Question 12 Marks
Determine the ratio of the volume of a cube to that of a sphere which will exactly fit inside thecube.
AnswerLet edge of the cube $=a$
volume of the cube $=a \times a \times a=a^3$
The sphere, which exactly fits in the cube, has radius $=\frac{a}{2}$
Therefore, volume of sphere
$=\frac{4}{3} \pi r^3=\frac{4}{3} \times \frac{22}{7}\left(\frac{a}{2}\right)^3=\frac{4}{3} \times \frac{22}{7} \times \frac{a^3}{8}=\frac{11}{21} a^3$
Volume of cube : volume of sphere
$=a^3: \frac{11}{21} a^3 $
$=1: \frac{11}{21} $
$ =21: 11$
View full question & answer→Question 22 Marks
A solid is in the form of a cone standing one hemi-sphere with both their radii being equal to 8cm and the height of cone is equal to its radius. Find, in terms of $\pi,$ the volume of the solid.
AnswerVolume of the solid
$=\frac{1}{3} \pi r^2 r+\frac{2}{3} \pi r^3$
$ =\frac{1}{3} \times \pi \times 8^3+\frac{2}{3} \times \pi \times 8^3 $
$ =\pi 8^3$
$ =512 \pi cm ^3$
View full question & answer→Question 32 Marks
A largest sphere is to be carved out of a right circular cylinder of radius $7$ cm and height $14$ cm.Find the volume of the sphere.
AnswerRadius of largest sphere that can be formed inside the cylinder should be equal to the radius ofthe cylinder.
Radius of the largest sphere $= 7 cm$
Volume of sphere
$=\frac{4}{3} \pi 7^3$
$=\frac{4}{3} \times \frac{22}{7} \times 7 \times 7$
$=\frac{4312}{3} $
$=1437 cm ^3$
View full question & answer→Question 42 Marks
The given figure shows the cross-section of a cone, a cylinder and a hemisphere all with the same diameter $10$ cm and the other dimensions are as shown. Calculate: the density of the material if its total weight is $1.7$ kg
Answer
Total weight of the solid $= 1.7 kg$
Total volume of solid $= 1.12 gm/cm^3$
$\therefore \text { Density }=\frac{1.7 \times 1000}{1519.0476} gm / cm ^3=1.119 gm / cm ^3$
$ \Rightarrow \text { Density }=1.12 gm / cm ^3$ View full question & answer→Question 52 Marks
A solid metallic hemisphere of diameter $28$ cm is melted and recast into a number of identicalsolid cones, each of diameter $14$ cm and height $8$ cm. find the number of cones so formed.
AnswerGiven -diameter $14\ cm;$ height $8\ cm; n =$ number of cones
Volume of hemisphere $= n \times $ Volume of $1$ cone
$\therefore \frac{2}{3} \pi R ^3= n \times \frac{1}{3} \pi r ^2 h $
$ \therefore \frac{2}{\not 3} \not R ^3= n \times \frac{1}{\not 3} \pi^2 r ^2 h$
$\therefore 2(14)^3= n (7)^2(8)$
$ \therefore \frac{2 \times 14 \times 14 \times 14}{7 \times 7 \times 8}= n$
$\therefore n =14$
View full question & answer→Question 62 Marks
Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r cm.
Answer
For the volume of cone to be largest, h = r cm
Volume of the cone
$\frac{1}{3} \pi r^2 h$
$ =\frac{1}{3} \pi \times r^2 \times r$
$ =\frac{1}{3} \pi r^3$ View full question & answer→Question 72 Marks
A hemi-spherical bowl has negligible thickness and the length of its circumference is $198\ cm.$ Find the capacity of the bowl.
AnswerLet r be the radius of the bowl.
$\therefore 2pir=198$
$\Rightarrow r=\frac{198 \times 7}{2 \times 22}$
$\Rightarrow r=31.5 cm$
Capacity of the bowl = $\frac{2}{3} \pi r^3$
$=\frac{2}{3} \times \frac{22}{7} \times(31.5)^3$
$=65488.5 cm ^3$
View full question & answer→Question 82 Marks
A cone and hemisphere have the same base and the same height. Find the ratio between their volumes.
