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Solids — PYPs

4 solved Solids previous year questions (PYQs) from past year papers — attempt each and check the answer.

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Q1:jipmat 2025QA › SolidsHardQA · MCQ

Consider the following statements, which of them is/are correct? A. If the height of cylinder is doubled, the area of curved surface is doubled. B. If the radius of a hemispherical solid is doubled, its total surface area becomes fourfold. C. If a hemisphere and cone have equal bases and equal heights, then the ratio of curved surface area is

\sqrt{2}:1

AB and C only
BA and C only
CA and B only
DA, B and C

Pick an option to attempt

The Setup: We are fact-checking 3D geometry claims. Let's run the formulas for each statement to see who is speaking straight facts and who is capping. Step 1: Check Statement A. Curved Surface Area (CSA) of a cylinder

=2\pi rh

. If

h

becomes

2h

, the new CSA

=2\pi r(2h)=2(2\pi rh)

. It perfectly doubles. Statement A is a W. Step 2: Check Statement B. Total Surface Area (TSA) of a hemisphere

=3\pi r^2

. If

r

becomes

2r

, the new TSA

=3\pi(2r)^2=3\pi(4r^2)=4(3\pi r^2)

. It scales by a factor of 4 (fourfold). Statement B is a W. Step 3: Check Statement C. Hemisphere CSA

=2\pi r^2

. For the cone, if it has the same base and height as the hemisphere, its radius is

r

and its height is also

r

(since a hemisphere's height equals its radius). Cone slant height (

l

l=\sqrt{r^2+h^2}=\sqrt{r^2+r^2}=\sqrt{2r^2}=r\sqrt{2}

. Cone CSA

=\pi rl=\pi r(r\sqrt{2})=\pi r^2\sqrt{2}

. Ratio of Hemisphere CSA to Cone CSA:

\frac{2\pi r^2}{\pi r^2\sqrt{2}}=\frac{2}{\sqrt{2}}=\frac{\sqrt{2}}{1}

The ratio is exactly

\sqrt{2}:1

. Statement C is a W. Step 4: Since all three are legit, option D is the one. Final Answer: A, B and C

Q2:ipmat indore 2019QA › SolidsMediumSA · TITA

The number of whole metallic tiles that can be produced by melting and recasting a circular metallic plate, if each of the tiles has a shape of a right-angled isosceles triangle and the circular plate has a radius equal in length to the longest side of the tile (Assume that the tiles and plate are of uniform thickness, and there is no loss of material in the melting and recasting process) is

Enter your answer to attempt

The Setup: Conservation of material dressed up as a 3D melting problem. Because the thickness is uniform and nothing is lost, the volume ratio equals the area ratio, and the whole thing reduces to plane geometry. Step 1: Total source material. With the plate's radius

R

\text{Plate area}=\pi R^2

Step 2: Area of one tile. The tile is a right-angled isosceles triangle, whose longest side is the hypotenuse, and the stem sets that hypotenuse equal to

R

. Letting each of the two equal legs be

x

, Pythagoras gives:

x^2+x^2=R^2 \implies 2x^2=R^2 \implies x^2=\frac{R^2}{2}

The legs are perpendicular, so they serve directly as base and height:

\text{Tile area}=\frac{1}{2}\cdot x\cdot x=\frac{x^2}{2}=\frac{1}{2}\cdot\frac{R^2}{2}=\frac{R^2}{4}

Step 3: Divide. Note

R

cancels completely - the answer does not depend on the plate's size, only on the stated relationship between the two:

n=\frac{\pi R^2}{\frac{R^2}{4}}=4\pi

Step 4: Filter for whole tiles. With

\pi\approx 3.14159

n\approx 4\times 3.14159=12.566

The stem asks for whole tiles, and you cannot cast a fraction of one, so the count floors to 12 - with material worth about

0.57

of a tile left unused. Rounding to 13 here is the trap: that would require metal that does not exist. Final Answer: 12

