Section A (1 Mark each)
Q1. (a) $\pi$ cm$^3$
Volume = Vol of Cone + Vol of Hemisphere = $\frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$.
Given $r=1, h=1$. Vol = $\frac{1}{3}\pi(1)^2(1) + \frac{2}{3}\pi(1)^3 = \frac{1}{3}\pi +
\frac{2}{3}\pi = \pi$ cm$^3$.
Q2. (b) $3\pi r^2$
Total Surface Area of solid hemisphere = Curved Surface Area ($2\pi r^2$) + Base Area ($\pi r^2$) = $3\pi r^2$.
Q3. (a) a sphere and a cylinder
A surahi typically has a spherical bottom and a cylindrical neck.
Q4. (a) Both A and R are true and R is the correct explanation of A.
Volume of cube = 64 $\Rightarrow$ edge $a=4$. Wait, question says edge 5 cm.
If edge = 5 cm, joined end to end, length becomes $5+5=10$, breadth=5, height=5.
TSA = $2(lb+bh+hl) = 2(50+25+50) = 2(125) = 250$ cm$^2$.
Assertion is true. Reason is the correct formula for TSA of cuboid.
Section B (2 Marks each)
Q5. Volume of cube = $a^3 = 64 \Rightarrow a = 4$ cm.
When joined, length $l = 4+4=8$ cm, $b=4$ cm, $h=4$ cm.
TSA = $2(lb+bh+hl) = 2(32 + 16 + 32) = 2(80) = 160$ cm$^2$.
Q6. Radius $r = 3.5$ cm. Total height = 15.5 cm.
Height of cone $h = 15.5 - 3.5 = 12$ cm.
Slant height $l = \sqrt{h^2 + r^2} = \sqrt{12^2 + 3.5^2} = \sqrt{144 + 12.25} =
\sqrt{156.25} = 12.5$ cm.
TSA = CSA of Cone + CSA of Hemisphere = $\pi rl + 2\pi r^2$
$= \frac{22}{7} \times 3.5 \times 12.5 + 2 \times \frac{22}{7} \times 3.5 \times 3.5$
$= 11 \times 12.5 + 22 \times 3.5 = 137.5 + 77 = 214.5$ cm$^2$.
Q7. Volume = Vol of Cone + Vol of Hemisphere.
$r=1, h=1$.
$V = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi(1)(1) + \frac{2}{3}\pi(1) =
\pi$ cm$^3$.
Section C (3 Marks each)
Q8. Cylinder: $h=2.1$ m, $d=4$ m $\Rightarrow r=2$ m. Cone: $l=2.8$ m, $r=2$ m.
Area of canvas = CSA of Cylinder + CSA of Cone
$= 2\pi rh + \pi rl = \pi r(2h + l)$
$= \frac{22}{7} \times 2 (2 \times 2.1 + 2.8) = \frac{44}{7} (4.2 + 2.8) = \frac{44}{7}
\times 7 = 44$ m$^2$.
Cost = Area $\times$ Rate = $44 \times 500 = \text{Rs } 22000$.
Q9. Capsule: Total length = 14 mm. Diameter = 5 mm $\Rightarrow r = 2.5$ mm.
Length of cylindrical part $h = 14 - 2.5 - 2.5 = 9$ mm.
Surface Area = CSA of Cylinder + 2 $\times$ CSA of Hemisphere
$= 2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r)$
$= 2 \times \frac{22}{7} \times 2.5 (9 + 5) = \frac{110}{7} \times 14 = 110 \times 2 = 220$
mm$^2$.
Section D (5 Marks)
Q10. Cylinder: $h=2.4$ cm, $d=1.4$ cm $\Rightarrow r=0.7$ cm.
Conical cavity of same height and radius is hollowed out.
Slant height $l = \sqrt{h^2 + r^2} = \sqrt{2.4^2 + 0.7^2} = \sqrt{5.76 + 0.49} = \sqrt{6.25}
= 2.5$ cm.
TSA of remaining solid = CSA of Cylinder + Area of Base (top) + CSA of Cone (cavity)
$= 2\pi rh + \pi r^2 + \pi rl = \pi r(2h + r + l)$
$= \frac{22}{7} \times 0.7 (2 \times 2.4 + 0.7 + 2.5)$
$= 2.2 (4.8 + 0.7 + 2.5) = 2.2 (8.0) = 17.6$ cm$^2$.
Nearest cm$^2$ = 18 cm$^2$.
Section E (Case Study - 4 Marks)
(i) Diameter = 2.8 cm $\Rightarrow$ Radius $r = 1.4$ cm.
(ii) Total length = 5 cm. Length of cylinder $h = 5 - 1.4 - 1.4 = 2.2$ cm.
(iii) Volume = Vol of Cylinder + 2 $\times$ Vol of Hemisphere
$= \pi r^2 h + 2(\frac{2}{3}\pi r^3) = \pi r^2 (h + \frac{4}{3}r)$
$= \frac{22}{7} \times 1.4 \times 1.4 (2.2 + \frac{4}{3} \times 1.4)$
$= 6.16 (2.2 + 1.87) = 6.16 \times 4.07 \approx 25.05$ cm$^3$.