Question 513 Marks
How many spherical lead shots of diameter 4cm can be made out of a solid cube of lead whose edge measures 44cm.
AnswerDiameter of the spherical lead shots = 4cm
Edge length of the solid cube (a) = 44cm.
Let n be the number of spherical lead shots made out of the solid cube.
n × Volume of the spherical lead shots = volume of the solid cube
$\Rightarrow\frac{\text{volume of the solid cube}}{\text{volume of the spherical lead shots}}=\text{n}$
$\Rightarrow\frac{\text{a}^3}{\frac{4}{3}\pi\text{r}^3}=\text{n}$
$\Rightarrow\frac{44^3}{\frac{4}{3}\pi\times\Big(\frac{4}{2}\Big)^3}=\text{n}$
$\Rightarrow2541=\text{n}$
Hence, 2541 spherical lead shot can be made.
View full question & answer→Question 523 Marks
A circus tent is in the shape of cylinder surmounted by a conical top of same diameter. If their common diameter is 56m, the height of the cylindrical part is 6m and the total height of the tent above the ground is 27m, find the area of the canvas used in making the tent.
AnswerWe have, diameter of base of cylinder = d = 56m
Radius of base of cylinder = $\text{r}=\Big(\frac{\text{d}}{2}\Big)=\Big(\frac{52}{2}\Big)=28\text{m}$
Height of tent = 27m
Height of cylinder = 6m
Height of conical portion = 27 – 6 = 21m
Radius of conical portion, r = 28m
Now, slant height of cone $=\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$=\sqrt{28^2+21^2}=35\text{m}$
Now, Area of canvas used = curved surface area of cylinder + curve surface area of cone.
$=2\pi\text{rh}+\pi\text{rl}$
$=\frac{22}{7}\times28(12+35)$
$=4136\text{m}^2$ View full question & answer→Question 533 Marks
A hemispherical bowl of internal radius 15cm contains a liquid. The liquid is to be filled into cylindrical-shaped bottles of diameter 5cm and height 6cm. How many bottles are necessary to empty the bowl?
AnswerInternal radius of hemispherical bowl (r) = 15cm
$\therefore$ Volume of liquid filled in it $=\frac{2}{3}\pi\text{r}^3$
$=\frac{2}{3}\pi(15)^3\text{cm}^3$
$\frac{2}{3}\pi\times15\times15\times15\text{cm}^3=2250\pi\ \text{cm}^3$
Diameter of cylindrical bottle = 5cm
$\therefore$ Radius ($r_1$) $=\frac{5}{2}\text{cm}$
and height (h) = 6cm
$\therefore$ Volume of one bottle $=\pi\text{r}^2\text{h}$
$=\pi\Big(\frac{5}{2}\Big)^2\times6\text{cm}^3$
$=\pi\times\frac{25}{4}\times6=\frac{75}{2}\pi\ \text{cm}^3$
$\therefore$ Number of bottles so filled $=2250\pi\div\frac{75}{2}\pi$
$=\frac{2250\pi\times2}{75\pi}=60$
View full question & answer→Question 543 Marks
A toy is in the form of a cone mounted on a hemisphere with the same radius. The diameter of the base of the conical portion is 6cm and its height is 4cm. Determine the surface area of the toy. $(\text{Use}\ \pi=3.14)$
AnswerRadius of hemisphere and the cone are the same.
So, r = 3cm
Surface area of the cone
$=\pi\text{rl}$
$=3.14\times3\times\sqrt{3^2+4^2}$
$=47.1\text{cm}^2$
Surface area of the hemisphere
$=2\pi\text{r}^2$
$=2\times3.14\times9$
$=56.52\text{cm}^2$
Total surface area of the toy = surface area of the cone + surface area of the hemisphere
$= 47.1 + 56.52cm^2$
$= 103.62cm^2$
View full question & answer→Question 553 Marks
Sushant has a vessel, of the form of an inverted cone, open at the top, of height 11cm and radius of top as 2.5cm and is full of water. Metallic spherical balls each of diameter 0.5cm are put in the vessel due to which $\Big(\frac{2}{5}\Big)^\text{th}$ of the water in the vessel flows out. Find how many balls were put in the vessel. Sushant made the arrangement so that the water that flows out irrigates the flower beds. What value has been shown by Sushant?
AnswerLet the number of the balls be n.
Volume of water flows out = Volume of n spherical bolls
$\Rightarrow\frac{2}{5}\times\frac{1}{3}\pi\text{r}^2\text{h}=\text{n}\times\frac{4}{3}\pi\text{r}^3$
$\Rightarrow\frac{2}{5}\times(2.5)^2\times11=\text{n}\times4\Big(\frac{0.5}{2}\Big)^3$
$\Rightarrow27.5=\frac{0.5}{8}\text{n}$
$\Rightarrow\text{n}=440$
View full question & answer→Question 563 Marks
The surface area of a solid metallic sphere is $616cm^2$. It is melted and recast into a cone of height 28cm. Find the diameter of the base of the cone so formed $(\text{use}\pi=\frac{22}{7}).$
AnswerThe surface area of the metallic sphere is 616 square cm. Let the radius of the metallic sphere is r. Therefore, we have
$\Rightarrow4\pi\text{r}^2=616$
$4\times\frac{22}{7}\times\text{r}^2=616$
$\text{r}^2=\frac{6.16\times7}{22\times4}$
$\Rightarrow\text{r}^2=7\times7$
$\Rightarrow\text{r}=7$
Therefore, the radius of the metallic sphere is 7cm and the volume of the sphere is $\text{V}_1=\frac{4}{3}\pi\times(7)^3\text{cm}^3$
The sphere is melted to recast a cone of height 28cm. Let the radius of the cone is R cm. Therefore, the volume of the cone is $\text{V}_2=\frac{1}{3}\pi\times(\text{R})^2\times28\text{cm}^3$
Since, the volumes of the sphere and the cone are same; we have
$V_1 = V_2$
$\Rightarrow\frac{4}{3}\pi\times(7)^3=\frac{1}{3}\pi\times(\text{R})^2\times28$
$\Rightarrow\text{R}^2=\frac{4\times(7)^3}{28}$
$\Rightarrow\text{R}^2=7^2$
$\Rightarrow\text{R}=7$
Hence, the diameter of the base of the cone so formed is two times its radius, which is 14cm.
