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41. The external and internal diameters of a hemispherical bowl are 10 cm and 8 cm respectively. What is the total surface area of the bowl ?

A. 257 cm²

B. 286 cm²

C. 292 cm²

D. 302 cm²

Discuss

Solution:

Internal radius, r = 4 cm
External radius, R = 5 cm
Total surface area :

\begin{aligned} = 2 π R^{2} + 2 π r^{2} + π (R^{2} - r^{2}) \\ = 3 π R^{2} + π r^{2} \\ = [π (3 \times 25 + 16)] c m^{2} \\ = (\frac{22}{7} \times 91) c m^{2} \\ = 286 c m^{2} \end{aligned}

42. A pyramid has an equilateral triangle as its base of which each side is 1 m. Its slant edge is 3 m. The whole surface are of the pyramid is equal to :

A. $\frac{\sqrt{3} + 2 \sqrt{13}}{4} s q . m$

B. $\frac{\sqrt{3} + 3 \sqrt{13}}{4} s q . m$

C. $\frac{\sqrt{3} + 3 \sqrt{35}}{4} s q . m$

D. $\frac{\sqrt{3} + 2 \sqrt{35}}{4} s q . m$

Discuss

Solution:

Area of base :

\begin{aligned} = (\frac{\sqrt{3}}{4} \times 1^{2}) m^{2} \\ = \frac{\sqrt{3}}{4} m^{2} \end{aligned}

Clearly, the pyramid has 3 triangular faces each with sides 3m, 3m and 1 m
So, area of each lateral face :

\begin{aligned} = \sqrt{\frac{7}{2} \times (\frac{7}{2} - 3) (\frac{7}{2} - 3) (\frac{7}{2} - 1)} m^{2} \\ [∵ s = \frac{3 + 3 + 1}{2} = \frac{7}{2}] \\ = \sqrt{\frac{7}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{5}{2}} m^{2} \\ = \frac{\sqrt{35}}{4} m^{2} \end{aligned}

∴ Whole surface area of the pyramid :

\begin{aligned} = (\frac{\sqrt{3}}{4} + 3 \times \frac{\sqrt{35}}{4}) m^{2} \\ = \frac{\sqrt{3} + 3 \sqrt{35}}{4} m^{2} \end{aligned}

43. A hemisphere and a cone have equal bases. If their heights are also equal, then the ratio of their curved surface will be :

A. $\sqrt{2} : 1$

B. $1 : \sqrt{2}$

C. 2 : 1

D. 1 : 2

Discuss

Solution:

Let,

\begin{aligned} O P = O Q = O R = r \\ ∴ O R = h = r \end{aligned}

∴ Curved surface area of the hemisphere =

2 π r^{2}

Curved surface area of a cone =

π r l

Where,

\begin{aligned} l = \sqrt{h^{2} + r^{2}} \\ = \sqrt{r^{2} + r^{2}} \\ = r \sqrt{2} \end{aligned}

∴ Required ratio :

\begin{aligned} = \frac{2 π r^{2}}{π r l} \\ = \frac{2 π r^{2}}{π r \times r \sqrt{2}} \\ = \frac{2}{\sqrt{2}} \\ = \frac{2 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} \\ = \frac{2 \sqrt{2}}{2} \\ = \frac{\sqrt{2}}{1} O r \sqrt{2} : 1 \end{aligned}

44. A swimming pool 9 m wide and 12 m long and 1 m deep on the shallow side and 4 m deep on the deeper side. Its volume is :

A. 360 m³

B. 270 m³

C. 420 m³

D. None of these

Discuss

Solution:

Given length of width of swimming pool is 9 m and 12 m respectively
Volume of swimming pool :

\begin{aligned} = 9 \times 12 \times (\frac{1 + 4}{2}) \\ = 9 \times 12 \times \frac{5}{2} \\ = 270 cu. metre \end{aligned}