AnswerLet the radius of base be 'r' and the height be 'h'
Volume of cone, $V_c$
$=\frac{1}{3} \pi r^2 h$
Volume of hemisphere, $V_h$
$=\frac{2}{3} \pi r^2 h$
$\frac{V_c}{V_h}=\frac{\frac{1}{3} \pi r^2 r}{\frac{2}{3} \pi r^3} $
$ \Rightarrow \frac{V_c}{V_h}=\frac{1}{2}$
View full question & answer→Question 92 Marks
The total area of a solid metallic sphere is $1256 cm^2$. It is melted and recast into solid right circular cones of radius $2.5 cm$ and height $8 cm$. Calculate: the number of cones recasted [π = 3.14]
Answer$ \therefore r=10 $
Volume of sphere $=\frac{4}{3} \pi r^3=\frac{4}{3} \times \frac{22}{7} \times 10 \times 10 \times 10=\frac{88000}{21} cm ^3$
volume of right circular cone $=$
$ \frac{1}{3} \pi r^2 h=\frac{1}{3} \times \frac{22}{7} \times(2.5)^2 \times 8=\frac{1100}{21} cm ^3 $
Number of cones
$=\frac{88000}{21} \div \frac{1100}{21} $
$=\frac{88000}{21} \times \frac{21}{1100} $
$ =80$
View full question & answer→Question 102 Marks
The total area of a solid metallic sphere is $1256 cm^2$. It is melted and recast into solid right circular cones of radius $2.5 cm$ and height $8 cm$. Calculate: the radius of the solid sphere.
AnswerTotal area of solid metallic sphere $=1256 cm ^2$
Let radius of the sphere is $r$ then
$4 \pi r^2=1256 $
$\Rightarrow 4 \times \frac{22}{7} r^2=1256 $
$ \Rightarrow r^2=\frac{1256 \times 7}{4 \times 22}$
$ \Rightarrow r^2=\frac{157 \times 7}{11} $
$ \Rightarrow r^2=\frac{1099}{11} $
$\Rightarrow r=\frac{1099}{11}$
$ \Rightarrow r=\sqrt{99.909} $
$ \Rightarrow r=9.995 cm \approx 10 cm $
View full question & answer→Question 112 Marks
The surface area of a solid metallic sphere is $2464 cm^2.$ It is melted and recast into solid right circular cones of radius $3.5 cm$ and height $7 cm.$ Calculate: the radius of the sphere $($Take $\pi = 22/7)$
AnswerTotal surface area of the sphere $= 4\pi r^2,$ where r is the radius of the sphere.
Thus,
$4 \pi r^2=2464 cm ^2$
$\Rightarrow 4 \times \frac{22}{7} \times r^2=2464$
$\Rightarrow r^2=196$
$\Rightarrow r=14 cm $
View full question & answer→Question 122 Marks
The internal and external diameters of a hollow hemi-spherical vessel are $21$ cm and $28$ cm respectively. Find volume of material of the vessel.
AnswerExternal radius $(R)=14 cm$
Internal radius $(r)=\frac{21}{2} cm$
$\frac{2}{3} \pi\left(R-r^3\right) $
$=\frac{2}{3} \times \frac{22}{7}\left((14)^3-\left(\frac{21}{2}\right)^3\right) $
$ \frac{44}{21}(2744-1157.625)$
$ =\frac{44}{21} \times 1586.375$
$ =3323.83 cm ^3$
View full question & answer→Question 132 Marks
The internal and external diameters of a hollow hemi-spherical vessel are $21$ cm and $28$ cm respectively. Find: external curved surface area .
AnswerExternal radius $(R) = 14\ cm$
Internal radius $(r)=\frac{21}{2}\ cm$
External curved surface area $=$
$2 \pi R^2$
$ =2 \times \frac{22}{7} \times 14 \times 14$
$ =1232 cm ^2$
View full question & answer→Question 142 Marks
The internal and external diameter of a hollow hemispherical vessel are $21$ cm and $28$ cm
respectively Find: internal curved surface area
AnswerExternal radius $(R) = 14 cm$
Internal radius $(r)=\frac{21}{2} cm$
Internal curved surface area $=$
$2 \pi r^2 $
$=2 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} $
$=693 cm ^3$
View full question & answer→Question 152 Marks
A solid metal sphere is cut through its center into $2$ equal parts. If the diameter of the sphere is $3 \frac{1}{2}$ cm, find the total surface area of each part correct to two decimal places.
