Maths Class X Question Bank

Question 1 of 45

1. Question
1 point(s)
The barrel of a fountain pen, cylindrical in shape is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen is used up on writing 3300 words on an average. How many words can be written in a bottle of ink containing one fifth of the litre?
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Hint

Volume of barrel $=\dfrac{22}{7}\times0.25\times0.25\times7=1.375\ cm^3$
Volume of ink in the bottle
$=\dfrac{1}{5}\times1000=200\ cm^3$
Total no. of barrels $=\dfrac{200}{1.375}$
Required no. of words $=\dfrac{200}{1.375}\times330=48000$
Question 2 of 45

2. Question
1 point(s)
A conical vessel of radius 6 cm and height 8 cm is completely filled with water. A sphere is lowered in the water and its size is such that when it touches the sides, it is just emerged. Whose fraction of water overflows?
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Hint

Volume of conical vessel $=\dfrac{1}{3}\pi (6)^2\times8=96\pi \ cm^3\\ \\ l=\sqrt{36+64}=10\ cm\\ \\ =\dfrac{6}{10}=\dfrac{r}{8-r}\\ \\ \Rightarrow 8r=24\ \Rightarrow r=3\ cm$
Volume of sphere $=\dfrac{4}{3}\pi\ (3)^3=36\pi\ cm^3$
Fraction $=\dfrac{36\pi}{96\pi}=\dfrac{3}{8}$
Question 3 of 45

3. Question
1 point(s)
A building is in the form of a right circular cylinder surmounted by a hemispherical dome both having the same base radii. The base diameter of the dome is equal to $\dfrac{2}{3} $ of the total height of the building. Find the height of the building, if it contains $67\dfrac{1}{21}m^3 $ of air.
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Hint

$2r=\dfrac{2}{3}\times(h+r)\ \Rightarrow 6r=2h+2r\\ \\ \Rightarrow 2r=h$
A/Q
$67\dfrac{1}{21}=\pi r^2h+\dfrac{2}{3}\pi r^3\\ \\ \Rightarrow \dfrac{1408}{21}=\pi r^2\ (h+\dfrac{2}{3}r)\\ \\ \Rightarrow \dfrac{1408}{21}=\pi r^2\left(\dfrac{8}{3}r\right)\\ \\ \Rightarrow \dfrac{1408}{7}=8\times\dfrac{22}{7}\times r^3\Rightarrow r=2m\\ \\ \Rightarrow h=4m$
Height of building $=r+h=6m$
Question 4 of 45

4. Question
1 point(s)
A shuttle cock used for playing Badminton has the shape of a frustum of a cone mounted on a hemisphere (see figure). The diameters of the ends of the frustum are 5 cm and 2 cm, the height of the entire shuttle cock is 7 cm. Find the external surface area. $\left(use\ \pi=\dfrac{22}{7}\right) $
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Hint

External surface area of the cock = Surface area of frustum + surface area of hemisphere
$=\pi(R+r)+2\pi r^2\\ \\ =\dfrac{22}{7}\times3.5\times6.1+2\times\dfrac{22}{7}\times1\times1\\ \\ =67.1+6.28\\ \\ =73.38\ cm^2$
Question 5 of 45

5. Question
1 point(s)
Decorative block shown in the figure is made of two solids, a cube and a hemisphere. The base of the block is a cube with edge 5 cm and hemisphere fixed on the top has a diameter 4.2 cm. Find the total surface area of the block $\left(Take\ \pi=\dfrac{22}{7}\right) $
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Hint

Total surface area of the block
= TSA of cube – base area of hemisphere + CSA of hemisphere
$=(6a^2-\pi r^2+2\pi r^2)\\ \\ =(25\times6+\pi r^2)\\ \\ =150+\dfrac{22}{7}\times(2.1)^2=163.86\ cm^2$
Question 6 of 45

6. Question
1 point(s)
If the radii of the ends of a bucket 45 cm high are 28 cm and 7 cm. Find its capacity and surface area.
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Hint

