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

S Chand Class 10 ICSE Maths Solutions Chapter 15 Three Dimensional Solids Ex 15(d)

Question 1. Find the volume and curved surface of the sphere if \( \left(\pi=\frac{22}{7}\right) \)
(i) radius = 1 cm
(ii) radius = 14 cm
(iii) radius = 28 cm
(iv) radius = \( 5\frac { 1 }{ 4 } \) cm
(v) Diameter = 14 cm
(vi) diameter = 35 cm
Answer:
(i) Radius of sphere (r) = 1 cm
The volume of the sphere is calculated using the formula \( \frac{4}{3}\pi r^3 \).
So, volume \( = \frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times 1 \times 1 \times 1 \text{ cm}^3 = \frac { 88 }{ 21 } \text{ cm}^3 \).
The curved surface area is found using \( 4\pi r^2 \).
So, curved surface area \( = 4 \times \frac { 22 }{ 7 } \times 1 \times 1 = \frac { 88 }{ 7 } \text{ cm}^2 \).
(ii) Radius (r) = 14 cm
Next, we calculate the volume with a radius of 14 cm.
Volume \( = \frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times 14 \times 14 \times 14 \text{ cm}^3 = \frac { 34496 }{ 3 } = 11498\frac { 2 }{ 3 } \text{ cm}^3 \).
Curved surface area \( = 4 \times \frac { 22 }{ 7 } \times 14 \times 14 \text{ cm}^2 = 2464 \text{ cm}^2 \).
(iii) Radius (r) = 28 cm
For a radius of 28 cm, the calculations are as follows.
Volume \( = \frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times 28 \times 28 \times 28 \text{ cm}^3 = \frac { 275968 }{ 3 } = 91989\frac { 1 }{ 3 } \text{ cm}^3 \).
Curved surface area \( = 4 \times \frac { 22 }{ 7 } \times 28 \times 28 \text{ cm}^3 = 9856 \text{ cm}^2 \).
(iv) Radius (r) \( = 5\frac { 1 }{ 4 } \text{ cm} = \frac { 21 }{ 4 } \text{ cm} \)
When the radius is a mixed fraction, convert it to an improper fraction first.
Volume \( = \frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times \frac { 21 }{ 4 } \times \frac { 21 }{ 4 } \times \frac { 21 }{ 4 } \text{ cm}^3 = \frac { 4851 }{ 8 } = 606\frac { 3 }{ 8 } \text{ cm}^3 \).
Curved surface area \( = 4 \times \frac { 22 }{ 7 } \times \frac { 21 }{ 4 } \times \frac { 21 }{ 4 } \text{ cm}^2 = \frac { 693 }{ 2 } = 346\frac { 1 }{ 2 } \text{ cm}^2 \).
(v) Diameter = 14 cm
First, find the radius from the given diameter.
Radius (r) \( = \frac { 14 }{ 2 } = 7 \text{ cm} \).
Volume \( = \frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times 7 \times 7 \times 7 \text{ cm}^3 = \frac { 4312 }{ 3 } = 1437\frac { 1 }{ 3 } \text{ cm}^3 = 1437.3 \text{ cm}^3 \).
Curved surface area \( = 4 \times \frac { 22 }{ 7 } \times 7 \times 7 \text{ cm}^2 = 616 \text{ cm}^2 \).
(vi) Diameter = 35 cm
Convert the diameter to radius for the last calculation.
Radius (r) \( = \frac { 35 }{ 2 } \text{ cm} \).
Volume \( = \frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times \frac { 35 }{ 2 } \times \frac { 35 }{ 2 } \times \frac { 35 }{ 2 } \text{ cm}^3 = \frac { 67375 }{ 3 } = 22458\frac { 1 }{ 3 } \text{ cm}^3 \).
Curved surface area \( = 4 \times \frac { 22 }{ 7 } \times \frac { 35 }{ 2 } \times \frac { 35 }{ 2 } \text{ cm}^2 = 3850 \text{ cm}^2 \).
In simple words: To find the volume, use \( \frac{4}{3}\pi r^3 \), and for the curved surface area, use \( 4\pi r^2 \). Remember to calculate the radius first if only the diameter is given.

🎯 Exam Tip: Always write down the formulas for volume and surface area of a sphere before substituting values to avoid errors and ensure full marks.

Question 2. Find the radius of the sphere whose surface area is equal to its volume.
Answer:
Let \( r \) be the radius of a sphere.
The formula for surface area is \( 4\pi r^2 \).
The formula for volume is \( \frac { 4 }{ 3 } \pi r^3 \).
Given that the surface area equals the volume:
\( 4\pi r^2 = \frac { 4 }{ 3 } \pi r^3 \)
Divide both sides by \( 4\pi r^2 \) (assuming \( r \ne 0 \)):
\( 1 = \frac { 1 }{ 3 } r \)
Multiply both sides by 3 to solve for \( r \).
\( \implies r = 3 \)
So, the radius is 3 units.
In simple words: When a sphere's surface area (the outside part) is the same number as its volume (the space it holds), its radius is always 3 units.

🎯 Exam Tip: When dealing with problems where volume and surface area are equal, remember to cancel common terms like \( \pi \) and \( r^2 \) to simplify the equation quickly.

