OP Malhotra Class 10 Maths Solutions Chapter 15 Three Dimensional Solids Exercise 15 (E)

Question 1. The diameter of a metallic sphere is 6 cm. The sphere is melted and drawn into a wire of uniform circular cross section. If the length of the wire is 36 cm, find its radius.
Answer: The diameter of the metallic sphere is 6 cm.
So, its radius \( (R) = \frac{6}{2} = 3 \) cm.
The volume of the sphere is \( \frac{4}{3}\pi R^3 = \frac{4}{3} \times \frac{22}{7} \times 3 \times 3 \times 3 \) cm³ \( = \frac{792}{7} \) cm³.
When the sphere is melted and made into a wire, its volume stays the same. The wire is a cylinder with length \( (h) = 36 \) cm. Let \( r \) be the radius of this circular wire.
The volume of the cylindrical wire is \( \pi r^2 h \).
So, \( \pi r^2 h = \frac{792}{7} \).
\( \implies \frac{22}{7} r^2 \times 36 = \frac{792}{7} \)
\( \implies r^2 = \frac{792 \times 7}{7 \times 22 \times 36} \)
\( \implies r^2 = 1 \)
\( \implies r = 1 \) cm. This means the radius of the wire is 1 cm.
In simple words: We find the volume of the sphere first. Then, we know this same volume will be used to make a wire. Since the wire is like a long cylinder, we use the cylinder volume formula and the given wire length to find its radius.

🎯 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 mensuration problems.

Question 2. How many lead balls, each of radius 1 cm, can be made from a sphere whose radius is 8 cm?
Answer: The radius of the large sphere is \( (R) = 8 \) cm.
The volume of the large sphere is \( \frac{4}{3}\pi R^3 = \frac{4}{3} \times \frac{22}{7} \times 8 \times 8 \times 8 \) cm³ \( = \frac{45056}{21} \) cm³.
Each small lead ball has a radius \( (r) = 1 \) cm.
The volume of one small lead ball is \( \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 1 \times 1 \times 1 \) cm³ \( = \frac{88}{21} \) cm³.
To find the number of small lead balls, divide the volume of the large sphere by the volume of one small lead ball.
Number of balls \( = \frac{\text{Volume of large sphere}}{\text{Volume of one small lead ball}} \)
\( \implies \) Number of balls \( = \frac{45056}{21} \div \frac{88}{21} \)
\( \implies \) Number of balls \( = \frac{45056}{21} \times \frac{21}{88} \)
\( \implies \) Number of balls \( = \frac{45056}{88} = 512 \). So, 512 small lead balls can be made.
In simple words: First, calculate how much space the big sphere takes up. Then, calculate how much space one tiny lead ball takes up. Divide the big space by the small space to find out how many small balls fit inside.

🎯 Exam Tip: When dividing volumes, ensure all dimensions are in the same units. If you see common factors like \( \frac{4}{3}\pi \) in both volumes, you can cancel them out before calculation to simplify the process.

Question 3. A solid metallic sphere of diameter 21 cm is melted and recast into a number of smaller cones, each of diameter 7 cm and height 3 cm. Find the number of cones so formed.
Answer: The diameter of the solid metallic sphere is 21 cm.
Its radius \( (R) = \frac{21}{2} \) cm.
The volume of the sphere is \( \frac{4}{3}\pi R^3 = \frac{4}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2} \) cm³ \( = 4851 \) cm³.
Each smaller cone has a diameter of 7 cm.
Its radius \( (r) = \frac{7}{2} \) cm and height \( (h) = 3 \) cm.
The volume of one cone is \( \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 3 \) cm³ \( = \frac{77}{2} \) cm³.
To find the number of cones, divide the volume of the sphere by the volume of one cone.
Number of cones formed \( = \frac{\text{Volume of sphere}}{\text{Volume of one cone}} \)
\( \implies \) Number of cones formed \( = 4851 \div \frac{77}{2} \)
\( \implies \) Number of cones formed \( = 4851 \times \frac{2}{77} \)
\( \implies \) Number of cones formed \( = 63 \times 2 = 126 \) cones.
In simple words: We calculate the total amount of metal from the big sphere. Then we find out how much metal is needed for one small cone. Dividing the total metal by the metal for one cone gives us how many cones can be made.

🎯 Exam Tip: Ensure that you correctly use the radius (half of the diameter) for calculations, and remember the formulas for the volume of a sphere and a cone.

