GSEB Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.3

Question 1. 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:
In the case of the sphere, the radius \( r = 4.2 \) cm.
The volume of the sphere is \( \frac {4}{3}\pi r^3 = \frac {4}{3}\pi (4.2)^3 \).
For the cylinder, the radius \( r = 6 \) cm and height \( h = ? \).
The volume of the cylinder is \( \pi r^2 h = \pi \times 6^2 \times h \).
Since the sphere is melted and recast into a cylinder, their volumes must be equal.
Volume of cylinder = Volume of sphere
\( \pi \times 6^2 \times h = \frac {4}{3}\pi (4.2)^3 \)
\( h = \frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 6 \times 6} \)
\( h = 4 \times 1.4 \times 0.7 \times 0.7 \)
\( h = 2.74 \) cm
Therefore, the height of the resulting cylinder is \( 2.74 \) cm.
In simple words: We find the sphere's volume first. Then, we use that same volume for the new cylinder. Knowing the cylinder's radius, we can calculate its height.

Exam Tip: Remember that when a solid is melted and recast into another shape, its volume remains constant. This is a key principle in such problems.

Question 2. Metal 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:
The radii of the solid spheres are:
\( r_1 = 6 \) cm
\( r_2 = 8 \) cm
\( r_3 = 10 \) cm
Let \( r \) be the radius of the resulting single sphere.
The total volume of the three smaller spheres will be equal to the volume of the single large sphere.
Volume of three spheres = Volume of resulting sphere
\( \frac {4}{3}\pi r_1^3 + \frac {4}{3}\pi r_2^3 + \frac {4}{3}\pi r_3^3 = \frac {4}{3}\pi r^3 \)
We can factor out \( \frac {4}{3}\pi \) from both sides:
\( \frac {4}{3}\pi [r_1^3 + r_2^3 + r_3^3] = \frac {4}{3}\pi r^3 \)
\( \implies r_1^3 + r_2^3 + r_3^3 = r^3 \)
Substitute the given radii:
\( 6^3 + 8^3 + 10^3 = r^3 \)
\( 216 + 512 + 1000 = r^3 \)
\( r^3 = 1728 \)
To find \( r \), take the cube root of 1728:
\( r = \sqrt[3]{1728} \)
\( r = 12 \) cm
Hence, the radius of the resulting solid sphere is \( 12 \) cm.
In simple words: When you combine smaller spheres, their total amount of material stays the same. We add up the volumes of the small spheres to get the volume of the big sphere, then find its radius.

Exam Tip: Always remember the formula for the volume of a sphere: \( V = \frac{4}{3}\pi r^3 \). When multiple objects are melted and reshaped, their combined volume is conserved.

Question 3. 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. [NCERT 20121]
Answer:
For the cylindrical well:
The diameter is \( 7 \) m, so the radius \( r = \frac {7}{2} \) m.
The depth (or height) of the well \( h = 20 \) m.
The volume of earth dug out from the well is the volume of the cylinder:
\( \text{Volume of well} = \pi r^2 h \)
\( = \frac {22}{7} \times \frac {7}{2} \times \frac {7}{2} \times 20 \)
\( = 11 \times 7 \times 10 \)
\( = 770 \text{ m}^3 \)
For the cuboidal platform:
The length \( l = 22 \) m.
The breadth \( b = 14 \) m.
Let the height of the platform be \( H \) m.
The volume of earth spread out in the form of the cuboidal platform is equal to the volume of earth dug out from the well.
Volume of cuboidal platform = Volume of earth dug out
\( l \times b \times H = 770 \)
\( 22 \times 14 \times H = 770 \)
\( H = \frac {770}{22 \times 14} \)
\( H = \frac {770}{308} \)
\( H = 2.5 \) m
Therefore, the height of the platform is \( 2.5 \) m.
In simple words: We calculate how much dirt comes out of the round well. Then, we spread that same amount of dirt to make a rectangular platform. Knowing the platform's length and width, we can find its height.

Exam Tip: In problems involving reshaping materials, the volume of the material remains constant. Ensure you use the correct volume formulas for cylinders and cuboids.