View full question & answer→Question 573 Marks
A canal is 300cm wide and 120cm deep. The water in the canal is flowing with a speed of 20km/h. How much area will it irrigate in 20 minutes if 8cm of standing water is desired?
AnswerVolume of water flows in the canal in one hour = width of the canal × depth of the canal × speed of the canal water
$= 3 x 1.2 \times 20 \times 1000m^3 = 72000m^3$
In 20 minutes the volume of water
$=\frac{72000\times20}{60}\text{m}^3=24000\text{m}^3$
Area irrigated in 20 minutes, if 8cm,
i.e., 0.08m standing water is required
Valume of water in canal = Area of field $\times\frac{8}{100}$
$24000=\text{A}\times\frac{8}{100}$
$\frac{24000\times100}{8}=\text{A}$
$\text{A}=300000\text{ m}^2$
$=30\text{ hectares.}$
View full question & answer→Question 583 Marks
Two cones have their heights in the ratio 1 : 3 and radii 3 : 1. What is the ratio of their volumes?
AnswerLet the radius of cones be 3r and r then the height of the cone be h and 3h
Then,
the volume of first cone
$\text{V}_1=\frac{1}{3}\pi\text{r}^2_1\text{h}_1$
$=\frac{1}{3}\pi(3\text{r})^2\times\text{h}$
$=\frac{1}{3}\pi\times9\text{r}^2\times\text{h}$
$=3\pi\text{r}^2\text{h}$
the volume of secound cone
$\text{V}_2=\frac{1}{3}\pi\text{r}^2_2\text{h}_2$
$\frac{1}{3}\pi\text{r}^2\times(3\text{h})$
$=\pi\text{hr}^2$
Therefore, ratio of their volume
$\Rightarrow\frac{\text{v}_1}{\text{v}_2}=\frac{3\pi\text{r}^2\text{h}}{\pi\text{hr}^2}$
$\Rightarrow\text{v}_1:\text{v}_2=3:1$
Hence, the required ratio are 3 : 1
View full question & answer→Question 593 Marks
A copper sphere of radius 3cm is melted and recast into a right circular cone of height 3cm. Find the radius of the base of the cone.
AnswerGiven radius of sphere = 3cm.
Volume of a sphere $=\frac{4}{3}\pi\text{r}^3$
$=\frac{4}{3}\times\pi\times3^3\text{cm}^3\ ...(\text{i})$
Given sphere is melted and recast into a right circular cone.
Given Hieght of circular cone = 3cm.
Volume of right circular cone $=\pi\text{r}^2\text{h}\times\frac{1}{3}$
$=\frac{\pi}{3}(\text{r})^2\times3\text{cm}^2\ ...(\text{ii})$
Equating 1 and 2 we get
$\frac{4}{3}\pi\times3^3=\frac{1}{3\pi}(\text{r})^2\times3$
$\text{r}^2=\frac{\frac{4}{3}\pi\times3^3}{\pi}$
$\text{r}^2=36\text{cm}$
$\therefore$ Radius of base of cone (r) = 6cm.
View full question & answer→Question 603 Marks
If the radii of the circular ends of a bucket 24cm high are 5cm and 15cm respectively, find the surface area of the bucket.
AnswerGiven height of a bucket (R) = 24cm Radius of circular ends of bucket 5cm and 15cm
$r_1 = 5cm; r_2 = 15cm$
Let 'l' be slant height of bucket
$\text{l}=\sqrt{\big(\text{r}_1-\text{r}_2\big)^2+\text{h}^2}$
$\Rightarrow\text{l}=\sqrt{\big(15-5\big)^2+24^2}$
$\Rightarrow\text{l}=\sqrt{100+576}=\sqrt{676}$
$\text{l}=26\text{cm}$
cueved surface area of bucket $= \pi\big(\text{r}_1+\text{r}_2\big)\text{l}+\pi\text{r}^2_2$
$=\pi (5+15)26+\pi(15)^2$
$=\pi (20)26+\pi(15)^2$
$=\pi (520-225)$
$= 745\pi \text{cm}^2$
$\therefore$ curved surface area of bucket $=745\pi\text{cm}^2$
View full question & answer→Question 613 Marks
The radii of the circular ends of a solid frustum of a cone are 33cm and 27cm and its slant height is 10cm. Find its total surface area.
AnswerThe slant height of the frustum of a cone is = 10cm. The radii of the upper and lower circles of the bucket are $r_1 = 33cm and r_2 = 27$cm respectively.
The total surface area of the frustum of the cone is
$\text{s}_1=\pi (\text{r}_1+\text{r}_2)\times\text{l}+\pi\text{r}^2_1+\pi\text{r}^2_2$
$=\pi\times(33+27)\times10+\pi\times33^2+\pi\times27^2$
$=600\pi+1089\pi+729\pi$
$=7599.42\text{cm}^2$
Hence total surface area is $7599.42cm^2$
View full question & answer→Question 623 Marks
A cone, a hemisphere and a cylinder stand on equal bases and have the same height. What is the ratio of their volumes?
AnswerLet r be the radius of the base.
And h is the height.
Here, h = r.
Now,
The ratio of their volumes will be
Volume of cone: volume of hemisphere: volume of a cylinder
$\frac{1}{3}\pi\text{r}^2\text{h}:\frac{2}{3}\pi\text{r}^3:\pi\text{r}^2\text{h}$
$\text{V}_1:\text{V}_2:\text{V}_3=\frac{1}{3}\pi\text{r}^3:\frac{2}{3}\pi\text{r}^3:\pi\text{r}^3$
Hence, $V_1: V_2: V_3 = 1: 2: 3$ View full question & answer→Question 633 Marks
A metallic sphere 1dm in diameter is beaten into a circular sheet of uniform thickness equal to 1mm. Find the radius of the sheet.
