CBSE Class 10 Mathematics Surface Areas And Volume VBQs Set D

SHORT ANSWER Type Questions

Question. A solid metallic hemisphere of radius 8 cm is melted and recast into a right circular cone of base radius 6 cm. Determine the height of the cone. 
Answer: For hemisphere, radius, \(r = 8\text{ cm}\)
Volume of hemisphere \( = \frac{2}{3}\pi r^3 = \frac{2}{3} \times \pi \times (8)^3 = \frac{1024\pi}{3}\text{ cm}^3\)
For cone, that is recasted from hemisphere base radius, \(r = 6\text{ cm}\)
Volume of cone \( = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (6)^2 h = 12\pi h\text{ cm}^3\)
As volume remains the same, when a body is reformed to another body
\(\therefore \text{Volume of hemisphere} = \text{Volume of cone}\)
\(\frac{1024\pi}{3} = 12\pi h\)
\(\Rightarrow h = \frac{1024\pi}{3 \times 12\pi} = \frac{256}{9} = 28.44\text{ cm}\)
Hence, the height of cone \( = 28.44\text{ cm}\).

Question. A girl empties a cylindrical bucket full of sand, of base radius 18 cm and height 32 cm on the floor to form a conical heap of sand. If the height of this conical heap is 24 cm, find its slant height and correct it to one place of the decimal. 
Answer: Base radius of the cylindrical bucket, \(r = 18\text{ cm}\)
Height of the cylindrical bucket, \(h = 32\text{ cm}\)
Height of the conical heap, \(H = 24\text{ cm}\)
Let the base radius of the conical heap be \(R\).
Volume of the cylindrical bucket \( = \) Volume of the conical heap
[Since, the bucket of sand is emptied in the form of a conical heap]
\(\therefore \pi r^2 h = \frac{1}{3} \pi R^2 H\)
\(\Rightarrow 18 \times 18 \times 32 = \frac{1}{3} \times R^2 \times 24\)
\(\Rightarrow R^2 = \frac{18 \times 18 \times 32}{8} = 18 \times 18 \times 4\)
\(\Rightarrow R = \sqrt{18 \times 18 \times 2 \times 2} = 18 \times 2 = 36\text{ cm}\)
Slant height, \(l = \sqrt{R^2 + H^2} = \sqrt{(36)^2 + (24)^2} = \sqrt{1296 + 576} = \sqrt{1872} = 43.27 \approx 43.3\text{ cm}\)
Hence, the slant height of the conical heap is 43.3 cm.

Question. How many cubic centimetres of iron is required to construct an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm, provided the thickness of the iron is 1.5 cm? If one cubic cm of iron weighs 7.5 g, find the weight of the box. 
Answer: Given: For open box
External length, \(l = 36\text{ cm}\)
External breadth, \(b = 25\text{ cm}\)
External height, \(h = 16.5\text{ cm}\)
\(\therefore \text{Volume of external box} = lbh = 36 \times 25 \times 16.5 = 14850\text{ cm}^3\)
Given, thickness of iron, \(x = 1.5\text{ cm}\) and box is open from top.
For internal box,
length, \(l' = l - 2x = 36 - (2 \times 1.5) = 36 - 3 = 33\text{ cm}\)
breadth, \(b' = b - 2x = 25 - (2 \times 1.5) = 25 - 3 = 22\text{ cm}\)
height, \(h' = h - 1x = 16.5 - 1.5 = 15\text{ cm}\)
\(\therefore \text{Volume of internal box} = l'b'h' = 33 \times 22 \times 15 = 10890\text{ cm}^3\)
Volume of metal used in box \( = \text{Volume of external box} - \text{Volume of internal box} = 14850 - 10890 = 3960\text{ cm}^3\)
Given: weight of \(1\text{ cm}^3\) of iron \( = 7.5\text{ gm}\)
\(\therefore \text{Weight of } 3960\text{ cm}^3 \text{ of iron} = 3960 \times 7.5\text{ gm} = \frac{3960 \times 7.5}{1000}\text{ kg} = 29.7\text{ kg}\)
Hence, the weight of the box is 29.7 kg and the volume of the metal \( = 3960\text{ cm}^3\).