Capacity of the bucket
$=\dfrac{1}{3}\pi h\ (r_1^2+r^2_2+r_1\ r_2)\\ \\ =\dfrac{1}{3}\times\dfrac{22}{7}\times45(28^2+7^2+28\times7)\\ \\ =330\times147=48510\ cm^3$
Question 7 of 45

7. Question
1 point(s)
A bucket made up of a metal sheet is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of the bucket if the cost of metal sheet used is Rs 15 per 100 $cm^2 $. (use $\pi=3.14 $)
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Hint

Surface area of bucket
$=\pi l(r_1+r_2)+\pi r^2_1\\ \\ =3.14\times20\times28+3.14\times64\\ \\ =1959.36\ cm^2$
Cost of $100\ cm^2$ metal $=15$
Cost of $1\ cm^2$ metal $=0.15$
Cost of $1959.36\ cm^2$ metal
$=293.90$
Question 8 of 45

8. Question
1 point(s)
From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of height 8 cm and of base radius 6 cm, is hollowed out. Find the volume of the remaining solid. (Take $\pi=3.1416 $. Also find the slant height of the cone.
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Hint

Volume of solid cylinder $=3.14\times6^2\times8=904.32$
Volume of conical cavity $=\dfrac{3.14\times6^2\times8}{3}=301.44$
Volume of resulting solid $=904.32-301.44=602.88\ cm^3$
Question 9 of 45

9. Question
1 point(s)
A container open from the top, made up of a metal sheet is in the form of a frustum of a cone of height 8 cm with radii of its lower and upper ends as 4 cm and 10 cm respectively. Find the cost of oil which can completely fill the container at the rate of Rs 20 per litre. Also, find the cost of metal used, if it costs Rs 8 per 100 $cm^2 $. (use $\pi=3.14 $).
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Hint

$l=\sqrt{16^2+(20-8)^2}=\sqrt{256+144}\\ \\ =\sqrt{400}=20\ cm$
Volume of metallic sheet
$=\dfrac{1}{3}\pi h\ (R^2+r^2+Rr)\\ \\ =\dfrac{1}{3}\pi\times 16 (20^2+8^2+20\times8)\\ \\ =\dfrac{1}{3}\times3.14\times16\times624\\ \\ =10449.92\ cm^3\\ \\ =\dfrac{10449.92}{1000}\ \text{litre}$
Cost of 1 litre of milk $=20$
Cost of 10.45 litre of milk $=20\times10.45=209$
Cost of metal sheet $=\pi(R+r)+\pi r^2$
$=\pi(20\times(20+8)+8^2)\\ \\ =1959.36\ cm^2$
Cost $=8\times\dfrac{1959.36}{100}=156.75$
Question 10 of 45

10. Question
1 point(s)
Water is flowing at the rate of 3 km/h through a circular pipe of 20 cm internal diameter into a cylindrical custern of diameter 10 m and depth 2 m. In how much time will the cistern be filled ? $\left(use\ \pi=\dfrac{22}{7}\right) $
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Hint

Volume of water that flows in the cistern in $x$ hours $=\pi r^2h$
$=\dfrac{22}{7}\times\dfrac{1}{10}\times\dfrac{1}{10}\times300x$
Volume of cistern $=\left(\dfrac{22}{7}\times5\times5\times2\right)m^3$
A/Q $\dfrac{22}{7}\times\dfrac{1}{10}\times\dfrac{1}{10}\times3000x=\dfrac{22}{7}\times5\times5\times2\\ \\ \Rightarrow x=\left(\dfrac{5\times5\times2\times10\times10}{3000}\right)\ hrs\\ \\ =1\ hr\ 40\ min$
Question 11 of 45

11. Question
1 point(s)
A solid toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius. The height of the cone is 2 cm and diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. [Take latex]\pi=3.14 [/latex]]
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Hint

Volume of toy $=\dfrac{1}{3}\pi r^2\ H+\dfrac{2}{3}\pi r^3$
$=\dfrac{1}{3}\pi r^2\ (H+2r)\\ \\ =\dfrac{1}{3}\times\dfrac{22}{7}\times2\times2\ (2+2\times2)=25.12\ cm^3$
Volume of cylinder $=\pi r^2h$
$=\dfrac{22}{7}\times2\times2\ (2+2)\\ \\ =50.24\ cm^3$
Difference $=50.24-25.12=25.12\ cm^3$
Question 12 of 45