Question 4. A metallic disc, in the shape of a right circular cylinder, is of height 2.5 mm, and base radius 12 cm. Metallic disc is melted and made into a sphere. Calculate the radius of the sphere.
Answer: The metallic disc is a right circular cylinder.
Its radius \( (R) = 12 \) cm.
Its height \( (H) = 2.5 \) mm. We need to convert this to cm: \( 2.5 \) mm \( = \frac{2.5}{10} = 0.25 \) cm.
The volume of the disc (cylinder) is \( \pi R^2 H = \frac{22}{7} \times 12 \times 12 \times 0.25 \) cm³.
\( = \frac{22}{7} \times 144 \times \frac{1}{4} = \frac{22}{7} \times 36 = \frac{792}{7} \) cm³.
When the disc is melted and made into a sphere, its volume remains the same. Let \( r \) be the radius of the new sphere.
The volume of the sphere is \( \frac{4}{3}\pi r^3 \).
So, \( \frac{4}{3}\pi r^3 = \frac{792}{7} \).
\( \implies \frac{4}{3} \times \frac{22}{7} r^3 = \frac{792}{7} \)
\( \implies r^3 = \frac{792}{7} \times \frac{3 \times 7}{4 \times 22} \)
\( \implies r^3 = 9 \times 3 = 27 \)
\( \implies r^3 = (3)^3 \)
\( \implies r = 3 \) cm. Thus, the radius of the sphere is 3 cm.
In simple words: First, convert all units to be the same. Then, calculate the volume of the metallic disc, which is shaped like a cylinder. Since this melted metal forms a sphere, the volume of the sphere will be the same as the disc. Use the sphere's volume formula to find its radius.

🎯 Exam Tip: Always pay close attention to units and convert them to be consistent (e.g., mm to cm) before starting calculations to avoid errors.

Question 5. A hollow sphere of internal and external diameters 6 cm and 10 cm respectively is melted and recast into a cone of base diameter 14 cm. Find the height of the cone.
Answer: The hollow sphere has an external diameter of 10 cm, so its external radius \( (R) = \frac{10}{2} = 5 \) cm.
Its internal diameter is 6 cm, so its internal radius \( (r) = \frac{6}{2} = 3 \) cm.
The volume of the metal used in the hollow sphere is the difference between the external and internal volumes: \( \frac{4}{3}\pi (R^3 - r^3) \).
\( = \frac{4}{3} \times \frac{22}{7} (5^3 - 3^3) \) cm³.
\( = \frac{88}{21} (125 - 27) = \frac{88}{21} \times 98 \) cm³.
\( = \frac{88 \times 14}{3} = \frac{1232}{3} \) cm³.
This metal is recast into a cone. The cone has a base diameter of 14 cm.
Its radius \( (r_1) = \frac{14}{2} = 7 \) cm. Let \( h \) be the height of the cone.
The volume of the cone is \( \frac{1}{3}\pi r_1^2 h \).
Since the volume of the metal remains constant, we have:
\( \frac{1}{3}\pi r_1^2 h = \frac{1232}{3} \).
\( \implies \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times h = \frac{1232}{3} \)
\( \implies \frac{154}{3} h = \frac{1232}{3} \)
\( \implies h = \frac{1232 \times 3}{3 \times 154} \)
\( \implies h = 8 \) cm. Therefore, the height of the cone is 8 cm.
In simple words: First, calculate the volume of metal in the hollow sphere by subtracting the inner empty space from the total outer volume. This metal is then shaped into a cone. Since the amount of metal is the same, set the volume of the cone equal to the volume of the metal, and then find the height of the cone.

🎯 Exam Tip: For hollow objects, remember to calculate the volume of the material itself (external volume minus internal volume) rather than just the external volume. This ensures you're using the actual amount of melted metal.