Question 3. Find the diameter of the sphere whose surface area is equal to the area of a circle of diameter 2.8 cm.
Answer:
Let \( r \) be the radius of the sphere.
The surface area of a sphere is \( 4\pi r^2 \).
The diameter of the circle is 2.8 cm.
The radius of the circle is \( \frac{2.8}{2} = 1.4 \text{ cm} \).
The area of the circle is \( \pi \times (\text{radius of circle})^2 = \pi \times (1.4)^2 = 1.96\pi \text{ cm}^2 \).
We are given that the surface area of the sphere equals the area of the circle.
So, \( 4\pi r^2 = 1.96\pi \).
Divide both sides by \( \pi \):
\( 4r^2 = 1.96 \)
Divide by 4:
\( r^2 = \frac{1.96}{4} = 0.49 \)
Take the square root of both sides:
\( r = \sqrt{0.49} = 0.7 \text{ cm} \).
The diameter of the sphere is \( 2r = 2 \times 0.7 = 1.4 \text{ cm} \).
In simple words: We want a sphere whose outside area is the same as a certain circle's area. First, find the radius of the circle and its area. Then, use that area to figure out the sphere's radius, and finally, double it to get the sphere's diameter.

🎯 Exam Tip: Pay close attention to whether the question asks for radius or diameter in the final answer, and ensure all units are consistent.

Question 4. Find the radius of the sphere whose volume is \( \frac { 32 }{ 3 }π cm³ \).
Answer:
Let \( r \) be the radius of the sphere.
The volume of a sphere is given by the formula \( \frac { 4 }{ 3 }πr³ \).
We are given that the volume of the sphere is \( \frac { 32 }{ 3 }π \text{ cm}^3 \).
So, we set the two expressions for volume equal to each other:
\( \frac { 4 }{ 3 }πr³ = \frac { 32 }{ 3 }π \)
Multiply both sides by 3:
\( 4\pi r^3 = 32\pi \)
Divide both sides by \( 4\pi \):
\( r^3 = \frac { 32\pi }{ 4\pi } \)
\( \implies r^3 = 8 \)
To find \( r \), take the cube root of 8.
\( r = 2 \text{ cm} \).
In simple words: We know the total space inside a sphere. We use the formula for sphere volume, then solve it like a puzzle to find its radius.

🎯 Exam Tip: Remember to isolate \( r^3 \) before taking the cube root, ensuring all constants are correctly moved to the other side of the equation.

Question 5. The volume of a sphere is 4.851 cm³. Find its surface area.
Answer:
Let \( r \) be the radius of the sphere.
The volume of a sphere is \( \frac { 4 }{ 3 }πr³ \).
We are given the volume is 4.851 cm³.
So, \( \frac { 4 }{ 3 }πr³ = 4.851 \)
Substitute \( \pi = \frac{22}{7} \):
\( \frac { 4 }{ 3 } \times \frac { 22 }{ 7 } r^3 = 4.851 \)
\( \frac { 88 }{ 21 } r^3 = 4.851 \)
To find \( r^3 \), multiply by \( \frac { 21 }{ 88 } \):
\( r^3 = 4.851 \times \frac { 21 }{ 88 } \)
\( \implies r^3 = 1.157625 \)
Taking the cube root of 1.157625:
\( r = \sqrt[3]{1.157625} = 1.05 \text{ cm} \).
Now we need to find the surface area of the sphere, using the formula \( 4\pi r^2 \).
Surface area \( = 4 \times \frac { 22 }{ 7 } \times (1.05)^2 \text{ cm}^2 \)
\( = 4 \times \frac { 22 }{ 7 } \times 1.1025 \)
\( = 13.86 \text{ cm}^2 \).
In simple words: First, we use the given volume to find the sphere's radius. Once we have the radius, we can use it to calculate the surface area.

🎯 Exam Tip: When a problem involves both volume and surface area, you usually need to find the radius from one given value before calculating the other. Always check the required \( \pi \) value if specified.

Question 6. Find the volume of a hollow sphere whose outer diameter is 10 cm and the thickness of the material of which it is made is 1 cm.
Answer:
The outer diameter of the hollow sphere is 10 cm.
This means the outer radius (R) \( = \frac { 10 }{ 2 } = 5 \text{ cm} \).
The thickness of the material is 1 cm.
To find the inner radius (r), subtract the thickness from the outer radius:
Inner radius (r) \( = 5 - 1 = 4 \text{ cm} \).
The volume of a hollow sphere is given by the formula \( \frac { 4 }{ 3 }π (R^3 – r^3) \).
Volume \( = \frac { 4 }{ 3 } \times \frac { 22 }{ 7 } [(5)^3 – (4)^3] \)
\( = \frac { 88 }{ 21 } [125 – 64] \)
\( = \frac { 88 }{ 21 } \times 61 \text{ cm}^3 \)
\( = \frac { 5368 }{ 21 } \approx 255.619 \text{ cm}^3 \).
Rounding to one decimal place, the volume is 255.6 cm³.
In simple words: For a hollow ball, you find the volume of the whole ball and then subtract the volume of the empty space inside. This gives you the volume of the material itself.

🎯 Exam Tip: For hollow solids, remember to calculate both outer and inner radii correctly before applying the volume formula for the material (Outer Volume - Inner Volume).