12. Question
1 point(s)
A tent is of the shape of a right circular cylinder upto a height of 3 metres and conical above it. The total height of the tent is 13.5 metres above the ground. Calculate the cost of painting the inner side of the tent at the rate of Rs 2 per square metre, if the radius of the base is 14 metres.
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Hint

$l=\sqrt{14^2+(10.5)^2}=\sqrt{306.25}=17.5\ m$
CSA of conical part $=\pi rl$
$=\dfrac{22}{7}\times14\times17.5=770\ m^2$
Total area $=(264+770)m^2=1034\ m^2$
Cost of painting $=2\times1034=2068$
Question 13 of 45

13. Question
1 point(s)
A solid is in the form of a right circular cylinder with hemispherical ends. The total height of the solid is 19 cm and the diameter of the cylinder and the hemispheres is 7 cm. Find the volume and total surface area of the solid.
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Hint

$h=19-2\times\dfrac{7}{2}=12\ cm$
Volume of solid $=\pi r^2h+2\times\dfrac{2}{3}\pi r^3$
$=\pi\times\left(\dfrac{7}{2}\right)^2\times12+\dfrac{4}{3}\pi\left(\dfrac{7}{2}\right)^3=641.66\ cm^3\\ \\ TSA=2\pi rh+2(2\pi r^2)\\ \\ =2\pi r(h+2r)\\ \\ =2\times\dfrac{22}{7}\times\dfrac{7}{2}\ (12+7)\\ \\ =418\ cm^2$
Question 14 of 45

14. Question
1 point(s)
A solid is composed of a cylinder with hemispherical ends. If the whole height of the solid is 100 cm and the diameter of cylindrical part and the hemispherical ends is 28 cm, find the cost of polishing the surface of the solid at the rate of 5 paise per sq. cm. $\left(use \ \pi=\dfrac{22}{7}\right) $
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Hint

$TSA=2\pi rh+2\times2\pi r^2\\ \\ =2\pi r\ (h+2r)\\ \\ =2\times\dfrac{22}{7}\times14\ (72+2\times14)\\ \\ =8800\ cm^2$
Cost $=8800\times\dfrac{5}{100}=440$
Question 15 of 45

15. Question
1 point(s)
A milk container is made of a metal sheet in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk which the container can hold when fully filled at Rs 20 per litre.
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Hint

Volume of frustum $=\dfrac{\pi h}{3}\ (R^2+r^2+Rr)$
$=\dfrac{\pi(16)}{3}\ (20^2+8^2+20\times8)\\ \\ =10459.428\ cm^3\\ \\ =10.459\ l$
Cost $=10.459\times20=209.18$
Question 16 of 45

16. Question
1 point(s)
A sphere and cube have same surface. Show that the ratio of the volume of sphere to that of the cube is $\sqrt6:\sqrt{\pi} $.
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Hint

$4\pi r^2=6a^2\\ \\ \Rightarrow \left(\dfrac{r}{a}\right)^2=\dfrac{3}{2\pi}\Rightarrow \dfrac{r}{a}=\sqrt{\dfrac{3}{2\pi}}\\ \\ \dfrac{Volume\ of\ sphere}{Volume\ of\ cube}=\dfrac{\dfrac{4}{3}\pi r^3}{a^3}\\ \\ =\dfrac{4\pi }{3}\ \left(\dfrac{\sqrt3}{2\pi}\right)^3\\ \\ =\sqrt{\dfrac{4\times3}{2\pi}}\\ \\ =\sqrt{\dfrac{6}{11}}$
Question 17 of 45

17. Question
1 point(s)
Find the mass of a 3.5 m long lead pipe, if the external diameter of pipe is 2.4 cm, thickness of the metal is 2 mm and $1\ cm^3 $ of lead weighs 11.4 kg.
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Hint