Question. The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen is used up on writing 3300 words on an average. How many words can be written in a bottle of ink containing one fifth of a litre? 
Answer: Given: Shape of barrel of fountain pen is cylinder.
Length of barrel i.e., \(h = 7\text{ cm}\)
diameter of barrel i.e., \(d = 5\text{ mm} = \frac{5}{10}\text{ cm} = \frac{1}{2}\text{ cm}\)
\(\therefore \text{radius, } r = \frac{1}{2 \times 2} = \frac{1}{4} = 0.25\text{ cm}\)
\(\Rightarrow \text{Volume of barrel} = \pi r^2 h = \frac{22}{7} \times (0.25)^2 \times 7 = 22 \times 0.0625 = 1.375\text{ cm}^3\)
According to question, \(1.375\text{ cm}^3\) of ink can write 3300 words.
\(\frac{1}{5}^{th}\) of a litre \( = \frac{1}{5} \times 1000\text{ cm}^3 = 200\text{ cm}^3\).
\(\therefore 200\text{ cm}^3 \text{ of ink can write} = \frac{3300}{1.375} \times 200 = 480000\text{ words}\)
Hence, \(\frac{1}{5}^{th}\) of a litre of ink can write 480000 words on an average.

Question. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that \(1\text{ cm}^3\) of iron has approximately 8 gm mass. (Use \(\pi = 3.14\)) 
Answer: In the given figure, In the small cylinder: radius, \(r_1 = 8\text{ cm}\), height, \(h_1 = 60\text{ cm}\).
In the big cylinder: radius, \(r_2 = \frac{24}{2} = 12\text{ cm}\), height, \(h_2 = 220\text{ cm}\).
Volume of the pole \( = \pi r_1^2 h_1 + \pi r_2^2 h_2\)
\( = \pi[(8)^2 \times 60 + (12)^2 \times 220]\)
\( = 3.14[64 \times 60 + 144 \times 220]\)
\( = 3.14 \times [3840 + 31,680]\)
\( = 111532.8\text{ cm}^3\)
Since \(1\text{ cm}^3\) of iron \( = 8\text{ gm}\) mass
\(\therefore 111532.8\text{ cm}^3 \text{ of iron} = 11532.8 \times 8\text{ g} = \frac{892262.4}{1000}\text{ kg} = 892.2624\text{ kg}\)
Hence, the mass of the pole is 892.2624 kg.

Question. A solid metallic sphere, 3 cm in radius, is melted and recast into three spherical balls with radii 1.5 cm, 2 cm and x cm. Find the value of x.
Answer: Radius of a solid metallic sphere \((R) = 3\text{ cm}\).
Radius of 3 spherical balls are:
First ball \(r_1 = 1.5\text{ cm}\)
Second ball \(r_2 = 2\text{ cm}\)
Third ball \(r_3 = x\text{ cm}\)
According to the question volume of solid metallic sphere \( = \) Volume of 3 balls
\(\frac{4}{3} \pi R^3 = \frac{4}{3} \pi r_1^3 + \frac{4}{3} \pi r_2^3 + \frac{4}{3} \pi r_3^3\)
\(R^3 = r_1^3 + r_2^3 + r_3^3\)
\((3)^3 = (1.5)^3 + (2)^3 + x^3\)
\(27 = 3.375 + 8 + x^3\)
\(x^3 = 27 - 11.375\)
\(x^3 = 15.625\)
\(x = \sqrt[3]{\frac{15625}{1000}} = \sqrt[3]{\frac{25 \times 25 \times 25}{10 \times 10 \times 10}} = \frac{25}{10} = 2.5\text{ cm}\)
Hence, the value of \(x\) is 2.5 cm.