Question 6. A hollow metallic cylindrical tube has an internal radius of 3 cm and height 21 cm. The thickness of the metal tube is \( \frac{1}{2} \) cm. The tube is melted and cast into a right circular cone of height 7 cm. Find the radius of cone correct to 1 decimal place. \( \left(\text{Take } \pi=\frac{22}{7}\right) \)
Answer: For the hollow metallic cylindrical tube:
Internal radius \( (r) = 3 \) cm.
Height \( (H) = 21 \) cm.
Thickness of metal \( = \frac{1}{2} \) cm \( = 0.5 \) cm.
Outer radius \( (R) = \text{Internal radius} + \text{Thickness} = 3 + 0.5 = 3.5 \) cm.
The volume of the metal in the tube is \( \pi R^2 H - \pi r^2 H = \pi (R^2 - r^2)H \).
\( = \frac{22}{7} \left( (3.5)^2 - 3^2 \right) \times 21 \) cm³.
\( = \frac{22}{7} \left( \frac{49}{4} - 9 \right) \times 21 \) cm³.
\( = \frac{22}{7} \left( \frac{49 - 36}{4} \right) \times 21 \) cm³.
\( = \frac{22}{7} \times \frac{13}{4} \times 21 \) cm³.
\( = 22 \times \frac{13}{4} \times 3 = 11 \times 13 \times \frac{3}{2} = \frac{429}{2} \) cm³.
This metal is cast into a right circular cone. Let \( r_c \) be the radius of the cone.
Height of cone \( (h_c) = 7 \) cm.
Volume of cone \( = \frac{1}{3}\pi r_c^2 h_c \).
Since the volume of metal is conserved:
\( \frac{1}{3}\pi r_c^2 h_c = \frac{429}{2} \).
\( \implies \frac{1}{3} \times \frac{22}{7} \times r_c^2 \times 7 = \frac{429}{2} \)
\( \implies \frac{22}{3} r_c^2 = \frac{429}{2} \)
\( \implies r_c^2 = \frac{429}{2} \times \frac{3}{22} \)
\( \implies r_c^2 = \frac{39 \times 3}{4} = \frac{117}{4} \).
\( \implies r_c = \sqrt{\frac{117}{4}} = \frac{\sqrt{117}}{2} \).
\( \implies r_c = \frac{10.816}{2} \approx 5.408 \) cm.
Rounding to one decimal place, the radius of the cone is 5.4 cm.
In simple words: First, calculate the volume of metal in the cylindrical tube by finding the difference between the outer and inner cylinder volumes. This metal is then melted to form a cone. By setting the volume of the cone equal to the volume of the metal, we can calculate the cone's radius. Remember to round the final answer to one decimal place.

🎯 Exam Tip: When dealing with hollow objects, ensure you correctly determine the outer and inner radii. Also, don't forget to convert units if necessary and pay attention to rounding instructions for the final answer.

Question 7. Three solid glass balls of radius 1, 6 and 8 cm, respectively are melted into a solid sphere. Find its radius.
Answer: Let the radii of the three solid glass balls be \( r_1 = 1 \) cm, \( r_2 = 6 \) cm, and \( r_3 = 8 \) cm.
The volume of the first ball is \( \frac{4}{3}\pi r_1^3 = \frac{4}{3}\pi (1)^3 \) cm³.
The volume of the second ball is \( \frac{4}{3}\pi r_2^3 = \frac{4}{3}\pi (6)^3 \) cm³.
The volume of the third ball is \( \frac{4}{3}\pi r_3^3 = \frac{4}{3}\pi (8)^3 \) cm³.
When these three balls are melted and combined, the total volume remains the same.
Total volume \( V = \frac{4}{3}\pi (r_1^3 + r_2^3 + r_3^3) \).
\( V = \frac{4}{3}\pi (1^3 + 6^3 + 8^3) \) cm³.
\( V = \frac{4}{3}\pi (1 + 216 + 512) \) cm³.
\( V = \frac{4}{3}\pi (729) \) cm³.
Let \( R \) be the radius of the new solid sphere formed.
The volume of the new sphere is \( \frac{4}{3}\pi R^3 \).
So, \( \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (729) \).
We can cancel \( \frac{4}{3}\pi \) from both sides.
\( \implies R^3 = 729 \).
Since \( 9^3 = 9 \times 9 \times 9 = 81 \times 9 = 729 \),
\( \implies R = 9 \) cm. Therefore, the radius of the new solid sphere is 9 cm.
In simple words: Calculate the volume of each small glass ball. Add these volumes together to get the total volume of glass. This total volume will be the same as the volume of the new, larger sphere. Use the sphere's volume formula to find the radius of this new sphere.

🎯 Exam Tip: When combining multiple spheres (or any solids) into one, the total volume is conserved. You can often factor out \( \frac{4}{3}\pi \) to simplify calculations if all original solids are spheres.

Question 8. A vessel in the form of an inverted cone is filled with water to the brim. Its height is 20 cm and diameter is 16.8 cm. Two equal solid cones are dropped in it so that they are fully submerged. As a result, one-third of the water in the original cone overflows. What is the volume of each of the solid copes submerged?