Volume of pipe $=\pi (r^1_2-r^2_1)\times h\\ \\ =\dfrac{22}{7}\ ((1.2)^2-(1.2-0.2)^2)\times350\\ \\ =1100\ (1.44-1)\\ \\ =484\ cm^3$
Weight $=484\times11.4=5.518\ kg$
Question 18 of 45

18. Question
1 point(s)
The difference between the outer and inner curved surface areas of a hollow right circular cylinder $14\ cm $ long, is $88\ cm^2 $. If the volume of metal used in making the cylinder is $176\ cm^3 $, find the outer and inner diameter of the cylinder.
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Hint

$2\pi r(R-r)=88\\ \\ \Rightarrow R-r=\dfrac{88}{2\times\dfrac{22}{7}\times14}\\ \\ \Rightarrow R-r=1$ …(1)
Also
$\pi h(R^2-r^2)=176\\ \\ \pi \times14(R+r)=176\\ \\ R+r=4$ ..(2)
Adding eqn (1) and (2) we get
$R=\dfrac{5}{2}\Rightarrow D=5$ and $r=\dfrac{3}{2},d=3$
Question 19 of 45

19. Question
1 point(s)
A bucket is in the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20 cm respectively. Find the capacity and surface area of the bucket. Also find the cost of milk which can completely fill the container at the rate of Rs 25 per litre. (use $\pi=3.14 $).
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Hint

Capacity of bucket $=\dfrac{\pi h}{3}\ (r^2_1+r^2_2+r_1r_2)$
$=3.14\times\dfrac{30}{3}\ (20^2+10^2+20\times10)\\ \\ =21980\ cm^3\\ \\ =21.980\ litres$
Cost of 1 litre = 25 Rs.
Cost of 21.980 litres $=25\times21.980=549.50$
Surface area of the bucket
$=\pi l(r_1+r_2)+\pi\ r^2\\ \\ =3.14\times 31.62\ (20+10)+3.14\times10^2\\ \\ =3292.6\ cm^2$
Question 20 of 45

20. Question
1 point(s)
A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 such gulab jamuns, each shaped like a cylinder with two hemispherical ends with total length 5 cm and diameter 2.8 cm.
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Hint

Volume of gulab jamun
$=3\pi r^2h+\dfrac{4}{3}\pi r^3\\ \\ =\dfrac{22}{7}\times1.4\times1.4\times\left(\dfrac{3\times2.2+4\times1.4}{3}\right)\\ \\ =\dfrac{75.152}{3}=25.05\ cm^3$
There are 45 gulab jamun
Volume $=45\times25.05=1127.28$
Volume $=1127.28\times\dfrac{30}{100}=338.184\ cm^3$
Question 21 of 45

21. Question
1 point(s)
A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has 8 gm mass.
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Hint

Volume of pole $=\pi r^2_1h_1+\pi r^2_2\ h_2$
$=(\pi\ (12)^2\times220)+(\pi(8)^2\times60)\\ \\ =3.14\ (31680+3840)\\ \\ =111532.8\ cm^3$
Mass $=111532.8\times8$
$=892262.4\ g$
Question 22 of 45

22. Question
1 point(s)
The perimeters of the ends of a frustum of a right circular cone are 44 cm and 33 cm. If the height of frustum is 16 cm, find its volume and total surface area.
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Hint

$2\pi R_1=44\Rightarrow R_1=7\ cm\\ \\ 2\pi R_2=33\ cm\Rightarrow R_2=5.25\ cm$
Volume $=\dfrac{1}{3}\pi h\ (R_1^2+R^2_2+R_1R_2)$
$=\dfrac{1}{3}\times\dfrac{22}{7}\times16\ (7^2+(5.25)^2+7\times5.25)=1900\ cm^3\\ \\ l=\sqrt{(7-5.25)^2+16^2}=16.10\ cm\\ \\ CSA=\pi(R_1+R_2)l\\ \\ =\dfrac{22}{7}\times16.10(7+5.25)=619.85\ cm^2\\ \\ TSA=CSA+\pi\ (R^2_1+R^2_2)\\ \\ =619.85+\dfrac{22}{7}\ (7^2+(5.25)^2)=860.475\ cm^2$
Question 23 of 45

