CBSE Class 10 Surface Areas and Volumes Sure Shot Questions Set F

Choose the correct answer from the given options:

Question.  A cylindrical pencil sharpened at one edge is the combination of
(a) a cone and a cylinder
(b) frustum of a cone and a cylinder
(c) a hemisphere and a cylinder
(d) two cylinders
Answer: a

Question. A surahi is the combination of
(a) a sphere and a cylinder
(b) a hemisphere and a cylinder
(c) two hemispheres
(d) a cylinder and a cone
Answer: a

Question. In a right circular cone, the cross-section made by a plane parallel to the base is a
(a) Triangle
(b) Circle
(c) Square
(d) None of these
Answer: b

Question. If two solid hemispheres of same base radius r are joined together along their bases, then curved surface area of this new solid is
(a) \( 4\pi r^2 \)
(b) \( 3\pi r^2 \)
(c) \( 2\pi r^2 \)
(d) \( \pi r^2 \)
Answer: a

Assertion-Reason Type Questions:
In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Assertion (A): The number of coins 1.75 cm in diameter and 2 mm thick is formed from a melted cuboid 10 cm \( \times \) 5.5 cm \( \times \) 3.5 cm is 400.
Reason (R): Volume of a cylinder \( = \pi r^2 h \) cubic units and volume of cuboid \( = (l \times b \times h) \) cubic units.
Answer: a

Question. Assertion (A): Number of spherical balls that can be made out of a solid cube of lead whose edge is 44 cm, each ball being 4 cm in diameter is 2541.
Reason (R): Number of balls \( = \frac{\text{Volume of one ball}}{\text{Volume of lead}} \).
Answer: c

Short Answer Type Questions

Question. A toy is in the form of a cone mounted on a hemisphere of radius 3.5 cm. The total height of the toy is 15.5 cm. Find the total surface area of the toy. 
Answer: Here, \( r = 3.5 \) cm. Height of cone \( h = (15.5 - 3.5) \text{ cm} = 12.0 \text{ cm} \).
Slant height \( l = \sqrt{12^2 + 3.5^2} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5 \text{ cm} \).
Total surface area of the toy \( = \pi rl + 2\pi r^2 = \pi r(l + 2r) \)
\( = \frac{22}{7} \times 3.5 \times (12.5 + 2 \times 3.5) = 11 \times (12.5 + 7) = 11 \times 19.5 = 214.5 \text{ cm}^2 \).

Question. A solid is in the form of a right circular cylinder with hemispherical ends. The total height of the solid is 58 cm and the diameter of the cylinder is 28 cm. Find the total surface area of the solid. [Use \( \pi = \frac{22}{7} \)] 
Answer: Radius \( r = 14 \) cm. Height of cylindrical part \( h = 58 - (14 + 14) = 58 - 28 = 30 \) cm.
Total surface area \( = \text{Curved surface area of cylindrical portion} + 2(\text{Curved surface area of hemispherical portion}) \)
\( = 2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r) = 2 \times \frac{22}{7} \times 14 \times (30 + 2 \times 14) \)
\( = 88 \times (30 + 28) = 88 \times 58 = 5104 \text{ cm}^2 \).

Short Answer Type Questions

Question. A toy is in the form of a cone mounted on a hemisphere with the same radius. The diameter of the base of the conical portion is 6 cm and its height is 4 cm. Determine the surface area of the toy. [Take \( \pi = 3.14 \)]
Answer: \( r = 3 \) cm, \( h = 4 \) cm \( \Rightarrow l = \sqrt{3^2 + 4^2} = 5 \) cm.
Surface area \( = 2\pi r^2 + \pi rl = \pi r(2r + l) = 3.14 \times 3 \times (2 \times 3 + 5) = 3.14 \times 3 \times 11 = 103.62 \text{ cm}^2 \).