AnswerDiameter of sphere $=3 \frac{1}{2} cm =\frac{7}{2} cm$
Therefore, radius of sphere $=\frac{7}{4} cm$
Total curved surface area of each hemispheres $=$
$2 \pi r^2+\pi r^2=3 \pi r^2$
$ =3 \times \frac{22}{7} \times \frac{7}{4} \times \frac{7}{4} $
$ =28.88 cm ^2$
View full question & answer→Question 162 Marks
If the number of square centimeters on the surface of a sphere is equal to the number of cubiccentimeters in its volume, what is the diameter of the sphere?
AnswerLet r be the radius of the sphere.
Surface area $=4 \pi r^2$ and volume $=\frac{4}{3} \pi r^3$
According to the condition:
$4 \pi r^2=\frac{4}{3} \pi r^3$
$ \Rightarrow \frac{r^3}{r^2}=4 \pi \times \frac{3}{4 \pi} $
$ \Rightarrow r =3 cm$
Diameter of sphere $= 2 \times 3 cm = 6 cm$
View full question & answer→Question 172 Marks
How many balls each of radius $1 cm$ can be made by melting a bigger ball whose diameter is $8cm.$
AnswerDiameter of bigger ball $= 8 cm$
Therefore, Radius of bigger ball $= 4 cm$
$\text { Volume }=\frac{4}{3} \pi r^3=\frac{4}{3} \pi \times 4 \times 4 \times 4=\frac{265 \pi}{3} cm ^3$
$ \text { Number of balls }=\frac{\frac{256 \pi}{3}}{\frac{4 \pi}{3}}=\frac{256 \pi}{3} \times \frac{3}{4 \pi}=64$
View full question & answer→Question 182 Marks
A spherical ball of lead has been melted and made into identical smaller balls with radius equalto half the radius of the original one. How many such balls can be made?
AnswerLet the radius of spherical ball = r
$\therefore$ Volume $=\frac{4}{3} \pi r^3$
Radius of smaller ball $=\frac{r}{2}$
$\therefore$ Volume of smaller ball $=\frac{4}{3} \pi\left(\frac{r}{2}\right)^3=\frac{4}{3} \pi \frac{r^3}{8}=\frac{\pi r^3}{6}$
Therefore, number of smaller balls made out of the given ball =
$\frac{\frac{4}{3} \pi r^3}{\pi \frac{r^3}{6}}=\frac{4}{3} \times 6=8$
View full question & answer→Question 192 Marks
The volume of a sphere is $38808 cm ^3$ find its diameter and the surface area.
AnswerVolume of the sphere $=38808 cm ^3$
Let radius of sphere $=r$
$\therefore \frac{4}{3} \pi r^3=38808 $
$ \Rightarrow \frac{4}{3} \times \frac{22}{7} \times r^3=38808$
$ \Rightarrow r^3=\frac{38808 \times 7 \times 3}{4 \times 22}=9261$
$\Rightarrow r=21 cm$
$\therefore \text { diameter }=2 r=21 \times c m=42 cm $
$\text { Surface area }=4 \pi r^2=4 \times \frac{22}{7} \times 21 \times 21 cm ^3=5544 cm ^3$
View full question & answer→Question 202 Marks
The circumference of the base of a $12$ m high conical tent is $66$ m. Find the volume of the aircontained in it.
AnswerCircumference of the conical tent $= 66 m$
and height (h) $= 12 m$
$\therefore$ Radius $=\frac{c}{2 \pi}=\frac{66 \times 7}{2 \times 22}=10.5 m$
Therefore, volume of air contained in it $=\frac{1}{3} \pi r^2 h$
$=\frac{1}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times 12 m ^3 $
$ =1386 m ^3$
View full question & answer→Question 212 Marks
The volume of a conical tent is $1232 m^3$ and the area of the base floor is $154 m^2. $ Calculate the: length of the canvas required to cover this conical tent if its width is $2 m.$
AnswerLet l be the slant height of the conical tent, then
$=l=\sqrt{h^2+r^2}$
$\therefore l=\sqrt{h^2+r^2}=\sqrt{(24)^2+(7)^2}=\sqrt{576+49}=\sqrt{625}=25 m$
The area of the canvas required to make the tent $=\pi r l m^2$
$\therefore \pi r l=\frac{22}{7} \times 7 \times 25 m^2=550 m^2$
Length of the canvas required to cover the conical tent of its width $2 m =$ $\frac{550}{2}=275 m$
View full question & answer→Question 222 Marks
The volume of a conical tent is $1232 m^3$ and the area of the base floor is $154 m^2$. Calculate the: height of the tent.