AnswerDiameter of a sphere = 1dm = 10cm
$\therefore$ Radius (r) $=\Big(\frac{10}{2}\Big)=5\text{cm}$
Volume of metal used in the sphere $=\Big(\frac{4}{3}\Big)\pi\text{r}^3$
$=\frac{4}{3}\times\pi(5)^3\text{cm}^3$
$=\frac{4}{3}\pi\times1256=\frac{500}{3}\pi\ \text{cm}^3$
Thickness of the sheet (h) $=1\text{mm}=\frac{1}{10}\text{cm}$
Let r be the radius of the circular sheet, then
$\pi\text{r}_1^2\text{h}=\frac{500}{3}\pi\Rightarrow\text{r}_1^2\text{h}=\frac{500}{3}$
$\Rightarrow\text{r}_1^2\times\frac{1}{10}=\frac{500}{3}$
$\Rightarrow\text{r}_1^2=\frac{500\times10}{3}=\frac{5000}{3}$
$\Rightarrow\text{r}=\sqrt{\frac{5000}{3}}\text{cm}^2\sqrt{16.67}=4.08\text{cm}$
$\therefore$ radius 4.08cm
View full question & answer→Question 643 Marks
A milk container of height 16cm is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as 8cm and 20cm respectively. Find the cost of milk at the rate of ₹ 44 per litre which the container can hold.
AnswerGiven that, height of milk container (h) = 16cm
Radius of lower end of milk container (r) = 8cm
and radius of upper end of milk container (R) = 20cm
$\therefore$ Volume of the milk container made of metal sheet in the form of a frustum of a cone
$=\frac{22\times16}{21}(400+64+160)$
$=\frac{22\times16\times624}{21}=\frac{219648}{21}$
$= 10459.42\text{cm}^3=10.45942\ \text{L}$
So, volume of the maik container is $10459.42cm^3$
$\because$ Cost of 1L milk = ₹ 44
$\therefore$ Cost of 10.45942L milk = 44 × 10.45942 = 460.24
Hence, the required cost of milk is ₹ 460.24 View full question & answer→Question 653 Marks
Lead spheres of diameter 6cm are dropped into a cylindrical beaker containing some water and are fully submerged. If the diameter of the beaker is 18cm and water rises by 40cm. find the number of lead spheres dropped in the water.
AnswerDiameter of cylindrical diameter = 18cm $\therefore$ Radius (R) $=\frac{18}{2}=9\text{cm}$ Height of water level (h) = 40cm $\therefore$ Volume of water $=\pi\text{R}^2\text{h}=\pi\times(9)^2\times40\text{cm}^3$ $=\pi\times28\times40=3240\pi\ \text{cm}^3$ Diameter of lead sphere = 6cm $\therefore$ Radius (r) $=\frac{6}{2}=3\text{cm}$ and volume $=\frac{4}{3}\pi\text{r}^3=\frac{4}{3}\pi(3)^3\text{cm}^3$ $=\frac{4}{3}\times27\pi=4\times9\pi\ \text{cm}^3$ $=36\pi\ \text{cm}^3$$\therefore$ Number of lead sphere $=\frac{3240\pi}{36\pi}=90$
View full question & answer→Question 663 Marks
A cone of maximum size is carved out from a cube of edge 14cm. Find the surface area of the cone and of the remaining solid left out after the cone carved out.
AnswerThe cone of maximum size that is carved out from a cube of edge 14cm will be of base radius 7cm and the height 14cm.
Surface area of the cone $=\pi\text{rl}+\pi\text{r}^2$
$=\frac{22}{7}\times7\times\sqrt{7^2+14^2}+\frac{22}{7}\times(7)^2$
$=\frac{22}{7}\times7\times\sqrt{245}+154$
$=(154\sqrt{5}+154)\text{cm}^2=154\sqrt{5}+1)\text{cm}^2$
Surface area of the cube $=6\times(14)^2$
$=6\times196=1176\text{cm}^2$
So, surface area of the rremaining solid left out after the cone is carved out
$=(1176-154+154\sqrt{5})\text{cm}^2$
$=(1022+154\sqrt{5})\text{cm}^2$
View full question & answer→Question 673 Marks
A 16m deep well with diameter 3.5m is dug up and the earth from it is spread evenly to form a platform 27.5m by 7m. Find the height of the platform.
AnswerAssume the well as a solid right circular cylinder.
Then, the radius of the solid right circular cylinder is$\text{r}=\frac{3.5}{2}=1.75\text{m}$
The well is 16m deep. Thus, the height of the solid right circular cylinder is 16m.
Therefore, the volume of the solid right circular cylinder is $\text{V}_1=\pi\text{r}^2\text{h}$
$=\frac{22}{7}\times(1.75)^2\times16\ \text{cubic meter}$
Let the height of the platform formed be x m.
The length and the breadth of the platform are l = 27.5m and b = 7m respectively.
Therefore, the volume of the platform is $V_2 = lbx = 27.5 \times 7 \times x = 192.5x$ cubic meters Since,
the well is spread to form the platform; the volume of the
well is equal to the volume of the platform. Hence,
we have $V_1 = V_2$
$\Rightarrow\frac{22}{7}\times(1.75)^2\times16=192.5\text{x}$
$\Rightarrow\ \text{x}=\frac{22}{7\times192.5}\times(1.75)^2\times16$
$\Rightarrow\frac{22\times3.0625\times16}{7\times192.5}$
$\Rightarrow0.8$
Hence, the height of the platform is 0.8m = 80cm.
View full question & answer→Question 683 Marks
A circus tent is cylindrical to a height of 3 metres and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53m, calculate the length of the canvas 5m wide to make the required tent.
AnswerDiameter of the tent = 105m $\therefore$ Radius (r) $=\Big(\frac{105}{2}\Big)\text{m}$
Height of cylindrical part ($h_1$) = 3m Slant height of conical part ($h_2$) = 53m
$\therefore$ Total surface area of the tent = curved surface area of the conical part + curved surface area of the cylindrical area $=\pi\text{rl}+2\pi\text{rh}_1$
$=\pi\text{r}(\text{l}+2\text{h}_1)=\frac{22}{7}\times\frac{105}{2}\ [53+2\times3]^2\text{m}^3$
$= 165 \times 599735m^2 $
Width of canvas = 5m
$\therefore$ Length of canvas = 9735 ÷ 5 = 1947m View full question & answer→Question 693 Marks
A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of the base of the cylinder or the cone is $24\ m$. The height of the cylinder is $11\ m$. If the vertex of the cone is 16m above the ground, find the area of the canvas required for making the tent. $(\text{Use}\ \pi=\frac{22}{7})$
AnswerSurface area of cylindrical part
$=2\pi\text{rh}$
$=2\times\frac{22}{7}\times\frac{24}{2}\times11$
$=829.71\text{m}^2$
Height of cone
$= 16 - 11$
$= 5m$
Surface area of conical part
$=\pi\text{rl}$
$\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$=\sqrt{(5)^2+(12)^2}$
$=13\text{m}$
S.A. of conical part
$=\frac{22}{7}\times12\times13$
$=490.29\text{m}^2$
Total area
$= 490.29 + 829.71$
$= 1319.99$
$= 1320m^2$
So canvas required will be $1320m^2$
View full question & answer→Question 703 Marks
A bucket is in the form of a frustum of a cone and holds $28.490$ litres of water. The radii of the top and bottom are $28\ cm$ and $21\ cm$ respectively. Find the height of the bucket.