Question. From a solid cylinder of height 12 cm and diameter of the base 10 cm a conical cavity of the same height and same diameter is hollowed out. Find the surface area of the remaining solid.
Answer: \( r = 5 \) cm, \( h = 12 \) cm \( \Rightarrow l = \sqrt{5^2 + 12^2} = 13 \) cm.
Surface area \( = \text{C.S.A of cylinder} + \text{Base area of cylinder} + \text{C.S.A of conical cavity} \)
\( = 2\pi rh + \pi r^2 + \pi rl = \pi r(2h + r + l) = \pi \times 5 \times (2 \times 12 + 5 + 13) = \pi \times 5 \times 42 = 210\pi \)
\( = 210 \times \frac{22}{7} = 660 \text{ cm}^2 \).

Question. A gulab jamun when completely ready for eating contains sugar syrup upto about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns shaped like a cylinder with two hemispherical ends, if the complete length of each of the gulab jamuns is 5 cm and its diameter is 2.8 cm. 
Answer: Radius \( r = 1.4 \) cm. Height of cylinder \( h = 5 - 2(1.4) = 2.2 \) cm.
Volume of 1 gulab jamun \( = \pi r^2 h + \frac{4}{3}\pi r^3 = \pi r^2 (h + \frac{4}{3}r) = \frac{22}{7} \times 1.4^2 \times (2.2 + \frac{4}{3} \times 1.4) \approx 25.05 \text{ cm}^3 \).
Volume of 45 gulab jamuns \( \approx 45 \times 25.05 = 1127.25 \text{ cm}^3 \).
Syrup volume \( = 30\% \text{ of } 1127.25 \approx 338.18 \text{ cm}^3 \). (Calculated approx: 337.88 \( \text{cm}^3 \)).

Question. A juice seller was serving his customers using glasses. The inner diameter of the cylindrical glass was 5 cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of the glass was 10 cm, find its actual capacity and its apparent capacity. (Use \( \pi = 3.14 \)) 
Answer: \( r = 2.5 \) cm, \( h = 10 \) cm.
Apparent capacity \( = \pi r^2 h = 3.14 \times 2.5 \times 2.5 \times 10 = 196.25 \text{ cm}^3 \).
Actual capacity \( = \text{Apparent capacity} - \text{Vol. of hemisphere} = 196.25 - \frac{2}{3} \times 3.14 \times 2.5^3 \)
\( = 196.25 - 32.71 = 163.54 \text{ cm}^3 \).

Long Answer Type Questions 

Question. A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of cylinder. The diameter and height of cylinder are 6 cm and 12 cm, respectively. If the slant height of the conical portion is 5 cm, then find the total surface area and volume of rocket. (Use \( \pi = 3.14 \)) 
Answer: \( r = 3 \) cm, \( H_{\text{cyl}} = 12 \) cm, \( l = 5 \) cm \( \Rightarrow h_{\text{cone}} = \sqrt{5^2 - 3^2} = 4 \) cm.
Volume \( = \pi r^2 H + \frac{1}{3}\pi r^2 h = \pi r^2 (H + \frac{h}{3}) = 3.14 \times 9 \times (12 + \frac{4}{3}) = 3.14 \times 3 \times 40 = 376.8 \text{ cm}^3 \).
TSA \( = \text{C.S.A cyl} + \text{C.S.A cone} + \text{Base area cyl} = 2\pi rH + \pi rl + \pi r^2 = \pi r(2H + l + r) = 3.14 \times 3 \times (24 + 5 + 3) = 301.44 \text{ cm}^2 \).

Question. A solid right circular cone of height 120 cm and radius 60 cm is placed in a right circular cylinder full of water of height 180 cm such that it touches the bottom. Find the volume of water left in cylinder, if the radius of the cylinder is equal to the radius to the cone. 
Answer: \( r = 60 \) cm, \( H_{\text{cyl}} = 180 \) cm, \( h_{\text{cone}} = 120 \) cm.
Vol. water left \( = \text{Vol. cylinder} - \text{Vol. cone} = \pi r^2 H - \frac{1}{3}\pi r^2 h = \pi r^2(H - \frac{h}{3}) \)
\( = \frac{22}{7} \times 60 \times 60 \times (180 - \frac{120}{3}) = \frac{22}{7} \times 3600 \times 140 = 22 \times 3600 \times 20 = 1,584,000 \text{ cm}^3 = 1.584 \text{ m}^3 \).