Question 4. 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. [CBSE 20121]
Answer:
For the cylindrical well:
The diameter is \( 3 \) m, so the inner radius \( r_1 = \frac {3}{2} \) m.
The depth (or height) of the well \( h = 14 \) m.
The volume of earth dug out from the well is \( \pi r_1^2 h = \pi \left(\frac {3}{2}\right)^2 \times 14 \).
For the circular embankment (ring shape):
The inner radius of the embankment is \( r_1 = \frac {3}{2} \) m.
The width of the embankment is \( 4 \) m.
So, the outer radius of the embankment \( R = r_1 + 4 = \frac {3}{2} + 4 = \frac {3+8}{2} = \frac {11}{2} \) m.
Let the height of the embankment be \( H \).
The volume of the embankment is the volume of the outer cylinder minus the volume of the inner cylinder (hole):
Volume of embankment \( = \pi R^2 H - \pi r_1^2 H = \pi (R^2 - r_1^2)H \)
Since the volume of the embankment is equal to the volume of earth dug out:
\( \pi (R^2 - r_1^2)H = \pi r_1^2 h \)
\( \pi \left[ \left(\frac{11}{2}\right)^2 - \left(\frac{3}{2}\right)^2 \right] H = \pi \left(\frac{3}{2}\right)^2 \times 14 \)
Divide both sides by \( \pi \):
\( \left[ \frac{121}{4} - \frac{9}{4} \right] H = \frac{9}{4} \times 14 \)
\( \left[ \frac{121-9}{4} \right] H = \frac{9}{4} \times 14 \)
\( \frac{112}{4} H = \frac{9}{4} \times 14 \)
\( 112 H = 9 \times 14 \)
\( H = \frac{9 \times 14}{112} \)
\( H = \frac{126}{112} \)
\( H = 1.125 \) m
Hence, the height of the embankment is \( 1.125 \) m.
In simple words: First, we find the volume of dirt removed from the well. This dirt is used to make a raised circular wall around the well. We calculate the volume of this wall using its inner and outer radii and set it equal to the dirt's volume to find the wall's height.

Exam Tip: When forming an embankment, the volume of the earth removed from the well is precisely the volume of the embankment. For a circular ring (embankment), its volume is found by subtracting the volume of the inner cylinder from the volume of the outer cylinder.

Question 5. A container shaped like a right circular cylinder having a diameter 12 cm and a 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:
For the right circular cylinder:
Diameter is \( 12 \) cm, so radius \( r_1 = 6 \) cm.
Height \( h_1 = 15 \) cm.
Volume of ice cream in the right circular cylinder is \( \pi r_1^2 h_1 = \pi \times 6^2 \times 15 = 540\pi \text{ cm}^3 \).
For each cone (with a hemispherical top):
Diameter is \( 6 \) cm, so radius \( r_2 = 3 \) cm.
Height of the conical part \( h_2 = 12 \) cm.
Volume of the cone part \( = \frac {1}{3} \pi r_2^2 h_2 = \frac {1}{3} \pi \times 3^2 \times 12 = 36\pi \text{ cm}^3 \).
The hemispherical part has a radius of \( r_2 = 3 \) cm.
Volume of the hemisphere part \( = \frac {2}{3} \pi r_2^3 = \frac {2}{3} \pi (3)^3 = \frac{2}{3}\pi \times 27 = 18\pi \text{ cm}^3 \).
Total volume of one ice cream cone (cone + hemisphere) \( = 36\pi + 18\pi = 54\pi \text{ cm}^3 \).
Number of cones filled with ice cream \( = \frac{\text{Volume of cylinder}}{\text{Total volume of one cone and hemisphere}} \)
\( = \frac{540\pi}{54\pi} \)
\( = 10 \)
Thus, \( 10 \) such cones can be filled with ice cream.
In simple words: First, we figure out how much ice cream is in the big cylinder. Then, we find out how much ice cream each small cone (with its dome-shaped top) can hold. Finally, we divide the total ice cream by the amount each cone holds to get the number of cones.

Exam Tip: Remember to calculate the volume of both parts of the ice cream serving (cone and hemisphere) and add them together before dividing the total volume of ice cream in the cylinder.