AnswerLet h be the height of the conical tent, then the volume $=$
$\frac{1}{3} \pi r^2 h m^3$
$\therefore \frac{1}{3} \pi r^2 h=1232$
$ \Rightarrow \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times h$
$ \Rightarrow h=\frac{1232 \times 3}{22 \times 7}=24$
Hence, radius of the base of the conical tent i.e. the floor $= 7 m$
View full question & answer→Question 232 Marks
The volume of a conical tent is $1232 m^3$ and the area of the base floor is $154 m^2.$ Calculate the: radius of the floor.
AnswerLet $r$ be the radius of the base of the conical tent, then area of the base floor $= πr^2m^2$
$\pi r^2=154$
$\frac{22}{7} \times r^2=154$
$\Rightarrow r^2=\frac{154 \times 7}{22}=49$
$\Rightarrow r=7$
Hence, radius of the base of the conical tent i.e. the floor $= 7m$
View full question & answer→Question 242 Marks
A cylindrical container with internal radius of its base $10 \cm,$ contains water up to a height of $7\ cm.$ find the area of wetted surface of the cylinder.
AnswerInternal radius of the cylinderical container $=10 cm$
Height of water $=7 cm$
Therefore, surface area of the wetted surface $=2 \pi r h+\pi r^2$
$=\pi(2 h+r)$
$=\frac{22}{7} \times 10 \times(2 \times 7+10)$
$=\frac{220}{7} \times 24$
$=754.29 cm ^2$
View full question & answer→Question 252 Marks
A cylinder has a diameter of $20$ cm. The area of curved surface is $100$ sq cm. Find: the volume of the cylinder correct to one decimal place.
AnswerDiameter of the cylinder $= 20$ cm
Hence, Radius (r) $= 10$ cm
Height $= h$ cm
$\text { Volime of the cylinder }=\pi r^2 h$
$=\frac{22}{7} \times 10 \times 10 \times 1.6$
$=502.9 cm ^3$
or
$=\frac{22}{7} \times 10 \times 10 \times \frac{35}{22}$
$=500 cm ^3$
View full question & answer→Question 262 Marks
A cylinder of circumference 8 cm and length 21 cm rolls without slideing $4 \frac{1}{2}$ for seconds at the rate of 9 complete round per second.
find,distance travelling by the cylinder in $4 \frac{1}{2}$ second,
AnswerCircumference of cylinder = 8 cm
Therefore, radius $=\frac{c}{2 \pi}=\frac{8 \times 7}{2 \times 22}=\frac{14}{11} cm$
Length of the cylinder $( h )=21 cm$
If distance covered in one revolution is 8 cm, then distance convered in 9 revolution
$=9 \times 8=72$
Therefore, distance converd in $4 \frac{1}{2}$ seconds $=72 \times \frac{9}{2} cm =324 cm$ View full question & answer→Question 272 Marks
The inner radius of a pipe is $2.1\ cm.$ How much water can $12\ m$ of this pipe hold?
AnswerInner radius of pipe $= 2.1 cm$
Length of the pipe $= 12 m = 1200 cm$
$\therefore \text { Volume }=\pi r^2 h$
$=\frac{22}{7} \times 2.1 \times 2.1 \times 1200 cm ^3$
$= 16632$ cm$^3$
View full question & answer→Question 282 Marks
A metal cube of side $11 cm$ is completely submerged in water contained in a cylindrical vessel with diameter $28 cm.$ Find the rise in the level of water.
AnswerClearly, Volume of cube of side $11 cm =$ Volume of water displaced in the cylinder
$\Rightarrow(11)^3=\pi r ^2 h $
$\Rightarrow 11 \times 11 \times 11=\frac{22}{7} \times \frac{28}{2} \times \frac{28}{2} \times h$
$\Rightarrow h =\frac{11 \times 11 \times 11 \times 7 \times 2 \times 2}{22 \times 28 \times 28}=\frac{121}{56}=2.16 cm$
View full question & answer→Question 292 Marks
The height of a circular cylinder is $20$ cm and the radius of its base is $7$ cm. Find : thetotal surface area.