Q3:jipmat 2025QA › SolidsMediumQA · MCQ

A solid brass sphere of radius 21 cm is converted into a right circular cylindrical rod of length 28 cm. The ratio of total surface areas of the rod to sphere is :

A3:1
B7:6
C7:3
D3:7

Pick an option to attempt

The Setup: Melting one 3D shape into another means their volumes are identical. We'll use the volume equality to find the missing radius of the cylinder, then compare their Total Surface Areas (TSA). Step 1: Equate the volumes to find the cylinder's radius (

R

). Sphere Volume = Cylinder Volume

\frac{4}{3}\pi r^3=\pi R^2h

Plug in the given specs:

r=21

and

h=28

\frac{4}{3} \times (21)^3=R^2 \times 28

\frac{4}{3} \times 9261=R^2 \times 28

4 \times 3087=R^2 \times 28

Divide by 28:

R^2=\frac{4 \times 3087}{28}=\frac{3087}{7}=441

R=\sqrt{441}=21\text{ cm}

So, the cylinder's radius matches the sphere's radius! Step 2: Calculate the Total Surface Area (TSA) of the cylinder (rod).

\text{TSA}_{\text{rod}}=2\pi R(R+h)

\text{TSA}_{\text{rod}}=2\pi(21)(21+28)=2\pi(21)(49)

Step 3: Calculate the Total Surface Area (TSA) of the sphere.

\text{TSA}_{\text{sphere}}=4\pi r^2=4\pi(21)^2

Step 4: Find the ratio.

\text{Ratio}=\frac{2\pi(21)(49)}{4\pi(21)(21)}

Cancel the

\pi

and one

21

\text{Ratio}=\frac{2 \times 49}{4 \times 21}=\frac{98}{84}

Divide top and bottom by 14:

\text{Ratio}=\frac{7}{6}

Final Answer: 7:6

Q4:ipmat indore 2019QA › SolidsHardMCQ · MCQ

Three cubes with integer edge lengths are given. It is known that the sum of their surface areas is

564 \ \text{cm}^2

. Then the possible values of the sum of their volumes are

A $764 \ \text{cm}^3$ and $586 \ \text{cm}^3$
B $586 \ \text{cm}^3$ and $564 \ \text{cm}^3$
C $764 \ \text{cm}^3$ and $564 \ \text{cm}^3$
D $586 \ \text{cm}^3$ and $786 \ \text{cm}^3$

Pick an option to attempt

The Setup: A Diophantine equation dressed as geometry. Convert the surface-area condition into a sum of three squares, find every integer solution, then convert back to volumes. Step 1: Reduce the surface-area equation. A cube of edge

a

has surface area

6a^2

, so with edges

a,b,c

6a^2+6b^2+6c^2=564 \implies a^2+b^2+c^2=94

Step 2: Bound the search before starting it. Order the edges

a\leq b\leq c

, with each at least 1 (an edge of 0 is not a cube). Then

c

is the largest, so

3c^2\geq 94

and

c^2\leq 94

\frac{94}{3}\leq c^2\leq 94 \implies 31.3\leq c^2\leq 94 \implies 6\leq c\leq 9

That leaves exactly four cases to test, which is what makes the search genuinely exhaustive rather than a lucky hunt. Step 3: Test all four.

$c$	$c^2$	$a^2+b^2$ needed	Solutions with $a\leq b\leq c$
9	81	13	$4+9$ -> edges $\{2,3,9\}$
8	64	30	none ( $30=1+29,4+26,9+21,16+14,25+5$ - no square pairs)
7	49	45	$9+36$ -> edges $\{3,6,7\}$
6	36	58	none (needs both $\leq 36$ ; $58-36=22$ and $58-25=33$ are not squares)

So there are exactly two admissible triples, and no more. Step 4: Convert to volumes.

\{2,3,9\}: \quad 2^3+3^3+9^3=8+27+729=764

\{3,6,7\}: \quad 3^3+6^3+7^3=27+216+343=586

Both triples do give surface area

6(94)=564

✓, so both are genuine and the answer must list both values. Final Answer:

764 \ \text{cm}^3

and

586 \ \text{cm}^3

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