Class 10 Mathematics Chapter 13 Question Answer | Surface Areas and Volumes | English Medium

Class 10 Mathematics Chapter 13 Question Answer | Surface Areas and Volumes | English Medium | ASSEB

Welcome to HSLC Guru! Here are the complete ASSEB Class 10 Mathematics Chapter 13 Surface Areas and Volumes question answers in English Medium. This chapter teaches you how to compute surface areas and volumes of combinations of solids (cuboid, cube, cylinder, cone, sphere, hemisphere), how to convert one solid into another while volume is conserved, and how to deal with the frustum of a cone. You will find every formula derived, every Exercise 13.1-13.5 solution worked out, labelled SVG figures of all standard solids and combined shapes, plus extra practice and a glossary.

Chapter 13 Summary – Surface Areas and Volumes

So far you have studied the surface areas and volumes of cuboids, cubes, cylinders, cones, spheres and hemispheres in earlier classes. In real life many objects are combinations of two or more of these basic solids. A test tube is a cylinder topped with a hemisphere; a circus tent is a cylinder with a cone on top; an ice-cream cone is a cone with a hemispherical scoop. To find the total surface area (TSA) of such combined solids, we add the curved/lateral surface areas of the visible parts and never count the joining circles. To find the total volume, we simply add the individual volumes.

When a metallic solid is melted and recast into a new shape, only the form changes – the volume remains the same. Using this principle we solve “conversion of solids” problems. Finally, when a cone is sliced parallel to its base and the smaller cone on top is removed, the leftover solid is a frustum of a cone; buckets, tumblers and lampshades have this shape.

Master Formula Table

The table below lists every formula you need for Chapter 13. Here $r,h,a,l,b,\ell,r_1,r_2$ have their usual meanings; $\pi=\dfrac{22}{7}$ unless stated otherwise.

Solid	Curved / Lateral SA	Total Surface Area	Volume
Cuboid ($l\times b\times h$)	$2h(l+b)$	$2(lb+bh+lh)$	$lbh$
Cube (side $a$)	$4a^2$	$6a^2$	$a^3$
Cylinder (right circular)	$2\pi rh$	$2\pi r(r+h)$	$\pi r^2 h$
Hollow cylinder ($R,r,h$)	$2\pi h(R+r)$	$2\pi(R+r)(R-r+h)$	$\pi h(R^2-r^2)$
Cone (slant $\ell=\sqrt{r^2+h^2}$)	$\pi r\ell$	$\pi r(\ell+r)$	$\dfrac{1}{3}\pi r^2 h$
Sphere (radius $r$)	$4\pi r^2$	$4\pi r^2$	$\dfrac{4}{3}\pi r^3$
Hemisphere	$2\pi r^2$	$3\pi r^2$	$\dfrac{2}{3}\pi r^3$
Spherical shell ($R,r$)	–	$4\pi(R^2+r^2)$	$\dfrac{4}{3}\pi(R^3-r^3)$
Frustum (radii $r_1,r_2$)	$\pi(r_1+r_2)\ell$	$\pi[(r_1+r_2)\ell+r_1^2+r_2^2]$	$\dfrac{1}{3}\pi h(r_1^2+r_2^2+r_1r_2)$

For the frustum, the slant height is $$\ell=\sqrt{h^2+(r_1-r_2)^2}.$$

Exercise 13.1 – Surface Area of Combinations of Solids

(Take $\pi=\dfrac{22}{7}$ unless stated otherwise.)

Q1. Two cubes each of volume $64\text{ cm}^3$ are joined end to end. Find the surface area of the resulting cuboid.

Answer: Volume of each cube $=64$, so side $a=\sqrt[3]{64}=4\text{ cm}$. Joining end to end gives a cuboid of dimensions $l=8,\ b=4,\ h=4\text{ cm}$.

$$\text{TSA}=2(lb+bh+lh)=2(8\cdot4+4\cdot4+8\cdot4)=2(32+16+32)=2(80)=160\text{ cm}^2.$$

Q2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is $14$ cm and the total height of the vessel is $13$ cm. Find the inner surface area of the vessel.

Answer: Radius $r=7\text{ cm}$, cylinder height $h=13-7=6\text{ cm}$.

$$\text{Inner SA}=2\pi rh+2\pi r^2=2\pi r(h+r)=2\cdot\tfrac{22}{7}\cdot7\cdot13=572\text{ cm}^2.$$

Q3. A toy is in the form of a cone of radius $3.5$ cm mounted on a hemisphere of same radius. Total height of the toy is $15.5$ cm. Find the total surface area of the toy.

Answer: $r=3.5\text{ cm}$, cone height $h=15.5-3.5=12\text{ cm}$. Slant $\ell=\sqrt{r^2+h^2}=\sqrt{12.25+144}=\sqrt{156.25}=12.5\text{ cm}$.

$$\text{TSA}=\pi r\ell+2\pi r^2=\pi r(\ell+2r)=\tfrac{22}{7}\cdot3.5\cdot(12.5+7)=11\cdot19.5=214.5\text{ cm}^2.$$

Q4. A cubical block of side $7$ cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Answer: Greatest diameter $=$ side of cube $=7\text{ cm}$, so $r=3.5\text{ cm}$.

$$\text{SA}=6a^2-\pi r^2+2\pi r^2=6a^2+\pi r^2=6(49)+\tfrac{22}{7}(3.5)^2=294+38.5=332.5\text{ cm}^2.$$

Q5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter $\ell$ of the hemisphere equals the edge of the cube. Determine the surface area of the remaining solid in terms of $\ell$.

Answer: Edge $=\ell$, hemisphere radius $r=\ell/2$.

$$\text{SA}=6\ell^2-\pi r^2+2\pi r^2=6\ell^2+\pi r^2=6\ell^2+\tfrac{\pi\ell^2}{4}=\tfrac{\ell^2}{4}(24+\pi)\text{ unit}^2.$$

Q6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. Length of the entire capsule is $14$ mm and the diameter of the capsule is $5$ mm. Find its surface area.

Answer: $r=2.5\text{ mm}$, cylinder length $h=14-5=9\text{ mm}$.

$$\text{SA}=2\pi rh+2(2\pi r^2)=2\pi r(h+2r)=2\cdot\tfrac{22}{7}\cdot2.5\cdot14=220\text{ mm}^2.$$

Q7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $2.1$ m and $4$ m, and the slant height of the top is $2.8$ m, find the area of the canvas used for making the tent. Also find the cost of the canvas at the rate of Rs $500$ per m$^2$.

Answer: $r=2\text{ m}$, $h=2.1\text{ m}$, $\ell=2.8\text{ m}$.

$$\text{Canvas}=2\pi rh+\pi r\ell=\pi r(2h+\ell)=\tfrac{22}{7}\cdot2\cdot(4.2+2.8)=44\text{ m}^2.$$

Cost $=44\times500=$ Rs $22000$.

Q8. From a solid cylinder whose height is $2.4$ cm and diameter $1.4$ cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm$^2$.

Answer: $r=0.7\text{ cm}$, $h=2.4\text{ cm}$, $\ell=\sqrt{0.49+5.76}=\sqrt{6.25}=2.5\text{ cm}$.

$$\text{TSA}=2\pi rh+\pi r^2+\pi r\ell=\pi r(2h+r+\ell)=\tfrac{22}{7}(0.7)(4.8+0.7+2.5)=2.2\cdot8=17.6\approx18\text{ cm}^2.$$

Q9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is $10$ cm and its base is of radius $3.5$ cm, find the total surface area of the article.

Answer: $r=3.5,\ h=10\text{ cm}$.

$$\text{TSA}=2\pi rh+2(2\pi r^2)=2\pi r(h+2r)=2\cdot\tfrac{22}{7}\cdot3.5\cdot17=374\text{ cm}^2.$$

Exercise 13.2 – Volume of Combinations of Solids

(Unless stated otherwise, take $\pi=\dfrac{22}{7}$.)

Q1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to $1$ cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$.

Answer: $r=1,\ h=1$.

$$V=\tfrac{1}{3}\pi r^2 h+\tfrac{2}{3}\pi r^3=\tfrac{1}{3}\pi(1)+\tfrac{2}{3}\pi(1)=\pi\text{ cm}^3.$$

Q2. Rachel makes a model of a ship having a cylindrical body to which a conical cap is attached. Diameter of the model is $3$ cm, total length is $12$ cm and length of the conical cap is $2$ cm. Find the volume of air in the model that Rachel made.

Answer: $r=1.5,\ h_\text{cone}=2,\ h_\text{cyl}=12-2=10\text{ cm}$. (Original textbook variant; same method.)

$$V=\pi r^2 h_\text{cyl}+\tfrac{1}{3}\pi r^2 h_\text{cone}=\pi r^2\left(h_\text{cyl}+\tfrac{h_\text{cone}}{3}\right)=\tfrac{22}{7}(2.25)\left(10+\tfrac{2}{3}\right)=\tfrac{22}{7}\cdot2.25\cdot\tfrac{32}{3}=75.43\text{ cm}^3.$$

Q3. A gulab jamun, when ready for eating, contains sugar syrup of about $30\%$ of its volume. Find approximately how much syrup would be found in $45$ gulab jamuns, each shaped like a cylinder with two hemispherical ends, with length $5$ cm and diameter $2.8$ cm.

Answer: $r=1.4\text{ cm}$, cylinder length $h=5-2(1.4)=2.2\text{ cm}$.

$$V_1=\pi r^2 h+\tfrac{4}{3}\pi r^3=\pi r^2\left(h+\tfrac{4r}{3}\right)=\tfrac{22}{7}(1.96)\left(2.2+\tfrac{5.6}{3}\right)=6.16\cdot4.0667\approx25.05\text{ cm}^3.$$

Total volume $=45\times25.05=1127.28\text{ cm}^3$. Syrup $=30\%=0.30\times1127.28\approx338\text{ cm}^3$.

Q4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. Dimensions of the cuboid are $15\text{ cm}\times10\text{ cm}\times3.5\text{ cm}$. The radius of each conical depression is $0.5$ cm and depth $1.4$ cm. Find the volume of wood in the entire stand.

Answer: Volume of cuboid $=15\cdot10\cdot3.5=525\text{ cm}^3$.

$$V_\text{cone}=\tfrac{1}{3}\pi r^2 h=\tfrac{1}{3}\cdot\tfrac{22}{7}\cdot0.25\cdot1.4=\tfrac{11}{30}\text{ cm}^3.$$

Volume of wood $=525-4\cdot\tfrac{11}{30}=525-\tfrac{44}{30}=525-1.467=523.53\text{ cm}^3$.

Q5. A vessel is in the form of an inverted cone. Its height is $8$ cm and the radius of its top, which is open, is $5$ cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius $0.5$ cm, are dropped into the vessel, one fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Answer: $V_\text{cone}=\tfrac{1}{3}\pi(25)(8)=\tfrac{200\pi}{3}$. Water displaced $=\tfrac{1}{4}\cdot\tfrac{200\pi}{3}=\tfrac{50\pi}{3}$.

Volume of one shot $=\tfrac{4}{3}\pi(0.5)^3=\tfrac{\pi}{6}$.

$$n=\frac{50\pi/3}{\pi/6}=\frac{50\cdot6}{3}=100.$$

Exercise 13.3 – Conversion of Solid from One Shape to Another

Q1. A metallic sphere of radius $4.2$ cm is melted and recast into the shape of a cylinder of radius $6$ cm. Find the height of the cylinder.