Question. A solid is in the shape of a cone surmounted on a hemisphere. The radius of each of them being 3.5 cm and the total height of the solid is 9.5 cm. Find the volume of the solid.
Answer: \( r = 3.5 \) cm, \( h_{\text{cone}} = 9.5 - 3.5 = 6 \) cm.
Volume \( = \text{Vol. hemisphere} + \text{Vol. cone} = \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2(2r + h) \)
\( = \frac{1}{3} \times \frac{22}{7} \times 3.5^2 \times (7 + 6) = \frac{1}{3} \times 38.5 \times 13 \approx 166.83 \text{ cm}^3 \).

Case Study Based Questions

I. The Great Stupa at Sanchi is one of the oldest stone structures in India, and an important monument of Indian Architecture. It was originally commissioned by the emperor Ashoka in the 3rd century BCE. Its nucleus is a simple hemispherical brick structure built over the relics of the Buddha. It is a perfect example of combination of solid figures. A big hemispherical dome with a cuboidal structure mounted on it. (Take \( \pi = \frac{22}{7} \))

Question. The volume of the hemispherical dome if the height of the dome is 21 m, is
(a) 19404 cu. m
(b) 2000 cu. m
(c) 15000 cu. m
(d) 19000 cu. m
Answer: a

Question. The formula to find the volume of sphere is
(a) \( \frac{2}{3}\pi r^3 \)
(b) \( \frac{4}{3}\pi r^3 \)
(c) \( 4\pi r^2 \)
(d) \( 2\pi r^2 \)
Answer: b

Question. The cloth required to cover the hemispherical dome if the radius of its base is 14 m is
(a) 1222 sq. m
(b) 1232 sq. m
(c) 1200 sq. m
(d) 1400 sq. m
Answer: b

Question. The total surface area of the combined figure, i.e. hemispherical dome with radius 14 m and cuboidal shaped top with dimensions 8 m \( \times \) 6 m \( \times \) 4 m is
(a) 1200 sq. m
(b) 1232 sq. m
(c) 1392 sq. m
(d) 1932 sq. m
Answer: c

Question. The volume of the cuboidal shaped top with dimensions mentioned in question 4, is
(a) 182.45 m\(^3\)
(b) 282.45 m\(^3\)
(c) 292 m\(^3\)
(d) 192 m\(^3\)
Answer: d

Question. Find approximately how much syrup would be found in 45 gulab jamuns shaped like a cylinder with two hemispherical ends, if the complete length of each of the gulab jamuns is 5 cm and its diameter is 2.8 cm.
Answer: Radius (r) = 1.4 cm; length (h) of cylindrical part = 5 cm - 2 x 1.5 cm = 2.2 cm. Volume of each gulab jamun = \( \pi(1.4)^2 \times 2.2 + 2 \times \frac{2}{3}\pi \times (1.4)^3 \) cm\(^3\). Volume of syrup found in 45 gulab jamuns = 45 x 30% of volume of each gulab jamun = 337.88 cm\(^3\).

Question. Since the inner diameter of the glass = 5 cm and height = 10 cm, the apparent capacity of the glass = \( \pi r^2 h = 3.14 \times 2.5 \times 2.5 \times 10 \text{ cm}^3 = 196.25 \text{ cm}^3 \). But the actual capacity of the glass is less by the volume of the hemisphere at the base of the glass. i.e., it is less by \( \frac{2}{3}\pi r^3 \). Find the actual capacity.
Answer: Volume of hemisphere = \( \frac{2}{3} \times 3.14 \times 2.5 \times 2.5 \times 2.5 \text{ cm}^3 = 32.71 \text{ cm}^3 \). So, the actual capacity of the glass = apparent capacity of glass - volume of the hemisphere = (196.25 - 32.71) = 163.54 cm\(^3\).