AnswerFor a circular cylinder ,
Height $= h = 20$ cm
Radius of the base $= r = 7$ cm
Total surface area of a cylinder $= 2\pi r (h + r)$
$=2 \times \frac{22}{7} \times 7(20+7) cm ^2$
$=2 \times 22 \times 27 cm ^2$
$= 1188 cm^2$
View full question & answer→Question 302 Marks
The height of a circular cylinder is $20\ cm$ and the radius of its base is $7\ cm$. Find : the volume
AnswerFor circular cylinder ,
Height = h $= 20\ cm$
Radius of the base = r $= 7\ cm$
Volume of a cylinder $= \pi r^2 h$
$=\frac{22}{7} \times 7 \times 7 \times 20 cm ^3$
$= 3080\ cm^3$
View full question & answer→Question 312 Marks
A solid is in the form of a cone standing one hemi$-$sphere with both their radii being equal to $8 \ cm$ and the height of cone is equal to its radius. Find, in terms of $\pi$ , the volume of the solid.
AnswerVolume of the solid
$=\frac{1}{3} \pi r^2 r+\frac{2}{3} \pi r^3$
$=\frac{1}{3} \times \pi \times 8^3+\frac{2}{3} \times \pi \times 8^3$
$=\pi 8^3$
$=512 \pi \ cm ^3$
View full question & answer→Question 322 Marks
A largest sphere is to be carved out of a right circular cylinder of radius $7 \ cm$ and height $14 \ cm.$ Find the volume of the sphere.
AnswerRadius of largest sphere that can be formed inside the cylinder should be equal to the radius ofthe cylinder.
Radius of the largest sphere $= 7 \ cm$
Volume of sphere
$=\frac{4}{3} \pi 7^3$
$=\frac{4}{3} \times \frac{22}{7} \times 7 \times 7$
$=\frac{4312}{3}$
$=1437 \ cm ^3$
View full question & answer→Question 332 Marks
The given figure shows the cross$-$section of a cone, a cylinder and a hemisphere all with the same diameter $10 \ cm$ and the other dimensions are as shown. Calculate: the density of the material if its total weight is $1.7 \ kg$
Answer
Total weight of the solid $= 1.7 \ kg$
Total volume of solid $= 1.12 \ gm/cm^3$
$\therefore \text { Density }=\frac{1.7 \times 1000}{1519.0476} \ gm / \ cm ^3=1.119 \ gm / \ cm ^3$
$\Rightarrow \text { Density }=1.12 \ gm / \ cm ^3$ View full question & answer→Question 342 Marks
A solid metallic hemisphere of diameter $28 \ cm$ is melted and recast into a number of identical solid cones, each of diameter $14 \ cm$ and height $8 \ cm.$ find the number of cones so formed.
AnswerGiven -diameter $14 \ cm;$ height $8 \ cm; n =$ number of cones
Volume of hemisphere $= n \times$ Volume of $1$ cone
$\therefore \frac{2}{3} \pi R ^3= n \times \frac{1}{3} \pi r ^2 h$
$\therefore \frac{2}{\not 3} \not R ^3= n \times \frac{1}{\not 3} \pi^2 r ^2 h$
$\therefore 2(14)^3= n (7)^2(8)$
$\therefore \frac{2 \times 14 \times 14 \times 14}{7 \times 7 \times 8}= n$
$\therefore n =14$
View full question & answer→Question 352 Marks
Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius $r \ cm.$
Answer
For the volume of cone to be largest, $h = r \ cm$
Volume of the cone
$\frac{1}{3} \pi r^2 h$
$=\frac{1}{3} \pi \times r^2 \times r$
$=\frac{1}{3} \pi r^3$ View full question & answer→Question 362 Marks
A hemi$-$spherical bowl has negligible thickness and the length of its circumference is $198 \ cm.$ Find the capacity of the bowl.
AnswerLet $r$ be the radius of the bowl.