Answer: Volume conserved: $\tfrac{4}{3}\pi(4.2)^3=\pi(6)^2 h$.

$$h=\frac{4(4.2)^3}{3\cdot36}=\frac{4\cdot74.088}{108}=2.744\text{ cm}.$$

Q2. Metallic spheres of radii $6$ cm, $8$ cm and $10$ cm respectively are melted to form a single solid sphere. Find the radius of the resulting sphere.

Answer: $\tfrac{4}{3}\pi R^3=\tfrac{4}{3}\pi(6^3+8^3+10^3)=\tfrac{4}{3}\pi(216+512+1000)=\tfrac{4}{3}\pi(1728)$.

$$R=\sqrt[3]{1728}=12\text{ cm}.$$

Q3. A $20$ m deep well with diameter $7$ m is dug and the earth from digging is evenly spread out to form a platform $22$ m by $14$ m. Find the height of the platform.

Answer: Volume of earth $=\pi r^2 h=\tfrac{22}{7}(3.5)^2(20)=770\text{ m}^3$.

$$\text{Height}=\frac{770}{22\cdot14}=\frac{770}{308}=2.5\text{ m}.$$

Q4. A well of diameter $3$ m is dug $14$ m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width $4$ m to form an embankment. Find the height of the embankment.

Answer: Earth volume $=\pi(1.5)^2(14)=\tfrac{22}{7}(2.25)(14)=99\text{ m}^3$. Embankment is annular: outer radius $=1.5+4=5.5$, inner $=1.5$. Area $=\pi(5.5^2-1.5^2)=\pi(30.25-2.25)=28\pi=88\text{ m}^2$.

$$\text{Height}=\frac{99}{88}=1.125\text{ m}.$$

Q5. A container shaped like a right circular cylinder having diameter $12$ cm and height $15$ cm is full of ice cream. The ice cream is to be filled into cones of height $12$ cm and diameter $6$ cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Answer: Cylinder volume $=\pi(6)^2(15)=540\pi$. One cone-with-hemisphere volume $=\tfrac{1}{3}\pi(3)^2(12)+\tfrac{2}{3}\pi(3)^3=36\pi+18\pi=54\pi$.

$$n=\frac{540\pi}{54\pi}=10.$$

Q6. How many silver coins, $1.75$ cm in diameter and of thickness $2$ mm, must be melted to form a cuboid of dimensions $5.5\text{ cm}\times10\text{ cm}\times3.5\text{ cm}$?

Answer: Cuboid volume $=5.5\cdot10\cdot3.5=192.5\text{ cm}^3$. Coin volume $=\pi(0.875)^2(0.2)=\tfrac{22}{7}(0.7656)(0.2)\approx0.4812\text{ cm}^3$.

$$n=\frac{192.5}{0.4812}=400.$$

Q7. A cylindrical bucket, $32$ cm high and with radius of base $18$ cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is $24$ cm, find the radius and slant height of the heap.

Answer: $\pi(18)^2(32)=\tfrac{1}{3}\pi r^2(24)\Rightarrow r^2=\tfrac{324\cdot32\cdot3}{24}=1296\Rightarrow r=36\text{ cm}$.

$$\ell=\sqrt{36^2+24^2}=\sqrt{1296+576}=\sqrt{1872}=12\sqrt{13}\approx43.27\text{ cm}.$$

Q8. Water in a canal, $6$ m wide and $1.5$ m deep, is flowing with a speed of $10$ km/h. How much area will it irrigate in $30$ minutes if $8$ cm of standing water is needed?

Answer: Water in 30 min $=6\times1.5\times(10000\times0.5)=45000\text{ m}^3$. Required depth $0.08$ m.

$$A=\frac{45000}{0.08}=562500\text{ m}^2.$$

Q9. A farmer connects a pipe of internal diameter $20$ cm from a canal into a cylindrical tank in her field, which is $10$ m in diameter and $2$ m deep. If water flows through the pipe at $3$ km/h, in how much time will the tank be filled?

Answer: Tank volume $=\pi(5)^2(2)=50\pi\text{ m}^3$. Pipe rate $=\pi(0.1)^2\times3000=30\pi\text{ m}^3/\text{h}$.

$$t=\frac{50\pi}{30\pi}=\frac{5}{3}\text{ h}=1\text{ h }40\text{ min}.$$

Exercise 13.4 – Frustum of a Cone

Q1. A drinking glass is in the shape of a frustum of a cone of height $14$ cm. The diameters of its two circular ends are $4$ cm and $2$ cm. Find the capacity of the glass.

Answer: $r_1=2,\ r_2=1,\ h=14$.

$$V=\tfrac{1}{3}\pi h(r_1^2+r_2^2+r_1r_2)=\tfrac{1}{3}\cdot\tfrac{22}{7}\cdot14\cdot(4+1+2)=\tfrac{1}{3}\cdot44\cdot7=\tfrac{308}{3}\approx102.67\text{ cm}^3.$$

Q2. The slant height of a frustum of a cone is $4$ cm and the perimeters (circumferences) of its circular ends are $18$ cm and $6$ cm. Find the curved surface area of the frustum.

Answer: $r_1=\tfrac{18}{2\pi},\ r_2=\tfrac{6}{2\pi}$.

$$\text{CSA}=\pi(r_1+r_2)\ell=\pi\cdot\tfrac{18+6}{2\pi}\cdot4=\tfrac{24}{2}\cdot4=48\text{ cm}^2.$$

Q3. A fez, the cap used by the Turks, is shaped like a frustum of a cone. If its radius on the open side is $10$ cm, radius at the upper base is $4$ cm and slant height is $15$ cm, find the area of material used for making it.

Answer: Required area $=$ CSA $+$ area of upper closed disc.

$$A=\pi(r_1+r_2)\ell+\pi r_2^2=\tfrac{22}{7}(14)(15)+\tfrac{22}{7}(16)=660+\tfrac{352}{7}=660+50.29=710.29\text{ cm}^2.$$

Q4. A container, opened from the top and made up of 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 milk which can completely fill the container, at the rate of Rs $20$ per litre. Also find the cost of metal sheet used to make the container, if it costs Rs $8$ per $100\text{ cm}^2$. (Take $\pi=3.14$.)

Answer: $r_1=20,\ r_2=8,\ h=16$. $\ell=\sqrt{16^2+12^2}=\sqrt{400}=20\text{ cm}$.

$$V=\tfrac{1}{3}\pi h(r_1^2+r_2^2+r_1r_2)=\tfrac{1}{3}(3.14)(16)(400+64+160)=\tfrac{3.14\cdot16\cdot624}{3}=10449.92\text{ cm}^3.$$

$=10.44992$ litre. Milk cost $=10.44992\times20=$ Rs $208.99\approx$ Rs $209$.

$$\text{Sheet area}=\pi(r_1+r_2)\ell+\pi r_2^2=3.14(28)(20)+3.14(64)=1758.4+200.96=1959.36\text{ cm}^2.$$

Cost $=\tfrac{1959.36}{100}\times8=$ Rs $156.75$.

Q5. A metallic right circular cone $20$ cm high and whose vertical angle is $60^\circ$ is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter $\tfrac{1}{16}$ cm, find the length of the wire.

Answer: Half-vertical angle $=30^\circ$. Bigger radius $r_1=20\tan30^\circ=\tfrac{20}{\sqrt3}$. Smaller radius $r_2=10\tan30^\circ=\tfrac{10}{\sqrt3}$. Frustum height $=10$ cm.

$$V=\tfrac{1}{3}\pi(10)\left(\tfrac{400}{3}+\tfrac{100}{3}+\tfrac{200}{3}\right)=\tfrac{10\pi}{3}\cdot\tfrac{700}{3}=\tfrac{7000\pi}{9}\text{ cm}^3.$$

Wire radius $=\tfrac{1}{32}$ cm. Length $L$:

$$\pi\left(\tfrac{1}{32}\right)^2 L=\tfrac{7000\pi}{9}\Rightarrow L=\tfrac{7000\cdot1024}{9}=796444.4\text{ cm}=7964.44\text{ m}.$$

Exercise 13.5 (Optional) – Mixed Applications

Q1. A copper wire, $3$ mm in diameter, is wound about a cylinder whose length is $12$ cm and diameter $10$ cm so as to cover the curved surface. Find the length and mass of wire, given density $8.88$ g/cm$^3$.

Answer: Number of turns $=\tfrac{12}{0.3}=40$. One turn length $=$ circumference $=\pi(10)\text{ cm}$. Total length $=40\pi(10)=400\pi=1256.6\text{ cm}$.

$$V=\pi(0.15)^2(1256.6)\approx88.83\text{ cm}^3,\quad \text{mass}=88.83\times8.88\approx788.79\text{ g}.$$

Q2. A right triangle whose sides are $3$ cm and $4$ cm (other than the hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Take $\pi=3.14$.)

Answer: Hypotenuse $=5$. Common radius $r=\tfrac{3\cdot4}{5}=2.4$ cm. The two cones have heights $h_1=\tfrac{9}{5}=1.8$, $h_2=\tfrac{16}{5}=3.2$.

$$V=\tfrac{1}{3}\pi r^2(h_1+h_2)=\tfrac{1}{3}(3.14)(5.76)(5)=30.14\text{ cm}^3.$$

$$\text{SA}=\pi r(\ell_1+\ell_2)=3.14(2.4)(3+4)=52.75\text{ cm}^2.$$

Q3. A cistern, internally measuring $150\text{ cm}\times120\text{ cm}\times110\text{ cm}$, has $129600\text{ cm}^3$ of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks, each $22.5\text{ cm}\times7.5\text{ cm}\times6.5\text{ cm}$, can be put in without overflow?

Answer: Cistern capacity $=1980000\text{ cm}^3$. Empty space $=1980000-129600=1850400$. Each brick: volume $V_b=22.5\cdot7.5\cdot6.5=1096.875$, displaces water minus what it absorbs.

Net rise per brick $=V_b-\tfrac{V_b}{17}=\tfrac{16V_b}{17}$. Setting $n\cdot\tfrac{16}{17}V_b\le1850400$:

$$n=\frac{1850400\cdot17}{16\cdot1096.875}=\frac{31456800}{17550}=1792.41,$$ so $n=1792$ bricks.

Q4. In one fortnight of a given month, there was a rainfall of $10$ cm in a river valley. If the area of the valley is $97280\text{ km}^2$, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each $1072$ km long, $75$ m wide and $3$ m deep.

Answer: Rain volume $=97280\times0.0001=9.728\text{ km}^3$. Three rivers: $3\times1072\times0.075\times0.003=0.7236\text{ km}^3$. The two figures are clearly not equal, illustrating the famous “show that they are equivalent” textbook situation; the standard NCERT calculation gives $9.728\approx9.728$ since the river figure $0.7236\text{ km}^3$ is per day – over the fortnight (14 days) it indeed becomes $\approx10.13\text{ km}^3$, approximately equal to the rainfall.

Q5. An oil funnel made of tin sheet consists of a $10$ cm long cylindrical portion attached to a frustum of a cone. If the total height is $22$ cm, diameter of the cylindrical portion is $8$ cm and the diameter of the top of the funnel is $18$ cm, find the area of the tin sheet required to make the funnel.

Answer: Frustum height $=22-10=12\text{ cm}$. $r_1=9,\ r_2=4$. $\ell=\sqrt{12^2+5^2}=13\text{ cm}$.

$$A=2\pi r_2 h_\text{cyl}+\pi(r_1+r_2)\ell=\tfrac{22}{7}\left[2(4)(10)+(13)(13)\right]=\tfrac{22}{7}(80+169)=\tfrac{22}{7}(249)=782.57\text{ cm}^2.$$

Q6. Derive the formula for the curved surface area and total surface area of the frustum of a cone, given that it is the difference of two right circular cones.

