Solutions: Surface Areas and Volumes - Test 2

Q1. (a) 2.74 cm

Vol of Sphere = Vol of Cylinder $\Rightarrow \frac{4}{3}\pi (4.2)^3 = \pi (6)^2 h$.
$h = \frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 36} = \frac{4 \times 74.088}{108} = 2.744$ cm.

Q2. (a) 6 cm

$R^3 = r_1^3 + r_2^3 + r_3^3 = 3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216$.
$R = \sqrt[3]{216} = 6$ cm.

Q3. (c) remain unaltered

When a solid is converted from one shape to another, its volume remains conserved.

Q4. (a) Both A and R are true and R is the correct explanation of A.

Ratio of volumes = $(R/r)^3 = (4/1)^3 = 64$.

Q5. Rod: $r=0.5$ cm, $h=8$ cm. Wire: $L=1800$ cm.

$\pi (0.5)^2 (8) = \pi r^2 (1800) \Rightarrow 0.25 \times 8 = 1800 r^2$.
$2 = 1800 r^2 \Rightarrow r^2 = \frac{1}{900} \Rightarrow r = \frac{1}{30}$ cm.
Thickness (diameter) = $\frac{2}{30} = \frac{1}{15}$ cm (approx 0.67 mm).

Q6. Sphere $r=6$ cm. Wire $d=2$ mm $\Rightarrow r=0.1$ cm.

$\frac{4}{3}\pi (6)^3 = \pi (0.1)^2 L$.
$\frac{4}{3} \times 216 = 0.01 L \Rightarrow 288 = 0.01 L$.
$L = 28800$ cm = 288 m.

Q7. Coin: $r=0.875$ cm ($7/8$ cm), $h=0.2$ cm ($1/5$ cm).

Vol of 1 coin = $\pi (\frac{7}{8})^2 (\frac{1}{5}) = \frac{22}{7} \times \frac{49}{64} \times \frac{1}{5} = \frac{77}{160}$ cm$^3$.
Vol of Cuboid = $5.5 \times 10 \times 3.5 = 192.5$ cm$^3$.
Number of coins = $\frac{192.5}{77/160} = \frac{192.5 \times 160}{77} = 400$.

Q8. Well: $r=1.5$ m, $h=14$ m. Embankment: Inner $r=1.5$, Outer $R=1.5+4=5.5$ m.

Vol of earth = $\pi (1.5)^2 (14) = 31.5\pi$ m$^3$.
Area of embankment ring = $\pi (R^2 - r^2) = \pi (5.5^2 - 1.5^2) = \pi (30.25 - 2.25) = 28\pi$ m$^2$.
Height = $\frac{\text{Vol}}{\text{Area}} = \frac{31.5\pi}{28\pi} = 1.125$ m.

Q9. Canal: $6 \times 1.5$ m$^2$. Speed 10 km/h = 10000 m/h.

Volume in 30 mins (0.5 h) = $6 \times 1.5 \times (10000 \times 0.5) = 9 \times 5000 = 45000$ m$^3$.
Area $\times$ Height = Volume $\Rightarrow A \times \frac{8}{100} = 45000$.
$A = \frac{45000 \times 100}{8} = 562500$ m$^2$.

Q10. Cylinder: $r=18, h=32$. Cone: $h=24$.

Vol Cylinder = Vol Cone $\Rightarrow \pi (18)^2 (32) = \frac{1}{3} \pi r^2 (24)$.
$324 \times 32 = 8 r^2 \Rightarrow r^2 = \frac{324 \times 32}{8} = 324 \times 4 = 1296$.
Radius $r = \sqrt{1296} = 36$ cm.
Slant height $l = \sqrt{r^2 + h^2} = \sqrt{36^2 + 24^2} = \sqrt{1296 + 576} = \sqrt{1872}$.
$l = \sqrt{144 \times 13} = 12\sqrt{13}$ cm.

(i) Vol of Sump = $l \times b \times h = 1.57 \times 1.44 \times 0.95$ m$^3$.
$= 2.14776$ m$^3$ (approx $2.15$ m$^3$).

(ii) Vol of Tank = $\pi r^2 h = 3.14 \times (0.6)^2 \times 0.95$ m$^3$.
$= 3.14 \times 0.36 \times 0.95 = 1.07388$ m$^3$ (approx $1.07$ m$^3$).

(iii) Ratio = $\frac{\text{Vol Tank}}{\text{Vol Sump}} = \frac{1.07388}{2.14776} = \frac{1}{2}$.
Capacity of tank is half the capacity of the sump.