AnswerGiven, volume of the frustum
$= 28.49 l = 28.49 \times 1000cm^3$ [$\because$ $1 L = 1000cm^3]$
$= 28490cm^3$
and radius of the top $ (r_1) = 28cm$
Radius of the bottom $ (r_2) = 21cm$
Let height of the bucket = h cm
Now, Volume of the buket
$=\frac{1}{3}\pi\text{h}(\text{r}_1^2+\text{r}_2^2+\text{r}_1\text{r}_2)=28490$ [given]
$=\frac{1}{3}\times\frac{22}{7}\times\text{h}(28^2+21^2+28\times21)=28490$
$\Rightarrow\text{h}(784+441+588)=\frac{28490\times3\times7}{22}$
$\Rightarrow1813\text{h}=1295\times21$
$\therefore\text{h}=\frac{1295\times21}{1813}=\frac{27195}{1813}=15\text{cm}$
View full question & answer→Question 713 Marks
A hemispherical bowl of internal radius 9cm is full of liquid. The liquid is to be filled into cylindrical shaped bottles each of radius 1.5cm and height 4cm. How many bottles are needed to empty the bowl?
AnswerGiven, radius of hemispherical bowl, r = 9cm
and radius of cylindrical bottles, R = 1.5cm
and height, h = 4cm
$\therefore$ Number of required cylindrical bottles
$\frac{\text{Volume of hemisphere bowl}}{\text{Volume of one cylindrical bottle} }$
$=\frac{\frac{2}{3}\pi\text{r}^3}{\pi\text{R}^2\text{h}}=\frac{\frac{2}{3}\pi\times9\times9\times9}{\pi\times1.5\times1.5\times4}=54$
View full question & answer→Question 723 Marks
The height of a solid cylinder is 15cm and the diameter of its base is 7cm. Two equal conical holes each of radius 3cm and height 4cm are cut off. Find the volume of the remaining solid.
Answer
The height of cylinder h = 15cm
Radius of cylinder r $=\frac72$
The volume of cylinder
$=\pi\text{r}^2\text{h}$
$=\pi\times\Big(\frac{7}{2}\Big)^2\times15\text{cm}^2$
$=183.75\pi$
The radius of conical holes = 3cm
Height of conical holes = 4cm.
The volume of conical holes
$=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\frac{1}{3}\pi\times9^3\times4$
$=12\pi\ \text{cm}^3$
Clearly,
The volume of remaining solid
= vol. of cylinder − 2 × vol. of cone
$=183.75\pi-24\pi$
$=501.6\text{cm}^3$ View full question & answer→Question 733 Marks
Find the volume of a solid in the form of a right circular cylinder with hemi-spherical ends whose total length is 2.7m and the diameter of each hemi-spherical end is 0.7m.
AnswerTotal height of solid = 2.7m
diameter of hemisphere = 0.7m
redius of hemisphere $=\frac{0.7}{2}\text{m}=0.35$
height of cylindrical = 2m.
So, that volume of the solid = 2 × volume of hemispherical part + volume of cylindrical part
$=2\times\frac{2}{3}\pi\text{r}^3+\pi\text{r}^2\text{h}$
$=\frac{4}{3}\times\frac{22}{7}\times0.35^3+\frac{22}{7}\times0.35^2\times2$
$=2\times\frac{22}{7}\times0.35\times0.35\times\Big[\frac{2\times0.35}{3}+1\Big]$
$=2\times22\times\frac{0.05}{100}\times\frac{0.35}{100}\Big[\frac{0.70+3}{3}\Big]$
$=2\times\frac{110\times35}{100\times100}\Big[\frac{3.70}{3}\Big]$
$2\times\frac{1.10\times35\times3.70}{300}$
$=0.9496$
$=0.955\text{m}^2$
View full question & answer→Question 743 Marks
A solid is in the form of a cylinder with hemispherical ends. Total height of the solid is $19\ cm$ and the diameter of the cylinder is $7\ cm$. Find the volume and total surface area of the solid.
AnswerVolume of cylinder $=\pi\text{r}^2\text{h}$
$=\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\times12$
$=462\text{cm}^3$
Volume of 2 hemisphere $=4\pi\text{r}^3$
$=\frac{4}{3}\times\frac{22}{2}\times\frac{7}{2}\times\frac{7}{2}\times\frac{7}{2}$
$=179.6\text{cm}^3$
Therefore,
Volume of solid $= 462 + 179.6$
$= 641.6cm^3$
Total surface area of the solid
$=2\pi\text{rh}+4\pi\text{r}^2$
$=2\pi\text{r}(\text{h}+2\text{r})$
$=2\times\frac{22}{7}\times\frac{7}{2}\Big(12+2\times\frac{7}{2}\Big)$
$=418\text{cm}^2$
View full question & answer→Question 753 Marks
Metal spheres, each of the radius $2\ cm$, are packed into a rectangular box of internal dimension $16\ cm \times 8\ cm \times 8\ cm$ when $16$ spheres are packed the box is filled with preservative liquid. Find the volume of this liquid.