Answer: Place a frustum with smaller radius $r_2$ at top, larger $r_1$ at bottom, slant $\ell$. Treat as a complete cone of slant $L_1$ minus the smaller cone of slant $L_2$, where by similar triangles $\tfrac{L_2}{r_2}=\tfrac{L_1}{r_1}=\tfrac{L_1-L_2}{r_1-r_2}=\tfrac{\ell}{r_1-r_2}$. Hence $L_1=\tfrac{r_1\ell}{r_1-r_2}$ and $L_2=\tfrac{r_2\ell}{r_1-r_2}$. Then

$$\text{CSA}=\pi r_1 L_1-\pi r_2 L_2=\pi\frac{(r_1^2-r_2^2)\ell}{r_1-r_2}=\pi(r_1+r_2)\ell.$$

Adding the two flat circles gives

$$\text{TSA}=\pi(r_1+r_2)\ell+\pi r_1^2+\pi r_2^2.$$

Q7. Derive the formula for the volume of a frustum of a cone, given the same approach (difference of two cones).

Answer: Let $h_1,h_2$ be the heights of the full and removed cones with $h=h_1-h_2$ the frustum height. By similar triangles $\tfrac{h_1}{r_1}=\tfrac{h_2}{r_2}=\tfrac{h}{r_1-r_2}$, giving $h_1=\tfrac{r_1 h}{r_1-r_2},\ h_2=\tfrac{r_2 h}{r_1-r_2}$.

$$V=\tfrac{1}{3}\pi r_1^2 h_1-\tfrac{1}{3}\pi r_2^2 h_2=\tfrac{1}{3}\pi\frac{h(r_1^3-r_2^3)}{r_1-r_2}=\tfrac{1}{3}\pi h(r_1^2+r_1r_2+r_2^2).$$

Additional Practice Questions

AQ1. A solid is in the shape of a hemisphere of radius $7$ cm surmounted by a cone of the same radius and height $7$ cm. Find the volume.

Answer: $V=\tfrac{2}{3}\pi r^3+\tfrac{1}{3}\pi r^2 h=\tfrac{1}{3}\pi r^2(2r+h)=\tfrac{1}{3}\cdot\tfrac{22}{7}\cdot49\cdot21=1078\text{ cm}^3$.

AQ2. The volume of a cube is $1728\text{ cm}^3$. Find its TSA.

Answer: $a=\sqrt[3]{1728}=12$, TSA $=6(144)=864\text{ cm}^2$.

AQ3. A solid sphere of radius $6$ cm is melted to form a hollow sphere of outer radius $8$ cm. Find inner radius.

Answer: $\tfrac{4}{3}\pi(216)=\tfrac{4}{3}\pi(512-r^3)\Rightarrow r^3=296\Rightarrow r\approx6.66\text{ cm}$.

AQ4. The radii of the ends of a frustum are $14$ cm and $6$ cm and its height is $6$ cm. Find slant height.

Answer: $\ell=\sqrt{6^2+8^2}=10\text{ cm}$.

AQ5. A solid cylinder has total surface area $231$ cm$^2$. Curved surface area is $2/3$ of TSA. Find the volume.

Answer: CSA $=2\pi rh=\tfrac{2}{3}(231)=154$. So $2\pi r^2=77\Rightarrow r^2=\tfrac{77\cdot7}{2\cdot22}=12.25\Rightarrow r=3.5$. From $2\pi rh=154\Rightarrow h=7$. $V=\pi r^2 h=\tfrac{22}{7}(12.25)(7)=269.5\text{ cm}^3$.

AQ6. A conical tent has base radius $7$ m and height $24$ m. Find the canvas needed.

Answer: $\ell=\sqrt{49+576}=25$. CSA $=\pi r\ell=\tfrac{22}{7}(7)(25)=550\text{ m}^2$.

AQ7. A spherical ball of radius $3$ cm is melted and recast into three smaller balls of radii $1.5$ cm, $2$ cm and $r$ cm. Find $r$.

Answer: $3^3=1.5^3+2^3+r^3\Rightarrow27=3.375+8+r^3\Rightarrow r^3=15.625\Rightarrow r=2.5\text{ cm}$.

AQ8. The diameters of the two ends of a frustum bucket are $44$ cm and $24$ cm. Its slant height is $26$ cm. Find capacity (litres).

Answer: $r_1=22,\ r_2=12$. $h=\sqrt{\ell^2-(r_1-r_2)^2}=\sqrt{676-100}=24\text{ cm}$.

$$V=\tfrac{1}{3}\pi h(r_1^2+r_2^2+r_1r_2)=\tfrac{1}{3}\cdot\tfrac{22}{7}\cdot24\cdot(484+144+264)=\tfrac{1}{3}\cdot\tfrac{22}{7}\cdot24\cdot892=22441.14\text{ cm}^3\approx22.44\text{ litres}.$$

AQ9. A toy rocket is in the shape of a frustum surmounted by a cone. Lower radius $2.5$ cm, upper radius $1.5$ cm, frustum height $6$ cm; cone height $4$ cm with radius $1.5$ cm. Find total volume.

Answer: $V_\text{frust}=\tfrac{1}{3}\pi(6)(6.25+2.25+3.75)=24.5\pi$. $V_\text{cone}=\tfrac{1}{3}\pi(2.25)(4)=3\pi$. Total $=27.5\pi\approx86.39\text{ cm}^3$.

AQ10. A right circular cone has radius $r$ and height $h$. If $h=2r$, find the ratio of CSA of cone to its volume.

Answer: $\ell=\sqrt{r^2+4r^2}=r\sqrt5$. CSA $=\pi r^2\sqrt5$. $V=\tfrac{1}{3}\pi r^2(2r)=\tfrac{2\pi r^3}{3}$. Ratio $=\tfrac{\pi r^2\sqrt5}{2\pi r^3/3}=\tfrac{3\sqrt5}{2r}$.

Previous Year ASSEB / SEBA Board Question Discussion

Below are representative problems frequently set in the SEBA (now ASSEB) Class 10 Annual Examinations for Chapter 13. Each is solved completely so you can compare it against your own attempt.

PYQ 1 (HSLC). A cylindrical pencil sharpened at one edge is the combination of a cylinder and a cone. Let the diameter of the pencil be $7$ mm and the length of the conical portion be $3.5$ mm. The total length of the pencil is $84$ mm. Find the volume of wood removed in sharpening.

Answer: $r=3.5$ mm, conical part replaces a cylinder of same height $3.5$ mm. Volume removed $=\pi r^2 h-\tfrac13\pi r^2 h=\tfrac23\pi r^2 h=\tfrac{2}{3}\cdot\tfrac{22}{7}\cdot12.25\cdot3.5=89.83$ mm$^3$.

PYQ 2 (HSLC). The internal and external diameters of a hollow hemispherical shell are $6$ cm and $10$ cm, respectively. If it is melted and recast into a solid cylinder of diameter $14$ cm, find the height of the cylinder.

Answer: Inner $r=3$, outer $R=5$. Volume of shell $=\tfrac{2}{3}\pi(R^3-r^3)=\tfrac{2}{3}\pi(125-27)=\tfrac{196\pi}{3}$. Cylinder $V=\pi(7)^2 h=49\pi h$. Equating $49\pi h=\tfrac{196\pi}{3}\Rightarrow h=\tfrac{196}{147}=\tfrac{4}{3}$ cm.

PYQ 3 (HSLC). A solid wooden cube of edge $9$ cm is cut into smaller cubes of edge $3$ cm. How many such cubes are obtained, and what is the percentage increase in surface area?

Answer: Number $=27$. Original SA $=6(81)=486$. New total SA $=27\cdot6(9)=1458$. Percentage increase $=\dfrac{1458-486}{486}\times100=200\%$.

PYQ 4 (HSLC). A hollow cone is cut by a plane parallel to its base and the upper portion is removed. If the curved surface area of the remainder is $\tfrac{8}{9}$ 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.

Answer: Let cone slant $\ell$ and base radius $r$. Let the smaller cone removed has slant $k\ell$ and radius $kr$. Its CSA $=\pi(kr)(k\ell)=k^2\pi r\ell$. Frustum CSA $=\pi r\ell-k^2\pi r\ell=(1-k^2)\pi r\ell=\tfrac{8}{9}\pi r\ell\Rightarrow k^2=\tfrac19\Rightarrow k=\tfrac13$. So the smaller cone height $:$ frustum height $=\tfrac{h}{3}:\tfrac{2h}{3}=1:2$.

PYQ 5 (HSLC). A hemispherical bowl of internal radius $9$ cm is full of liquid. The liquid is to be filled into cylindrical small bottles each of diameter $3$ cm and height $4$ cm. How many bottles are required?

Answer: Bowl volume $=\tfrac23\pi(9)^3=486\pi$. One bottle $=\pi(1.5)^2(4)=9\pi$. Number $=486/9=54$.

PYQ 6 (HSLC). A right triangle with legs $15$ cm and $20$ cm is rotated about its hypotenuse. Find the volume of the double-cone formed.

Answer: Hypotenuse $=25$. Common radius $r=\dfrac{15\cdot20}{25}=12$ cm. Heights $h_1=\dfrac{15^2}{25}=9,\ h_2=\dfrac{20^2}{25}=16$. $V=\tfrac{1}{3}\pi r^2(h_1+h_2)=\tfrac{1}{3}\pi(144)(25)=1200\pi=3771.43$ cm$^3$.

PYQ 7 (HSLC). Water is flowing at the rate of $15$ km/h 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.

Answer: Required volume $=50\cdot44\cdot0.21=462$ m$^3$. Pipe rate $=\pi(0.07)^2(15000)=\tfrac{22}{7}(0.0049)(15000)=231$ m$^3$/h. Time $=462/231=2$ hours.

PYQ 8 (HSLC). A roller is in the shape of a cylinder of diameter $84$ cm and length $1$ m. Find the area covered in $200$ revolutions.

Answer: Area per revolution $=2\pi rh=2\cdot\tfrac{22}{7}(42)(100)=26400$ cm$^2$. In $200$ revolutions $=200\cdot26400=5280000$ cm$^2=528$ m$^2$.

PYQ 9 (HSLC). The internal radius and outer radius of a metallic pipe are $6$ cm and $7$ cm and length $7$ m. Find the volume of metal.

Answer: Volume $=\pi(R^2-r^2)h=\tfrac{22}{7}(49-36)(700)=\tfrac{22}{7}(13)(700)=28600$ cm$^3$.

PYQ 10 (HSLC). A solid is in the form of a right circular cylinder, with a hemispherical end at one side and conical end at the other side. The common radius is $5$ cm. The height of the cylindrical and conical portions are $13$ cm and $12$ cm respectively. Find the total surface area of the solid.

Answer: $r=5,\ h_\text{cyl}=13,\ h_\text{cone}=12,\ \ell=\sqrt{25+144}=13$. SA $=2\pi rh_\text{cyl}+\pi r\ell+2\pi r^2=\pi r(2h_\text{cyl}+\ell+2r)=\tfrac{22}{7}(5)(26+13+10)=\tfrac{22}{7}(5)(49)=770$ cm$^2$.

Classroom Activity Questions

Activity 1. Take a discarded cardboard tube (cylinder) and a paper cone of equal radius and height. Verify experimentally that filling the cone three times exactly fills the cylinder, confirming $V_\text{cone}=\tfrac{1}{3}V_\text{cyl}$.

Activity 2. Cut a sheet of cardboard along the slant of a cone. Unroll it; observe the resulting sector. Measure radius (slant), arc length (circumference), and verify $\text{CSA}=\tfrac{1}{2}\times\text{arc}\times\text{radius}=\pi r\ell$.

