JIPMAT Solids PYQs

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

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: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

JIPMAT Solids — PYPs

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