SSC CGL 2025 Tier-I · 24 September 2025, Shift 2 · Quantitative Aptitude · Question 5 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 13 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 rectangular aquarium (100 cm L × 50 cm W × 60 cm H) contains water with an initial level of 20 cm. 10 identical metal spheres, each with a diameter of 10 cm, are gently lowered into the aquarium. Assuming the spheres are fully submerged, what is the rise in the water level?

(a)π⁄6 cm
(b)π⁄3 cm
(c)2π⁄3 cm
(d)π cm

Answer

Why

Correct — B. The water rises by the spheres' volume spread over the aquarium's base.

Radius of each sphere: 10 ÷ 2 = 5 cm
One sphere: 4⁄3 π × 5³ = 500π⁄3 cm³
Ten spheres: 10 × 500π⁄3 = 5000π⁄3 cm³

Base area: 100 × 50 = 5000 cm²
Rise: (5000π⁄3) ÷ 5000 = π⁄3 cm → option (b)

Why the others are wrong

(a)
π⁄6 cm — π⁄6 cm is half the true rise. It is what the hemisphere formula 2⁄3 πr³ gives, or five spheres instead of ten.
(c)
2π⁄3 cm — 2π⁄3 cm is double the true rise: the displacement of twenty such spheres, not ten.
(d)
π cm — π cm comes from dropping the ÷3 in 4⁄3 πr³: 10 × 4π × 125 ÷ 5000 = π. The sphere formula keeps the 3 in the denominator.

Concept

A fully submerged solid pushes aside its own volume of water. In a tank with vertical walls that volume spreads over the base, so rise = volume immersed ÷ base area.

The 60 cm height and the 20 cm starting level do not enter the formula. They only confirm the setup works: 20 cm of water covers a sphere 10 cm tall, and the level rises by about 1.05 cm, far below the rim.

Key facts

Rise in level = volume immersed ÷ base area of the vessel.
Volume of a sphere = 4⁄3 πr³, with r half the diameter.
π⁄3 cm ≈ 1.05 cm.

Study next

Immersing a solid in a cylinder: rise = volume ÷ πR²Finding how many solids raise the level by a given height

Common traps

Using the diameter as the radius. 10³ in place of 5³ makes the rise eight times too big, 8π⁄3 cm.
Dividing by the whole tank volume, 100 × 50 × 60, instead of the base area.

Displaced volume over base area also decides 12 Sep 2025, 16:00, Quant Q.24, where a hemispherical stone of radius 7 cm raises oil in a cylinder by 3 cm and the cylinder's radius must be found.

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 99 SSC CGL questions on it, every solved shift.