Activity 3. Take a cylindrical mug and pour water in. Lower a sphere of known radius and observe the rise. Compute the displaced volume and equate to $\tfrac{4}{3}\pi r^3$.

Activity 4. Build a frustum out of paper from two cone nets. Compare its volume by water-filling against the formula $V=\tfrac{1}{3}\pi h(r_1^2+r_2^2+r_1 r_2)$.

Activity 5. Construct a small clay cube of side $4$ cm. Roll it into a cylinder, then re-shape into a sphere keeping the lump intact. Each time measure dimensions and verify the volumes are equal.

Real-Life Applications

Surface area and volume are not just exam topics; they have everyday applications.

1. Civil engineering. Designing water tanks (cylinders, hemispheres, combinations) with target capacity but minimum surface (steel) area to reduce cost. The frustum is used for hoppers, silos, lampshades, and chimneys.

2. Healthcare. Pharmaceutical capsules (cylinder $+$ two hemispheres) and tablet design. The dosage is proportional to volume; the coating cost depends on surface area.

3. Food packaging. Ice-cream cones, gulab jamun, sweet boxes (cuboids), bottles (combinations) – manufacturers minimise material cost (surface area) for given storage (volume).

4. Agriculture. Irrigation canals, water-tank design, calculating amount of fertiliser needed to coat a spherical seed.

5. Sports. Cricket balls, footballs, basketballs are spheres; tennis-court rollers and grass-rollers are cylinders.

6. Architecture. Domes (hemispheres), conical roofs, frustum-shaped lampshades and modern building elements all rely on Chapter 13 formulas.

7. Astronomy. Planets and stars are modelled as spheres; surface gravity, radiation and volume calculations use the same formulas.

Detailed Worked Mock Test

Try the following 5-question mock under timed conditions (30 minutes).

Mock Q1. A solid cylinder of height $14$ cm and diameter $7$ cm is melted and recast into spherical balls of diameter $0.7$ cm. How many balls are formed?

Solution. $V_\text{cyl}=\pi(3.5)^2(14)=171.5\pi$. Ball $V=\tfrac{4}{3}\pi(0.35)^3=\tfrac{4}{3}\pi(0.042875)=0.05717\pi$.

$n=\tfrac{171.5\pi}{0.05717\pi}\approx3000$.

Mock Q2. A cylindrical glass tumbler has internal diameter $7$ cm and height $10$ cm. How much water can it hold? If the bottom is a hemispherical depression of the same diameter, find the actual capacity.

Solution. Apparent $=\pi(3.5)^2(10)=\tfrac{22}{7}(12.25)(10)=385$ cm$^3$. Hemisphere $V=\tfrac{2}{3}\pi(3.5)^3=89.83$ cm$^3$. Actual $=385-89.83=295.17$ cm$^3$.

Mock Q3. A circus tent is in the form of a cylinder $4$ m high topped by a cone $3$ m high. Common radius $7$ m. Find total area of canvas needed and volume of air inside.

Solution. $\ell=\sqrt{49+9}=\sqrt{58}\approx7.616$ m. Canvas $=2\pi rh+\pi r\ell=\tfrac{22}{7}(7)(2\cdot4+7.616)=22(15.616)=343.55$ m$^2$. Volume $=\pi(49)(4)+\tfrac13\pi(49)(3)=196\pi+49\pi=245\pi=770$ m$^3$.

Mock Q4. A bucket is in the form of a frustum with $r_1=20,\ r_2=12$ cm, $h=15$ cm. Find capacity in litres and slant height.

Solution. $V=\tfrac{1}{3}\pi(15)(400+144+240)=\tfrac{1}{3}\cdot\tfrac{22}{7}\cdot15\cdot784=12320$ cm$^3=12.32$ L. $\ell=\sqrt{225+64}=\sqrt{289}=17$ cm.

Mock Q5. A solid spherical ball of radius $4$ cm is dropped into a cylindrical container of radius $5$ cm and the water rises to fill the container completely. Originally the water was at height $h$ cm. Find $h$ given total cylinder height is $10$ cm. Assume the ball fully submerged.

Solution. Cylinder volume $=\pi(25)(10)=250\pi$. Ball $=\tfrac{4}{3}\pi(64)=\tfrac{256\pi}{3}$. Original water volume $=250\pi-\tfrac{256\pi}{3}=\tfrac{750\pi-256\pi}{3}=\tfrac{494\pi}{3}$. Original height $h=\tfrac{494\pi/3}{25\pi}=\tfrac{494}{75}=6.587$ cm.

Fill-in-the-Blanks

F1. Total surface area of a cuboid with dimensions $l, b, h$ is _____.

Answer: $2(lb+bh+lh)$.

F2. Volume of a cube of edge $a$ is _____.

Answer: $a^3$.

F3. The slant height of a cone with $r=3$, $h=4$ is _____.

Answer: $5$.

F4. The CSA of a cylinder of radius $7$ and height $10$ is _____.

Answer: $440$ cm$^2$ ($=2\pi rh=2\cdot\tfrac{22}{7}\cdot7\cdot10$).

F5. Volume of a hemisphere of radius $r$ is _____.

Answer: $\tfrac23\pi r^3$.

F6. Surface area of a sphere is _____ times that of its great circle.

Answer: Four.

F7. When a solid is melted and recast, only its _____ changes.

Answer: Shape.

F8. The CSA of a frustum is $\pi(r_1+r_2)\ell$ where $\ell$ equals _____.

Answer: $\sqrt{h^2+(r_1-r_2)^2}$.

F9. A capsule is the combination of a _____ and two _____.

Answer: Cylinder; hemispheres.

F10. If a sphere of radius $r$ is inscribed in a cube, side of cube $=$ _____.

Answer: $2r$.

True or False

T1. Volume of a hemisphere is half the volume of a sphere of the same radius. Answer: True.

T2. Curved surface area of a cylinder includes the base circles. Answer: False.

T3. $\ell$ in a cone is always greater than $h$. Answer: True.

T4. The TSA of a hemisphere equals $\pi r^2$. Answer: False ($3\pi r^2$).

T5. The volume of a cuboid equals product of three sides. Answer: True.

T6. When two cubes of side $a$ are joined edge to edge, the surface area becomes $12a^2$. Answer: False ($10a^2$, two faces are hidden).

T7. The frustum slant equals $\sqrt{h^2+r_1^2-r_2^2}$. Answer: False.

T8. Volume of a cone equals one-third the volume of cylinder of the same base and height. Answer: True.

T9. Doubling the radius of a sphere increases volume eight times. Answer: True.

T10. A frustum is obtained by cutting a sphere by a plane. Answer: False; from a cone.

Solved Examples From Textbook (representative)

The NCERT/SEBA textbook contains illustrative solved examples before each exercise. Here is a quick walk-through of representative examples.

Example A. Two solid cones $A$ and $B$ are placed in a cylindrical tube. The ratio of their capacities is $2:1$. Find heights and capacities of cones, given the tube’s diameter is $4$ cm and total height $14$ cm.

Solution. Cone radii both $=2$. Capacities ratio $\tfrac{V_A}{V_B}=\tfrac{h_A}{h_B}=\tfrac{2}{1}$ so $h_A=2h_B$. Also $h_A+h_B=14\Rightarrow3h_B=14\Rightarrow h_B=\tfrac{14}{3},\ h_A=\tfrac{28}{3}$. $V_A=\tfrac13\pi(4)\tfrac{28}{3}=\tfrac{112\pi}{9},\ V_B=\tfrac{56\pi}{9}$. Numerically $V_A\approx39.11,\ V_B\approx19.56$ cm$^3$.

Example B. From a solid cube of side $7$ cm, a conical cavity of height $7$ cm and base radius $3$ cm is hollowed out. Find the volume of the remaining solid.

Solution. $V=a^3-\tfrac13\pi r^2 h=343-\tfrac13\cdot\tfrac{22}{7}\cdot9\cdot7=343-66=277$ cm$^3$.

Example C. A wooden block of dimensions $11\text{ cm}\times7\text{ cm}\times7\text{ cm}$ has the largest possible cylinder cut out from it (axis along the longer side). Find the volume of wood remaining.

Solution. Largest cylinder has $r=3.5$, $h=11$. Cylinder $V=\pi(12.25)(11)=423.5$ cm$^3$. Block $V=11\cdot7\cdot7=539$. Wood remaining $=539-423.5=115.5$ cm$^3$.

Example D. A solid metallic sphere of radius $r$ is melted and recast into the shape of a cylinder of radius $r$. Find the height of the cylinder.

Solution. $\tfrac{4}{3}\pi r^3=\pi r^2 h\Rightarrow h=\tfrac{4r}{3}$.

Example E. A solid hemisphere has radius $r$. It is reshaped into a cone of base radius $r$. Find the height of the cone.

Solution. $\tfrac{2}{3}\pi r^3=\tfrac{1}{3}\pi r^2 h\Rightarrow h=2r$.

Example F. A roller is $1$ m long and $0.84$ m in diameter. It takes $750$ revolutions to flatten a playground. Find the area of the playground in m$^2$.

Solution. Area covered per revolution $=2\pi rh=2\cdot\tfrac{22}{7}(0.42)(1)=2.64$ m$^2$. Total $=750\cdot2.64=1980$ m$^2$.

Example G. The radii of internal and external surfaces of a hollow spherical shell are $3$ cm and $5$ cm. If it is melted and recast into a solid cylinder of height $\dfrac{8}{3}$ cm, find the radius of the cylinder.

Solution. Shell volume $=\tfrac{4}{3}\pi(125-27)=\tfrac{392\pi}{3}$. Cylinder $\pi r^2\cdot\tfrac{8}{3}=\tfrac{392\pi}{3}\Rightarrow r^2=49\Rightarrow r=7$ cm.

Example H. A drinking glass is in the shape of a cylinder of radius $3$ cm and height $11$ cm with a hemispherical raised portion at the bottom of radius $3$ cm. Find apparent and actual capacities.

Solution. Apparent $=\pi(9)(11)=99\pi=311.14$ cm$^3$. Hemisphere $=\tfrac{2}{3}\pi(27)=18\pi=56.57$ cm$^3$. Actual $=99\pi-18\pi=81\pi=254.57$ cm$^3$.

Example I. The diameter of a road roller of length $120$ cm is $84$ cm. If it takes $500$ complete revolutions to level a playground, find its area in $m^2$.

Solution. Lateral surface per turn $=2\pi rh=2\cdot\tfrac{22}{7}\cdot42\cdot120=31680$ cm$^2$. Total $=500\cdot31680=15840000$ cm$^2=1584$ m$^2$.

Example J. A solid metallic cylinder of radius $14$ cm and height $\tfrac{2}{3}$ cm is melted to form $n$ small cones of base radius $1$ cm and height $1$ cm. Find $n$.

Solution. $V_\text{cyl}=\pi(196)(\tfrac{2}{3})=\tfrac{392\pi}{3}$. Cone $V=\tfrac{1}{3}\pi(1)(1)=\tfrac{\pi}{3}$. $n=\tfrac{392\pi/3}{\pi/3}=392$.

Application Word Problems

App 1. A swimming pool $25$ m long, $15$ m wide is to be filled with water $1.6$ m deep. The hose pumps at $7$ kL/min. How many minutes to fill?

Answer: Volume needed $=25\cdot15\cdot1.6=600$ m$^3=600$ kL. Time $=600/7=85.71$ min $\approx1$ h $26$ min.

App 2. A spherical balloon of radius $r$ is inflated so its radius doubles. Find ratio of new SA to old SA.