45. A closed aquarium of dimensions 30 cm × 25 cm × 20 cm is made up entirely of glass plates held together with tapes. The total length of tape required to hold the plates together (ignore the overlapping tapes) is :

A. 75 cm

B. 120 cm

C. 150 cm

D. 300 cm

Discuss

Solution:

Total length of tape required :
= Sum of lengths of edges
= (30 × 4 + 25 × 4 + 20 × 4) cm
= 300 cm

46. A swimming pool 9 m wide and 12 m long is 1 m deep on the shallow side and 4 m deep on the deeper side. It volume is :

A. 208 m³

B. 270 m³

C. 360 m³

D. 408 m³

Discuss

Solution:

Volume :

\begin{aligned} = [12 \times 9 \times (\frac{1 + 4}{2})] m^{3} \\ = (12 \times 9 \times 2.5) m^{3} \\ = 270 m^{3} \end{aligned}

47. An aluminium sheet 27 cm long, 8 cm broad and 1 cm thick is melted into a cube. The difference in the surface areas of the two solids would be :

A. Nil

B. 284 cm²

C. 286 cm²

D. 296 cm²

Discuss

Solution:

Volume of cube = Volume of sheet = (27 × 8 × 1) cm³ = 216 cm³
Edge of cube :

\sqrt[3]{216} c m = 6 c m

Surface area of sheet :

\begin{aligned} = 2 (l b + b h + l h) \\ = 2 (27 \times 8 + 8 \times 1 + 27 \times 1) c m^{2} \\ = (216 + 8 + 27) c m^{2} \\ = 502 c m^{2} \end{aligned}

Surface area of cube :

\begin{aligned} = 6 a^{2} \\ = (6 \times 6^{2}) c m^{2} \\ = 216 c m^{2} \end{aligned}

∴ Required difference :

\begin{aligned} = (502 - 216) c m^{2} \\ = 286 c m^{2} \end{aligned}

48. The volumes of two cubes are in the ratio 8 : 27. The ratio of their surface areas is :

A. 2 : 3

B. 4 : 9

C. 12 : 9

D. None of these

Discuss

Solution:

Let their edges be a and b
Then,

\begin{aligned} \frac{a^{3}}{b^{3}} = \frac{8}{27} \\ \Rightarrow {(\frac{a}{b})}^{3} = {(\frac{2}{3})}^{3} \\ \Rightarrow \frac{a}{b} = \frac{2}{3} \\ \Rightarrow \frac{a^{2}}{b^{2}} = \frac{4}{9} \\ \Rightarrow \frac{6 a^{2}}{6 b^{2}} = \frac{4}{9} O r 4 : 9 \end{aligned}

49. The height of a right circular cylinder is 6 m. If three times the sum of the areas of its two circular faces is twice the area of the curved surface, then the radius of its base is :

A. 1 m

B. 2 m

C. 3 m

D. 4 m

Discuss

Solution:

\begin{aligned} 3 \times 2 π r^{2} = 2 \times 2 π r h \\ \Rightarrow 6 r = 4 h \\ \Rightarrow r = \frac{2}{3} h \\ \Rightarrow r = (\frac{2}{3} \times 6) m \\ \Rightarrow r = 4 m \end{aligned}

50. Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 metres per second into a cylindrical tank, the radius of whose base is 60 cm. By how much will the level of water rise in 30 minutes ?

A. 2 m

B. 3 m

C. 4 m

D. 5 m

Discuss

Solution:

Volume of water flown through the pipe in 30 min :

\begin{aligned} = [(π \times 0.01 \times 0.01 \times 6) \times 30 \times 60] m^{3} \\ = (1.08 π) m^{3} \end{aligned}

Let the rise in level of water be h metres
Then,

\begin{aligned} π \times 0.6 \times 0.6 \times h = 1.08 π \\ \Rightarrow h = (\frac{1.08}{0.6 \times 0.6}) \\ \Rightarrow h = 3 m \end{aligned}

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