CUET UG 2025 — General Test Geometry

Curved Surface Area of a cylinder $= 2\pi rh$=2πrh where $r$r is the radius and $h$h is the height.

For the original cylinder:

Height $= h$=h

Curved Surface Area $= 2\pi rh$=2πrh

When height is doubled:

New height $= 2h$=2h

New Curved Surface Area $= 2\pi r(2h)$=2πr(2h)

$$= 4\pi rh$$=4πrh

The ratio of new to original curved surface area:

$$\dfrac{4\pi rh}{2\pi rh} = 2$$2πrh4πrh​=2

The curved surface area is doubled.

Statement I is true.

---

For a hemispherical solid, the total surface area includes the curved surface and the flat circular base.

Total Surface Area $= 2\pi r^2 + \pi r^2 = 3\pi r^2$=2πr2+πr2=3πr2

For the original hemisphere:

Radius $= r$=r

Total Surface Area $= 3\pi r^2$=3πr2

When radius is doubled:

New radius $= 2r$=2r

New Total Surface Area $= 3\pi(2r)^2$=3π(2r)2

$$= 3\pi(4r^2)$$=3π(4r2)

$$= 12\pi r^2$$=12πr2

The ratio of new to original total surface area:

$$\dfrac{12\pi r^2}{3\pi r^2} = 4$$3πr212πr2​=4

The total surface area becomes fourfold.

Statement II is true.

---

Both Statement I and Statement II are true.