Answer: The diameter of the conical vessel is 16.8 cm, so its radius (R) is \( \frac{16.8}{2} = 8.4 \) cm. The height (H) is 20 cm.
The volume of water initially filling the cone is \( \frac{1}{3} \pi R^2 H \).
\( \implies \) Volume \( = \frac{1}{3} \pi \times (8.4)^2 \times 20 \) cm³
\( \implies \) Volume \( = \frac{1}{3} \pi \times 70.56 \times 20 = 470.4 \pi \) cm³.
When two solid cones are dropped, one-third of the water overflows. This means the volume of the two small solid cones is equal to the volume of the overflowing water.
Volume of overflowing water \( = \frac{1}{3} \times 470.4 \pi = 156.8 \pi \) cm³.
So, the volume of two small solid cones is \( 156.8 \pi \) cm³.
To find the volume of one solid cone, we divide this by 2.
Volume of one solid cone \( = \frac{156.8 \pi}{2} = 78.4 \pi \) cm³.
Using \( \pi = \frac{22}{7} \), the volume is \( 78.4 \times \frac{22}{7} = 11.2 \times 22 = 246.4 \) cm³.
In simple words: First, we find how much water the big cone can hold. When small cones are added, some water spills out. The amount of water that spills out is the same as the space the small cones take up. We then find the volume of one small cone by dividing the total displaced volume by two.

🎯 Exam Tip: Remember that when an object is submerged in a liquid, the volume of liquid displaced is equal to the volume of the object. Also, ensure consistent units throughout your calculation.

Question 9. The surface area of a solid metallic sphere is 616 cm². It is melted and recast into smaller spheres of diameter 3.5 cm. How many such spheres can be obtained?
Answer: The surface area of the large metallic sphere is given as 616 cm².
The formula for the surface area of a sphere is \( 4 \pi R^2 \), where R is its radius.
So, \( 4 \pi R^2 = 616 \).
Using \( \pi = \frac{22}{7} \), we have \( 4 \times \frac{22}{7} \times R^2 = 616 \).
\( \implies \frac{88}{7} R^2 = 616 \).
\( \implies R^2 = \frac{616 \times 7}{88} = \frac{4312}{88} = 49 \).
\( \implies R = \sqrt{49} = 7 \) cm.
Now, we find the volume of the large sphere: \( V_L = \frac{4}{3} \pi R^3 = \frac{4}{3} \times \frac{22}{7} \times (7)^3 \).
\( \implies V_L = \frac{4}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 = \frac{4 \times 22 \times 49}{3} = \frac{4312}{3} \) cm³.
For the smaller spheres, the diameter is 3.5 cm, so the radius (r) is \( \frac{3.5}{2} = 1.75 \) cm.
The volume of one small sphere is \( V_S = \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (1.75)^3 \).
\( \implies V_S = \frac{4}{3} \times \frac{22}{7} \times (1.75 \times 1.75 \times 1.75) = \frac{88}{21} \times 5.359375 \approx 22.47 \) cm³.
To find the number of small spheres, divide the volume of the large sphere by the volume of one small sphere.
Number of spheres \( = \frac{V_L}{V_S} = \frac{\frac{4312}{3}}{\frac{4}{3} \pi (1.75)^3} = \frac{4312}{4 \times \frac{22}{7} \times (1.75)^3} \).
\( = \frac{4312}{4 \times \frac{22}{7} \times 5.359375} = \frac{4312}{\frac{88}{7} \times 5.359375} = \frac{4312 \times 7}{88 \times 5.359375} \).
\( = \frac{30184}{471.625} = 64 \).
Alternatively, Number of spheres \( = \frac{\frac{4}{3} \pi R^3}{\frac{4}{3} \pi r^3} = \left( \frac{R}{r} \right)^3 = \left( \frac{7}{1.75} \right)^3 = (4)^3 = 64 \). This method is much faster.
In simple words: First, we find the radius of the big sphere using its surface area. Then, we find the volume of the big sphere and also the volume of one small sphere. The number of small spheres is found by dividing the total volume of the large sphere by the volume of one small sphere.

🎯 Exam Tip: When an object is melted and recast, its volume remains the same. For spheres, remember that the ratio of volumes is the cube of the ratio of their radii.

