SSC CGL 2025 Tier-I · 12 September 2025, Shift 3 · Quantitative Aptitude · Question 24 of 100

We transcribed this text from the pictures on a candidate's response sheet for this shift; the answer is the key SSC colour-marked on that sheet. Solution last reviewed 11 September 2026.

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SSC does not publish CGL question papers or a public answer key. From 2025 the candidate response sheet prints every question and option as a picture, so the text here is transcribed from those pictures for this shift and checked against them independently; where the layout carries meaning, the picture is shown as well. Each answer is the key SSC colour-marked on that sheet. SSC released a tentative key for challenge and later its final one, both only to candidates behind a login, and the sheet does not say which of the two it carries. The solutions are written by UnlockIAS.

A vertical cylindrical container is filled with oil. A solid hemispherical stone of radius 7 cm is immersed completely, and the oil level rises by 3 cm. What is the radius of the cylinder?

(a)6.29 cm
(b)7.43 cm
(c)8.73 cm
(d)11.41 cm

Answer

Why

Correct — C. The oil that rises takes up exactly the volume of the stone.

Hemisphere volume = ⅔πr³ = ⅔π × 343 = 686π⁄3 cm³
Risen oil = cylinder of radius R and height 3 = 3πR²

Equate: 3πR² = 686π⁄3 → R² = 686⁄9 ≈ 76.22
Take the root: R = √76.22 ≈ 8.73 cm → option (c)

Why the others are wrong

(a)
6.29 cm — With R = 6.29 cm the stone would lift the oil 228.67 ÷ 6.29² ≈ 5.78 cm, not 3 cm. That cylinder, 12.58 cm across, is also narrower than the stone's 14 cm flat face.
(b)
7.43 cm — With R = 7.43 cm the rise would be 228.67 ÷ 7.43² ≈ 4.14 cm. A narrower cylinder lifts the oil higher, so R must be larger to hold the rise to 3 cm.
(d)
11.41 cm — With R = 11.41 cm the rise would be only 228.67 ÷ 11.41² ≈ 1.76 cm. The cylinder is too wide for the stone's volume to lift the oil 3 cm.

Concept

A fully immersed solid pushes aside its own volume of liquid. In a vertical cylinder that volume shows up as a disc-shaped layer of oil: πR² × rise.

Set the two volumes equal and π cancels: 3R² = ⅔ × 7³.

A hemisphere is half a sphere, so its volume is ⅔πr³, half of (4⁄3)πr³.

The stem says the container is filled with oil, yet the level rises by 3 cm. Read it as holding oil with room above the surface, since a brim-full container would overflow instead.

Key facts

Volume of a hemisphere = ⅔πr³.
A fully immersed solid raises the liquid in a vertical cylinder by (volume of solid) ÷ πR².
Here ⅔π × 343 = 686π⁄3 cm³, so R² = 686⁄9 and R ≈ 8.73 cm.

Study next

Rise in level when spheres or balls are dropped into a cylinderMelting and recasting solids, where the total volume is conservedTotal surface area of a hemisphere, 3πr²

Common traps

Using the full sphere's volume, (4⁄3)πr³: that gives R ≈ 12.35 cm, which is not an option.
Equating the stone's volume to the whole cylinder of oil instead of the 3 cm layer that rises.

The hemisphere volume ⅔πr³ is asked on its own at 19 Sep 2025, 09:00, Quant Q.10 (radius 5 cm gives 250π⁄3 cm³).

Volume conservation drives 18 Sep 2025, 12:30, Quant Q.11, where hemispheres of radius 2 cm and 4 cm are recast into one with R³ = 8 + 64 = 72.

Related PYQs

No directly related past PYQ was found.

Topic: Right Prism, Right Circular Cone, Right Circular Cylinder, Sphere, Hemispheres, Rectangular Parallelepiped, Regular Right Pyramid with triangular or square base — all 51 SSC CGL questions on it, every solved shift.