AnswerThe radius of each of the metallic sphere is 2cm. Therefore, the volume of each metallic sphere is $\text{V}=\frac{4}{3}\pi\times (2)^3\text{cm}^3$
The total volume of the 16 spheres is
$\text{V}_1=16\times\frac{4}{3}\pi\times(2)^3\text{cm}^3$
The internal dimension of the rectangular box is 16cm × 8cm × 8cm. Therefore, the volume of the rectangular box is $V_2 = 16 \times 8 \times 8cm^3$
Therefore, the volume of the liquid is V2 - V1
$=16\times8\times8-16\times\frac{4}{3}\pi\times(2)^3$
$= 1024 - 536.03$
$= 488$
Hence volume of liquid is $488cm^3$
View full question & answer→Question 763 Marks
An iron pole consisting of a cylindrical portion $110\ cm$ high and of base diameter $12\ cm$ is surmounted by a cone $9\ cm$ high. Find the mass of the pole, given that $1\ cm^3$ of iron has 8 gram mass approximately. $\Big(\text{Use}:\ \pi=\frac{355}{115}\Big)$
AnswerDiameter of the base of the cylindrical pole = 12cm
$\therefore$ Radius (r) $=\Big(\frac{22}{2}\Big)=6\text{cm}$
Height of cylindrical portion $(h_1) = 110cm$
and height of conical portion $(h_2) = 9cm$
$\therefore$ Volume of the pole = Volume of cylindrical portion + volume of conical portion
$=\pi\text{r}^2\text{h}_1+\frac{1}{3}\pi\text{r}^2\text{h}_2$
$=\pi\text{r}^2\Big(\text{h}_1+\frac{1}{3}\text{h}_2\Big)=\frac{355}{115}(6)^2\Big(110+\frac{1}{3}\times9\Big)\text{cm}^3$
$=\frac{355}{115}\times36\times113\text{cm}^3$
$=\frac{71}{23}\times36\times113\text{cm}^3=\frac{288828}{23}\text{cm}^3$
Weight of $1cm^2 = 8gm$
$=\frac{288828\times8}{23\times1000}\text{kg}=100.46\text{kg}$
Note: If height of cone is 15cm instead of 9cm then the weight will be 102.24kg View full question & answer→Question 773 Marks
A solid is composed of a cylinder with hemispherical ends. If the length of the whole solid is $108\ cm$ and the diameter of the cylinder is $36\ cm$, find the cost of polishing the surface at the rate of $7$ paise per $cm^2$. $ (\text{Use}\ \pi=3.1416)$
AnswerHeight of the cylinder $= 108 - 36 = 72cm r = 18cm $C.S.A. of cylinder $=2\pi\text{rh}$ $=2\pi\times18\times72$ $=2\times3.14\times18\times72$$=8138.88\text{cm}^2$
C.S.A. of 2 hemisphere = surface of a sphere$=4\pi\text{r}^2$
Total surface $=2\pi\text{rh}+4\pi\text{r}^2$ $=2\pi\text{r}(\text{h}+2\text{r})$ $=2\times3.14\times18(72+2\times18)$$=12214.54\text{cm}^2$
Total cost $= 7 \times 12214.54 = 85502$ paise $= 855.02 Rs.$
View full question & answer→Question 783 Marks
A farmer runs a pipe of internal diameter 20cm from the canal into a cylindrical tank in his field which is 10m in diameter and 2m deep. If water flows through the pipe at the rate of 3km/h, in how much time will the tank be filled?
AnswerDiameter of pipe = 20cm
$\therefore$ Radius (r) $=\frac{20}{2}=10\text{cm}=\frac{10}{100}=\frac{1}{10}\text{m}$
Diameter of cylinderical tank = 10m
$\therefore$ Radius of tank (R) $=\frac{10}{2}=5\text{m}$
and depth {h} = 2m
$\therefore$ Volume of water filled in it $=\pi\text{r}^2\text{h}$
$=\frac{22}{7}\times\frac{1}{10}\times\frac{1}{10}\times\text{H}=\frac{1100}{7}$
$\text{H}=\frac{1100}{7}\times\frac{700}{22}=5000\text{m}=5\text{km}$
Speed of water = 3km/hr
$\therefore$ Time taken $=\frac{5}{3}\text{hr}=1\frac{2}{3}\text{hours}$
= 1 hr 40 minutes.
View full question & answer→Question 793 Marks
A conical vessel whose internal radius is $10\ cm$ and height 48cm is full of water. Find the volume of water. If this water is poured into a cylindrical vessel with internal radius $20\ cm$, find the height to which the water level rises in it.
AnswerInternal radius of the conical vessel $(r_1) = 10cm$
Height $(h_1) = 48cm$
$\therefore$ Volume of water in it $=\frac{1}{3}\pi\text{r}_1^2\text{h}_1=\frac{1}{3}\pi\times(10)^2\times48\text{cm}^3$
$=1600\pi\text{cm}^3=1600\times3.14=5024\text{cm}^2$
Now volume of water in the cylindrical vessel
$=1600\pi\ \text{cm}^3$
Internal radius $(r_2) = 20cm$
Let $h_1$ be the height of water level, then
$\pi\text{r}_2^2\text{h}_2=1600\pi\Rightarrow\pi\times(20)^2\text{h}_2=1600\pi$
$\Rightarrow400\pi\ \text{h}_2=1600\pi\Rightarrow\text{h}_2=\frac{1600\pi}{400\pi}=4$
Hence height $= 4cm$
View full question & answer→Question 803 Marks
If $r_1$ and $r_2$ be the radii of two solid metallic spheres and if they are melted into one solid sphere, prove that the radius of the new sphere is $(\text{r}_1^3+\text{r}_2^3).\frac{1}{3}$
AnswerLet $r_1$ and $r_2$ be the radii of two solid sphere$\therefore$ Volume of both the sphere
$=\frac{4}{3}\pi\text{r}_1^3+\frac{4}{3}\pi\text{r}_2^3=\frac{4}{3}\pi(\text{r}_1^3+\text{r}_2^3)$
Now volume of the new sphere so formed whose volume is
$=\frac{4}{3}\pi(\text{r}_1^3+\text{r}_2^3)$
$\therefore$ Radius of new sphere
$=\frac{\frac{4}{3}\pi\sqrt[3]{(\text{r}_1^3+\text{r}_2^3)}}{\frac{4}{3}\pi}=\sqrt[3]{\text{r}_1^3+\text{r}_1^3}$
$=(\text{r}_1^3+\text{r}_2^3)^\frac{1}{3}$
View full question & answer→Question 813 Marks
A factory manufactures 120,000 pencils daily. The pencils are cylindrical in shape each of length $25\ cm$ and circumference of base as $1.5\ cm$. Determine the cost of colouring the curved surfaces of the pencils manufactured in one day at ₹ $0.05$ per $dm^2.$
AnswerLength of the pencil, $h = 25\ cm$
circumference of the base $= 1.5\ cm$
Curved surface area of the pencil which needs to be painted will be
CSA = circumference × height
$= 1.5 \times 25cm^2$
$= 37.5cm^2$
$= 0.375dm^2$
Pencils manufactured in one day = 120000
So, the total area to be painted will be
$120000 \times 0.375dm^2 = 45000dm^2$
View full question & answer→Question 823 Marks
$16$ glass spheres each of radius $2\ cm$ are packed into a cuboidal box of internal dimensions $16\ cm \times 8\ cm \times 8\ cm$ and then the box is filled with water. Find the volume of water fdled in the box.