Question. Water flows at the rate of 10 m/minute through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the base is 40 cm and depth 24 cm? [
Answer: When water flows through a pipe of a certain area of cross-section A with velocity v, then, volume of water coming from pipe in time t \( = \text{Area of cross-section} \times \text{Length}\).
Given: speed of water flow \( = 10\text{ m/min} = 1000\text{ cm/min}\).
For pipe: Diameter, \(d = 5\text{ mm} = \frac{5}{10}\text{ cm}\). Radius, \(r = \frac{5}{10 \times 2}\text{ cm} = 0.25\text{ cm}\).
\(\therefore \text{Amount of water that flows out of cylindrical pipe in 1 minute} = \pi r^2 h = \pi \times (0.25)^2 \times 10 \times 100 = 62.5\pi\text{ cm}^3\).
Amount of water required to fill conical vessel \( = \) Volume of conical vessel \( = \frac{1}{3} \pi r^2 h\).
Here, radius of conical vessel \( = \frac{40}{2} = 20\text{ cm}\). Depth i.e., height of conical vessel \( = 24\text{ cm}\).
\(\Rightarrow \text{Volume} = \frac{1}{3} \pi \times (20)^2 \times 24 = 3200\pi\text{ cm}^3\).
Time required to fill the vessel \( = \frac{\text{Volume of conical vessel}}{\text{Volume of water that flows out in 1 minute}} = \frac{3200\pi}{62.5\pi} = \frac{32000}{625} = 51.2\text{ minutes}\).
\( = 51\text{ minutes} + \frac{2}{10} \times 60\text{ seconds} = 51\text{ minutes 12 seconds}\).
Hence, the time required is 51 minutes 12 seconds.

Question. The height of a cone is 30 cm. A small cone is cut off at the top, at some height above the base, by a plane parallel to the base. If its volume is \(\frac{1}{27}\) of the volume of the given cone, at what height above the base has the section been made? 
Answer: Given, the height of the cone \((H), OB = 30\text{ cm}\). Let the radius of the bigger cone be \(R\). Now, suppose a small cone is cut at height, ‘\(h\)’ from the top and its radius is ‘\(r\)’. In \(\Delta OAD\) and \(\Delta OBC\): \(\angle AOD = \angle BOC\) (common), \(\angle OAD = \angle OBC = 90^\circ\). \(\therefore \Delta OAD \sim \Delta OBC\) (by AA similarity). Then, \(\frac{OA}{OB} = \frac{AD}{BC}\) (Since the corresponding sides of similar triangles are proportional). \(\Rightarrow \frac{h}{30} = \frac{r}{R}\) ...(i).
Now, volume of the bigger cone, \(V_1 = \frac{1}{3} \pi R^2 H = \frac{1}{3} \pi R^2 \times 30\).
Volume of the smaller cone, \(V_2 = \frac{1}{3} \pi r^2 h\).
Then, \(V_2 = \frac{1}{27} V_1\) (given).
\(\frac{1}{3} \pi r^2 h = \frac{1}{27} \times \frac{1}{3} \pi R^2 \times 30\)
\(\Rightarrow r^2 h = R^2 \times \frac{30}{27}\)
\(\Rightarrow h = \frac{R^2}{r^2} \times \frac{10}{9}\)
\(\Rightarrow h = \left( \frac{30}{h} \right)^2 \times \frac{10}{9}\) (From (i))
\(\Rightarrow h^3 = \frac{30 \times 30 \times 10}{9}\)
\(\Rightarrow h = \sqrt[3]{10 \times 10 \times 10} = 10\text{ cm}\).
Then, the height above the base, \(AB = 30 - 10 = 20\text{ cm}\).
Hence, the height above the base where the section has been made is 20 cm.

Question. A factory manufactures 1,20,000 pencils daily. The pencils are cylindrical in shape, each of length 25 cm and circumference of base as 1.5 cm. Determine the cost of colouring the curved surfaces of the pencils manufactured in one day at \(Rs 0.05\) per \(dm^2\). 
Answer: Given: Shape of the pencils is cylindrical. Let radius of base be \(r\). length of pencil, \(h = 25\text{ cm}\). circumference of base \( = 1.5\text{ cm}\).
\(\Rightarrow 2\pi r = 1.5\text{ cm}\).
Curved surface area of 1 pencil \( = 2\pi rh = (2\pi r) \times h = 1.5 \times 25 = 37.5\text{ cm}^2\).
We know that \(1\text{ cm} = 0.1\text{ dm}\), \(1\text{ cm}^2 = 0.01\text{ dm}^2\).
\(37.5\text{ cm}^2 = 0.01 \times 37.5\text{ dm}^2 = 0.375\text{ dm}^2\).
\(\Rightarrow \text{Curved surface area of 120000 pencils} = 0.375 \times 120000 = 45000\text{ dm}^2\).
Given cost of colouring \(1\text{ dm}^2\) curved surface area of pencil \( =(Rs 0.05\).
\(\therefore \text{Cost of colouring 45000 dm}^2 \text{ CSA of pencil} = (Rs 0.05 \times 45000 = (Rs 2250\).