23. Question
1 point(s)
A milk container is made of metal sheet in the shape of frustum of cone whose volume is $10459\dfrac{3}{7}cm^3 $. The radii of its lower and upper circular ends are 8 cm and 20 cm. Find the cost of metal used in making the container at the rate of Rs 1.40 per square cm.
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Hint

Volume of bucket $=\dfrac{73216}{7}\ cm^3$
$h=\dfrac{3V}{\pi(R^2+r^2+R\times r)}\\ \\ =\dfrac{3\times\dfrac{73216}{7}}{\dfrac{22}{7}\times(20^2+8^2+20\times8)}\\ \\ =\dfrac{219648}{13728}=16 \ cm\\ \\ TSA=\pi\ (R+r)\ \sqrt{(R-r)^2+h^2+\pi R^2}\\ \\ =(20+8)\ \sqrt{(20-8)^2+16^2+20}\pi\\ \\ =\dfrac{21120}{7}\ cm^2$
Cost $1.40 \ per\ cm^2$
Total cost $=\dfrac{21120}{7}\times1.40=4224\ Rs$
Question 24 of 45

24. Question
1 point(s)
A solid wooden toy is in the form of a cone mounted on a hemisphere. If the radii of hemisphere and base of cone are 4.2 cm each and the total height of toy is 10.2 cm, find the volume of wood used in the toy. Also, find the total surface area of toy.
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Hint

Volume of toy = Volume of hemisphere + Volume of cone
$=\dfrac{2}{3}\times\dfrac{22}{7}\times(4.2)^3+\dfrac{1}{3}\times\dfrac{22}{7}\times(4.2)^2\times6\\ \\ =266.112\ cm^3$
Question 25 of 45

25. Question
1 point(s)
A circus tent is in the form of right circular cylinder and right circular cone above it. The diameter and height of cylindrical part of tent at 126 m and 5 m respectively. The total height of tent is 21 m. Find the total cost of tent if the canvas used costs Rs 12 per $m^2 $.
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Hint

$l=\sqrt{r^2+h^2}=\sqrt{63^2+16^2}\\ \\ =\sqrt{4225}=65\ m$
Total canvas $=2\pi rh+\pi rl$
$=2\times\dfrac{22}{7}\times63\times5+\dfrac{22}{7}\times63\times65\\ \\ =1980+12870=14850\ m^2$
cost $14850\times12=178200$
Question 26 of 45

26. Question
1 point(s)
A right triangles, whose sides other than hypotenuse, are 3 cm and 4 cm is made to revolve about its hypotenuse. Find the volume of the double cone so formed. (see Fig.)
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Hint

hypotenuse $=\sqrt{9+16}=\sqrt{25}=5\ cm$
Area of $\triangle =\dfrac{1}{2}\times5\times h=\dfrac{1}{2}\times4\times3$
$\Rightarrow r=\dfrac{12}{5}=2.4\ cm$
Volume of double cone $=\dfrac{1}{3}\pi r^2\ (h_1+h_2)$
$=\dfrac{1}{3}\times3.14\times(2.4)^2\times5=30.14\ cm^2$
Question 27 of 45

27. Question
1 point(s)
Wax cylinder of diameter 21 cm and height 21 cm is chipped off and shaped to form a cone of maximum volume. The chipped off wax is recast into a solid sphere. Find the diameter of the sphere.
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Hint

Volume of cylinder $=\dfrac{22}{7}\times\left(\dfrac{21}{2}\right)^2\times21=7276.5\ cm^3$
Volume of cone $=\dfrac{1}{3}\times\dfrac{22}{7}\times\left(\dfrac{21}{2}\right)^2\times 21=2425.5\ cm^3$
Remaining volume $=7276.5-2425.5=4851\ cm^3$
A/Q $\dfrac{4}{3}\times\dfrac{22}{7}\times r^3=4851\\ \\ \Rightarrow r=10.5\ cm$
Question 28 of 45