Question 7. The surface areas of two spheres are in the ratio 4 : 9. Find the ratio of their volumes.
Answer:
Let \( r_1 \) and \( r_2 \) be the radii of the two spheres.
The surface area of the first sphere is \( 4\pi r_1^2 \).
The surface area of the second sphere is \( 4\pi r_2^2 \).
We are given that their surface areas are in the ratio 4 : 9.
So, \( 4\pi r_1^2 : 4\pi r_2^2 = 4 : 9 \).
This simplifies to \( r_1^2 : r_2^2 = 4 : 9 \).
Taking the square root of both sides gives the ratio of their radii:
\( \frac{r_1}{r_2} = \sqrt{\frac{4}{9}} = \frac{2}{3} \).
Now, let's find the ratio of their volumes. The volume of a sphere is \( \frac { 4 }{ 3 }πr^3 \).
Volume of first sphere \( = \frac { 4 }{ 3 }πr_1^3 \).
Volume of second sphere \( = \frac { 4 }{ 3 }πr_2^3 \).
The ratio of their volumes will be:
\( \frac { \frac { 4 }{ 3 }πr_1^3 }{ \frac { 4 }{ 3 }πr_2^3 } = \frac{r_1^3}{r_2^3} = \left(\frac{r_1}{r_2}\right)^3 \)
Since \( \frac{r_1}{r_2} = \frac{2}{3} \), the ratio of volumes is:
\( \left(\frac{2}{3}\right)^3 = \frac{2^3}{3^3} = \frac{8}{27} \).
Thus, the ratio of the volumes of the two spheres is 8 : 27.
In simple words: If you know how the surface areas of two spheres compare, you can find how their radii compare by taking the square root. Then, to compare their volumes, you cube the ratio of their radii.

🎯 Exam Tip: Remember the fundamental relationships: the ratio of surface areas is the square of the ratio of radii, and the ratio of volumes is the cube of the ratio of radii. This shortcut is very useful.

Question 8. The volumes of two spheres are in the ratio 64 : 27. Find their radii if the sum of their radii is 21 cm.
Answer:
Let \( r_1 \) and \( r_2 \) be the radii of the two spheres.
The volume of the first sphere is \( V_1 = \frac { 4 }{ 3 }πr_1^3 \).
The volume of the second sphere is \( V_2 = \frac { 4 }{ 3 }πr_2^3 \).
We are given that the ratio of their volumes is 64 : 27.
So, \( \frac { V_1 }{ V_2 } = \frac { \frac { 4 }{ 3 }πr_1^3 }{ \frac { 4 }{ 3 }πr_2^3 } = \frac{r_1^3}{r_2^3} = \frac{64}{27} \).
Taking the cube root of both sides:
\( \frac{r_1}{r_2} = \sqrt[3]{\frac{64}{27}} = \frac{4}{3} \).
So, we have the ratio \( r_1 : r_2 = 4 : 3 \).
We can write \( r_1 = 4k \) and \( r_2 = 3k \) for some constant \( k \).
We are also given that the sum of their radii is 21 cm:
\( r_1 + r_2 = 21 \)
Substitute the expressions in terms of \( k \):
\( 4k + 3k = 21 \)
\( 7k = 21 \)
\( k = \frac{21}{7} = 3 \).
Now, find the individual radii:
\( r_1 = 4k = 4 \times 3 = 12 \text{ cm} \).
\( r_2 = 3k = 3 \times 3 = 9 \text{ cm} \).
The radii of the two spheres are 12 cm and 9 cm.
In simple words: We are given how the volumes of two spheres compare and what their radii add up to. First, we use the volume ratio to find the ratio of their radii. Then, we use this ratio along with their sum to find each radius separately.

🎯 Exam Tip: When given a ratio and a sum, representing the quantities with a common multiplier (like \( 4k \) and \( 3k \)) is an efficient way to solve for the individual values.

Question 9. Find the cost of painting a hemispherical dome of diameter 10 m at the rate of Rs. 1.40 per m².
Answer:
The diameter of the hemispherical dome is 10 m.
So, the radius (r) \( = \frac { 10 }{ 2 } = 5 \text{ m} \).
The curved surface area of a hemisphere is given by the formula \( 2\pi r^2 \).
Curved surface area \( = 2 \times \frac { 22 }{ 7 } \times 5 \times 5 \text{ m}^2 = \frac { 1100 }{ 7 } \text{ m}^2 \).
The rate of painting is Rs. 1.40 per m².
To find the total cost of painting, multiply the area by the rate.
Total cost \( = \frac { 1100 }{ 7 } \times 1.40 \)
\( = \frac { 1100 \times 1.40 }{ 7 } \)
\( = 1100 \times 0.2 \)
\( = 220 \).
The total cost of painting the dome is Rs. 220.
In simple words: First, find the radius from the diameter. Then, calculate the curved area of the dome using the correct formula. Finally, multiply this area by the cost per square meter to get the total painting cost.

🎯 Exam Tip: Remember that a dome is typically a curved surface, so use the curved surface area formula (\( 2\pi r^2 \)) for a hemisphere, not the total surface area which includes the base.