Question 6. 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 x 10 cm x 3.5 cm?
Answer:
For each silver coin (which is cylindrical):
The diameter is \( 1.75 \) cm, so the radius \( r_1 = \frac {1.75}{2} = 0.875 \) cm, which can also be written as \( \frac {7}{8} \) cm.
The thickness (height) \( h_1 = 2 \) mm \( = \frac {2}{10} \) cm \( = \frac {1}{5} \) cm.
The volume of a coin \( = \pi r_1^2 h_1 \)
\( = \frac {22}{7} \times \left(\frac {7}{8}\right)^2 \times \frac {1}{5} \)
\( = \frac {22}{7} \times \frac {49}{64} \times \frac {1}{5} \)
\( = \frac {11 \times 7}{32 \times 5} \text{ cm}^3 \)
\( = \frac {77}{160} \text{ cm}^3 \)
For the cuboid:
Length \( l = 5.5 \) cm.
Breadth \( b = 10 \) cm.
Height \( h = 3.5 \) cm.
The volume of the cuboid \( = l \times b \times h \)
\( = 5.5 \times 10 \times 3.5 \text{ cm}^3 \)
\( = 192.5 \text{ cm}^3 \)
The number of coins needed \( = \frac{\text{Volume of cuboid}}{\text{Volume of a coin}} \)
\( = \frac{192.5}{\frac{77}{160}} \)
\( = \frac{192.5 \times 160}{77} \)
\( = \frac{1925 \times 16}{77} \)
\( = 25 \times 16 \)
\( = 400 \)
Therefore, \( 400 \) silver coins must be melted.
In simple words: We find the volume of one small coin and the volume of the big cuboid we want to make. Then, we divide the big volume by the small volume to see how many coins are needed.

Exam Tip: Ensure all units are consistent (e.g., convert mm to cm) before performing calculations. The total volume of material remains constant when reshaping objects.

Question 7. 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:
For the cylindrical bucket:
The radius of the bucket \( r = 18 \) cm.
The height of the bucket \( h = 32 \) cm.
The volume of sand filled in the cylindrical bucket \( = \pi r^2 h \)
\( = \pi \times 18^2 \times 32 \)
\( = \pi \times 324 \times 32 \)
\( = 10368\pi \text{ cm}^3 \)
For the conical heap:
Let the radius of the conical heap be \( R \).
The height of the conical heap \( H = 24 \) cm.
The volume of the conical heap is equal to the volume of sand from the cylindrical bucket.
Volume of conical heap = Volume of cylindrical bucket
\( \frac {1}{3}\pi R^2 H = 10368\pi \)
Divide both sides by \( \pi \):
\( \frac {1}{3} R^2 \times 24 = 10368 \)
\( 8 R^2 = 10368 \)
\( R^2 = \frac{10368}{8} \)
\( R^2 = 1296 \)
\( R = \sqrt{1296} \)
\( R = 36 \) cm
So, the radius of the conical heap of sand is \( 36 \) cm.
Now, we find the slant height \( l \) of the conical heap using the formula \( l = \sqrt{R^2 + H^2} \):
\( l = \sqrt{36^2 + 24^2} \)
\( l = \sqrt{1296 + 576} \)
\( l = \sqrt{1872} \)
To simplify \( \sqrt{1872} \):
\( 1872 = 144 \times 13 \)
\( l = \sqrt{144 \times 13} \)
\( l = 12\sqrt{13} \) cm
Hence, the slant height of the sand heap is \( 12\sqrt{13} \) cm.
In simple words: First, we calculate the total sand in the cylindrical bucket. When this sand forms a cone, its volume stays the same. We use the cone's height and its volume to find its radius. Then, with the radius and height, we calculate the slant height.

Exam Tip: Remember the volume formulas for both cylinders and cones. For the slant height, the Pythagorean theorem \( l^2 = r^2 + h^2 \) is essential. Simplify square roots if possible to present the answer in its most precise form.