Question 10. The volume of a conical tent is 1232 m³ and the area of the base floor is 154 m². Calculate the: (i) radius of the floor; (ii) height of the tent; (iii) length of canvas required to cover this conical tent, if its width is 2 m.
Answer: A conical tent has a base that is a circle. The area of the base floor is \( 154 \) m².
(i) The area of a circle is \( \pi r^2 \). So, \( \pi r^2 = 154 \).
Using \( \pi = \frac{22}{7} \), we have \( \frac{22}{7} r^2 = 154 \).
\( \implies r^2 = \frac{154 \times 7}{22} = 7 \times 7 = 49 \).
\( \implies r = \sqrt{49} = 7 \) m.
So, the radius of the floor is 7 m.
(ii) The volume of a conical tent is \( \frac{1}{3} \pi r^2 h \). We know the volume is 1232 m³ and \( \pi r^2 = 154 \).
So, \( \frac{1}{3} \times 154 \times h = 1232 \).
\( \implies 154 h = 1232 \times 3 = 3696 \).
\( \implies h = \frac{3696}{154} = 24 \) m.
So, the height of the tent is 24 m.
(iii) To find the length of canvas needed, we first need the curved surface area (CSA) of the cone, which is \( \pi r l \), where \( l \) is the slant height. We need to find \( l \) using the radius \( r \) and height \( h \).
Slant height \( l = \sqrt{r^2 + h^2} = \sqrt{(7)^2 + (24)^2} \).
\( \implies l = \sqrt{49 + 576} = \sqrt{625} = 25 \) m.
Curved surface area (CSA) \( = \pi r l = \frac{22}{7} \times 7 \times 25 = 22 \times 25 = 550 \) m².
The canvas has a width of 2 m. To find the length, divide the total area by the width.
Length of canvas \( = \frac{\text{Area}}{\text{Width}} = \frac{550}{2} = 275 \) m.
In simple words: We first use the floor area to find the tent's base radius. Then, we use the volume and radius to find the tent's height. Finally, we calculate the slant height and curved area of the tent, then divide by the canvas width to find the required length.

🎯 Exam Tip: Remember the formulas for area of a circle, volume of a cone, and curved surface area of a cone. The Pythagorean theorem is crucial for finding the slant height when radius and height are known.

Question 11. The given figure represents a hemisphere surmounted by a conical block of wood. The diameter of their bases is 6 cm each and the slant height of the cone is 5 cm. Calculate : (i) the height of the cone ; (ii) the volume of the solid.

Answer: The solid is made of a hemisphere at the bottom and a cone on top.
The diameter of their bases is 6 cm, so the radius (r) for both the cone and the hemisphere is \( \frac{6}{2} = 3 \) cm.
The slant height (l) of the cone is 5 cm.
(i) To find the height (h) of the cone, we use the Pythagorean theorem: \( l^2 = r^2 + h^2 \).
So, \( 5^2 = 3^2 + h^2 \).
\( \implies 25 = 9 + h^2 \).
\( \implies h^2 = 25 - 9 = 16 \).
\( \implies h = \sqrt{16} = 4 \) cm.
The height of the cone is 4 cm.
(ii) To find the volume of the entire solid, we add the volume of the cone and the volume of the hemisphere.
Volume of cone \( = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (3)^2 (4) = \frac{1}{3} \pi \times 9 \times 4 = 12 \pi \) cm³.
Volume of hemisphere \( = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (3)^3 = \frac{2}{3} \pi \times 27 = 18 \pi \) cm³.
Total volume of the solid \( = 12 \pi + 18 \pi = 30 \pi \) cm³.
Using \( \pi = \frac{22}{7} \), Total volume \( = 30 \times \frac{22}{7} = \frac{660}{7} \approx 94.29 \) cm³.
This shape is common in real-world items like ice cream cones, combining two simpler solids. If using the formula \( \frac{1}{3} \pi r^2 (h + 2r) \), then \( \frac{1}{3} \pi (3)^2 (4 + 2 \times 3) = \frac{1}{3} \pi \times 9 \times (4+6) = 3 \pi \times 10 = 30 \pi \) cm³.
In simple words: First, we use the given slant height and base radius of the cone to calculate its vertical height. Then, we find the volume of the cone and the volume of the hemisphere separately. Adding these two volumes together gives the total volume of the entire wooden block.

🎯 Exam Tip: When dealing with combined solids, break the problem into finding the dimensions and volumes of the individual shapes, then add or subtract as needed. Always double-check calculations involving \( \pi \).