$\therefore 2\pi r=198$
$\Rightarrow r=\frac{198 \times 7}{2 \times 22}$
$\Rightarrow r=31.5 \ cm$
Capacity of the bowl $= \frac{2}{3} \pi r^3$
$=\frac{2}{3} \times \frac{22}{7} \times(31.5)^3$
$=65488.5 \ cm ^3$
View full question & answer→Question 372 Marks
A cone and hemisphere have the same base and the same height. Find the ratio between theirvolumes.
AnswerLet the radius of base be $'r\ '$ and the height be $'h\ '$
Volume of cone, $V_c$
$=\frac{1}{3} \pi r^2 h$
Volume of hemisphere, $V_h$
$=\frac{2}{3} \pi r^2 h$
$\frac{V_c}{V_h}=\frac{\frac{1}{3} \pi r^2 r}{\frac{2}{3} \pi r^3}$
$\Rightarrow \frac{V_c}{V_h}=\frac{1}{2}$
View full question & answer→Question 382 Marks
A hemi-spherical bowl of internal radius 9 cm is full of liquid. This liquid is to be filled into conical shaped small containers each of diameter 3 cm and height 4 cm. How manycontainersare necessary to empty the bowl?
AnswerRadius of hemispherical bowl = 9 cm
Volume $=\frac{1}{2} \times \frac{4}{3} \pi r ^3=\frac{2}{3} \pi 9^3=\frac{2}{3} \pi \times 729=486 \pi cm ^2$
Diameter each of cylindrical bottle = 3 cm
Radius $=\frac{3}{2} cm$, and height $=4 cm$
$\therefore$ Volume of bottle $=\frac{1}{3} \pi r ^2 h =\frac{1}{3} \pi \times\left(\frac{3}{2}\right)^2 \times 4=3 \pi$
$\therefore$ No. of. bottles $=\frac{486 \pi}{3 \pi}=162$
View full question & answer→Question 392 Marks
The surface area of a solid metallic sphere is $2464 \ cm^2.$ It is melted and recast into solid right circular cones of radius $ 3.5 \ cm$ and height $7 \ cm.$ Calculate: the radius of the sphere $($Take $\pi = 22/17)$
AnswerTotal surface area of the sphere $= 4\pi r^2,$ where r is the radius of the sphere.
Thus,
$4 \pi r^2=2464 \ cm ^2$
$\Rightarrow 4 \times \frac{22}{7} \times r^2=2464$
$\Rightarrow r^2=196$
$\Rightarrow r=14 \ cm $
View full question & answer→Question 402 Marks
The internal and external diameters of a hollow hemi$-$spherical vessel are $21 \ cm$ and $28 \ cm$ respectively. Find: external curved surface area .
AnswerExternal radius $(R) = 14 \ cm$
Internal radius $(r)=\frac{21}{2} \ cm$
External curved surface area $=2 \pi R^2$
$=2 \times \frac{22}{7} \times 14 \times 14$
$=1232 \ cm ^2$
View full question & answer→Question 412 Marks
The internal and external diameter of a hollow hemispherical vessel are $21 \ cm$ and $28 \ cm$
respectively Find: internal curved surface area
AnswerExternal radius $(R) = 14 \ cm$
Internal radius $(r)=\frac{21}{2} \ cm$
Internal curved surface area $=2 \pi r^2$
$=2 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}$
$=693 \ cm ^3$
View full question & answer→Question 422 Marks
A solid metal sphere is cut through its center into $2$ equal parts. If the diameter of the sphere is $3 \frac{1}{2} \ cm,$ find the total surface area of each part correct to two decimal places.
AnswerDiameter of sphere $=3 \frac{1}{2} \ cm =\frac{7}{2} \ cm$
Therefore, radius of sphere $=\frac{7}{4} \ cm$
Total curved surface area of each hemispheres $=2 \pi r^2+\pi r^2=3 \pi r^2$
$=3 \times \frac{22}{7} \times \frac{7}{4} \times \frac{7}{4}$
$=28.88 \ cm ^2$
View full question & answer→Question 432 Marks
How many balls each of radius $1 \ cm$ can be made by melting a bigger ball whose diameter is $8\ cm.$
AnswerDiameter of bigger ball $= 8 \ cm$
Therefore, Radius of bigger ball $= 4 \ cm$
$\text { Volume }=\frac{4}{3} \pi r^3=\frac{4}{3} \pi \times 4 \times 4 \times 4=\frac{265 \pi}{3} \ cm ^3$
$\text { Number of balls }=\frac{\frac{256 \pi}{3}}{\frac{4 \pi}{3}}=\frac{256 \pi}{3} \times \frac{3}{4 \pi}=64$
View full question & answer→Question 442 Marks
The diameter of two cones are equal. If their slant heights are in the ratio 5 : 4, find the ratio oftheir curved surface areas.