Question 10. A certain spherical ball of diameter 4 cm weighs 8 kg. Find the weight of a spherical ball of the same material whose inner and outer diameters are 8 cm and 10 cm respectively.
Answer:
For the first ball:
Diameter = 4 cm, so radius (r) \( = \frac { 4 }{ 2 } = 2 \text{ cm} \).
Volume of this ball \( = \frac { 4 }{ 3 }πr³ = \frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times 2 \times 2 \times 2 \text{ cm}^3 = \frac { 704 }{ 21 } \text{ cm}^3 \).
Weight of this ball = 8 kg.
So, the weight per cm³ \( = \frac { 8 \text{ kg} }{ \frac { 704 }{ 21 } \text{ cm}^3 } = \frac { 8 \times 21 }{ 704 } \text{ kg/cm}^3 \).
\( = \frac { 168 }{ 704 } \text{ kg/cm}^3 \approx 0.2386 \text{ kg/cm}^3 \).
For the second, hollow spherical ball:
Outer diameter = 10 cm, so outer radius (R) \( = \frac { 10 }{ 2 } = 5 \text{ cm} \).
Inner diameter = 8 cm, so inner radius (r) \( = \frac { 8 }{ 2 } = 4 \text{ cm} \).
Volume of the material of this hollow ball \( = \frac { 4 }{ 3 }π (R^3 – r^3) \).
\( = \frac { 4 }{ 3 } \times \frac { 22 }{ 7 } [(5)^3 – (4)^3] \)
\( = \frac { 88 }{ 21 } [125 – 64] \)
\( = \frac { 88 }{ 21 } \times 61 \text{ cm}^3 \)
\( = \frac { 5368 }{ 21 } \text{ cm}^3 \).
Now, calculate the weight of this hollow ball using the density from the first ball.
Weight \( = \text{Volume of material} \times \text{Weight per unit volume} \)
Weight \( = \frac { 5368 }{ 21 } \text{ cm}^3 \times \frac { 168 }{ 704 } \text{ kg/cm}^3 \)
\( = \frac { 5368 \times 168 }{ 21 \times 704 } \text{ kg} \)
\( = \frac { 901824 }{ 14784 } \text{ kg} = 61 \text{ kg} \).
The hollow spherical ball weighs 61 kg.
In simple words: First, find out how much 1 cubic centimeter of the material weighs using the information from the small solid ball. Then, calculate the total volume of material in the hollow ball (outer volume minus inner volume) and multiply it by the weight per cubic centimeter to find its total weight.

🎯 Exam Tip: For problems involving weight and volume of the same material, always find the density (weight per unit volume) first, as it remains constant for the same material.

Question 11. Prove that the surface area of a sphere of diameter d is \( \pi d² \) and the volume is \( \frac { 1 }{ 6 }πd³ \).
(i) Prove that the surface area of a sphere of diameter d is \( \pi d² \) and the volume is \( \frac { 1 }{ 6 }πd³ \).
(ii) The volumes and diameters of a cone and sphere are equal. Prove that the height of the cone is twice the diameter of the sphere.
Answer:
(i) Diameter of a sphere = \( d \)
If the diameter is \( d \), then the radius (r) \( = \frac { d }{ 2 } \).
The formula for the surface area of a sphere is \( 4\pi r^2 \).
Substitute \( r = \frac{d}{2} \) into the formula:
Surface area \( = 4\pi \left(\frac{d}{2}\right)^2 = 4\pi \frac{d^2}{4} = \pi d^2 \).
This proves the surface area formula in terms of diameter.
The formula for the volume of a sphere is \( \frac { 4 }{ 3 }πr³ \).
Substitute \( r = \frac{d}{2} \) into the formula:
Volume \( = \frac { 4 }{ 3 }π \left(\frac{d}{2}\right)^3 = \frac { 4 }{ 3 }π \frac{d^3}{8} = \frac { 4πd^3 }{ 24 } = \frac { 1 }{ 6 }πd³ \).
This proves the volume formula in terms of diameter.
(ii) Let the diameter of the cone and sphere be \( D \).
So, the radius of the cone is \( r_{cone} = \frac{D}{2} \).
The radius of the sphere is \( r_{sphere} = \frac{D}{2} \).
Let \( h \) be the height of the cone.
The volume of the cone is \( V_{cone} = \frac { 1 }{ 3 }π(r_{cone})^2 h = \frac { 1 }{ 3 }π \left(\frac{D}{2}\right)^2 h \).
The volume of the sphere is \( V_{sphere} = \frac { 4 }{ 3 }π(r_{sphere})^3 = \frac { 4 }{ 3 }π \left(\frac{D}{2}\right)^3 \).
We are given that their volumes are equal:
\( \frac { 1 }{ 3 }π \left(\frac{D}{2}\right)^2 h = \frac { 4 }{ 3 }π \left(\frac{D}{2}\right)^3 \)
Cancel \( \frac { 1 }{ 3 }π \) and \( \left(\frac{D}{2}\right)^2 \) from both sides:
\( h = 4 \left(\frac{D}{2}\right) \)
\( \implies h = 2D \).
This means the height of the cone is twice the diameter of the sphere. This proves the statement.
In simple words: For part (i), we replace the radius in the normal sphere formulas with half the diameter to show the formulas in terms of diameter. For part (ii), we set the volume formulas for a cone and a sphere equal, using their diameters. We then simplify to show that the cone's height must be twice the sphere's diameter.

🎯 Exam Tip: When proving formulas, clearly state your substitutions and simplification steps. For comparative problems, writing down both formulas and setting up the equation is crucial.

Question 12. The external diameter of a hollow metal sphere is 12 cm, and its thickness is 2 cm. Find the radius of a solid sphere made of the same material and having the same weight as the hollow sphere.
Answer:
For the hollow metal sphere:
External diameter = 12 cm, so outer radius (R) \( = \frac { 12 }{ 2 } = 6 \text{ cm} \).
Thickness of the metal = 2 cm.
Internal radius (r) \( = \text{Outer radius} - \text{Thickness} = 6 - 2 = 4 \text{ cm} \).
Volume of the metal in the hollow sphere \( = \frac { 4 }{ 3 }π (R^3 – r^3) \).
\( = \frac { 4 }{ 3 }π [(6)^3 – (4)^3] \)
\( = \frac { 4 }{ 3 }π [216 – 64] \)
\( = \frac { 4 }{ 3 }π (152) \text{ cm}^3 \)
\( = \frac { 608 }{ 3 } π \text{ cm}^3 \).