Question 12. A hemispherical bowl of diameter 7.2 cm is filled completely with chocolate sauce. This sauce is poured into an inverted cone of radius 4.8 cm. Find the height of the cone.
Answer: The hemispherical bowl has a diameter of 7.2 cm, so its radius (r) is \( \frac{7.2}{2} = 3.6 \) cm.
The volume of chocolate sauce is equal to the volume of the hemispherical bowl.
Volume of hemisphere \( = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (3.6)^3 \).
\( \implies \) Volume \( = \frac{2}{3} \pi \times 46.656 = 31.104 \pi \) cm³.
This chocolate sauce is then poured into an inverted cone that has a radius of 4.8 cm. Let the height of this cone be h.
The volume of the cone is \( \frac{1}{3} \pi R^2 h \), where R is the cone's radius.
Since the sauce fills the cone, its volume will be the same as the sauce from the hemisphere.
So, \( \frac{1}{3} \pi (4.8)^2 h = 31.104 \pi \).
We can cancel \( \pi \) from both sides.
\( \frac{1}{3} (4.8)^2 h = 31.104 \).
\( \implies \frac{1}{3} \times 23.04 \times h = 31.104 \).
\( \implies 7.68 h = 31.104 \).
\( \implies h = \frac{31.104}{7.68} = 4.05 \) cm.
The height of the cone is 4.05 cm. Understanding that the volume stays constant helps solve these problems.
In simple words: First, we find how much chocolate sauce is in the hemispherical bowl. Then, we know this same amount of sauce goes into the cone. Using the volume of the cone formula and the given radius, we can work backward to find the height of the cone.

🎯 Exam Tip: When a substance is transferred from one container to another, its volume remains constant. This principle is key to solving problems where shapes are melted and recast or liquids are poured.

Question 13. A solid cone of radius 5 cm and height 8 cm is melted and made into small spheres of radius 0.5 cm. Find the number of spheres formed.
Answer: First, calculate the volume of the solid cone.
Radius of the cone (R) = 5 cm, height of the cone (H) = 8 cm.
Volume of cone \( V_C = \frac{1}{3} \pi R^2 H = \frac{1}{3} \pi (5)^2 (8) \).
\( \implies V_C = \frac{1}{3} \pi \times 25 \times 8 = \frac{200 \pi}{3} \) cm³.
Next, calculate the volume of one small sphere.
Radius of one small sphere (r) = 0.5 cm.
Volume of sphere \( V_S = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (0.5)^3 \).
\( \implies V_S = \frac{4}{3} \pi \times 0.125 = \frac{0.5 \pi}{3} \) cm³.
To find the number of spheres formed, divide the total volume of the cone by the volume of one small sphere.
Number of spheres \( = \frac{V_C}{V_S} = \frac{\frac{200 \pi}{3}}{\frac{0.5 \pi}{3}} \).
\( \implies \) Number of spheres \( = \frac{200 \pi}{3} \times \frac{3}{0.5 \pi} = \frac{200}{0.5} = 400 \).
Therefore, 400 small spheres can be formed from the melted cone. This shows how a larger volume can be broken down into many smaller, identical volumes.
In simple words: We find the total amount of material in the cone by calculating its volume. Then, we find the amount of material in just one tiny sphere. Dividing the total material from the cone by the material for one sphere tells us how many small spheres can be made.

🎯 Exam Tip: Always remember that when a solid is melted and recast into other solids, the total volume remains conserved. This principle is fundamental for solving such problems efficiently.

Question 14. A hollow sphere of internal and external radii 6 cm and 8 cm respectively is melted and recast into small cones of base radius 2 cm and height 8 cm. Find the number of cones.
Answer: First, calculate the volume of metal in the hollow sphere.
External radius (R) = 8 cm, internal radius (r) = 6 cm.
Volume of metal in hollow sphere \( V_H = \frac{4}{3} \pi (R^3 - r^3) \).
\( \implies V_H = \frac{4}{3} \pi (8^3 - 6^3) = \frac{4}{3} \pi (512 - 216) \).
\( \implies V_H = \frac{4}{3} \pi \times 296 = \frac{1184 \pi}{3} \) cm³.
Next, calculate the volume of one small cone.
Base radius of cone (\( r_c \)) = 2 cm, height of cone (h) = 8 cm.
Volume of one cone \( V_C = \frac{1}{3} \pi r_c^2 h = \frac{1}{3} \pi (2)^2 (8) \).
\( \implies V_C = \frac{1}{3} \pi \times 4 \times 8 = \frac{32 \pi}{3} \) cm³.
To find the number of cones (n), divide the total volume of metal from the hollow sphere by the volume of one cone.
Number of cones \( n = \frac{V_H}{V_C} = \frac{\frac{1184 \pi}{3}}{\frac{32 \pi}{3}} \).
\( \implies n = \frac{1184}{32} = 37 \).
Thus, 37 small cones can be formed from the hollow sphere. This problem demonstrates the conservation of volume in melting and recasting processes.
In simple words: We first find the total amount of metal in the hollow sphere by subtracting the inner empty space from the total outer volume. Then, we calculate the volume of one small cone. Finally, we divide the total metal volume by the volume of one cone to find out how many cones can be made.