Short Answer Type Questions 

Question. 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 of sphere = Volume of cylinder
\( \frac{4}{3}\pi (4.2)^3 = \pi (6)^2 \times h \)
\( \frac{4}{3} \times 4.2 \times 4.2 \times 4.2 = 36 \times h \)
\( 4 \times 1.4 \times 4.2 \times 4.2 = 36h \)
\( h = \frac{5.6 \times 17.64}{36} = 2.744 \) cm.

Question. Three solid metal cubes of edges 6 cm, 8 cm and 10 cm are melted and recasted into a single solid cube. Find the length of the edge of the cube so obtained.
Answer: Volume of new cube = \( 6^3 + 8^3 + 10^3 = 216 + 512 + 1000 = 1728 \text{ cm}^3 \).
Edge \( = \sqrt[3]{1728} = 12 \) cm.

Question. A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire. 
Answer: Volume of rod \( = \pi (0.5)^2 \times 8 = 2\pi \text{ cm}^3 \).
Volume of wire \( = \pi r^2 \times 1800 \text{ cm} \).
\( 1800\pi r^2 = 2\pi \Rightarrow r^2 = \frac{1}{900} \Rightarrow r = \frac{1}{30} \) cm.
Thickness (diameter) \( = 2r = \frac{1}{15} \text{ cm} \approx 0.67 \text{ mm} \).

Question. Two spheres of same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the new sphere.
Answer: Since weights are in ratio 1:7, volumes are in ratio 1:7.
Total volume = Volume of (smaller + larger) = \( 1V + 7V = 8V \).
Volume of small sphere \( = \frac{4}{3}\pi (3)^3 = 36\pi \).
Volume of new big sphere \( = 8 \times 36\pi = 288\pi \).
\( \frac{4}{3}\pi R^3 = 288\pi \Rightarrow R^3 = 216 \Rightarrow R = 6 \) cm.
Diameter = 12 cm.

Question. 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 \( = \pi (\frac{7}{2})^2 \times 20 = \frac{22}{7} \times \frac{49}{4} \times 20 = 770 \text{ m}^3 \).
Volume of platform \( = 22 \times 14 \times H = 770 \).
\( H = \frac{770}{22 \times 14} = 2.5 \) m.

Question. A cone and a cylinder have the same radii but the height of the cone is 3 times that of the cylinder. Find the ratio of their volumes. 
Answer: \( V_{\text{cone}} = \frac{1}{3}\pi r^2 (3h) = \pi r^2 h \).
\( V_{\text{cylinder}} = \pi r^2 h \).
Ratio \( = 1:1 \).

Question. The volume of a right circular cylinder with its height equal to the radius is \( 25 \frac{1}{7} \text{ cm}^3 \). Find the height of the cylinder. [Use \( \pi = \frac{22}{7} \)] 
Answer: Volume \( = \pi r^2 h = \pi h^3 \) (since \( r=h \)).
\( \frac{22}{7} h^3 = \frac{176}{7} \Rightarrow 22 h^3 = 176 \Rightarrow h^3 = 8 \Rightarrow h = 2 \) cm.

Short Answer Type Questions

Question. A solid sphere is melted and recasted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 4 cm, its height 24 cm and thickness 2 cm; find the radius of the sphere. 
Answer: External radius \( R = 4 \) cm, Thickness = 2 cm \( \Rightarrow \) Internal radius \( r = 4-2 = 2 \) cm.
Volume of hollow cylinder \( = \pi(R^2 - r^2)h = \pi(4^2 - 2^2) \times 24 = \pi(16-4) \times 24 = 288\pi \text{ cm}^3 \).
Volume of sphere \( = \frac{4}{3}\pi R_{\text{sphere}}^3 = 288\pi \)
\( R_{\text{sphere}}^3 = \frac{288 \times 3}{4} = 72 \times 3 = 216 \Rightarrow R_{\text{sphere}} = 6 \) cm.