AnswerLet radius of each cone = r
Ratio between their slant heights = 5:4
Let slant height of the first cone = 5x
and slant height of second cone = 4x
Therefore, curved surface area of the first cone $=\pi r \times(5 x)=5 \pi r x$
curved surface area of the second cone $=\pi r l=\pi r \times(4 x)=4 \pi r x$ Hence, ratio between them $=5 \pi r x: 4 \pi r x=5: 4$
View full question & answer→Question 452 Marks
The circumference of the base of a $12 m$ high conical tent is $66 m.$ Find the volume of the aircontained in it.
AnswerCircumference of the conical tent $= 66 m$
and height $(h) = 12 m$
$\therefore$ Radius $=\frac{c}{2 \pi}=\frac{66 \times 7}{2 \times 22}=10.5 m$
Therefore, volume of air contained in it $=\frac{1}{3} \pi r^2 h$
$=\frac{1}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times 12 m ^3$
$=1386 m ^3$
View full question & answer→Question 462 Marks
The volume of a conical tent is $1232 m^3$ and the area of the base floor is $154 m^2.$ Calculate the: height of the tent.
AnswerLet h be the height of the conical tent, then the volume $=$ $\frac{1}{3} \pi r^2 h m^3$
$\therefore \frac{1}{3} \pi r^2 h=1232$
$\Rightarrow \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times h$
$\Rightarrow h=\frac{1232 \times 3}{22 \times 7}=24$
Hence, radius of the base of the conical tent i.e. the floor $= 7 m$
View full question & answer→Question 472 Marks
The volume of a conical tent is $1232 m^3$ and the area of the base floor is $154 m^2.$ Calculate the: radius of the floor.
AnswerLet $r$ be the radius of the base of the conical tent, then area of the base floor $= \pi r^2m^2$
$\pi r^2=154$
$\frac{22}{7} \times r^2=154$
$\Rightarrow r^2=\frac{154 \times 7}{22}=49$
$\Rightarrow r=7$
Hence, radius of the base of the conical tent
i.e. the floor $= 7m$
View full question & answer→Question 482 Marks
A cylindrical container with internal radius of its base $10 \ cm ,$ contains water up to a height of $7 \ cm.$ find the area of wetted surface of the cylinder.
AnswerInternal radius of the cylinderical container $=10 \ cm$
Height of water $=7 \ cm$
Therefore, surface area of the wetted surface $=2 \pi r h+\pi r^2$
$=\pi(2 h+r)$
$=\frac{22}{7} \times 10 \times(2 \times 7+10)$
$=\frac{220}{7} \times 24$
$=754.29 \ cm ^2$
View full question & answer→Question 492 Marks
The inner radius of a pipe is $2.1 \ cm.$ How much water can $12 m$ of this pipe hold?
AnswerInner radius of pipe $= 2.1 \ cm$
Length of the pipe $= 12 m = 1200 \ cm$
$\therefore \text { Volume }=\pi r^2 h$
$=\frac{22}{7} \times 2.1 \times 2.1 \times 1200 \ cm ^3$
$= 16632 \ cm^3$
View full question & answer→Question 502 Marks
A metal cube of side $11 \ cm$ is completely submerged in water contained in a cylindrical vessel with diameter $28 \ cm.$ Find the rise in the level of water.
AnswerClearly, Volume of cube of side $11 \ cm =$ Volume of water displaced in the cylinder
$\Rightarrow(11)^3=\pi r ^2 h$
$\Rightarrow 11 \times 11 \times 11$
$=\frac{22}{7} \times \frac{28}{2} \times \frac{28}{2} \times h $
$\Rightarrow h =\frac{11 \times 11 \times 11 \times 7 \times 2 \times 2}{22 \times 28 \times 28}$
$=\frac{121}{56}$
$=2.16 \ cm$
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