🎯 Exam Tip: For hollow objects, the volume of material is the difference between the external and internal volumes. Always ensure you subtract the volumes correctly before proceeding to division.

Question 15. A solid sphere of radius 15 cm is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate the number of cones recast.
Answer: First, calculate the volume of the large solid sphere.
Radius of sphere (R) = 15 cm.
Volume of sphere \( V_S = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (15)^3 \).
\( \implies V_S = \frac{4}{3} \pi \times 3375 = 4 \pi \times 1125 = 4500 \pi \) cm³.
Next, calculate the volume of one small right circular cone.
Radius of cone (r) = 2.5 cm, height of cone (h) = 8 cm.
Volume of cone \( V_C = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (2.5)^2 (8) \).
\( \implies V_C = \frac{1}{3} \pi \times 6.25 \times 8 = \frac{50 \pi}{3} \) cm³.
To find the number of cones, divide the total volume of the sphere by the volume of one cone.
Number of cones \( = \frac{V_S}{V_C} = \frac{4500 \pi}{\frac{50 \pi}{3}} \).
\( \implies \) Number of cones \( = 4500 \pi \times \frac{3}{50 \pi} = \frac{4500 \times 3}{50} = 90 \times 3 = 270 \).
Thus, 270 cones can be recast from the solid sphere. This again illustrates that volume is conserved during transformation of solids.
In simple words: We first find the volume of the large sphere. Then, we find the volume of a single small cone. To know how many cones can be made, we divide the volume of the large sphere by the volume of one small cone.

🎯 Exam Tip: Always use the full volume formulas, including \( \pi \), and simplify by canceling \( \pi \) if it appears in both the numerator and denominator, which often saves calculation steps.

Question 16. The surface area of a solid metallic sphere is 2464 cm². It is melted and recast into solid right circular cones of radius 3.5 cm and height 7 cm. Calculate: (i) the radius of the sphere. (ii) the number of cones recast. (Take \( \pi = 22/7 \))
Answer: (i) The surface area of the solid metallic sphere is 2464 cm².
The formula for the surface area of a sphere is \( 4 \pi R^2 \).
So, \( 4 \pi R^2 = 2464 \).
Using \( \pi = \frac{22}{7} \), we have \( 4 \times \frac{22}{7} \times R^2 = 2464 \).
\( \implies \frac{88}{7} R^2 = 2464 \).
\( \implies R^2 = \frac{2464 \times 7}{88} = \frac{17248}{88} = 196 \).
\( \implies R = \sqrt{196} = 14 \) cm.
The radius of the sphere is 14 cm.
(ii) First, calculate the volume of the large sphere.
Volume of sphere \( V_S = \frac{4}{3} \pi R^3 = \frac{4}{3} \times \frac{22}{7} \times (14)^3 \).
\( \implies V_S = \frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14 = \frac{4 \times 22 \times 2 \times 14 \times 14}{3} = \frac{34496}{3} \) cm³.
Next, calculate the volume of one small cone.
Radius of cone (r) = 3.5 cm, height of cone (h) = 7 cm.
Volume of cone \( V_C = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times 7 \).
\( \implies V_C = \frac{1}{3} \times \frac{22}{7} \times (3.5 \times 3.5) \times 7 = \frac{1}{3} \times 22 \times 3.5 \times 3.5 = \frac{269.5}{3} \) cm³.
To find the number of cones recast, divide the volume of the sphere by the volume of one cone.
Number of cones \( = \frac{V_S}{V_C} = \frac{\frac{34496}{3}}{\frac{269.5}{3}} \).
\( \implies \) Number of cones \( = \frac{34496}{269.5} = 128 \).
A quicker way is to express 3.5 as 7/2. Then \( V_C = \frac{1}{3} \pi (\frac{7}{2})^2 (7) = \frac{1}{3} \pi \frac{49}{4} \times 7 = \frac{343 \pi}{12} \).
Number of cones \( = \frac{\frac{4}{3} \pi (14)^3}{\frac{343 \pi}{12}} = \frac{4 \times 14^3}{3} \times \frac{12}{343} = \frac{4 \times 2744}{1} \times \frac{4}{343} = 4 \times 8 \times 4 = 128 \).
In simple words: First, we use the sphere's surface area to find its radius. Then, we calculate the total volume of the sphere. After that, we find the volume of one small cone. Finally, we divide the total sphere volume by the volume of one cone to find out how many cones can be made.

