universal Antarctic thresholds.

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

Revise: Lesson E4 (Fossils & Climate).

🏁 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.

Know another parent preparing too?

Send them this free diagnostic — it shows any child exactly where they're strong and where to focus. Pure help, nothing to buy.

Want to turn the red and amber bars green? The full Junior Science Olympiad Y7-8 course teaches all the areas above from the ground up — full coverage of Physics, Chemistry, Biology and Earth Science for the ASI Junior Science Olympiad.

▶ Try the first 3 lessons free Unlock the full course — A$299

Studying a second Science course too? Use code SCIENCE20 at checkout for 20% off.