Question. A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap? 
Answer: \( r = 12 \) m, \( h = 3.5 \) m.
Volume \( = \frac{1}{3} \times \frac{22}{7} \times 12 \times 12 \times 3.5 = 22 \times 4 \times 12 \times 0.5 = 528 \text{ m}^3 \).
Slant height \( l = \sqrt{12^2 + 3.5^2} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5 \) m.
Canvas area \( = \pi rl = \frac{22}{7} \times 12 \times 12.5 = 471.42 \text{ m}^2 \).

Question. A cone of height 24 cm and radius of base 6 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere 
Answer: Volume of cone = Volume of sphere.
\( \frac{1}{3}\pi \times 6 \times 6 \times 24 = \frac{4}{3}\pi R^3 \)
\( 36 \times 24 = 4 R^3 \Rightarrow R^3 = 36 \times 6 = 216 \Rightarrow R = 6 \) cm.

Question. 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 cm \( \times \) 10 cm \( \times \) 3.5 cm?
Answer: Vol. of one coin \( = \pi (\frac{1.75}{2})^2 \times 0.2 = \frac{22}{7} \times \frac{1.75 \times 1.75}{4} \times 0.2 \).
Vol. of cuboid \( = 5.5 \times 10 \times 3.5 = 192.5 \text{ cm}^3 \).
Number of coins \( = \frac{192.5 \times 7 \times 4}{22 \times 1.75 \times 1.75 \times 0.2} = 400 \).

Question. The \( \frac{3}{4} \)th part of a conical vessel of internal radius 5 cm and height 24 cm is full of water. The water is emptied into a cylindrical vessel with internal radius 10 cm. Find the height of water in cylindrical vessel. 
Answer: Volume of water \( = \frac{3}{4} \times (\frac{1}{3}\pi \times 5^2 \times 24) = 150\pi \text{ cm}^3 \).
Vol. in cylinder \( = \pi \times 10^2 \times h = 100\pi h \).
\( 100\pi h = 150\pi \Rightarrow h = 1.5 \) cm.

Question. 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 standing water is needed? 
Answer: Volume of water in 30 mins \( = \text{width} \times \text{depth} \times (\text{speed} \times \text{time}) \)
\( = 6 \times 1.5 \times (10000 \times \frac{30}{60}) = 9 \times 5000 = 45000 \text{ m}^3 \).
Irrigated Area \( \times 0.08 \text{ m} = 45000 \text{ m}^3 \).
Area \( = \frac{45000}{0.08} = 562500 \text{ m}^2 = 56.25 \text{ hectares} \).

Question. In a hospital, used water is collected in a cylindrical tank of diameter 2 m and height 5 m. After recycling, this water is used to irrigate a park of hospital whose length is 25 m and breadth is 20 m. If tank is filled completely then what will be the height of standing water used for irrigating the park? 
Answer: Vol. of tank \( = \pi \times 1^2 \times 5 = 5\pi \text{ m}^3 \approx 15.7 \text{ m}^3 \).
Area of park \( = 25 \times 20 = 500 \text{ m}^2 \).
Height of water \( = \frac{\text{Volume}}{\text{Area}} = \frac{15.7}{500} = 0.0314 \text{ m} = 3.14 \text{ cm} \).

Question. In a rain-water harvesting system, the rain-water from a roof of 22 m \( \times \) 20 m drains into a cylindrical tank having diameter of base 2 m and height 3.5 m. If the tank is full, find the rainfall in cm.
Answer: Vol. of tank \( = \frac{22}{7} \times 1^2 \times 3.5 = 11 \text{ m}^3 \).
Area of roof \( = 22 \times 20 = 440 \text{ m}^2 \).
Rainfall height \( = \frac{11}{440} = 0.025 \text{ m} = 2.5 \text{ cm} \).

