Volume and surface area Questions and answers

Topics

General Knowledge
General Science
General English
Aptitude
General Computer Science
General Intellingence and Reasoning
Current Affairs
Exams
Elementary Mathematics
English Literature

Elementary Mathematics

21). The curved surface area of a right circular cone of radius 14 cm is 440 sq cm. What is the slant height of the cone?

A). 10 cm

B). 11cm

C). 12 cm

D). 13 cm

View Answer

22). The diameter of base of a right circular cone is 7 cm and slant height is 10 cm, then what is its lateral surface area?

A). 110 sq cm

B). 100 sq cm

C). 70 sq cm

D). 49 sq cm

View Answer

Correct Answer: 110 sq cm

Given that.

Diameter of a right circular cone = 7 cm

Radius of a right circular cone = \( \Large \frac{7}{2} \) cm

and slant height of a right circular cone (l) = 10 cm

Therefore, Lateral surface area of a cone = \( \Large \pi rl \)

= \( \Large \frac{22}{7} \times \frac{7}{2} \times 10 \)

= \( \Large 11 \times 10 = \) = 110 \( \Large cm^{2} \)

23). What is the whole surface area of a cone of base radius 7 cm and height 24 cm?

A). 654 sq cm

B). 704 sq cm

C). 724 sq cm

D). 964 sq cm

View Answer

Correct Answer: 704 sq cm

Slant height, \( \Large l = \sqrt{h^{2} + r^{2}} \)

= \( \Large \sqrt{ \left(24\right)^{2} + \left(7\right)^{2} } = \sqrt{576 + 49} \)

\( \Large \sqrt{625} = 25 \)

Total surface area = \( \Large \pi r \left(l + r\right) \)

= \( \Large \frac{22}{7} \times 7 \left(25 + 7\right) \)

= \( \Large \frac{22}{7} \times 7 \times 32 = 704 sq.cm. \)

24). The volume of a right circular cone is 100 \( \Large \pi cm^{3} \) and its height is 12 cm. Find its slant height.

A). 13 cm

B). 16 cm

C). 9 cm

D). 26 cm

View Answer

Correct Answer: 13 cm

Volume = \( \Large \frac{1}{3} \pi r^{2}h \)

According to the question,

\( \Large \frac{1}{3} \pi r^{2}h = 100 \pi => \frac{1}{3} \pi r^{2} \times 12 = 100 \pi \)

=> \( \Large r^{2} = 25 \)

Therefore, r = \( \Large \sqrt{25} \) = 5 cm

Therefore, Slant height (l) = \( \Large \sqrt{h^{2}+r^{2}} = \sqrt{12^{2}+5^{2}} \)

= \( \Large \sqrt{169} \) = 13 cm

25). The radius of the base of a right circular cone is doubled. To keep the volume fixed, the height of the cone will be

A). half of the previous height

B). one-third of the previous height

C). one-fourth of the previous height

D). \( \Large \frac{1}{\sqrt{2}} \) times of the previous height

View Answer

Correct Answer: one-fourth of the previous height

In first situation,

Radius = \( \Large r_{1} \), height = \( \Large h_{1} \) and volume = \( \Large v_{1} \)

In second situation,

Radius = 2 \( \Large r_{1} \), height = \( \Large h_{2} \) and volume = \( \Large v_{2} \)

In the volume of fixed, then

\( \Large v_{1} \) = \( \Large v_{2} \)

\( \Large \frac{1}{3} \pi r_{1}^{2}h_{1} = \frac{1}{3} \pi \left(2r_{1}\right)^{2}h_{2} \)

\(h_{1} \) = 4 \( \Large h_{2} \)

\( \Large h_{2} = \frac{h_{1}}{4} \)

Therefore, height of the cone will be one-fourth of the previous height.

26). A cone of radius r cm and height h cm 15 divided into two parts by drawing a plane through the middle point of its height and parallel to the base. What is the who of the volume of the original cone to the volume of the smaller cone?

A). 4 : 1

B). 8 : 1

C). 2 : 1

D). 6 : 1

View Answer

Correct Answer: 8 : 1

Let the cone is divided into two parts line l.