For the solid sphere:
Let its radius be \( x \).
Volume of the solid sphere \( = \frac { 4 }{ 3 }πx³ \).
Since the solid sphere is made of the same material and has the same weight as the hollow sphere, their volumes must be equal.
\( \frac { 4 }{ 3 }πx³ = \frac { 608 }{ 3 } π \)
Cancel \( \frac { 4 }{ 3 }π \) from both sides:
\( x³ = \frac { 608 }{ 4 } \)
\( \implies x³ = 152 \).
To find \( x \), take the cube root of 152.
\( x = \sqrt[3]{152} \approx 5.3368 \text{ cm} \).
Rounding to two decimal places, the radius of the solid sphere is 5.34 cm.
In simple words: First, calculate how much metal is actually in the hollow sphere by finding its volume. Since the new solid sphere is made of the same metal and has the same weight, its volume must be equal to the volume of metal in the hollow sphere. Then, use this volume to find the radius of the solid sphere.

🎯 Exam Tip: Remember that "same weight" implies "same volume" if the material (and thus density) is the same. This is a key insight for solving such problems efficiently.

Question 13. A hollow copper ball has an external diameter of 12 cm, and a thickness of 0.1 cm. Find
(i) the outer surface area of the ball;
(ii) the weight of the ball if 1 cm³ of copper weighs 8.88 g. (Take \( \pi \) to be 3.14).
Answer:
The external diameter of the hollow spherical ball is 12 cm.
So, the outer radius (R) \( = \frac { 12 }{ 2 } = 6 \text{ cm} \).
The thickness of the metal used is 0.1 cm.
Internal radius (r) \( = \text{Outer radius} - \text{Thickness} = 6 - 0.1 = 5.9 \text{ cm} \).

(i) Outer surface area of the ball:
Formula for surface area is \( 4\pi R^2 \).
Outer surface area \( = 4 \times 3.14 \times (6)^2 \text{ cm}^2 \)
\( = 4 \times 3.14 \times 36 \)
\( = 452.16 \text{ cm}^2 \).
(ii) Volume of the metal used:
Formula for volume of hollow sphere material \( = \frac { 4 }{ 3 }π (R^3 – r^3) \).
Volume \( = \frac { 4 }{ 3 } \times 3.14 \times [(6)^3 – (5.9)^3] \text{ cm}^3 \)
\( = \frac { 4 }{ 3 } \times 3.14 \times [216 – 205.379] \text{ cm}^3 \)
\( = \frac { 4 }{ 3 } \times 3.14 \times (10.621) \text{ cm}^3 \)
\( \approx 44.467 \text{ cm}^3 \).
The weight of 1 cm³ of copper is 8.88 g.
Total weight of the ball \( = \text{Volume of metal} \times \text{Weight per cm}^3 \).
Total weight \( = 44.467 \times 8.88 \text{ g} \)
\( \approx 394.87 \text{ g} \).
In simple words: First, calculate the outer surface area using the outer radius. Then, find the volume of just the copper material by subtracting the inner empty space volume from the total outer volume. Finally, multiply this material volume by the given weight per cubic centimeter to get the total weight of the ball.

🎯 Exam Tip: Remember to use the correct radius (outer or inner) for each calculation. Use the given value of \( \pi \) (here, 3.14) if specified in the question, otherwise use \( \frac{22}{7} \).

Question 14. Marbles of diameter 1.4 cm, are dropped into a beaker containing some water and are placed into it. The diameter of the beaker is 7 cm. Find how many marbles have been dropped in it if the water rises by 5.6 cm. \( \left(\pi=\frac{22}{7}\right) \)
Answer:
For each marble:
Diameter = 1.4 cm, so radius (r) \( = \frac { 1.4 }{ 2 } = 0.7 \text{ cm} \).
Volume of one marble \( = \frac { 4 }{ 3 }πr³ = \frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times (0.7)^3 \text{ cm}^3 \)
\( = \frac { 88 }{ 21 } \times 0.343 \text{ cm}^3 \)
\( \approx 1.437333 \text{ cm}^3 \).
For the beaker (a cylinder):
Diameter of beaker = 7 cm, so radius of the base (r₁) \( = \frac { 7 }{ 2 } = 3.5 \text{ cm} \).
The water level rises by 5.6 cm. This rise in water level means the volume of displaced water is equal to the volume of the marbles.
Volume of water displaced \( = \pi r_1^2 h \), where \( h \) is the rise in water level.
Volume of water displaced \( = \frac { 22 }{ 7 } \times (3.5)^2 \times 5.6 \text{ cm}^3 \)
\( = \frac { 22 }{ 7 } \times 12.25 \times 5.6 \text{ cm}^3 \)
\( = 22 \times 1.75 \times 5.6 \text{ cm}^3 \)
\( = 215.6 \text{ cm}^3 \).
Number of marbles \( = \frac { \text{Volume of water displaced} }{ \text{Volume of one marble} } \)
\( = \frac { 215.6 \text{ cm}^3 }{ 1.437333 \text{ cm}^3 } \)
\( \approx 150 \).
Therefore, 150 marbles were dropped into the vessel.
In simple words: First, calculate the volume of one marble. Then, figure out the total volume of water that was pushed up in the beaker. Finally, divide the total volume of displaced water by the volume of one marble to find out how many marbles caused that rise.