28. Question
1 point(s)
The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be $ \dfrac{1}{27}$ th volume of the cone, at what height above the base is the section made?
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Hint

$V_2=\dfrac{1}{27}.V_1\\ \\ \Rightarrow \dfrac{V_2}{V_1}=\dfrac{1}{27}\\ \\ \Rightarrow \dfrac{\dfrac{1}{3}\pi r^2h}{10\pi r^2_1}=\dfrac{1}{27}\\ \\ \Rightarrow \left(\dfrac{r_2}{r_1}\right)^2\times\dfrac{h}{30}=\dfrac{1}{27}$
But $\dfrac{r_2}{r_1}=\dfrac{h}{30}$
So $h=10\ cm$
Thus at a height 20 cm above the base a small cone is cut.
Question 29 of 45

29. Question
1 point(s)
A container shaped like a circular cylinder having diameter 12 cm and height 15 cm is full of icecream. The ice-cream is filled into cones of height 12 cm and diameter 6 cm each having a hemispherical shape on the top. Find the number of such cones which can be filled with ice-cream.
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Hint

$\pi r^2_1h_1=n\left(\dfrac{1}{3}\pi r^2_2h_2+\dfrac{2}{3}\pi r^3_2\right)\\ \\ \Rightarrow \pi \times 6^2\times15=n\left(\dfrac{1}{3}\pi \times3^2\times12+\dfrac{2}{3}\pi\times3^2\right)\\ \\ \Rightarrow n=\dfrac{36\times15}{\dfrac{1}{3}\times9\times12+\dfrac{2}{3}\times27}=10$
Question 30 of 45

30. Question
1 point(s)
A metallic bucket is in the shape of a frustum of a cone. If the diameter of two circular ends of the bucket are 45 cm and 25 cm, respectively and the total vertical height is 24 cm, find the area of the metallic sheet used to make the bucket. Also find the volume of water it can hold.
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Hint

$l=\sqrt{24^2+(22.5-12.5)^2}=26\ cm$
Metal sheet needed $=\pi(R+r)l+\pi r^2$
$=\dfrac{22}{7}\times(22.5+12.5)\times26+\dfrac{22}{7}\times12.5\times12.5=3351.07\ cm^2$
Volume of water $=\dfrac{\pi h}{3}\ (R^2+r^2+Rr)$
$=\dfrac{22}{7}\times\dfrac{1}{3}\times24[(22.5)^2-(12.5^2+22.5\times12.5)]=23728\ cm^3$
Question 31 of 45

31. Question
1 point(s)
A spherical copper shell, of external diameter 18 cm, is melted and recast into a solid cone of base radius 14 cm and height $4\dfrac{3}{7} $ cm. Find the inner diameter of the shell.
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Hint

$\dfrac{4}{3}\pi (r^3_c-r_c)^3=\dfrac{1}{3}\pi r^2_c\times h_c\\ \\ \Rightarrow 4(9^3-r^3_1)=14^2\times\dfrac{31}{7}\\ \\ \Rightarrow 729-r^3_c=217\Rightarrow r^3_c=512\Rightarrow r_c=8\ cm$
Internal diameter $=16\ cm$
Question 32 of 45

32. Question
1 point(s)
A bucket is in the form of a frustum of a cone with a capacity of $12308.8\ cm^3$. The radii of the top and bottom circular ends of the bucket are 20 cm and 12 cm respectively. Find the height of the bucket and also the area of metal sheet used in making it. (Use $ \pi=3.14$).
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Hint

$12308.8=3.14\times\dfrac{h}{3}(20^2+12^2+20\times12)\\ \\ \Rightarrow h=15\ cm$
Surface area $=\pi r^2_2+\pi (r_1+r_2)l$
$l=\sqrt{15^2+8^2}=17\ cm$
Surface area $=3.14\times12^2+\dfrac{22}{7}\times32\times17=2160.32\ cm^2$
Question 33 of 45

33. Question
1 point(s)
Water is flowing at the rate of 15 km per hour through a pipe of diameter 14 cm into a rectangular tank which is 50 m long and 44 m wide. Find the time in which the level of water in the tank will rise by 21 cm.
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Hint