Answer: $\dfrac{4\pi(2r)^2}{4\pi r^2}=4:1$.

App 3. A cylindrical drum of radius $7$ cm and height $20$ cm is filled with $2/3$ water. A solid cone of radius $5$ cm and height $9$ cm is dropped into it. By how much does the water level rise (assume drum doesn’t overflow)?

Answer: Cone volume $=\tfrac{1}{3}\pi(25)(9)=75\pi=235.71$ cm$^3$. Initial water $=\tfrac{2}{3}\pi(49)(20)=2053.33$ cm$^3$. Cylinder cross-section $=49\pi=154$ cm$^2$. Rise $=235.71/154\approx1.53$ cm.

App 4. An open metallic bucket is in the shape of a frustum mounted on a hollow cylindrical base of radius $r$. If the bucket has $r_1=21$ cm, $r_2=14$ cm, height of frustum $40$ cm, base cylinder height $10$ cm and radius $14$ cm, find the volume of water it can hold.

Answer: $V=\tfrac{1}{3}\pi(40)(441+196+294)+\pi(196)(10)=\tfrac{40\pi}{3}(931)+1960\pi=12413.33\pi+1960\pi=14373.33\pi=45172.38$ cm$^3=45.17$ L.

App 5. A toy is made by joining a hemisphere of diameter $7$ cm on top of a cube of edge $7$ cm. Find total cost of painting the surface at Rs $5$ per cm$^2$.

Answer: SA $=6a^2-\pi r^2+2\pi r^2=6(49)+\pi(3.5)^2=294+\tfrac{22}{7}(12.25)=294+38.5=332.5$ cm$^2$. Cost $=332.5\cdot5=$ Rs $1662.50$.

App 6. A right circular hollow tube has internal radius $r$, external $R$, height $h$. Show its volume is $\pi h(R-r)(R+r)$.

Answer: $V=\pi R^2 h-\pi r^2 h=\pi h(R^2-r^2)=\pi h(R-r)(R+r)$.

App 7. A circus tent is a cylinder of height $5$ m mounted by a cone of slant $13$ m, base radius $12$ m. If the tent is to be made from $1.5$ m wide canvas, find length needed.

Answer: Canvas area $=2\pi rh+\pi r\ell=\tfrac{22}{7}(12)(2\cdot5+13)=\tfrac{22}{7}(12)(23)=868.57$ m$^2$. Length $=868.57/1.5=579.05$ m.

App 8. A solid wooden right circular cone has radius $7$ cm and height $24$ cm. From the larger end a hemisphere is hollowed out. Find the surface area of the remaining solid.

Answer: $\ell=\sqrt{49+576}=25$. CSA cone $=\pi(7)(25)=175\pi$. Removed circle $\pi(49)$ replaced by hemisphere curved area $2\pi(49)$. SA $=175\pi+2\pi(49)=175\pi+98\pi=273\pi=858$ cm$^2$.

App 9. A wooden cylinder of base radius $14$ cm and length $30$ cm is to be carved into a cone of same base radius and height $30$ cm. Find volume of wood wasted.

Answer: $V_\text{cyl}=\pi(196)(30)=5880\pi$. $V_\text{cone}=\tfrac13\pi(196)(30)=1960\pi$. Waste $=3920\pi=12320$ cm$^3$.

App 10. A solid cuboid of dimensions $9\times8\times6$ cm is melted to form a cube. Find the side of the cube.

Answer: $V=432$. $a=\sqrt[3]{432}\approx7.56$ cm.

Conceptual Questions

C1. Why does the lateral surface area of a cube equal $4a^2$?

Answer: A cube has $6$ equal faces of area $a^2$ each. The “lateral” faces (sides) are the four vertical faces, excluding top and bottom; hence $4a^2$.

C2. If two cones have the same volume, what can you say about their dimensions?

Answer: $\tfrac{1}{3}\pi r_1^2 h_1=\tfrac{1}{3}\pi r_2^2 h_2\Rightarrow r_1^2 h_1=r_2^2 h_2$. They need not be congruent; one can be wide and short, the other narrow and tall.

C3. Why is the volume of a sphere $\dfrac{4}{3}\pi r^3$ and not $\dfrac{2}{3}\pi r^3$?

Answer: $\tfrac{2}{3}\pi r^3$ is the volume of a hemisphere; the sphere is two hemispheres joined back to back, doubling to $\tfrac{4}{3}\pi r^3$.

C4. When can the slant height of a frustum equal its actual height?

Answer: Only if $r_1=r_2$, but then the frustum is a cylinder. So in a true frustum, $\ell>h$ always.

C5. Two cubes of side $a$ are joined to form a cuboid. Why is the surface area $10a^2$ rather than $12a^2$?

Answer: Two of the faces become internal (joining face) and are hidden, removing $2a^2$ from the total $12a^2$.

C6. Why must the height $h$ in the cone formula be the perpendicular height and not the slant height?

Answer: Volume formula derives from stacking discs; the perpendicular height measures how many discs fit. Slant height does not represent vertical thickness.

C7. Identify the shape of an Indian “matka” (water pot). Why is the standard frustum formula not directly used?

Answer: A matka is a curved sphere-like shape; it is not a true frustum (which has straight slant sides). Its volume is found by integration in higher classes.

C8. State Cavalieri’s principle and how it relates to volume.

Answer: Two solids of equal heights with cross-sections of equal area at every level have equal volumes. It justifies the Pyramid-Prism rule and is the conceptual basis for $V=\tfrac{1}{3}\text{base}\times h$ for cones/pyramids.

C9. Why is sand from a cylindrical bucket usually heaped into a cone, not a hemisphere?

Answer: The angle of repose of dry sand is around $30^\circ$, which corresponds naturally to a cone shape. A hemisphere shape would require sand to defy gravity at $90^\circ$.

C10. If a frustum’s bigger radius equals twice the smaller, derive a simple ratio of volumes of cone-removed to frustum.

Answer: Take $r_1=2k,\ r_2=k$. Heights via similar triangles: $h_1=2h_2$. So smaller cone height $h_2$, full cone height $2h_2$, frustum height $h=h_2$. $V_\text{small}=\tfrac{1}{3}\pi k^2 h_2$. $V_\text{full}=\tfrac{1}{3}\pi(4k^2)(2h_2)=\tfrac{8}{3}\pi k^2 h_2$. $V_\text{frustum}=V_\text{full}-V_\text{small}=\tfrac{7}{3}\pi k^2 h_2$. Ratio $V_\text{small}:V_\text{frustum}=1:7$.

Extra Problems for Self-Practice

No.	Problem	Final Answer
E1	SA of sphere with radius $14$ cm.	$2464$ cm$^2$
E2	Volume of cylinder $r=21$, $h=20$.	$27720$ cm$^3$
E3	Slant of cone $r=9$, $h=12$.	$15$ cm
E4	TSA of hemisphere $r=7$.	$462$ cm$^2$
E5	Volume of cube of edge $5$ cm.	$125$ cm$^3$
E6	Volume of frustum $r_1=10,\ r_2=4,\ h=15$.	$\approx2495.71$ cm$^3$
E7	How many cubes of $1$ cm side fit in a $10$ cm box?	$1000$
E8	CSA of cone $r=7,\ \ell=25$.	$550$ cm$^2$
E9	Diameter of sphere with volume $\dfrac{4}{3}\pi$.	$2$
E10	Volume of cone $r=3.5,\ h=12$.	$154$ cm$^3$
E11	SA of cuboid $5\times4\times3$.	$94$
E12	Volume of hemisphere $r=21$.	$19404$ cm$^3$
E13	Slant of frustum $r_1=8,\ r_2=4,\ h=3$.	$5$
E14	SA of cylinder $r=2,\ h=10$, both ends.	$48\pi$
E15	Volume of sphere $r=10.5$.	$4851$ cm$^3$

Quick-Recall Formula Strip

Quantity	Cuboid $l,b,h$	Cube $a$	Cylinder	Cone	Sphere	Hemisphere	Frustum
CSA / LSA	$2h(l+b)$	$4a^2$	$2\pi rh$	$\pi r\ell$	$4\pi r^2$	$2\pi r^2$	$\pi(r_1+r_2)\ell$
TSA	$2(lb+bh+lh)$	$6a^2$	$2\pi r(r+h)$	$\pi r(\ell+r)$	$4\pi r^2$	$3\pi r^2$	$\pi[(r_1+r_2)\ell+r_1^2+r_2^2]$
Volume	$lbh$	$a^3$	$\pi r^2 h$	$\tfrac{1}{3}\pi r^2 h$	$\tfrac{4}{3}\pi r^3$	$\tfrac{2}{3}\pi r^3$	$\tfrac{1}{3}\pi h(r_1^2+r_2^2+r_1r_2)$

Glossary

Term	Meaning
Surface Area (SA)	Sum of areas of all faces of a 3-D solid; measured in unit$^2$.
Curved Surface Area (CSA)	Area of curved/lateral face only, excluding flat ends.
Total Surface Area (TSA)	Curved surface area plus area of all flat faces.
Volume (V)	Capacity or 3-D space occupied; unit$^3$.
Cuboid	Rectangular box with three mutually perpendicular pairs of equal faces.
Cube	Cuboid in which all edges are equal.
Right circular cylinder	Solid generated by revolving a rectangle about one of its sides.
Right circular cone	Solid generated by revolving a right triangle about one of its legs.
Slant height ($\ell$)	Distance from the apex of a cone to any point on the base circumference; $\ell=\sqrt{r^2+h^2}$.
Sphere	Set of all points in space at fixed distance $r$ from a fixed centre.
Hemisphere	Half of a sphere bounded by a great circle (flat disc).
Spherical shell	Region between two concentric spheres.
Frustum of a cone	Portion of a cone between its base and a cutting plane parallel to the base.
Conversion of solids	Recasting one shape into another; volume is conserved.
Combination of solids	Solid formed by joining two or more standard solids; SA omits hidden faces, V is additive.
$\pi$	Constant ratio of circumference to diameter; $\pi\approx\tfrac{22}{7}\approx3.14$.

Deep-Dive Derivations

Although ASSEB Class 10 expects you to use the formulas, knowing how they are derived strengthens your problem-solving instinct. We summarise the four most-asked derivations.

Derivation 1 – Curved Surface Area of a Cylinder

Slit a right circular cylinder of radius $r$ and height $h$ vertically along one edge and unroll it. The flat shape is a rectangle of width equal to the circumference $2\pi r$ and height $h$.

$$\text{CSA}=\text{rectangle area}=2\pi r\times h=2\pi rh.$$

Adding the two circular ends (each of area $\pi r^2$) gives the total surface area: $\text{TSA}=2\pi rh+2\pi r^2=2\pi r(h+r)$.

Derivation 2 – Volume of a Cylinder

Approximate a cylinder by stacking $n$ thin discs each of negligible thickness $h/n$. Each disc has area $\pi r^2$ and tiny volume $\pi r^2(h/n)$. Summing all $n$ discs:

$$V=n\cdot\pi r^2\cdot\frac{h}{n}=\pi r^2 h.$$

This is the prism rule: V = (base area) $\times$ height.

Derivation 3 – Curved Surface Area of a Cone

Cut a right cone of slant height $\ell$ and radius $r$ along a slant edge and unroll. You get a sector of radius $\ell$ and arc length $2\pi r$ (the original circumference).

$$\text{Sector area}=\tfrac{1}{2}\times\text{arc}\times\text{radius}=\tfrac{1}{2}\cdot2\pi r\cdot\ell=\pi r\ell.$$

Adding the base circle gives $\text{TSA}=\pi r\ell+\pi r^2=\pi r(\ell+r)$.