Question. In a rectangular park of dimensions 50 m \(\times\) 40 m, a rectangular pond is constructed so that the area of a grass strip of uniform width surrounding the pond would be 1184 m2. Find the length and breadth of the pond. 
Answer: Let ABCD be a rectangular park of dimensions 50 m \(\times\) 40 m, and EFGH be a rectangular pond. Let ‘\(x\)’ be the width of the grassy area around the pond. Now, length of lawn \( = 50\text{ m}\). width of the lawn \( = 40\text{ m}\).
\(\Rightarrow \text{length of the pond} = (50 - 2x)\text{m}\), \(\text{width of the pond} = (40 - 2x)\text{m}\).
Also given, Area of grass surrounding the pond \( = 1184\text{ m}^2\).
\(\Rightarrow \text{Area of rectangular lawn} - \text{Area of pond} = 1184\).
\(\Rightarrow 50 \times 40 - (50 - 2x) \times (40 - 2x) = 1184\).
\(\Rightarrow -4x^2 + 180x - 1184 = 0\).
\(\Rightarrow x^2 - 45x + 296 = 0\).
\(\Rightarrow x^2 - 37x - 8x + 296 = 0\) (on splitting the middle term).
\(\Rightarrow (x - 37) (x - 8) = 0\).
\(\Rightarrow x = 37 \text{ or } 8\).
But \(x = 37\) is not possible, as in the case the length of the pond becomes \(50 - 2 \times 37 = -29\) (not possible). \(\therefore x = 8\).
Then, length of the pond \( = 50 - 2 \times 8 = 34\text{ m}\), breadth of the pond \( = 40 - 2 \times 8 = 24\text{ m}\).
Hence, the length and breadth of the pond are 34 m and 24 m respectively.

Question. From a rectangular block of wood, having dimensions 15 cm \(\times\) 10 cm \(\times\) 3.5 cm, a pen stand is made by making four conical depressions. The radius of each one of the depression is 0.5 cm and the depth is 2.1 cm. Find the volume of wood left in the pen-stand.
Answer: Dimensions of the rectangular block of wood : length \((l) = 15\text{ cm}\), breadth \((b) = 10\text{ cm}\), height \((h) = 3.5\text{ cm}\).
Dimension of conical depression radius of cone \((r) = 0.5\text{ m}\), height of cone \((h) = 2.1\text{ cm}\).
Volume of wood left \( = \text{Volume of cuboid} - \text{Total Volume of four conical depression}\).
\( = l \times b \times h - 4 \times \frac{1}{3} \pi r^2 h\)
\( = 15 \times 10 \times 3.5 - 4 \times \frac{1}{3} \times \frac{22}{7} \times 0.5 \times 0.5 \times 2.1\)
\( = 525 - 2.2 = 522.8\text{ cm}^3\).
Hence, the volume of wood left in the pen stand is \(522.8\text{ cm}^3\).