Question. Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted to form a solid sphere. Find the radius of the resulting sphere.
Answer: \( \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (6^3 + 8^3 + 10^3) \Rightarrow R^3 = 216 + 512 + 1000 = 1728 \Rightarrow R = 12 \) cm.

Long Answer Type Questions 

Question. A well, whose diameter is 3 m, has been dug 21 m deep and the earth dug out is used to form an embankment 4 m wide around it. Find the height of the embankment.
Answer: Vol. of well \( = \pi \times (1.5)^2 \times 21 \).
Embankment radii: \( r = 1.5 \) m, \( R = 1.5 + 4 = 5.5 \) m.
Area of embankment base \( = \pi(5.5^2 - 1.5^2) = \pi(30.25 - 2.25) = 28\pi \).
Height \( H = \frac{\pi \times 1.5 \times 1.5 \times 21}{28\pi} = \frac{2.25 \times 21}{28} = \frac{2.25 \times 3}{4} = 1.6875 \) m.

Question. A farmer connects a pipe of internal diameter 20 cm from the canal into a cylindrical tank in his field which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled? 
Answer: Vol. of tank \( = \pi \times 5^2 \times 2 = 50\pi \text{ m}^3 \).
Radius of pipe \( = 0.1 \) m. Speed \( = 3000 \text{ m/h} \).
Vol. of water per hour \( = \pi \times (0.1)^2 \times 3000 = 30\pi \text{ m}^3 \).
Time \( = \frac{50\pi}{30\pi} = \frac{5}{3} \text{ hours} = 1 \text{ hour } 40 \text{ minutes} = 100 \text{ minutes} \).

Question. Selvi’s house has an overhead tank in the shape of a cylinder. This is filled by pumping water from an underground tank which is in the shape of a cuboid. The underground tank has dimensions 1.57 m \( \times \) 1.44 m \( \times \) 95 cm. The overhead tank has its radius of 60 cm and its height is 95 cm. Find the height of the water left in the underground tank after the overhead tank has been completely filled with water which had been full. Compare the capacity of the overhead tank with that of the underground tank. [Use \( \pi = 3.14 \)] 
Answer: Vol. of underground tank \( = 157 \times 144 \times 95 \text{ cm}^3 \).
Vol. of overhead tank \( = 3.14 \times 60 \times 60 \times 95 \text{ cm}^3 \).
Water left vol. \( = (157 \times 144 \times 95) - (3.14 \times 3600 \times 95) = 95(22608 - 11304) = 95 \times 11304 \text{ cm}^3 \).
Height of water left \( = \frac{95 \times 11304}{157 \times 144} = \frac{95}{2} = 47.5 \) cm.
Capacity ratio \( = \frac{11304 \times 95}{22608 \times 95} = \frac{1}{2} \).

Question. In a cylindrical vessel of radius 10 cm, containing some water, 9000 small spherical balls are dropped which are completely immersed in water which raises the water level. If each spherical ball is of radius 0.5 cm, then find the rise in the level of water in the vessel. 
Answer: Vol. of 9000 balls \( = 9000 \times \frac{4}{3}\pi (0.5)^3 = 3000 \times 4\pi \times 0.125 = 1500\pi \text{ cm}^3 \).
Rise in level \( h = \frac{\text{Volume}}{\text{Base Area}} = \frac{1500\pi}{\pi \times 10^2} = 15 \) cm.

Question. Water is flowing through a cylindrical pipe of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm at the rate of 0.7 m/sec. By how much will the water rise in the tank in half an hour? 
Answer: Pipe radius \( r = 1 \) cm \( = 0.01 \) m. Time = 1800 sec.
Vol. of water in 30 mins \( = \pi \times (0.01)^2 \times (0.7 \times 1800) = \pi \times 0.0001 \times 1260 = 0.126\pi \text{ m}^3 \).
Tank base area \( = \pi \times (0.4)^2 = 0.16\pi \text{ m}^2 \).
Rise in level \( H = \frac{0.126\pi}{0.16\pi} = 0.7875 \text{ m} = 78.75 \) cm.