🎯 Exam Tip: The volume of water displaced by submerged objects is always equal to the total volume of the objects themselves. This is a fundamental principle of buoyancy.

Question 15. A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed into the tub. If the radius of the hemisphere is 3.5 cm and height of the cone outside the hemisphere is 5 cm, find the volume of water left in the tub. (Take \( \pi = 22/7 \))
Answer:
For the cylindrical tub:
Radius (r) = 5 cm, height (h) = 9.8 cm.
Volume of water initially in the tub \( = \pi r^2 h \).
\( = \frac { 22 }{ 7 } \times 5 \times 5 \times 9.8 \text{ cm}^3 \)
\( = 22 \times 25 \times 1.4 \text{ cm}^3 \)
\( = 770 \text{ cm}^3 \).

For the solid (cone mounted on a hemisphere):
Radius of hemisphere (r₁) = 3.5 cm.
Height of cone (h₁) = 5 cm.
The total volume of the solid is the sum of the volume of the hemisphere and the volume of the cone.
Volume of hemisphere \( = \frac { 2 }{ 3 }πr_1^3 \).
Volume of cone \( = \frac { 1 }{ 3 }πr_1^2 h_1 \). (Note: The cone has the same base radius as the hemisphere.)
Volume of solid \( = \frac { 2 }{ 3 }πr_1^3 + \frac { 1 }{ 3 }πr_1^2 h_1 \)
\( = \frac { 1 }{ 3 }πr_1^2 (2r_1 + h_1) \)
\( = \frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times (3.5)^2 (2 \times 3.5 + 5) \)
\( = \frac { 22 }{ 21 } \times 12.25 (7 + 5) \)
\( = \frac { 22 }{ 21 } \times 12.25 \times 12 \)
\( = 22 \times 0.5833 \times 12 \)
\( \approx 154 \text{ cm}^3 \).

The volume of water left in the tub \( = \text{Initial volume of water} - \text{Volume of solid} \).
Volume left \( = 770 - 154 = 616 \text{ cm}^3 \).
In simple words: First, calculate the total water in the tub. Then, calculate the volume of the object being put in (a cone on top of a hemisphere). The amount of water left is the starting water volume minus the object's volume.

🎯 Exam Tip: Break down complex shapes into simpler geometric forms (like hemisphere and cone) to calculate their individual volumes, then sum them up for the total volume of the immersed object.

Question 16. 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.
Answer:
The total height of the toy is 15.5 cm.
The radius of the hemispherical part (r) = 3.5 cm.
The height of the conical part (h) \( = \text{Total height} - \text{Radius of hemisphere} \).
\( h = 15.5 - 3.5 = 12 \text{ cm} \).

Next, calculate the slant height (l) of the cone using the Pythagorean theorem.
\( l = \sqrt{r^2+h^2} = \sqrt{(3.5)^2+(12)^2} \)
\( = \sqrt{12.25+144} = \sqrt{156.25} \)
\( l = 12.5 \text{ cm} \).
The total surface area of the toy is the sum of the curved surface area of the cone and the curved surface area of the hemisphere.
Total surface area \( = \pi rl + 2\pi r^2 = \pi r(l + 2r) \).
\( = \frac { 22 }{ 7 } \times 3.5 (12.5 + 2 \times 3.5) \text{ cm}^2 \)
\( = 11 (12.5 + 7) \)
\( = 11 (19.5) \)
\( = 214.5 \text{ cm}^2 \).
In simple words: First, find the cone's height by subtracting the hemisphere's radius from the toy's total height. Then, calculate the slant height of the cone. Finally, add the curved surface area of the cone to the curved surface area of the hemisphere to get the toy's total outside area.

🎯 Exam Tip: When finding the surface area of composite solids, remember that the contact surface between the parts (e.g., the circular base of the cone and hemisphere) is *not* exposed and should not be included in the total surface area.

Question 17. The largest sphere is carved out of a cube of side 7 cm. Find the volume of the sphere.
Answer:
The side of the cube is 7 cm.
When the largest sphere is carved out of a cube, its diameter will be equal to the side of the cube.
So, the diameter of the sphere = 7 cm.
The radius of the sphere (r) \( = \frac { 7 }{ 2 } \text{ cm} \).

The volume of a sphere is given by the formula \( \frac { 4 }{ 3 }πr³ \).
Volume \( = \frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times \left(\frac{7}{2}\right)^3 \text{ cm}^3 \)
\( = \frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times \frac { 343 }{ 8 } \text{ cm}^3 \)
\( = \frac { 88 }{ 21 } \times \frac { 343 }{ 8 } \text{ cm}^3 \)
\( = \frac { 11 \times 49 }{ 3 } \text{ cm}^3 \)
\( = \frac { 539 }{ 3 } \text{ cm}^3 \)
\( \approx 179.666 \text{ cm}^3 \).
Rounding to one decimal place, the volume is 179.7 cm³.
In simple words: When a sphere is cut from a cube, its diameter will be the same as the cube's side length. Use this diameter to find the sphere's radius, and then calculate its volume.

🎯 Exam Tip: Always visualize the geometric relationship between the carved solid and its container to correctly determine the dimensions (e.g., diameter of sphere equals side of cube).

Question 18. A cylindrical bucket, whose base has a radius of 15 cm, is filled with water up to a height of 20 cm. A heavy iron spherical ball of a radius 10 cm is dropped to submerge completely in water in the bucket. Find the increase in the level of water.
Answer:
For the cylindrical bucket:
Radius of base (R) = 15 cm.
Initial height of water (H) = 20 cm.