AnswerGiven, dimensions of the cuboidal
$= 16cm \times 8cm \times 8cm$
$\therefore$ Volume of the cuboidal $= 16 \times 8 \times 8$
$= 1024cm^3$
Also, given radius of one glass sphere = 2cm
$\therefore$ Volume of one glass sphere $=\Big(\frac{4}{3}\Big)\pi\text{r}^3$
$=\Big(\frac{4}{3}\times\frac{22}{7}\times(2)^3\Big)$
$=\Big(\frac{704}{21}\Big)=33.523\text{cm}^3$
Now, volume of 16 glass spheres $= 16 \times 33.523 = 536.37cm$
$\therefore$ Required volume of water = Volume of cuboidal – Volume of 16 glass spheres
$= 1024 - 536.37 = 487.6cm^3$
View full question & answer→Question 833 Marks
A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is $104\ cm$ and the radius of each of the hemispherical ends is $7\ cm$, find the cost of polishing its surface at the rate of Rs $10$ per $dm^2.$
AnswerWe have a solid composed of cylinder with hemispherical ends.
Radius of the two curved surfaces (r) = 7cm
Height of cylinder is h.
Total height of the body (h + 2r) = 104cm
So, total surface area is given by,
Total surface area = curved surface area of cylinder + 2 (curves surface area of hemisphere)
$=2\pi\text{rh}+2(2\pi\text{r}^2)$
$=2\pi\text{r}(\text{h}+2\text{r})$
$=2(3.14)(7)(104)\text{cm}^2$
$=4571.84\text{cm}^2$
Change the units of curved surface area as,
Total surface area $=\frac{4571.84}{100}\text{dm}^2$
$= 45.7184dm^2$
Cost of polishing the surface is Rs 10 per $dm^2.$
So total cost,
$= Rs (45.7184)(10)$
$= Rs 457.18$
View full question & answer→Question 843 Marks
Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere?
AnswerLet the radius of the hemisphere be r units.
Volume of a hemisphere = Surface area of the hemisphere
$\Rightarrow\frac{2}{3}\pi\text{r}^3=3\pi\text{r}^2$
$\Rightarrow\frac{2}{3}\text{r}=3$
$\Rightarrow\text{r}=\frac{9}{2}$
$\Rightarrow\text{d}=9\text{units}$
Hence, diameter of the hemisphere is equal to 9 units.
View full question & answer→Question 853 Marks
Find the mass of a $3.5\ m$ long lead pipe, if the external diameter of the pipe is $2.4\ cm$, thickness of the metal is $2\ mm$ and the mass of $1\ cm^3$ of lead is $11$.4g.
AnswerExternal diameter of a cylindrical pipe = 2.4cm
Radius (R) $=\Big(\frac{2.4}{2}\Big)=1.2\text{cm}$
Thickness of the pipe $=2\text{mm}=\Big(\frac{2}{10}\Big)=0.2\text{cm}$
Inner radius $(r) = 1.2 – 0.2 = 1.0cm$
Height (length) of the pipe (h) = 3.5m
$= 350cm$
Volume of the mass of the pipe $=\pi\text{h}(\text{R}^2-\text{r}^2)$
$=\frac{22}{7}\times350(1.2^2-1^2)\text{cm}^3$
$=1100(1.44-1)\text{cm}^3$
$=1100\times0.44=484\text{cm}^3$
weight of $1\ cm^3$ of lead $= 11$.4gm
total wweight of thew pipe = 484 × 11.4gm
$=5517.6\text{gm}$
$=\frac{5517.6}{1000}=5.5176\text{kg}$
$=5.518\text{kg}$
View full question & answer→Question 863 Marks
Marbles of diameter 1.4cm are dropped into a cylindrical beaker of diameter 7cm containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6cm.
AnswerDiameter of the marbles = 1.4cm
Radius of the marbles, $\text{r}=\frac{1.4}{2}=0.7\text{cm}$
Diameter of the cylindrical beaker = 7cm
Radius of the beaker $=\frac{7}{2}=3.5\text{cm}$
Rise in the level of water = 5.6cm = Height of cylindar
let the volume of marbles be n.
n × volume of the marbles = volume of the water raise
$\Rightarrow\text{n}=\frac{\text{volume of the water risen}}{\text{volume of the marble}}$
$\Rightarrow\text{n}=\frac{\pi\Big(\frac{7}{2}\Big)^2(5.6)}{\frac{4}{3}\pi(0.7)^3}=150$
Hence, the number marbbles will be 150.
View full question & answer→Question 873 Marks
A right circular cylinder having diameter $12\ cm$ and height $15\ cm$ is full ice-cream. The ice-cream is to be filled in cones of height $12\ cm$ and diameter 6cm having a hemispherical shape on the top. Find the number of such cones which can be filled with ice-cream.
AnswerWe have to find the number of cones which can be filled using the ice cream in the cylindrical vessel.
Radius of the cylinder $(r_1) = 6cm$
Height of cylinder $(h) = 15cm$
Radius of cone and the hemisphere on it $(r_2) = 3cm$
Height of cone $(l) = 12cm$
Let ‘n’ number of cones filled. So we can write it as,
n (Volume of each cone) = Volume of cylinder
So,
$(\text{n})\Big(\frac{1}{3}\pi\text{r}^2_2\text{l}+\frac{2}{3}\pi\text{r}^3_2\Big)=\pi\text{r}^2_1\text{h}$
$(\text{n})\Big(\frac{\text{r}^2_2(\text{l}+2\text{r}_2)}{3}\Big)=\text{r}^2_1\text{h}$
Now put the values to get,
$(\text{n})\Big(\frac{9(12+6)}{3}\Big)=36(15)$
$54\text{n}=540$
Therefore, $n = 10$
View full question & answer→Question 883 Marks
If the radii of the circular ends of a conical bucket which is $45\ cm$ high be 28cm and $7\ cm$, find the capacity of the bucket. $(\text{use}\ \pi=\frac{22}{7}).$
AnswerThe height of the conical bucket is $h = 45\ cm$. The radii of the bottom and top circles are $r_1 = 28cm$ and $r_2 = 7cm$ respectively.
