Class 10 Maths: Ch 12 – Surface Areas and Volumes – Extra Questions with Answers

Ans: C

Solution:
Let the radii be 3r and 4r, and the heights be 2h and 5h.
Volume of cylinder 1: π(3r)^2 * 2h = 18πr^2h
Volume of cylinder 2: π(4r)^2 * 5h = 80πr^2h
Ratio of volumes: (18πr^2h) / (80πr^2h) = 18/80 = 9/40

Answer: 9:40

Ans: C

Solution:
The lateral surface area of a frustum of a cone is given by πl(R + r), where l is the slant height, R is the radius of the larger base, and r is the radius of the smaller base.
In this case, l = 10 cm, R = 12 cm, and r = 6 cm.
Area = π * 10 * (12 + 6) = 180π cm²

Answer: 180π cm²

Ans: C

Solution:
Volume of cylinder = πr²h = π(3²)(8) = 72π cm³
Volume of 6 spheres = 72π cm³
Volume of 1 sphere = 72π/6 = 12π cm³
(4/3)πr³ = 12π
r³ = 12 * (3/4) = 9
r = ∛9 cm

Answer: ∛9 cm

Ans: B

Solution:
Volume of sphere = (4/3)πr³ = (4/3)π(3³) = 36π cm³
Volume of cone = (1/3)πr²h = (1/3)π(6²)h = 12πh cm³
Since the volume remains constant, 36π = 12πh
h = 3 cm

Answer: 3 cm

Ans: C

Solution:
The curved surface area of a hemisphere is 2πr². The curved surface area of a cone is πrl.
2πr² = πrl
2r = l
r/l = 1/2

Answer: 1/2

Ans: C

Solution:
Volume of sphere = $\frac{4}{3}\pi(2r)^3 = \frac{32}{3}\pi r^3$
Volume of cone = $\frac{1}{3}\pi R^2 (4r) = \frac{4}{3}\pi R^2 r$
Equating the volumes:
$\frac{32}{3}\pi r^3 = \frac{4}{3}\pi R^2 r$
$32r^2 = 4R^2$
$R^2 = 8r^2$
$R = 2\sqrt{2}r$
Answer: $2\sqrt{2}r$

Ans: B

Solution:
The volume of the original sphere is (4/3)π(6^3) = 288π cm³.
The volumes of the two smaller spheres are (4/3)π(3^3) = 36π cm³ and (4/3)π(4^3) = 256/3π cm³.
Let r be the radius of the third sphere. Its volume is (4/3)πr³.
The sum of the volumes of the three smaller spheres equals the original volume: 36π + 256/3π + (4/3)πr³ = 288π.
Simplifying, (4/3)πr³ = 288π – 36π – 256/3π = (864 – 108 – 256)/3π = 500/3π.
r³ = 500/4 = 125.
r = 5 cm.

Answer: 5 cm

Ans: D

Solution:
The tank is a cone. The ratio of radius to depth is constant. The radius at depth 12 cm is 3 cm. Thus, r/h = 3/12 = 1/4. The volume V of a cone is (1/3)πr²h. For the tank, r=h/4. So, V = (1/3)π(h/4)²h = (π/48)h³. The total depth is h = 12, so V = (π/48) * 12³ = 36π.
Answer: 36π cm³

Ans: C

Solution:
1. **Convert units:**
* Pipe diameter: 6 mm = 0.6 cm
* Pipe radius: 0.3 cm
* Water flow rate: 12 m/min = 1200 cm/min
* Conical vessel base diameter: 30 cm
* Conical vessel base radius: 15 cm
2. **Calculate the flow rate of the pipe (volume per minute):**
* Area of pipe: π * (0.3 cm)^2 = 0.09π cm^2
* Volume flow rate: 0.09π cm^2 * 1200 cm/min = 108π cm^3/min
3. **Calculate the volume of the conical vessel:**
* Volume of cone: (1/3) * π * (15 cm)^2 * 20 cm = 1500π cm^3
4. **Calculate the time to fill the vessel:**
* Time = Volume of cone / Flow rate of pipe
* Time = 1500π cm^3 / (108π cm^3/min) ≈ 13.89 minutes

Answer: 13.89 minutes

Ans: B

Solution:
Let $r$ be the radius of the base of the cone and $h$ be the height of the cone. The radius of the hemisphere is 5.
From the geometry, we have $r^2 + (h-5)^2 = 5^2$, which simplifies to $r^2 = 10h – h^2$.
The volume of the cone is $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (10h – h^2)h = \frac{1}{3}\pi(10h^2 – h^3)$.
To maximize the volume, we take the derivative with respect to $h$ and set it to 0:
$\frac{dV}{dh} = \frac{1}{3}\pi(20h – 3h^2) = 0$, so $h(20 – 3h) = 0$.
Since $h \neq 0$, we have $h = \frac{20}{3}$.
Then $r^2 = 10(\frac{20}{3}) – (\frac{20}{3})^2 = \frac{200}{3} – \frac{400}{9} = \frac{600-400}{9} = \frac{200}{9}$.
Thus, $r = \frac{10\sqrt{2}}{3}$.
The maximum volume is $V = \frac{1}{3}\pi(\frac{200}{9})(\frac{20}{3}) = \frac{4000\pi}{81}$.

Answer: $\frac{4000\pi}{81}$