Section A (1 Mark each)
Q1. (b) $2r$
Volume of Sphere = Volume of Cone $\Rightarrow \frac{4}{3}\pi r^3 = \frac{1}{3}\pi R^2
r$.
$4r^2 = R^2 \Rightarrow R = 2r$.
Q2. (a) 12 cm
Volume = $6^3 + 8^3 + 10^3 = 216 + 512 + 1000 = 1728$ cm$^3$.
Edge $a = \sqrt[3]{1728} = 12$ cm.
Q3. (c) two cones and a cylinder
A gilli has two conical ends and a cylindrical middle part.
Q4. (a) Both A and R are true and R is the correct explanation of A.
Diameter of inscribed sphere = Side of cube $a$. Radius = $a/2$.
Section B (2 Marks each)
Q5. Rise in water level.
Volume displaced by 500 persons = $500 \times 0.04 = 20$ m$^3$.
Area of pond = $80 \times 50 = 4000$ m$^2$.
Rise in level $h = \frac{\text{Volume}}{\text{Area}} = \frac{20}{4000} = \frac{1}{200}$ m =
0.5 cm.
Q6. Number of cones.
Volume of Sphere = $n \times$ Volume of Cone.
$\frac{4}{3}\pi (10.5)^3 = n \times \frac{1}{3}\pi (3.5)^2 (3)$.
$4 \times 10.5 \times 10.5 \times 10.5 = n \times 3.5 \times 3.5 \times 3$.
$n = \frac{4 \times 1157.625}{36.75} = \frac{4630.5}{36.75} = 126$.
Q7. Number of bricks.
Volume of wall = $270 \times 300 \times 350 = 28350000$ cm$^3$.
Volume occupied by bricks = $\frac{7}{8} \times 28350000 = 24806250$ cm$^3$.
Volume of 1 brick = $22.5 \times 11.25 \times 8.75 = 2214.84375$ cm$^3$.
Number of bricks = $\frac{24806250}{2214.84375} = 11200$.
Section C (3 Marks each)
Q8. Internal diameter of pipe.
Rate = 2.52 km/h = 2520 m/h. Time = 0.5 h. Length of water column $h = 1260$ m.
Volume in tank = $\pi R^2 H = \pi (0.4)^2 (3.15)$ (using meters).
Volume from pipe = $\pi r^2 h = \pi r^2 (1260)$.
$\pi r^2 (1260) = \pi (0.16)(3.15) \Rightarrow r^2 = \frac{0.504}{1260} = 0.0004$.
$r = 0.02$ m = 2 cm. Diameter = 4 cm.
Q9. Area to colour (TSA of toy).
Radius $r = 3.5/2 = 1.75$ cm. Total height = 5 cm.
Height of cone $h = 5 - 1.75 = 3.25$ cm.
Slant height $l = \sqrt{3.25^2 + 1.75^2} = \sqrt{10.5625 + 3.0625} = \sqrt{13.625} \approx
3.7$ cm.
TSA = CSA Hemisphere + CSA Cone = $2\pi r^2 + \pi rl = \pi r(2r + l)$.
$= \frac{22}{7} \times 1.75 (3.5 + 3.7) = 5.5 \times 7.2 = 39.6$ cm$^2$.
Section D (5 Marks)
Q10. Number of lead shots.
Volume of Cone = $\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (5)^2 (8) = \frac{200\pi}{3}$
cm$^3$.
Water flowed out = $\frac{1}{4} \times \frac{200\pi}{3} = \frac{50\pi}{3}$ cm$^3$.
Volume of 1 lead shot = $\frac{4}{3}\pi (0.5)^3 = \frac{4}{3}\pi (0.125) = \frac{0.5\pi}{3}
= \frac{\pi}{6}$ cm$^3$.
Number of shots $n \times \frac{\pi}{6} = \frac{50\pi}{3} \Rightarrow n = \frac{50\pi}{3}
\times \frac{6}{\pi} = 100$.
Section E (Case Study - 4 Marks)
(i) Apparent Capacity = Volume of Cylinder = $\pi r^2 h$.
$= 3.14 \times (2.5)^2 \times 10 = 3.14 \times 6.25 \times 10 = 196.25$ cm$^3$.
(ii) Volume of Hemisphere = $\frac{2}{3}\pi r^3 = \frac{2}{3} \times 3.14
\times (2.5)^3 = \frac{2}{3} \times 3.14 \times 15.625 \approx 32.71$ cm$^3$.
Actual Capacity = $196.25 - 32.71 = 163.54$ cm$^3$.
(iii) Volume of hemispherical portion is approx $32.71$ cm$^3$.