The volume/ capacity of the conical bucket is,
$\text{V}=\frac{1}{3}\pi\big(\text{r}^2_1+\text{r}_1\text{r}_2+\text{r}^2_2\big)\times \text{h}$
$=\frac{1}{3}\pi\big(28^2+28\times7+7^2\big)\times45$
$=\frac{1}{3}\times\frac{22}{7}\times1029\times45$
$=22\times147\times15$
$= 48510\text{cm}^3$
Hence, volume $= 48510\ cm^3$
View full question & answer→Question 893 Marks
In a cylindrical vessel of diameter $24\ cm$, filled up with sufficient quantity of water, a solid spherical ball of radius 6cm is completely immersed. Find the increase in height of water level.
AnswerRadius of spherical ball $r = 6cm$, radius of cylindrical vessel $r_1 = 12cm$
Since, the ball completely immersed into the vessel, the water level is increased.
Let the height of increased level.
Therefore,
The volume of increase water level = volume of ball
$\pi\times(12)^2\times\text{x}=\frac{4}{3}\pi\times(6)^3$
$144\text{x}=\frac{4}{3}\times216$
$144\text{x}=4\times72$
$\text{x}=\frac{4\times72}{144}$
$\text{x}=2\text{cm}$
Hence, the level of water increased by 2cm. View full question & answer→Question 903 Marks
A hemispherical depression is cut out from one face of a cubical wooden block of edge $21\ cm$, such that the diameter of the hemisphere is equal to the edge of the cube. Determine the volume and total surface area of the remaining block.
AnswerGiven edge of wooden block $(a) = 21cm$
Given diameter of hemisphere = edge of cube
Radius $=\frac{21}{2}=10.5\text{cm}$
Volume of remaining block = Volume of box - Volume of hemisphere
$=\text{a}^3-\frac{2}{3}\pi\text{r}^3$
$=(21)^3-\frac{2}{3}\pi(10.5)^3$
$=6835.5\text{cm}^3$
Surface area of box $= 6a^2$
Curved surface area of hemisphere $=\pi\text{r}^2$
So remaining surface area of box = (1) - (2) + (3)
$=6\text{a}^2-\pi\text{r}^2+2\pi\text{r}^2$
$=6(21)^2-\pi(10.5)+2\pi(10.5)^2$
$=2992.5\text{cm}^2$
$\therefore$ Remaining surface area of box $= 2992.5cm^2$
Volume of remaining block $= 6835.5cm^3$
View full question & answer→Question 913 Marks
Rain water, which falls on a flat rectangular surface of length $6\ m$ and breadth $4\ m$ is transferred into a cylindrical vessel of internal radius $20\ cm$. What will be the height of water in the cylindrical vessel if a rainfall of 1cm has fallen?
AnswerThe fallen rains are in the form of a cuboid of height 1cm, length 6m = 600cm and breadth 4m = 400 cm. Therefore, the volume of the fallen rains is
$V = 600 \times 400 \times 1 = 240000cm^3$
The fallen rains are transferred into a cylindrical vessel of internal radius $r_1 = 20cm$. Let, the height of the water in the cylindrical vessel is $h_1cm$. Then, the volume of the water in the cylinder is
$\text{V}_1=\pi\text{r}^2_1\text{h}_1=\frac{22}{7}\times(20)^2\times\text{h}_1$
Since, the volume of the water in the cylinder is same as the volume of the rainfalls, we have
$V_{1 =}V$
$\Rightarrow\frac{22}{7}\times(20)^2\times\text{h}_1=240000$
$\Rightarrow\text{h}_1=\frac{240000\times7}{(20)^2\times22}$
$\Rightarrow190.9$
Therefore, the height of the water in the cylinder is 190.9cm.
View full question & answer→Question 923 Marks
A petrol tank is a cylinder of base diameter $21\ cm$ and length 18cm fitted with conical ends each of axis length 9cm. Determine the capacity of the tank.
AnswerDiameter of cylindrical part = $21cm$
Radius (r) $=\Big(\frac{21}{2}\Big)\text{cm}$
Height of cylindrical part $(h_1) = 18cm$
and height of each conical part $(h_2) = 9cm$
Total volume of the tank $=2\times\frac{1}{3}\pi\text{r}^2\text{h}_2+\pi\text{r}^2\text{h}_1$
$=\pi\text{r}^2\Big(\frac{2}{3}\text{h}_2+\text{h}_1\Big)$
$=\frac{22}{7}\Big(\frac{21}{2}\Big)^2\Big(\frac{2}{3}\times9+18\Big)\text{cm}^3$
$=\frac{22}{7}\times\frac{441}{4}(6+18)\text{cm}^3$
$=\frac{11\times63}{2}\times24\text{cm}^3=8316\text{cm}^3$ View full question & answer→Question 933 Marks
How many spherical bullets each of $5\ cm$ in diameter can be cast from a rectangular block of metal $11dm \times 1m \times 5dm$?
AnswerWe are given a metallic block of dimension $= 11dm \times 1m \times 5dm.$
We know that, 1dm $= 10^{-1}m$
So, the volume of the given metallic block is $= 11 \times 10^{-1}\times 1 \times 5 \times 10^{-1} = 55 \times 10^{-2}m^3$
We want to know how many spherical bullets can be formed from this volume of the metallic block. It is given that the diameter of each bullet should be 5 cm.
We know,
$\text{Volume of a sphere}=\frac{4}{3}\pi(\text{r})^3$
Here, $r = 25 \times 10^{-3}m$
Let the no. of bullets formed be n.
We know that the sum of the volumes of the bullets formed should be equal to the volume of the metallic block.
$\Rightarrow55\times10^{-2}=\text{n}\times\frac{4}{3}\times\frac{22}{7}\times(25\times10^{-3})^3$
$\text{n}=\frac{55\times3\times7\times10^{-2}}{4\times22\times25\times25\times25\times10^{-9}}$
$=\frac{21\times10^7}{(2\times5)^3\times25}$
$=8400$
Hence the no. of bullets that can be formed is 8400.
View full question & answer→Question 943 Marks
A well of diameter 3m is dug 14m deep. The earth taken out of it is spread evenly all around it to a width of 4m to form an embankment. Find the height of the embankment.