Derivation 4 – Volume of a Cone (Cavalieri / experiment)

By experiment – filling a hollow cone with sand and pouring three times into a cylinder of the same base and height fills the cylinder exactly – we find $V_\text{cone}=\tfrac13 V_\text{cyl}$:

$$V_\text{cone}=\tfrac{1}{3}\pi r^2 h.$$

Derivation 5 – Surface Area and Volume of a Sphere

Archimedes’ classical result states that a sphere of radius $r$ has the same curved area as the side of the cylinder of radius $r$ and height $2r$ that just encloses it: $\text{SA}=2\pi r\cdot2r=4\pi r^2$. The volume is two-thirds of the enclosing cylinder: $V=\tfrac{2}{3}\pi r^2(2r)=\tfrac{4}{3}\pi r^3$.

Derivation 6 – Frustum Volume by Subtraction

A frustum with radii $r_1>r_2$ and height $h$ comes from a big cone of radius $r_1$ and height $h_1$ minus a similar small cone of radius $r_2$ and height $h_2$, where $h=h_1-h_2$. Similar triangles give $\dfrac{h_1}{r_1}=\dfrac{h_2}{r_2}=\dfrac{h}{r_1-r_2}$, so $h_1=\dfrac{r_1 h}{r_1-r_2}$ and $h_2=\dfrac{r_2 h}{r_1-r_2}$.

$$V=\tfrac{1}{3}\pi r_1^2 h_1-\tfrac{1}{3}\pi r_2^2 h_2=\tfrac{1}{3}\pi\frac{h(r_1^3-r_2^3)}{r_1-r_2}=\tfrac{1}{3}\pi h(r_1^2+r_1r_2+r_2^2).$$

Worked-Example Variants

The exam often varies the shape, units or unknown. Practising the same situation in different guises makes you fluent. Here are eight extended worked examples.

Variant A. A hemispherical dome of radius $63$ dm is to be painted from outside at the cost of Rs $2$ per dm$^2$. Find the cost.

Answer: Curved area only $=2\pi r^2=2\cdot\tfrac{22}{7}\cdot63^2=24948\text{ dm}^2$. Cost $=24948\times2=$ Rs $49896$.

Variant B. A spherical iron ball of radius $4.2$ cm is dropped into a cylindrical jar of radius $7$ cm filled partly with water. Find the rise in the water level (assume ball fully submerged).

Answer: Volume of ball $=\tfrac{4}{3}\pi(4.2)^3=\tfrac{4}{3}\pi(74.088)=310.46\pi/3$ approx; equivalently $\tfrac{4}{3}\cdot\tfrac{22}{7}\cdot74.088\approx310.46$ cm$^3$. Rise $=\dfrac{310.46}{\pi(7)^2}=\dfrac{310.46}{154}\approx2.016$ cm.

Variant C. A solid metallic sphere of radius $10.5$ cm is melted and recast into smaller cones each of radius $3.5$ cm and height $3$ cm. Find the number of cones.

Answer: Sphere volume $=\tfrac{4}{3}\pi(10.5)^3=\tfrac{4}{3}\pi(1157.625)=4851\pi/\dots$ Numerically: $\tfrac{4}{3}\cdot\tfrac{22}{7}\cdot1157.625=4851\text{ cm}^3$. Cone volume $=\tfrac{1}{3}\pi(12.25)(3)=12.25\pi=38.5$ cm$^3$. So $n=4851/38.5=126$.

Variant D. The diameters of the lower and upper ends of a bucket in the form of a frustum are $20$ cm and $40$ cm. Its height is $20$ cm. Find capacity in litres.

Answer: $r_1=20,\ r_2=10,\ h=20$. $V=\tfrac{1}{3}\pi h(r_1^2+r_2^2+r_1r_2)=\tfrac{1}{3}\cdot\tfrac{22}{7}\cdot20\cdot700=\tfrac{22\cdot20\cdot700}{21}=14666.67$ cm$^3=14.667$ L.

Variant E. A toy is in the shape of a hemisphere surmounted by a right cone. The radius of the hemisphere is $7$ cm and total height $21$ cm. Find the surface area.

Answer: $r=7$, cone height $h=21-7=14$ cm. $\ell=\sqrt{49+196}=\sqrt{245}=7\sqrt5$ cm.

$$\text{SA}=\pi r\ell+2\pi r^2=\tfrac{22}{7}(7)(7\sqrt5)+2\cdot\tfrac{22}{7}(49)=22\cdot7\sqrt5+308=154\sqrt5+308\approx344.36+308=652.36\text{ cm}^2.$$

Variant F. A right triangle ABC with sides $5,12,13$ revolves about the side $12$. Find the volume of the cone produced and its CSA.

Answer: The cone has radius $5$ and height $12$, slant $13$. $V=\tfrac{1}{3}\pi(25)(12)=100\pi\approx314.29\text{ cm}^3$. $\text{CSA}=\pi(5)(13)=65\pi\approx204.29$ cm$^2$.

Variant G. The diameters of the ends of a frustum are $35$ cm and $30$ cm and its height is $14$ cm. Find the volume in litres.

Answer: $r_1=17.5,\ r_2=15,\ h=14$.

$$V=\tfrac{1}{3}\cdot\tfrac{22}{7}\cdot14\cdot(306.25+225+262.5)=\tfrac{1}{3}\cdot\tfrac{22}{7}\cdot14\cdot793.75=11645.83\text{ cm}^3=11.646\text{ L}.$$

Variant H. A solid is in the shape of a cone surmounted by a hemisphere of equal radius $r$ and equal heights. Show that the total surface area $=\pi r^2(\sqrt5+2)$.

Answer: Cone height $h=r$, so $\ell=\sqrt{r^2+r^2}=r\sqrt2$. SA $=\pi r\ell+2\pi r^2=\pi r^2\sqrt2+2\pi r^2=\pi r^2(\sqrt2+2)$. (The original problem statement uses different proportions; verify with the figure given in your textbook.)

Cross-Sectional View of a Combined Solid

Multiple Choice Questions

MCQ1. The total surface area of a hemisphere of radius $r$ is: (a) $2\pi r^2$ (b) $3\pi r^2$ (c) $4\pi r^2$ (d) $\pi r^2$.

Answer: (b) $3\pi r^2$. The curved part contributes $2\pi r^2$ and the flat circular base contributes $\pi r^2$, hence $3\pi r^2$.

MCQ2. If a cone of radius $r$ and height $h$ has slant height $\ell$, then $\ell^2$ equals: (a) $r^2-h^2$ (b) $r^2+h^2$ (c) $rh$ (d) $2rh$.

Answer: (b) $r^2+h^2$ by Pythagoras.

MCQ3. The volume of a sphere of radius $r$ is: (a) $\tfrac{4}{3}\pi r^3$ (b) $\tfrac{2}{3}\pi r^3$ (c) $4\pi r^2$ (d) $\pi r^3$.

Answer: (a).

MCQ4. A solid sphere of radius $r$ is melted and recast into $8$ smaller spheres of equal radius $r’$. Then $r’/r=$ (a) $1/2$ (b) $1/4$ (c) $1/3$ (d) $1/8$.

Answer: (a) Equating volumes: $\tfrac{4}{3}\pi r^3=8\cdot\tfrac{4}{3}\pi r’^3\Rightarrow r’^3=r^3/8\Rightarrow r’=r/2$.

MCQ5. The volume of the frustum with $r_1=14,\ r_2=7,\ h=12$ is closest to: (a) $4312\text{ cm}^3$ (b) $3850\text{ cm}^3$ (c) $4620\text{ cm}^3$ (d) $5544\text{ cm}^3$.

Answer: (a). $V=\tfrac{1}{3}\pi h(r_1^2+r_2^2+r_1 r_2)=\tfrac{1}{3}\cdot\tfrac{22}{7}\cdot12\cdot(196+49+98)=\tfrac{264\cdot343}{21}=4312\text{ cm}^3$.

MCQ6. The total surface area of a solid hemisphere whose volume is $\tfrac{2}{3}\pi$ is: (a) $3\pi$ (b) $4\pi$ (c) $\pi$ (d) $2\pi$.

Answer: (a). $V=\tfrac{2}{3}\pi r^3=\tfrac{2}{3}\pi\Rightarrow r=1$, hence TSA $=3\pi(1)^2=3\pi$.

MCQ7. If the radii of two right circular cylinders are in the ratio $2:3$ and their heights are in the ratio $5:4$, ratio of their volumes is: (a) $5:9$ (b) $5:6$ (c) $4:9$ (d) $1:1$.

Answer: (a). $\dfrac{V_1}{V_2}=\dfrac{r_1^2 h_1}{r_2^2 h_2}=\dfrac{4\cdot5}{9\cdot4}=\dfrac{20}{36}=\dfrac{5}{9}$.

MCQ8. A circus tent is cylindrical to a height of $3$ m and conical above it. Common radius $52.5$ m, slant of cone $53$ m. Canvas required is closest to: (a) $9735$ m$^2$ (b) $7735$ m$^2$ (c) $8735$ m$^2$ (d) $11825$ m$^2$.

Answer: (a). $A=2\pi rh+\pi r\ell=\pi r(2h+\ell)=\tfrac{22}{7}(52.5)(59)=9735\text{ m}^2$.

MCQ9. A cylinder, a cone and a hemisphere have equal base and equal height. The ratio of their volumes is: (a) $3:1:2$ (b) $1:2:3$ (c) $2:1:3$ (d) $1:3:2$.

Answer: (a). For cylinder $V=\pi r^2 h$, cone $\tfrac{1}{3}\pi r^2 h$, hemisphere $\tfrac{2}{3}\pi r^3=\tfrac{2}{3}\pi r^2 h$ (since $h=r$). Ratio $1:\tfrac13:\tfrac23=3:1:2$.

MCQ10. The slant height of a cone whose CSA is $308$ cm$^2$ and base radius $7$ cm is: (a) $14$ cm (b) $11$ cm (c) $13$ cm (d) $12$ cm.

Answer: (a). $\pi r\ell=308\Rightarrow\ell=\dfrac{308\cdot7}{22\cdot7}=\dfrac{308}{22}=14$ cm.

MCQ11. If the radius of a sphere is doubled, its surface area becomes: (a) doubled (b) tripled (c) four times (d) eight times.

Answer: (c). SA scales as $r^2$.

MCQ12. A right circular cone of height $h$ is cut into two equal halves by a plane through its midpoint parallel to base. Volume of upper part : volume of lower frustum = (a) $1:7$ (b) $1:4$ (c) $1:8$ (d) $7:1$.

Answer: (a). Upper cone is similar with linear ratio $1:2$, so volume ratio $1:8$. Frustum (lower) $=8-1=7$. Hence $1:7$.

Higher-Order Thinking Skills (HOTS)

HOTS 1. A solid metallic sphere of diameter $28$ cm is melted and recast into a number of smaller cones, each of radius $\tfrac{14}{3}$ cm and height $3$ cm. Find the number of cones formed.

Answer: Sphere volume $=\tfrac{4}{3}\pi(14)^3=\tfrac{10976\pi}{3}$. Cone volume $=\tfrac{1}{3}\pi\cdot\tfrac{196}{9}\cdot3=\tfrac{196\pi}{9}$.

$$n=\frac{10976\pi/3}{196\pi/9}=\frac{10976\cdot9}{3\cdot196}=\frac{10976\cdot3}{196}=\frac{32928}{196}=168.$$

HOTS 2. A bucket made of metal sheet is in the form of a frustum, $35.2$ cm high and the radii of its lower and upper ends are $16$ cm and $33$ cm, respectively. Find the area of the metal sheet used and the volume of water it can hold (Take $\pi=\tfrac{22}{7}$).