Question 8. 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:
For the canal:
Width of canal \( (b) = 6 \) m.
Depth of canal \( (h) = 1.5 \) m.
Speed of flow of water \( = 10 \) km/h.
Convert speed to m/minute:
\( = \frac{10 \times 1000 \text{ m}}{60 \text{ min}} = \frac{10000}{60} \text{ m/minute} = \frac{500}{3} \text{ m/minute} \).
The distance the water flows in 30 minutes (length \( l \)):
\( l = \text{speed} \times \text{time} = \frac{500}{3} \times 30 = 5000 \text{ m} \).
The volume of water flowed in 30 minutes \( = l \times b \times h \)
\( = 5000 \times 6 \times 1.5 \)
\( = 5000 \times 9 \)
\( = 45000 \text{ m}^3 \).
For the field to be irrigated:
The height of standing water needed \( = 8 \) cm \( = \frac{8}{100} \) m.
Let the area of the field be \( A \).
The volume of water in the field \( = \text{Area of field} \times \text{height of water} \).
Since the volume of water from the canal irrigates the field:
Volume of water in field = Volume of water flowed in 30 minutes
\( \text{Area of field} \times \frac{8}{100} = 45000 \)
\( \text{Area of field} = \frac{45000 \times 100}{8} \)
\( \text{Area of field} = 562500 \text{ m}^2 \).
To convert to hectares, we know \( 1 \text{ hectare} = 10000 \text{ m}^2 \).
Area of field irrigated \( = \frac{562500}{10000} = 56.25 \text{ hectares} \).
In simple words: First, we calculate how much water flows out of the canal in 30 minutes. Then, we use this volume of water to cover a field with a certain depth. We divide the total water volume by the required water depth to find the area of land that can be irrigated.

Exam Tip: Be careful with unit conversions (km/h to m/minute, cm to m) and always ensure consistency. The key principle is that the volume of water provided by the canal equals the volume of water spread over the field.

Question 9. 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 the rate of 3 km/h, in how much time will the tank be filled?
Answer:
For the cylindrical pipe:
Internal diameter \( = 20 \) cm, so radius \( r = 10 \) cm \( = \frac{10}{100} \) m \( = 0.1 \) m.
Speed of flow of water in the pipe \( = 3 \) km/h.
Convert speed to m/s:
\( = \frac{3 \times 1000 \text{ m}}{60 \times 60 \text{ s}} = \frac{3000}{3600} \text{ m/s} = \frac{5}{6} \text{ m/s} \).
Volume of water flowed per second through the pipe (this represents the height \( h \) of water flowing per second through a cylinder with radius \( r \))
\( = \pi r^2 \times \text{speed} \)
\( = \pi (0.1)^2 \times \frac{5}{6} \)
\( = \pi \times 0.01 \times \frac{5}{6} \)
\( = \frac{0.05\pi}{6} \text{ m}^3/\text{s} \).
For the cylindrical tank:
Diameter \( = 10 \) m, so radius \( R = 5 \) m.
Depth \( H = 2 \) m.
Volume of the tank \( = \pi R^2 H \)
\( = \pi \times 5^2 \times 2 \)
\( = \pi \times 25 \times 2 \)
\( = 50\pi \text{ m}^3 \).
Time taken to fill the tank \( = \frac{\text{Volume of tank}}{\text{Volume of water flowed in one second}} \)
\( = \frac{50\pi}{\frac{0.05\pi}{6}} \)
\( = \frac{50}{0.05} \times 6 \)
\( = 1000 \times 6 \)
\( = 6000 \) seconds.
Convert seconds to minutes and hours:
\( 6000 \text{ seconds} = \frac{6000}{60} \text{ minutes} = 100 \text{ minutes} \).
\( 100 \text{ minutes} = \frac{100}{60} \text{ hour} = 1 \text{ hour } 40 \text{ minutes} \).
Therefore, the tank will be filled in 1 hour and 40 minutes.
In simple words: We first calculate how much water the pipe delivers every second. Then, we find out how much water the tank can hold. By dividing the tank's total volume by the pipe's flow rate, we determine the time it will take to fill the tank.

Exam Tip: Be very careful with unit conversions (cm to m, km/h to m/s or m/minute). Ensure the units for volume and time are consistent throughout the calculation to avoid errors. The formula for the volume of water flowing through a pipe is essentially the volume of a cylinder whose length is the distance water travels in a given time.