Volume of water flowing out of the pipe in 1 hour $=\pi r^2h$
$=\dfrac{22}{7}\times\dfrac{7}{100}\times\dfrac{7}{100}\times15000=231 \ m^3$
Volume of water in t hours $=231\ t$
Volume of water in the cuboid $=50\times44\times\dfrac{21}{100}=462\ m^3$
A/Q $231t=462\Rightarrow t=2\ hours$
Question 34 of 45

34. Question
1 point(s)
A drinking glass open at the top is in the shape of a frustum of a cone of height 24 cm. The diameters of its top and bottom circular ends are 18 cm and 4 cm respectively. Find the capacity and total surface area of the glass.
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Hint

$h=24,\ R=9,\ r=2$
Volume $=\dfrac{1}{3}\times\dfrac{22}{7}\ (9^2+2^2+9\times2)\times24$
$=\dfrac{1}{3}\times\dfrac{22}{7}\times103\times24=2589.714\ cm^3\\ \\ l=\sqrt{576+49}=25$
Surface area $=\pi(R+r)l+\pi r^2$
$=\dfrac{22}{7}\times11\times25+\dfrac{22}{7}\times4\\ \\ =\dfrac{22}{7}\ (279)=876.857\ cm^2$
Question 35 of 45

35. Question
1 point(s)
A solid in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and height of the cone is equal to its radius. Find the volume and surface area of the solid.
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Hint

Volume of solid = Volume of cone + Volume of hemisphere
$=\dfrac{1}{3}\pi r^2_1\ h_1+\dfrac{2}{3}\pi r^3_2\\ \\ =\dfrac{1}{3}\pi \times1\times1\times1+\dfrac{2}{3}\pi\times 1^3=\pi\ cm^3$
Question 36 of 45

36. Question
1 point(s)
An iron pillar has some part in the form of a right circular cylinder and the remaining in the form of a right circular cone. The radius of the base of each of the cone and the cylinder is 8 cm. The cylindrical part is 240 cm high and conical part is 36 cm high. Find the weight of the pillar if 1 cu cm of iron weighs 7.5 grams.
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Hint

Volume of pillar $=\left(\dfrac{22}{7}\times8\times8\times240\right)+\left(\dfrac{1}{3}\times\dfrac{22}{7}\times8^2\times36\right)=50688\ cm^3$
Weight $=50688\times7.8=395.37\ kg$
Question 37 of 45

37. Question
1 point(s)
The internal radii of the ends of a bucket, full of milk and of internal height 16 cm, are 14 cm and 7 cm. If this milk is poured into a hemispherical vessel, the vessel is completely filled. Find the internal diameter of the hemispherical vessel.
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Hint

$\dfrac{1}{3}\pi h(r^2_1+r^2_2+r_1r_2)=\dfrac{2}{3}\pi R^3\\ \\ \Rightarrow \dfrac{1}{3}\times \pi\times16(14^2+7^2+14\times7)\\ \\ =\dfrac{2}{3}\pi\times R^3\\ \\ \Rightarrow R^3=\dfrac{16}{2}\times343=14\ cm$
Internal diameter $=2\times14=28\ cm$
Question 38 of 45

38. Question
1 point(s)
A right triangle, whose sides are 6 cm and 8 cm (other than hypothenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed.
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Hint

hypothenuse $=\sqrt{36+64}=\sqrt{100}=10\ cm$
$=\dfrac{1}{2}\times10\times r=24\\ \\ \Rightarrow r=4.8\ cm$
Volume $=\dfrac{1}{3}\pi r^2\ (h_1+h_2)=\dfrac{1}{3}\times\dfrac{22}{7}\times(4.8)^2\times10=241.3\ cm^3$
surface area $=\pi r(6+8)$
$=\dfrac{22}{7}\times4.8\times14=211.2\ cm^2$
Question 39 of 45

39. Question
1 point(s)
A vessel in the form of a hemispherical bowl is full of water. Its water is emptied into a cylinder. The internal radii of bowl and the cylinder are $10\dfrac{1}{2} $ cm and 7 cm respectively. Find the height of water in the cylinder.
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Hint