Answer: $\ell=\sqrt{(33-16)^2+35.2^2}=\sqrt{289+1239.04}=\sqrt{1528.04}\approx39.09$ cm.

Sheet $=\pi(r_1+r_2)\ell+\pi r_2^2=\tfrac{22}{7}[(49)(39.09)+(256)]=\tfrac{22}{7}(1915.41+256)=\tfrac{22}{7}(2171.41)=6824.43\text{ cm}^2$.

$$V=\tfrac{1}{3}\pi h(r_1^2+r_2^2+r_1r_2)=\tfrac{1}{3}\cdot\tfrac{22}{7}\cdot35.2\cdot(1089+256+528)=\tfrac{22\cdot35.2\cdot1873}{21}=69088.27\text{ cm}^3.$$

HOTS 3. A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere. If the radius of the hemisphere is $4.2$ cm and the total height of the toy is $10.2$ cm, find the volume of wood used in the toy.

Answer: $r=4.2,\ h_\text{cone}=10.2-4.2=6$.

$$V=\tfrac{2}{3}\pi r^3+\tfrac{1}{3}\pi r^2 h=\tfrac{1}{3}\pi r^2(2r+h)=\tfrac{1}{3}\cdot\tfrac{22}{7}\cdot17.64\cdot14.4=266.11\text{ cm}^3.$$

HOTS 4. The radii of two right circular cylinders are in the ratio $1:2$ and their heights in the ratio $5:3$. Calculate the ratio of their (a) curved surface areas, (b) volumes.

Answer: (a) $\tfrac{2\pi r_1 h_1}{2\pi r_2 h_2}=\tfrac{1\cdot5}{2\cdot3}=\tfrac{5}{6}$. (b) $\tfrac{\pi r_1^2 h_1}{\pi r_2^2 h_2}=\tfrac{1\cdot5}{4\cdot3}=\tfrac{5}{12}$.

HOTS 5. A solid is composed of a cylinder with two hemispherical ends. The total length is $108$ cm and the diameter of each hemisphere is $36$ cm. Find the cost of polishing the surface of the solid at the rate of $7$ paise per cm$^2$.

Answer: $r=18$, cylinder length $=108-36=72$ cm.

$$\text{SA}=2\pi rh+2(2\pi r^2)=2\pi r(h+2r)=2\cdot\tfrac{22}{7}\cdot18\cdot108=12219.43\text{ cm}^2.$$

Cost $=12219.43\times0.07=$ Rs $855.36$.

HOTS 6. A solid metallic right circular cone of height $20$ cm and base radius $5$ cm is melted and recast into spherical balls of diameter $1$ cm each. Find the number of balls.

Answer: Cone volume $=\tfrac{1}{3}\pi(25)(20)=\tfrac{500\pi}{3}$. Ball $=\tfrac{4}{3}\pi(0.5)^3=\tfrac{\pi}{6}$.

$$n=\frac{500\pi/3}{\pi/6}=\frac{500\cdot6}{3}=1000.$$

HOTS 7. A spherical glass vessel has a cylindrical neck $8$ cm long, $2$ cm in diameter; the diameter of the spherical part is $8.5$ cm. By measuring the amount of water it holds, a child finds its volume to be $345$ cm$^3$. Check his answer ($\pi=3.14$).

Answer: Cylindrical neck volume $=\pi(1)^2(8)=8\pi=25.13$. Spherical part $=\tfrac{4}{3}\pi(4.25)^3=\tfrac{4}{3}\pi(76.77)=321.39$. Total $=346.52\text{ cm}^3$. Hence the child’s value $345$ is incorrect (off by $\approx1.5$).

HOTS 8. A juice seller serves his customers using a glass which has a hemispherical raised portion at the bottom which reduces the apparent capacity. Inner diameter of cylindrical glass $=5$ cm, height $=10$ cm. Find the apparent and actual capacity.

Answer: $r=2.5,\ h=10$. Apparent $=\pi r^2 h=\tfrac{22}{7}(6.25)(10)=196.43\text{ cm}^3$. Hemisphere volume $=\tfrac{2}{3}\pi(2.5)^3=\tfrac{2}{3}\cdot\tfrac{22}{7}\cdot15.625=32.74\text{ cm}^3$. Actual $=196.43-32.74=163.69\text{ cm}^3$.

HOTS 9. A solid iron pole consists of a cylinder of height $220$ cm and base diameter $24$ cm, surmounted by another cylinder of height $60$ cm and radius $8$ cm. Find the mass of the pole, given that $1$ cm$^3$ of iron has $\approx 8$ g of mass.

Answer: $V=\pi(12)^2(220)+\pi(8)^2(60)=\pi[31680+3840]=35520\pi=111577.14\text{ cm}^3$. Mass $=111577.14\times8=892617.14$ g $\approx892.62$ kg.

HOTS 10. A hemispherical tank full of water is emptied by a pipe at the rate of $3\tfrac{4}{7}$ litres per second. How much time will it take to empty half the tank if it is $3$ m in diameter?

Answer: Tank radius $=1.5$ m. Volume $=\tfrac{2}{3}\pi(1.5)^3=\tfrac{2}{3}\cdot\tfrac{22}{7}\cdot3.375=7.071\text{ m}^3=7071$ litres. Half $=3535.5$ litres. Rate $=\tfrac{25}{7}$ L/s. Time $=\tfrac{3535.5\cdot7}{25}=989.94$ s $\approx16.5$ minutes.

Important Concepts and Tips

Tip 1. Always re-read the problem to identify which surface is required: inner surface of a hollow vessel does not include any outer wall; outer surface of a wooden article that has hemispherical scoops still includes the curved scoop area but not its flat circle (which has been removed).

Tip 2. When two solids are joined the joining circle disappears – this is true for both surfaces. So a cone-on-hemisphere has SA $=\pi r\ell+2\pi r^2$ (no $\pi r^2$ for the joining circle).

Tip 3. “Volume conserved” problems are easy if you set up the equation $V_\text{old}=V_\text{new}$ first, then substitute the relevant formula on each side.

Tip 4. For a frustum, before computing CSA always find $\ell=\sqrt{h^2+(r_1-r_2)^2}$. Do not confuse this with the slant of the original cone $L=\sqrt{h_1^2+r_1^2}$.

Tip 5. When a wire is drawn from a melted volume, equate $\pi r_\text{wire}^2 L = V_\text{melted}$ to find length $L$. Wire is treated as a cylinder.

Tip 6. A flowing canal/pipe converts speed to volume by writing the cross-section area times the length traversed in the chosen time interval.

Tip 7. In bricks-and-cistern absorption problems, displacement equals (brick volume) $\times$ (1 – absorption fraction). Use this net rise per brick to compute number of bricks before overflow.

Tip 8. Always check units: convert mm to cm or cm to m before substituting; the answer’s unit should match what the problem asked for.

Common Mistakes to Avoid

Mistake	Correction
Counting the hidden circle when joining a cone to a hemisphere.	Subtract or simply omit it; SA of cone-on-hemi $=\pi r\ell+2\pi r^2$.
Using cone slant for the frustum slant.	Frustum slant $=\sqrt{h^2+(r_1-r_2)^2}$, not $\sqrt{h^2+r_1^2}$.
Forgetting to convert mm to cm in capsule problems.	Stick to one unit throughout, usually cm.
Computing TSA when the question wants only canvas (CSA) for tents.	The base of a tent is open, so canvas $=$ CSA only.
Treating a hemispherical cavity as adding to SA only on top.	Both the curved cavity and removed flat circle change the SA; net effect $=+\pi r^2$ in most cases.
Using radius when the diameter is given (and vice versa).	Halve diameters first; double-check the labelled figure.
Forgetting the factor $\tfrac13$ for cone volume.	Cone $V=\tfrac13\pi r^2 h$. Always.
Mis-applying $\tfrac23\pi r^3$ where $\tfrac43\pi r^3$ is required.	Hemisphere is half a sphere ($\tfrac23$); a sphere is $\tfrac43$.

Frequently Asked Questions

FAQ 1. Is the formula for a hemisphere’s TSA always $3\pi r^2$?

Answer: Yes, when the hemisphere is solid. It comprises curved part $2\pi r^2$ plus circular base $\pi r^2$. A hollow hemispherical bowl, where the bottom circle is open, has only $2\pi r^2$ inside-curved area.

FAQ 2. How can I remember which formulas need a $\tfrac13$?

Answer: Pyramidal/conical shapes – cone, pyramid – are one-third of the prism with the same base and height. Spheres need $\tfrac43$. Hemispheres get $\tfrac23$. Cylinders, cuboids and cubes have no fraction.

FAQ 3. Why is $\pi$ approximated as $\tfrac{22}{7}$?

Answer: For convenient cancellation. NCERT/SEBA conventionally use $\tfrac{22}{7}$ unless told to use $3.14$. Final answers may differ slightly between the two.

FAQ 4. In conversion problems, does mass change?

Answer: No. Mass and density are conserved (only the shape changes). Volume too is conserved as long as the same material is used.

FAQ 5. What is the difference between curved surface area and lateral surface area?

Answer: “Curved” is used when the surface is curved (cylinder, cone, sphere); “lateral” is the prismatic equivalent (cuboid, cube). Both exclude the top and bottom flat faces.

FAQ 6. How do you find the slant height of a frustum given only the radii and the volume?

Answer: First solve $V=\tfrac13\pi h(r_1^2+r_2^2+r_1r_2)$ for $h$, then plug into $\ell=\sqrt{h^2+(r_1-r_2)^2}$.

FAQ 7. Why doesn’t the joining circle in combined solids appear in the SA?

Answer: It is hidden inside the solid; surface area counts only the visible exterior. Volumes, however, simply add because each solid contributes its own interior.

Assertion – Reason Questions

For each pair, choose: (A) Both A and R are true and R is the correct explanation of A; (B) Both A and R are true but R is NOT the correct explanation; (C) A is true, R is false; (D) A is false, R is true.

AR1. Assertion (A): The TSA of a hemisphere of radius $r$ is $3\pi r^2$. Reason (R): A hemisphere has both a curved surface and a flat circular base.

Answer: (A). Curved $2\pi r^2$ + base $\pi r^2$ $=3\pi r^2$, exactly because of R.

AR2. Assertion (A): If a sphere of radius $r$ is melted and recast into a cylinder of radius $r$, the cylinder’s height is $\tfrac{4r}{3}$. Reason (R): Volume is conserved when one solid is recast into another of the same material.

Answer: (A). $\tfrac{4}{3}\pi r^3=\pi r^2 h\Rightarrow h=\tfrac{4r}{3}$.

AR3. Assertion (A): The CSA of a cone with $r=7,\ \ell=10$ is $220$ cm$^2$. Reason (R): CSA $=\pi r\ell$.

Answer: (A). $\tfrac{22}{7}(7)(10)=220$.

AR4. Assertion (A): The volume of a frustum is always less than the volume of the corresponding full cone. Reason (R): A frustum is the full cone minus a smaller similar cone.

Answer: (A).

AR5. Assertion (A): When two cubes of equal side are joined face to face, the result is a cube. Reason (R): The sides of the joined object equal $a$, $a$, $2a$.

Answer: (D). The result is a cuboid (not a cube), but R is correct.

Case Study Questions

Case Study 1 – Water Tank Design. A village water-tank consists of a cylindrical body $7$ m diameter and $4$ m high topped by a hemispherical dome of the same diameter. The bottom is a flat closed circular plate.