In triangle AOB and ACD

By basic proportionality theorem,

\( \Large CD = \frac{r}{2}, since \ AC = \frac{h}{2} \)

Therefore, Ratio = \( \Large \frac{Volume \ of \ original \ cone}{Volme \ of \ smaller \ cone } \)

= \( \Large \frac{\frac{1}{3} \pi r^{2}h}{\frac{1}{3} \pi \left(\frac{r}{2}\right)^{2} \left(\frac{h}{2}\right) } = \frac{8}{1} \)

Therefore, Ratio = 8 : 1

27). If the volumes of two right circular cones are in the ratio 1 : 3 and their diameters are in the ratio 3 : 5, then the ratio of their heights is

A). 25 : 27

B). 1 : 5

C). 3 : 5

D). 5 : 27

View Answer

Correct Answer: 25 : 27

Let diameter, radius and height of first cone are \( \Large d_{1} \), \( \Large r_{1} \) and \( \Large h_{1} \) respectively and that of second cone are \( \Large d_{2} \), \( \Large r_{2} \) and \( \Large h_{2} \) respectively.

\( \Large \frac{r_{1}}{r_{2}} = \frac{d_{1}}{d_{2}} = \frac{3}{5}. \frac{h_{1}}{h_{2}} = ? \)

Given, \( \Large \frac{\frac{1}{3}\pi r_{1}^{2}h_{1}}{\frac{1}{3} \pi r_{2}^{2}h_{2}} = \frac{1}{3} => \left(\frac{r_{1}}{r_{2}}\right)^{2} \times \frac{h_{1}}{h_{2}} = \frac{1}{3} \)

=> \( \Large \left(\frac{3}{5}\right)^{2} \times \frac{h_{1}}{h_{2}} = \frac{1}{3} => \frac{h_{1}}{h_{2}} = \frac{1}{3} \times \frac{25}{9} = \frac{25}{27} \)

28). The diameters of two cones are equal, If their slant heights be in the ratio of 5 : 7 then find the ratio of their Curved surface areas.

A). 25 : 7

B). 25 : 49

C). 5 : 49

D). 5 : 7

View Answer

Correct Answer: 5 : 7

Given,

\( \Large \frac{l_{1}}{l_{2}} = \frac{5}{7} \)

Now, curved surface area of first cone

= \( \Large \pi rl_{1} \)

and curved surface area of second cone

= \( \Large \pi rl_{2} \)

Therefore, Ratio = \( \Large \frac{ \pi rl_{1}}{ \pi rl_{2}} = \frac{l_{1}}{l_{2}} = 5 : 7 \)

29). A conical cap has the base diameter 24 cm and height 16 cm. What is the cost of painting the surface of the cap at the rate of 70 paise per sq cm?

A). Rs.520

B). Rs.524

C). Rs.528

D). Rs.532

View Answer

Correct Answer: Rs.532

\( \Large l = \sqrt{h^{2} + r^{2}} \)

= \( \Large \sqrt{16^{2} + 12^{2}} = \sqrt{256+144} \)

=\( \Large \sqrt{400} - 20 cm \)

Curved surface area = \( \Large \pi rl \)

= \( \Large \frac{22}{7} \times 12 \times 20 \)

Therefore, Cost of painting = \( \Large \frac{22}{7} \times 12 \times 20 \times 0.70 \)

= Rs.528

30). The curved surface area and the total surface area of a cylinder are in the ratio 1 : 2. If the total surface area of the right cylinder is 616 \( \Large cm^{2} \), then its volume is

A). 1632 \( \Large cm^{3} \)

B). 1078 \( \Large cm^{3} \)

C). 1232 \( \Large cm^{3} \)

D). 1848 \( \Large cm^{3} \)

View Answer

Correct Answer: 1632 \( \Large cm^{3} \)

Curved surface area of cylinder

= \( \Large 2 \pi rh cm^{2} \)

Total surface area of cylinder

= \( \Large 2 \pi r \left(h + r\right) cm^{2} \)

Now,

According to the question,

\( \Large \frac{2 \pi rh}{2 \pi r \left(h + r\right) } = \frac{1}{2} \)

=> \( \Large \frac{h}{h+r} = \frac{1}{2} \)

=> 2h = h + r

=> h = r

Now,

Total surface area = 616 \( \Large cm^{2} \)

=> \( \Large 2 \pi r \left(h + r\right) = 616 \)

=> \( \Large 2 \pi r \left(r + r\right) = 616 \)

\( \Large \left(Because h = r \right) \)

=> \( \Large 2 \pi r \left(2r\right) = 616 \)

=> \( \Large 4 \pi r^{2} = 616\)

=> \( \Large r = \sqrt{\frac{616}{4 \pi }} = \sqrt{\frac{616 \times 7}{4 \times 22}} \)

= \( \Large \sqrt{7 \times 7} = 7 \)

Therefore, r = h = 7 cm

Therefore, Volume = \( \Large \pi r^{2}h \)

= \( \Large \frac{22}{7} \times \left(7 \times 7\right) \times 7 \)

= 1078 \( \Large cm^{3} \)

1	2	3	4	5	6	7

Go to :

Total Pages : 12