Question. In a hospital, used water is collected in a cylindrical tank diameter 2 m and height 5 m. After recycling, this water is used to irrigate a park of a hospital whose length is 25 m and breadth is 20 m. If the tank is filled completely then what will be the height of standing water used for irrigating the park. 
Answer: Diameter of the cylindrical tank \((d) = 2\text{ m}\). \(\therefore \text{Radius of the cylindrical tank } (r) = 1\text{ m}\). Height of the cylindrical tank \((h) = 5\text{ m}\).
Then, volume of the cylindrical tank \( = \pi r^2 h = \pi(1)^2 \times 5 = 5\pi\text{ m}^3\) ...(i).
Length of the park \((l) = 25\text{ m}\), breadth of the park \((b) = 20\text{ m}\). Let the height of standing water in the park be ‘\(h\)’ m. Then, the volume of water in the park \( = lbh = 25 \times 20 \times h\) ...(ii).
Now, water from the tank is used to irrigate the park. So, the volume of cylindrical tank \( = \) Volume of water in the park.
\(5\pi = 25 \times 20 \times h\) [from (i) & (ii)] \(\Rightarrow h = \frac{5\pi}{25 \times 20} = \frac{\pi}{100} = \frac{3.14}{100} = 0.0314\text{ m}\).
Hence, the height of the standing water in park is 0.0314m.
By recycling water, better use of nature resource occurs without wastage. It helps in reducing and preventing pollution. It helps in saving and conserving water also.

Question. In a rectangular park of dimensions 50 m \(\times\) 40 m, a rectangular pond is constructed so that the area of a grass strip of uniform width surrounding the pond would be 1184 m\(^2\). Find the length and breadth of the pond. 
Answer: Let ABCD be a rectangular park of dimensions 50 m \(\times\) 40 m, and EFGH be a rectangular pond. Let ‘x’ be the width of the grassy area around the pond. Now, length of lawn \( = 50 \text{ m} \), width of the lawn \( = 40 \text{ m} \).
\( \Rightarrow \text{length of the pond} = (50 - 2x) \text{ m} \), width of the pond \( = (40 - 2x) \text{ m} \).
Also given, Area of grass surrounding the pond \( = 1184 \text{ m}^2 \).
\( \Rightarrow \text{Area of rectangular lawn} - \text{Area of pond} = 1184 \)
\( \Rightarrow 50 \times 40 - (50 - 2x) \times (40 - 2x) = 1184 \)
\( \Rightarrow - 4x^2 + 180x - 1184 = 0 \)
\( \Rightarrow x^2 - 45x + 296 = 0 \)
\( \Rightarrow x^2 - 37x - 8x + 296 = 0 \) (on splitting the middle term).
\( \Rightarrow (x - 37) (x - 8) = 0 \Rightarrow x = 37 \text{ or } 8 \).
But \( x = 37 \) is not possible, as in the case the length of the pond becomes \( 50 - 2 \times 37 = -29 \) (not possible). \( \therefore x = 8 \).
Then, length of the pond \( = 50 - 2 \times 8 = 34 \text{ m} \), breadth of the pond \( = 40 - 2 \times 8 = 24 \text{ m} \).
Hence, the length and breadth of the pond are 34 m and 24 m respectively.

Question. From a rectangular block of wood, having dimensions 15 cm \(\times\) 10 cm \(\times\) 3.5 cm, a pen stand is made by making four conical depressions. The radius of each one of the depression is 0.5 cm and the depth is 2.1 cm. Find the volume of wood left in the pen-stand. 
Answer: Dimensions of the rectangular block of wood : length (l) \( = 15 \text{ cm} \), breadth (b) \( = 10 \text{ cm} \), height (h) \( = 3.5 \text{ cm} \).
Dimension of conical depression: radius of cone (r) \( = 0.5 \text{ m} \), height of cone (h) \( = 2.1 \text{ cm} \).
Volume of wood left \( = \text{Volume of cuboid} - \text{Total Volume of four conical depression} \)
\( = l \times b \times h - 4 \times \frac{1}{3}\pi r^2 h \)
\( = 15 \times 10 \times 3.5 - 4 \times \frac{1}{3} \times \frac{22}{7} \times 0.5 \times 0.5 \times 2.1 \)
\( = 525 - 2.2 = 522.8 \text{ cm}^3 \).
Hence, the volume of wood left in the pen stand is 522.8 cm\(^3\).