Q1. Find the volume of water the tank can hold.

Answer: $r=3.5$. $V=\pi r^2 h+\tfrac{2}{3}\pi r^3=\tfrac{22}{7}(12.25)(4)+\tfrac{2}{3}\cdot\tfrac{22}{7}(42.875)=154+89.83=243.83$ m$^3$.

Q2. Find the area of metallic sheet needed to make the tank.

Answer: Cylindrical wall $+$ bottom $+$ dome curved surface $=2\pi rh+\pi r^2+2\pi r^2=2\pi r(h+\tfrac{3r}{2})$. Numerically $=\tfrac{22}{7}\cdot7\cdot(8+10.5)\cdot$ Wait, let’s compute carefully: $2\pi rh=88$, $\pi r^2=38.5$, $2\pi r^2=77$. Total $=88+38.5+77=203.5$ m$^2$.

Q3. If sheet costs Rs $100$ per m$^2$, find total cost.

Answer: $203.5\times100=$ Rs $20350$.

Case Study 2 – Ice-Cream Factory. An ice-cream factory uses cones consisting of a conical body of radius $3$ cm, height $9$ cm, topped by a hemispherical scoop of the same radius. The factory has $36$ litres of ice cream available.

Q1. Find the volume of one ice-cream cone.

Answer: $V=\tfrac13\pi r^2 h+\tfrac23\pi r^3=\tfrac13\pi(9)(9)+\tfrac23\pi(27)=27\pi+18\pi=45\pi=141.43$ cm$^3$.

Q2. How many cones can be filled?

Answer: $36$ L $=36000$ cm$^3$. $n=\tfrac{36000}{141.43}\approx254.55$, so $254$ cones.

Q3. Each cone is sold for Rs $20$. Find total revenue.

Answer: $254\times20=$ Rs $5080$.

Case Study 3 – Cement Mixing. A cylindrical concrete column is being made of radius $0.5$ m and height $4$ m. The cement comes in $50$ kg bags, each producing $0.025$ m$^3$ of concrete.

Q1. Find the volume of the column.

Answer: $V=\pi(0.25)(4)=\pi=3.14$ m$^3$.

Q2. How many bags of cement are needed?

Answer: $n=3.14/0.025=125.6$, so $126$ bags.

Q3. If each bag costs Rs $400$, total cost?

Answer: $126\times400=$ Rs $50400$.

Case Study 4 – Hot-Air Balloon. A spherical hot-air balloon of radius $4$ m is being inflated. As gas is added, the radius increases at the rate of $0.05$ m/s.

Q1. Find the surface area at $r=4$.

Answer: $SA=4\pi(16)=64\pi\approx201.06$ m$^2$.

Q2. Find the volume at $r=4$.

Answer: $V=\tfrac{4}{3}\pi(64)=\tfrac{256\pi}{3}\approx268.08$ m$^3$.

Q3. If the radius increases to $4.05$, find the change in volume (approximately).

Answer: $\Delta V\approx4\pi r^2\,\Delta r=4\pi(16)(0.05)=10.05$ m$^3$.

Case Study 5 – Lampshade Manufacturing. A frustum-shaped lampshade of upper radius $5$ cm, lower radius $10$ cm, height $12$ cm is being manufactured. The shade has fabric only on the curved surface.

Q1. Find the slant height of the lampshade.

Answer: $\ell=\sqrt{144+25}=\sqrt{169}=13$ cm.

Q2. Find the area of fabric needed.

Answer: $\pi(r_1+r_2)\ell=\tfrac{22}{7}(15)(13)=612.86$ cm$^2$.

Q3. If fabric costs Rs $0.50$ per cm$^2$, total fabric cost?

Answer: $612.86\times0.50=$ Rs $306.43$.

Final Mock Test (Practice on Your Own)

Time: $1$ hour. Maximum marks: $40$. Each question carries $4$ marks except where noted.

FT1. A cone of base radius $7$ cm and height $24$ cm is melted to form a sphere. Find the radius of the sphere. ($\pi=\tfrac{22}{7}$)

FT2. A bucket made of metal sheet has the form of a frustum with $r_1=15$ cm, $r_2=5$ cm, height $24$ cm. Find the area of metal sheet needed and capacity in litres.

FT3. A children’s tent is in the shape of a cylinder of base radius $5$ m, height $3$ m, surmounted by a cone of slant height $8$ m. Find canvas needed.

FT4. A hollow cylindrical pipe is $21$ dm long. Its inner and outer radii are $5$ cm and $7$ cm. Find volume of the metal.

FT5. A solid is composed of a cylinder of radius $7$ cm and height $14$ cm with two hemispherical scoops at the two ends. Find the SA of the resulting solid.

FT6. A cubical wooden block of side $14$ cm is to have the largest possible cone scooped out of it (axis along a diagonal of one face). Find the volume of wood remaining.

FT7. (5 marks) A solid metallic right circular cone $84$ cm high with base radius $21$ cm is melted to form a uniform-thickness wire $7$ mm in diameter. Find length of wire.

FT8. (5 marks) A pole of total length $11$ m has a cylinder of length $9$ m and diameter $40$ cm and is topped by a hemisphere. Find total surface area.

Mock Test – Answer Key

FT1. Sphere $r$: $\tfrac{4}{3}\pi r^3=\tfrac{1}{3}\pi(49)(24)\Rightarrow r^3=294\Rightarrow r=\sqrt[3]{294}\approx6.65$ cm.

FT2. $\ell=\sqrt{576+100}=\sqrt{676}=26$ cm. Sheet $=\pi(r_1+r_2)\ell+\pi r_2^2=\tfrac{22}{7}(20)(26)+\tfrac{22}{7}(25)=1634.29+78.57=1712.86$ cm$^2$. $V=\tfrac{1}{3}\pi(24)(225+25+75)=\tfrac{1}{3}\cdot\tfrac{22}{7}\cdot24\cdot325=8171.43$ cm$^3=8.17$ L.

FT3. Canvas $=2\pi rh+\pi r\ell=\tfrac{22}{7}(5)(2\cdot3+8)=\tfrac{22}{7}(5)(14)=220$ m$^2$.

FT4. $V=\pi h(R^2-r^2)=\tfrac{22}{7}(210)(49-25)=\tfrac{22}{7}(210)(24)=15840$ cm$^3$.

FT5. SA $=2\pi rh-2\pi r^2+2\cdot2\pi r^2=2\pi rh+2\pi r^2$. Wait – careful: scooping out replaces the closing top circle with the curved hemisphere, so net add per scoop $=+\pi r^2$. SA $=2\pi rh+2\pi r^2=2\cdot\tfrac{22}{7}(7)(14+7)=924$ cm$^2$.

FT6. Largest cone: base radius $7$ cm, height $14$ cm. Cone $V=\tfrac{1}{3}\pi(49)(14)=\tfrac{2156\pi}{3}=718.67\cdot\pi/\pi=$. Numerically: $\tfrac{1}{3}\cdot\tfrac{22}{7}(49)(14)=718.67$ cm$^3$. Cube $V=2744$. Wood $=2744-718.67=2025.33$ cm$^3$.

FT7. Cone $V=\tfrac{1}{3}\pi(441)(84)=12348\pi$. Wire $\pi(0.35)^2 L=12348\pi\Rightarrow L=12348/0.1225=100800$ cm $=1008$ m.

FT8. $r=20$ cm, cylinder length $9$ m. Hemisphere length $20$ cm $=0.2$ m, so total length is $9.2$ m, not $11$ – so hemisphere with radius matching pole. Recompute with $h=11-0.2=10.8$ m. SA $=2\pi rh+2\pi r^2=2\pi(0.2)(10.8+0.2)=2\pi(0.2)(11)=4.4\pi=13.83$ m$^2$.

Summary Mind-Map

Concept	Key Idea
Surface Area of Combined Solids	Add the visible curved/flat areas; ignore hidden joining circles.
Volume of Combined Solids	Always additive: $V=V_1+V_2+\dots$
Conversion of Solids	Volume conserved when material is recast.
Frustum of a Cone	Difference of two similar cones; $\ell=\sqrt{h^2+(r_1-r_2)^2}$.
Real-Life Use	Tents, capsules, buckets, gulab jamun, water tanks, lampshades.
$\pi$ value	$\tfrac{22}{7}$ unless specified $3.14$.

Twenty Rapid-Fire Solved Numerical Drills

D1. Volume of cuboid $20\times15\times8$ cm. Answer: $V=2400$ cm$^3$.

D2. SA of cube of edge $9$ cm. Answer: $6(81)=486$ cm$^2$.

D3. Curved surface of cylinder $r=10,\ h=14$. Answer: $2\pi rh=880$ cm$^2$.

D4. Volume of cone $r=6,\ h=7$. Answer: $\tfrac{1}{3}\pi(36)(7)=264$ cm$^3$.

D5. SA of sphere $r=21$. Answer: $4\pi(441)=5544$ cm$^2$.

D6. Volume of hemisphere $r=14$. Answer: $\tfrac{2}{3}\pi(2744)=5749.33$ cm$^3$.

D7. Slant height of cone $r=8,\ h=15$. Answer: $\ell=17$.

D8. CSA of frustum $r_1=12,\ r_2=4,\ \ell=10$. Answer: $\pi(16)(10)=502.86$ cm$^2$.

D9. TSA of hemisphere $r=10.5$. Answer: $3\pi(110.25)=1039.5$ cm$^2$.

D10. Volume of sphere of diameter $14$ cm. Answer: $r=7$, $V=\tfrac{4}{3}\pi(343)=1437.33$ cm$^3$.

D11. A cube of side $a$ has the same volume as cylinder of radius $r$ and height $h$. If $a=4,\ r=2$, find $h$. Answer: $\pi(4)h=64\Rightarrow h=\tfrac{64}{4\pi}=5.09$ cm.

D12. A cone has volume $616$ cm$^3$ and height $12$. Find $r$. Answer: $r^2=\tfrac{616\cdot3}{12\pi}=49\Rightarrow r=7$.

D13. Twelve identical spheres of radius $1$ are melted into one sphere. New radius. Answer: $R^3=12\Rightarrow R=\sqrt[3]{12}\approx2.29$.

D14. A cube floats in water with $\tfrac{1}{4}$ of its volume submerged. If side $=10$ cm, displaced water $=$? Answer: $\tfrac{1000}{4}=250$ cm$^3$.

D15. CSA of cylinder $=44$ cm$^2$, $h=2$ cm. Find $r$. Answer: $r=\tfrac{44}{4\pi}=3.5$ cm.

D16. Cone of $r=5,\ h=12$ has slant $13$. CSA $=$? Answer: $\pi(5)(13)=204.29$ cm$^2$.

D17. Hemisphere $r=7$, find total volume of two such hemispheres joined. Answer: $V=2\cdot\tfrac{2}{3}\pi(343)=\tfrac{4}{3}\pi(343)=1437.33$ cm$^3$ (a sphere).

D18. Volume of frustum $r_1=8,\ r_2=6,\ h=10$. Answer: $V=\tfrac{1}{3}\pi(10)(64+36+48)=\tfrac{10\pi}{3}(148)=1549.52$ cm$^3$.

D19. Total surface of cuboid $5\times4\times6$. Answer: $2(20+24+30)=148$ cm$^2$.

D20. Volume increment if a cube’s edge is increased from $5$ to $6$. Answer: $216-125=91$ cm$^3$.

Bonus Visual: Combined Solid Decomposition

Keep practising every solved example, redraw each labelled figure and remember the golden rule: when faces are joined, do not count the hidden circles in surface area; for volumes, just add. Best of luck for your ASSEB Class 10 Mathematics examination!