$\pi r^2h=\dfrac{2}{3}\pi r^3\\ \\ \Rightarrow \pi\times4^2\times h=\dfrac{2}{3}\pi\times 6^3\\ \\ \Rightarrow h=9\ cm$
Question 40 of 45

40. Question
1 point(s)
A hollow cone is cut by a plane parallel to the base and the upper portion is removed. If the curved surface of the remainder is $\dfrac{8}{9} $ th of the of the curved surface of the whole cone, find the ratio of the line segments into which the cone’s altitude is divided by the plane.
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Hint

$\dfrac{h}{H}=\dfrac{r}{R}=\dfrac{1}{K}\\ \\ \dfrac{\pi rl}{\pi Rl}=\dfrac{l}{R^2}=\dfrac{1}{9}\\ \\ \Rightarrow R=3\\ \\ \Rightarrow \dfrac{h}{H}=\dfrac{1}{3}\\ \\ \dfrac{h}{H-h}=\dfrac{1}{3-1}=\dfrac{1}{2}$
Question 41 of 45

41. Question
1 point(s)
A sphere, of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by $3\dfrac{5}{9} $ cm. Find the diameter of the cylindrical vessel.
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Hint

$\pi r^2h=\dfrac{4}{3}\pi R^3\\ \\ r=\sqrt{\dfrac{4\times6^3\times9}{3\times32}}=9\ cm$
Diameter $=2\times9=18\ cm$
Question 42 of 45

42. Question
1 point(s)
Water flows out through a circular pipe whose internal radius is 1 cm, at the rate of 80 cm/ second into an empty cylindrical tank, the radius of whose base is 40 cm. By how much will the level of water rise in the tank in half an hour.
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Hint

Volume of cylinder pipe $=\pi r^2h$
$=\pi(1)^2\ (80)=251.4285\ cm^3/sec$
volume of water $=251.42\times30\times60=452571.5\ cm^3$
Volume of cylinder tank $=\pi (40)^2h=5028.57\times h$
$\Rightarrow h=\dfrac{452571.5}{5028.5714}=90\ cm$
Question 43 of 45

43. Question
1 point(s)
A tent consists of a frustum of a cone, surmounted by a cone. If the diameter of the upper and lower circular ends of the frustum be 14 m and 26 m respectively, the height of the frustum be 8 m and the slant height of the surmounted conical portion be 12 m, find the area of canvas required to make the tent. (Assume that the radii of the upper circular end of the frustum and the base of surmounted conical portion are equal.)
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Hint

$l=\sqrt{8^2+(13-7)^2}=\sqrt{8^2+6^2}=\sqrt{100}=10\ m\\ \\ CSA=\pi rl=\pi\ (13)\ (12+10)=286\ \pi \ m^2$
Question 44 of 45

44. Question
1 point(s)
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 20 cm and radius of the base is 3.5 cm, find the total surface area of the article.
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Hint

Volume of toy $=\pi r^2h-\left(2\times\dfrac{2}{3}\pi r^3\right)$
$=[\pi(3.5)^2\times10]-\left[2\times\dfrac{2}{3}\times \pi(3.5)^3\right]=205.251\ cm^3$
Question 45 of 45

45. Question
1 point(s)
A juice seller serves his customers using a glass as shown in figure. The inner diameter of the cylindrical glass is 5 cm, but the bottom of the glass has a hemispherical portion raised which reduces the capacity of the glass. If the height of the glass is 10 cm, find the apparent capacity of the glass and its actual capacity. (Use $\pi=3.14 $)
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Hint

Volume of cylinder $=3.14\times\left(\dfrac{5}{2}\right)^2\times10$
$=196.25\ cm^3$
Volume of hemisphere $=\dfrac{2}{3}\times3.14\times \left(\dfrac{5}{2}\right)^3=32.7\ cm^3$
Actual capacity $=196.25-32.7=163.55\ cm^3$
Apparent capacity $=196.25\ cm^3$