For the spherical ball:
Radius (r) = 10 cm.
Volume of the spherical ball \( = \frac { 4 }{ 3 }πr³ = \frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times (10)^3 \text{ cm}^3 \)
\( = \frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times 1000 \text{ cm}^3 \)
\( = \frac { 88000 }{ 21 } \text{ cm}^3 \).
When the spherical ball is dropped into the bucket, it displaces a volume of water equal to its own volume. This displaced water causes the water level to rise.
Let the increase in water level be \( h \text{ cm} \).
The volume of water rise will be \( \pi R^2 h \) (cylindrical volume).
So, \( \pi R^2 h = \text{Volume of spherical ball} \).
\( \frac { 22 }{ 7 } \times (15)^2 \times h = \frac { 88000 }{ 21 } \)
\( \frac { 22 }{ 7 } \times 225 \times h = \frac { 88000 }{ 21 } \)
\( \frac { 4950 }{ 7 } h = \frac { 88000 }{ 21 } \)
To find \( h \), rearrange the equation:
\( h = \frac { 88000 }{ 21 } \times \frac { 7 }{ 4950 } \)
\( h = \frac { 88000 }{ 3 \times 4950 } \)
\( h = \frac { 8800 }{ 1485 } \)
\( h = \frac { 1760 }{ 297 } \)
\( h \approx 5.9259 \text{ cm} \).
As a mixed fraction, \( h = 5\frac { 25 }{ 27 } \text{ cm} \).
The increase in water level is \( 5\frac { 25 }{ 27 } \text{ cm} \).
In simple words: When a ball is dropped into water, the water level rises because the ball pushes water out of its way. The amount of water pushed out is the same as the ball's volume. We use this displaced volume and the bucket's base area to figure out how much the water level went up.

🎯 Exam Tip: Remember that the volume of a submerged object is equal to the volume of the liquid it displaces. This displaced volume, when contained in a cylindrical vessel, forms a cylinder with the same base area and an increased height.

Question 19. The radius of the base of a cone and the radius of a sphere are the same, each being 8 cm. Given that the volumes of these two solids are also the same, calculate the slant height of the cone, correct to one place of decimal.
Answer:
For the cone and the sphere:
Radius (r) = 8 cm.
The volume of the sphere \( = \frac { 4 }{ 3 }πr³ \).
The volume of the cone \( = \frac { 1 }{ 3 }πr²h \), where \( h \) is the height of the cone.
We are given that their volumes are equal:
\( \frac { 1 }{ 3 }πr²h = \frac { 4 }{ 3 }πr³ \)
Cancel \( \frac { 1 }{ 3 }πr² \) from both sides:
\( h = 4r \).
Substitute \( r = 8 \text{ cm} \):
\( h = 4 \times 8 = 32 \text{ cm} \).
Now, we need to find the slant height (l) of the cone. The formula is \( l = \sqrt{r^2+h^2} \).
\( l = \sqrt{(8)^2+(32)^2} \)
\( = \sqrt{64+1024} \)
\( = \sqrt{1088} \)
\( l \approx 32.9848 \text{ cm} \).
Rounding to one decimal place, the slant height is 33.0 cm.
In simple words: We are told that a cone and a sphere have the same radius and the same volume. First, we use their volume formulas to find the cone's height. Once we have the cone's height and radius, we use the Pythagorean theorem to calculate its slant height.

🎯 Exam Tip: When given equal volumes or surface areas, set up the equations for both shapes and solve for the unknown dimension. For a cone, remember that the slant height involves the radius and vertical height.

Question 20. 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 rim. 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. \( \left(\text { Take } \pi \text { to be } \frac{22}{7}\right) \)
Answer:
For the conical vessel:
Radius (r) = 5 cm, height (h) = 8 cm.
Volume of water filled in it (which is the volume of the cone) \( = \frac { 1 }{ 3 }πr²h \).
\( = \frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 5 \times 5 \times 8 \text{ cm}^3 \)
\( = \frac { 4400 }{ 21 } \text{ cm}^3 \).

When lead shots are dropped, one-fourth of the water flows out.
Volume of water that flowed out \( = \frac { 1 }{ 4 } \times \text{Volume of cone} \)
\( = \frac { 1 }{ 4 } \times \frac { 4400 }{ 21 } \text{ cm}^3 = \frac { 1100 }{ 21 } \text{ cm}^3 \).
This volume of water that flowed out is equal to the total volume of the lead shots dropped.
For each lead shot (sphere):
Radius (r₁) = 0.5 cm.
Volume of one lead shot \( = \frac { 4 }{ 3 }πr_1^3 \)
\( = \frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times (0.5)^3 \text{ cm}^3 \)
\( = \frac { 88 }{ 21 } \times 0.125 \text{ cm}^3 \)
\( = \frac { 11 }{ 21 } \text{ cm}^3 \).
Number of lead shots \( = \frac { \text{Total volume of lead shots} }{ \text{Volume of one lead shot} } \)
\( = \frac { \frac { 1100 }{ 21 } \text{ cm}^3 }{ \frac { 11 }{ 21 } \text{ cm}^3 } \)
\( = \frac { 1100 }{ 11 } = 100 \).
Therefore, 100 lead shots were dropped in the vessel.
In simple words: First, find the total volume of water in the cone. Then, calculate one-fourth of this volume, which is the amount of water that spilled out. This spilled volume is equal to the total volume of all the lead shots. Finally, calculate the volume of one lead shot and divide the total volume of spilled water by the volume of one shot to find the number of shots.

