Vardaan Learning Institute
Chapter Solutions: Surface Areas and Volumes
SECTION A: OBJECTIVE TYPE QUESTIONS (1 Mark Each)
1. Two cubes (vol 64) joined. Surface Area is:
$a^3 = 64 \Rightarrow a = 4$ cm. Cuboid $L=8$, $B=4$, $H=4$.
$TSA = 2(LB+BH+HL)$
$= 2(32+16+32) = 2(80) = 160$ cm$^2$.
Correct Option: (b) 160 cm$^2$
2. Volume of Cone + Hemisphere (r=1, h=1):
Vol = $\frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$.
$= \frac{1}{3}\pi + \frac{2}{3}\pi = \pi$.
Correct Option: (a) $\pi$ cm$^3$
3. Recast Sphere (4.2cm) to Cylinder (6cm). Find h:
$\frac{4}{3}\pi R^3 = \pi r^2 h$
$\Rightarrow \frac{4}{3}(4.2)^3 = 6^2 h$.
$h = \frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 36} = 2.744$ cm.
Correct Option: (a) 2.74 cm
4. Ratio of volumes 64:27. SA Ratio?
$V_1/V_2 = r_1^3/r_2^3 = 64/27 \Rightarrow r_1/r_2 = 4/3$.
$A_1/A_2 = r_1^2/r_2^2 = 16/9$.
Correct Option: (d) 16:9
5. Diagonal of cube $6\sqrt{3}$. TSA?
$a\sqrt{3} = 6\sqrt{3} \Rightarrow a=6$.
TSA $= 6a^2 = 6(36) = 216$ cm$^2$.
Correct Option: (b) 216 cm$^2$
6. SA of sphere 616. radius?
$4\pi r^2 = 616 \Rightarrow 4 \times \frac{22}{7} \times r^2 = 616$.
$r^2 = \frac{616 \times 7}{88} = 49 \Rightarrow r=7$ cm.
Correct Option: (a) 7 cm
7. Cylinder (r=8,h=2) to Cone (h=6). Radius?
$\pi(8)^2(2) = \frac{1}{3}\pi r^2 (6)$.
$128 = 2r^2 \Rightarrow r^2=64 \Rightarrow r=8$.
Correct Option: (c) 8 cm
8. curved SA of 2 joined hemispheres.
Curved SA of one hemi = $2\pi r^2$. Since joined at bases, total curved SA is outer surface.
Area = $2\pi r^2 + 2\pi r^2 = 4\pi r^2$. (Assuming sphere formation).
Correct Option: (a) $4\pi r^2$
9. Spheres (d=6) from Cyl (h=45, d=4).
Vol Cyl = $\pi (2)^2 (45) = 180\pi$.
Vol Sphere = $\frac{4}{3}\pi (3)^3 = 36\pi$.
Number = $180\pi / 36\pi = 5$.
Correct Option: (b) 5
10. Assertion: TSA vs CSA.
TSA is $2\pi r(r+h)$. Assertion says $2\pi rh$ (which is CSA).
So A is False. R is True.
Correct Option: (d) A is false but R is true.
SECTION B: SHORT ANSWER TYPE QUESTIONS (2 Marks Each)
11. Capsule Surface Area.
SA = $2\pi r(h+2r) = 2 \times \frac{22}{7} \times 2.5 (9+5) = 220$ mm$^2$.
220 mm$^2$
12. Tent Canvas Area.
Area = $\frac{22}{7} \times 2 (4.2 + 2.8) = 44$ m$^2$.
44 m$^2$
13. TSA of remaining solid (Cylinder - Cone).
TSA = $2\pi rh + \pi rl + \pi r^2 = 18$ cm$^2$.
18 cm$^2$
14. 3 Cubes (3,4,5) -> Single Cube. Edge?
Total vol = $3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216$.
$a^3 = 216 \Rightarrow a = 6$ cm.
6 cm
SECTION C: SHORT ANSWER TYPE II QUESTIONS (3 Marks Each)
15. Wooden article (Cylinder - 2 Hemispheres). TSA.
TSA = $2\pi r(h+2r) = 22 \times 17 = 374$ cm$^2$.
374 cm$^2$
16. Number of lead shots dropped.
Number = $\frac{\text{Volume Flowed}}{\text{Volume 1 Shot}}$
$= \frac{50\pi/3}{\pi/6} = 100$.
100
17. Mass of solid iron pole.
Vol = $111532.8$ cm$^3$.
Mass = $892.26$ kg.
892.26 kg
18. Well embankment.
Vol of Earth = $\pi r^2 h = \pi (1.5)^2 (14) = 31.5\pi$.
Area of Embankment Ring = $\pi(R^2 - r^2) = \pi(5.5^2 - 1.5^2)$ (Width 4, so R=1.5+4=5.5).
$= \pi(30.25 - 2.25) = 28\pi$.
Height = $Volume / Area = 31.5\pi / 28\pi = 1.125$ m.
1.125 m
19. Canal irrigation.
Vol in 30 min (0.5 hr) = $6 \times 1.5 \times (10000 \times 0.5) = 45000$ m$^3$.
Area = $Volume / Height = 45000 / 0.08$ (8cm = 0.08m)
$= 562500$ m$^2$.
562500 m$^2$
20. Solid: Cyl + 2 Hemi ends. Total h=20, d=7. Volume.
$r=3.5$. Cyl h = $20 - 7 = 13$.
Vol = $\pi r^2 h + \frac{4}{3}\pi r^3 = \pi r^2 (h + \frac{4}{3}r)$
$= \frac{22}{7} \times 12.25 (13 + 4.67)$
$= 38.5 \times 17.67 \approx 680$ cm$^3$. (Exact: $38.5 \times (13 + 14/3) = 38.5 \times 53/3 = 2040.5/3
= 680.17$)
680.17 cm$^3$
SECTION D: LONG ANSWER TYPE QUESTIONS (5 Marks Each)
21. Gulab Jamun syrup volume.
Vol of 1 Jamun = Cyl Vol + 2 Hemi Vol.
Cyl length = $5 - 2.8 = 2.2$. $r=1.4$.
Vol = $\pi(1.4)^2(2.2) + \frac{4}{3}\pi(1.4)^3 = \pi(1.4)^2 [2.2 + 1.86] \approx 25.05$ cm$^3$.
45 Jamuns = $45 \times 25.05 \approx 1127.25$.
Syrup = $30\% \text{ of } 1127.25 \approx 338$ cm$^3$.
338 cm$^3$
22. Pipe to Tank time.
Pipe $r=10$cm = 0.1m. Speed = 3 km/h = 3000 m/h.
Vol per hour = $\pi (0.1)^2 (3000) = 30\pi$ m$^3$.
Tank Vol = $\pi (5)^2 (2) = 50\pi$ m$^3$.
Time = $50\pi / 30\pi = 5/3$ hrs = 1 hr 40 mins.
1 hour 40 minutes
SECTION E: CASE STUDY ANSWERS
23. Case Study: Top (Lattu)
(i) $h = 3.25$ cm.
(ii) $l \approx 3.7$ cm.
(iii) TSA = $39.6$ cm$^2$.
(i) 3.25 cm (ii) 3.7 cm (iii) 39.6 cm$^2$