AnswerWe have,
Radius of well $\text{r}=\frac{3}{2}\text{m}$
Depth of well = 14m
The volume of well
$=\pi\times\Big(\frac{3}{2}\Big)^2\times14$
$=\pi\times\frac{9}{4}\times14$
$=\frac{63}{2}\pi\ \text{cm}^3$
Therefore,
Volume of earth dugout = volume of well $=\frac{63}{2}\pi\ \text{m}^3$
Let h be the height of embankment.
Clearly, embankment form a cylindrical shell whose inner and outer radius are
$\frac{3}{2}\text{m}\ \text{and}\ \Big(\frac{3}{2}+4\Big)=\frac{11}{2}\text{m}$
Volume of embankment View full question & answer→Question 953 Marks
Prove that the surface area of a sphere is equal to the curved surface area of the circumscribed cylinder.
AnswerLet radius of a sphere be r
Curved surface area of sphere $=4\pi\text{r}^2$
$\text{S}_1=4\pi\text{r}^2$
let radius of cylinder be 'r' cm
Height of cylinder be '2r' cm
Curved surface area of cylinder $=2\pi\text{r}\text{h}$
$\text{S}_2=2\pi\text{r}(2\text{r})=4\pi\text{r}$
$S_1$ and $S_2$ are equal. Hence proved.
So curved surface area of sphere = surface area of cylinder.
View full question & answer→Question 963 Marks
The sum of the radius of base and height of a solid right circular cylinder is $37\ cm$. If the total surface area of the solid cylinder is $1628cm^2$, find the volume of cylinder.$\Big(\text{use}\ \pi=\frac{22}{7}\Big)$
AnswerLet the radius of the base of the cylinder be r cm
Let the height be h cm
Now given that r + h = 37cm
Total surface area $= 1628cm^2 ...(1)$
Total surface area of the cylinder $=2\pi\text{r}^2+2\pi\text{rh}$
$=2\pi\text{r}(\text{r}+\text{h})$
$=2\pi\text{r}\times37$
$=74\pi\text{r}\ ...(2)$
From equating (1) and (2) we get
$74\pi\text{r}=1628$
$\Rightarrow\text{r}=\frac{1628}{74\pi}=7\text{cm}^3$
Thus, the height will be 37 - 7 = 30cm
Thus, the volume of the cylinder $=\pi\text{r}^2\text{h}=\pi(7)^2\times30=4620\text{cm}^3$
Hence the volume is $4620cm^3$
View full question & answer→Question 973 Marks
If $r_1$ and $r_2$ denote the radii of the circular bases of the frustum of a cone such that $r_1 > r_2$, then write the ratio of the height of the cone of which the frustum is a part to the height fo the frustum.
Answer
Since, $\triangle\text{VO'B}\sim\triangle\text{VOA}$
Therefore,
In $\triangle\text{VO'B}\sim\triangle\text{VOA}$
$\frac{\text{O'B}}{\text{OA}}=\frac{\text{O'V}}{\text{OV}}$
$\frac{\text{r}_2}{\text{r}_1}=\frac{\text{h}-\text{h}_1}{\text{h}}$
$\frac{\text{r}_2}{\text{r}_1}=1-\frac{\text{h}_1}{\text{h}}$
$\frac{\text{h}_1}{\text{h}}=1-\frac{\text{r}_2}{\text{r}_1}$
$\frac{\text{r}_1-\text{r}_2}{\text{r}_1}$
Hence,
The ratio of the height of cone of which the frustum is a part to the height fo the frustum.
$\frac{\text{OV}}{OO'}=\frac{\text{h}}{\text{h}_1}=\frac{\text{r}_1}{\text{r}_1-\text{r}_2}$
Hence, $h : h_1 = r_1 : (r_1 - r_2)$ View full question & answer→Question 983 Marks
A wall 24m, 0.4m thick and 6m high is constructed with the bricks each of dimensions $25\ cm x 16\ cm x 10\ cm$. If the mortar occupies $\Big(\frac{1}{10}\Big)$ of the volume of the wall, then find the number of bricks used in constructing the wall.
AnswerDimensions of the wall are $24 m \times 0.4 m \times 6 m$
Volume of the wall $=24 m \times 0.4 m \times 6 m=57.6 m^3$
Dimensions of the bricks are $25 m \times 16 m \times 10 m$
Volume of each brick $=4000 cm^3=0.004 m^3$
Volume of mortor $=\frac{1}{10} \times$ Volume of the wall $=\frac{1}{10} \times 57.6=5.76 m^3$
Volume of all the bricks $=$ Volume of the wall - volume of motor $=57.6-5.76 m^3$
Let the number of bricks used in making the wall be $n$.
$\frac{\text{volume of all the bricks}}{\text{volume of each bricks}}=\text{n}$
$\Rightarrow\frac{51.84}{0.004}=\text{n}$
$\Rightarrow\text{n}=12960$
Hence, $12960$ bricks are used to make the wall.
View full question & answer→Question 993 Marks
A solid metallic sphere of radius 10.5cm is melted and recast into a number of smaller cones, each of radius 3.5cm and height 3cm. Find the number of cones so formed.
AnswerThe volume of the solid metallic sphere $=\frac{4}{3}\pi(10.5)^3\text{cm}^3$
Volume of a cone of radius 3.5cm and height 3cm $=\frac{1}{3}\pi(3.5)^2\times3\text{cm}^3$
Number of cones so formed.
The volume of the solid metallic sphere $=\frac{4}{3}\pi(10.5)^3\text{cm}^3$
Volume of a cone of radius 3.5cm and hieght.
$3\text{cm}=\frac{1}{3}\pi(3.5)^2\times3\text{cm}^3$
Number of cones so formed
$\frac{\frac{4}{3}\pi\times10.5\times10.5\times10.5}{\frac{1}{3}\pi\times3.5\times3.5\times3}=126$
View full question & answer→Question 1003 Marks
The largest sphere is carved out of a cube of side 10.5cm. Find the volume of the sphere.
Answer
The side of cube a = 0.5cm.
Since, a largest sphere is curved out of that cube i.e., radius of sphere,
$\text{r}=\frac{\text{a}}{2}$
$\text{r}=\frac{10.5}{2}\text{cm}$
$\text{r}=5.25\text{cm}$
The volume of sphere
$=\frac{4}{3}\pi(5.25)^3$
$=\frac{4}{3}\times\frac{22}{7}\times5.25\times5.25\times5.25$
$=22\times0.75\times1.75\times21$
$=606.375\text{cm}^3$ View full question & answer→