🎯 Exam Tip: Always pay attention to the value of \( \pi \) specified in the question (e.g., \( \frac{22}{7} \) or 3.14). Also, simplify fractions and powers before multiplying large numbers to avoid errors.

Question 17. Two solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed.
Answer: First, calculate the volume of the two solid spheres.
Radius of small sphere (\( r_1 \)) = 2 cm.
Volume of small sphere \( V_1 = \frac{4}{3} \pi r_1^3 = \frac{4}{3} \pi (2)^3 = \frac{4}{3} \pi \times 8 = \frac{32 \pi}{3} \) cm³.
Radius of big sphere (\( r_2 \)) = 4 cm.
Volume of big sphere \( V_2 = \frac{4}{3} \pi r_2^3 = \frac{4}{3} \pi (4)^3 = \frac{4}{3} \pi \times 64 = \frac{256 \pi}{3} \) cm³.
Total volume of metal from both spheres \( V_{total} = V_1 + V_2 = \frac{32 \pi}{3} + \frac{256 \pi}{3} = \frac{288 \pi}{3} = 96 \pi \) cm³.
This total volume is recast into a cone with a height (h) of 8 cm. Let the radius of this cone be R.
The volume of the cone \( V_C = \frac{1}{3} \pi R^2 h \).
Since the metal is recast, the volume remains the same: \( V_C = V_{total} \).
So, \( \frac{1}{3} \pi R^2 (8) = 96 \pi \).
We can cancel \( \pi \) from both sides.
\( \frac{8}{3} R^2 = 96 \).
\( \implies R^2 = \frac{96 \times 3}{8} = 12 \times 3 = 36 \).
\( \implies R = \sqrt{36} = 6 \) cm.
The radius of the cone formed is 6 cm. This process shows how material can be reshaped while keeping its total volume constant.
In simple words: We find the volume of each of the two spheres and add them together to get the total amount of metal. This total metal volume is then used to form a new cone. Knowing the cone's height and total volume, we can calculate its radius.

🎯 Exam Tip: Always sum up the volumes of all source objects before equating it to the volume of the target object. This ensures that all material is accounted for during the recasting process.

Question 18. A certain number of metallic cones, each of radius 2 cm and height 3 cm are melted and recast into a solid sphere of radius 6 cm. Find the number of cones.
Answer: First, calculate the volume of one metallic cone.
Radius of cone (\( r_c \)) = 2 cm, height of cone (h) = 3 cm.
Volume of one cone \( V_C = \frac{1}{3} \pi r_c^2 h = \frac{1}{3} \pi (2)^2 (3) \).
\( \implies V_C = \frac{1}{3} \pi \times 4 \times 3 = 4 \pi \) cm³.
Next, calculate the volume of the solid sphere.
Radius of sphere (\( r_s \)) = 6 cm.
Volume of sphere \( V_S = \frac{4}{3} \pi r_s^3 = \frac{4}{3} \pi (6)^3 \).
\( \implies V_S = \frac{4}{3} \pi \times 216 = 4 \pi \times 72 = 288 \pi \) cm³.
To find the number of cones (n) required to form the sphere, divide the volume of the sphere by the volume of one cone.
Number of cones \( n = \frac{V_S}{V_C} = \frac{288 \pi}{4 \pi} \).
\( \implies n = \frac{288}{4} = 72 \).
Therefore, 72 metallic cones are needed to form the solid sphere. This demonstrates the conservation of material volume during the process of melting and recasting solids.
In simple words: We figure out how much material is in one small cone. Then, we figure out how much material is in the large sphere. Dividing the total material of the sphere by the material in one cone tells us how many cones were needed to make the sphere.

🎯 Exam Tip: When multiple small objects are melted to form a single large object (or vice versa), the total volume of the smaller objects must equal the volume of the larger object. Set up the equation and solve for the unknown quantity (e.g., number of objects, radius, or height).