Question. Water is flowing at the rate of 15 km/h through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time will the level of water in the pond rise by 21 cm? 
Answer: Given: Speed of water \( = 15 \text{ km/h} = 15000 \text{ m/h} \).
Diameter of pipe, \( 2r = 14 \text{ cm} \Rightarrow r = 7 \text{ cm} \).
Shape of pipe is cylinder. Volume of water that flows out of pipe in 1 hour \( = \pi r^2 h = \frac{22}{7} \times \frac{7}{100} \times \frac{7}{100} \times 15000 \text{ m}^3 = 231 \text{ m}^3 \) ...(i)
It is given that for cuboidal pond: length, \( l = 50 \text{ m} \), breadth, \( b = 44 \text{ m} \), depth required, \( h = 21 \text{ cm} = \frac{21}{100} \text{ m} \).
\( \therefore \text{Volume of water in cuboidal pond} = lbh = 50 \times 44 \times \frac{21}{100} = 22 \times 21 = 462 \text{ m}^3 \).
From eqn (i), 231 m\(^3\) of water flows out of pipe in 1 hour. \( \Rightarrow 462 \text{ m}^3 \) of water flows out of pipe in \( \frac{1}{231} \times 462 = 2 \text{ hours} \).
Hence, the required time is 2 hours.

Question. In a hospital, used water is collected in a cylindrical tank diameter 2 m and height 5m. After recycling, this water is used to irrigate a park of a hospital whose length is 25 m and breadth is 20 m. If the tank is filled completely then what will be the height of standing water used for irrigating the park? 
Answer: Diameter of the cylindrical tank (d) \( = 2 \text{ m} \), \( \therefore \text{Radius of the cylindrical tank (r)} = 1 \text{ m} \), Height of the cylindrical tank (h) \( = 5 \text{ m} \).
Then, volume of the cylindrical tank \( = \pi r^2 h = \pi (1)^2 \times 5 = 5\pi \text{ m}^3 \) ...(i)
Length of the park (l) \( = 25 \text{ m} \), Breadth of the park (b) \( = 20 \text{ m} \). Let the height of standing water in the park be ‘h’ m.
Then, the volume of water in the park \( = lbh = 25 \times 20 \times h \) ...(ii)
Now, water from the tank is used to irrigate the park. So, the volume of cylindrical tank \( = \text{Volume of water in the park} \).
\( 5\pi = 25 \times 20 \times h \) [from (i) & (ii)] \( \Rightarrow h = \frac{5\pi}{25 \times 20} = \frac{\pi}{100} = \frac{3.14}{100} = 0.0314 \text{ m} \).
Hence, the height of the standing water in park is 0.0314 m.

Question. A cone of maximum size is carved out from a solid cube of side 14 cm. Find the total surface area of the remaining solid left out. 
Answer: Given, side of the cube, a \( = 14 \text{ cm} \). The maximum size cone carved out from cube would have the base radius (r) \( = \frac{14}{2} = 7 \text{ cm} \), height (h) \( = 14 \text{ cm} \).
Total surface of the remaining solid \( = \text{Surface Area of cube} + \text{Curved surface area of cone} - \text{Base area of cone} \) ...(i)
Then, the surface area of the cube \( = 6 (\text{side})^2 = 6 (14)^2 = 6 \times 196 = 1176 \text{ cm}^2 \) ...(ii)
Curved surface area of cone \( = \pi r l \), where, \( l = \sqrt{h^2 + r^2} = \sqrt{(14)^2 + 7^2} = \sqrt{196 + 49} = \sqrt{245} = 7\sqrt{5} \text{ cm} \).
\( \therefore \text{C.S.A. of cone} = \pi \times 7 \times 7\sqrt{5} = \frac{22}{7} \times 7 \times 7\sqrt{5} = 154\sqrt{5} \text{ cm}^2 \) ...(iii)
Base area of the cone \( = \pi r^2 = \frac{22}{7} \times 7 \times 7 = 154 \text{ cm}^2 \) ...(iv)
Put the values of (ii), (iii) and (iv) in (i), we get the Total surface area of the remaining solid: \( = 1176 + 154\sqrt{5} - 154 = (1022 + 154\sqrt{5}) \text{ cm}^2 \).
Hence, the total surface area of the remaining solid is \( (1022 + 154\sqrt{5}) \text{ cm}^2 \).