🎯 Exam Tip: The volume of water that overflows from a vessel when objects are immersed is exactly equal to the volume of the immersed objects. This is Archimedes' principle in action.

Question 21. A buoy is made in the form of a hemisphere surmounted by a right cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 metres and its volume is \( \frac { 2 }{ 3 } \) of the hemisphere. Calculate the height of the cone and the surface of the buoy correct to 2 places of decimals. \( \left(\text { Take } \pi=3 \frac{1}{7}\right) \)
Answer:
For the buoy:
Radius of cone (r) = 3.5 m.
This is also the radius of the hemisphere (r) = 3.5 m.
We are given \( \pi = 3\frac{1}{7} = \frac{22}{7} \).

(i) Height of the cone:
Volume of hemisphere \( = \frac { 2 }{ 3 }πr³ = \frac { 2 }{ 3 } \times \frac { 22 }{ 7 } \times (3.5)^3 \text{ m}^3 \).
\( = \frac { 44 }{ 21 } \times 42.875 \text{ m}^3 \)
\( = \frac { 539 }{ 6 } \text{ m}^3 \).
Volume of cone \( = \frac { 1 }{ 3 }πr^2 h \).
We are given that the volume of the cone is \( \frac { 2 }{ 3 } \) of the volume of the hemisphere.
So, \( \frac { 1 }{ 3 }πr^2 h = \frac { 2 }{ 3 } \times \frac { 539 }{ 6 } \).
\( \frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times (3.5)^2 h = \frac { 539 }{ 9 } \).
\( \frac { 22 }{ 21 } \times 12.25 h = \frac { 539 }{ 9 } \).
\( \frac { 22 \times 12.25 }{ 21 } h = \frac { 539 }{ 9 } \).
\( \frac { 269.5 }{ 21 } h = \frac { 539 }{ 9 } \).
\( 12.8333 h = 59.8889 \).
\( h = \frac { 539 }{ 9 } \times \frac { 21 }{ 269.5 } \)
\( h = \frac { 539 \times 7 }{ 3 \times 269.5 } = \frac { 3773 }{ 808.5 } \)
\( h = \frac { 14 }{ 3 } \text{ m} \approx 4.666 \text{ m} \).
So, the height of the cone is \( 4\frac { 2 }{ 3 } \text{ m} \approx 4.67 \text{ m} \).
(ii) Surface area of the buoy:
First, find the slant height (l) of the cone.
\( l = \sqrt{r^2+h^2} = \sqrt{(3.5)^2+\left(\frac{14}{3}\right)^2} \)
\( = \sqrt{\left(\frac{7}{2}\right)^2+\left(\frac{14}{3}\right)^2} \)
\( = \sqrt{\frac{49}{4}+\frac{196}{9}} \)
\( = \sqrt{\frac{441+784}{36}} = \sqrt{\frac{1225}{36}} \)
\( l = \frac{35}{6} \text{ m} \).
The surface area of the buoy is the sum of the curved surface area of the cone (\( \pi rl \)) and the curved surface area of the hemisphere (\( 2\pi r^2 \)).
Surface area \( = \pi rl + 2\pi r^2 = \pi r(l + 2r) \).
\( = \frac { 22 }{ 7 } \times 3.5 \left(\frac{35}{6} + 2 \times 3.5\right) \)
\( = 11 \left(\frac{35}{6} + 7\right) \)
\( = 11 \left(\frac{35+42}{6}\right) \)
\( = 11 \times \frac{77}{6} \)
\( = \frac{847}{6} \text{ m}^2 \)
\( \approx 141.166 \text{ m}^2 \).
Rounding to two decimal places, the surface area of the buoy is 141.17 m².
In simple words: For the buoy, we first use the given ratio of cone volume to hemisphere volume to calculate the height of the cone. Then, using this height and the radius, we find the cone's slant height. Finally, we add the curved surface areas of the cone and the hemisphere to get the total surface area of the buoy.

🎯 Exam Tip: When dealing with composite figures, clearly identify each component shape and its dimensions. Calculate intermediate values like cone height or slant height carefully before combining them for the final surface area.

Question 22. A cylinder, whose height is equal to its diameter, has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder, correct to 1 decimal place.
Answer: First, we are given that the height of the cylinder is equal to its diameter. This means if the cylinder's radius is \( r_c \), its height \( h \) will be \( 2r_c \). We are also told that the volume of this cylinder is the same as the volume of a sphere with a radius of 4 cm. The volume of a sphere is found using the formula \( \frac{4}{3}\pi r^3 \).
So, for the sphere:
Radius \( r_s = 4 \text{ cm} \)
Volume of sphere \( = \frac{4}{3}\pi (4)^3 \)
\( = \frac{4}{3}\pi (64) \)
\( = \frac{256}{3}\pi \text{ cm}^3 \)
Next, for the cylinder:
Let its radius be \( r_c \).
Height \( h = 2r_c \)
Volume of cylinder \( = \pi r_c^2 h \)
Substitute \( h = 2r_c \) into the cylinder's volume formula:
Volume of cylinder \( = \pi r_c^2 (2r_c) \)
\( = 2\pi r_c^3 \)
Now, we know that the volume of the cylinder is equal to the volume of the sphere:
\( 2\pi r_c^3 = \frac{256}{3}\pi \)
To find \( r_c \), we can divide both sides by \( 2\pi \):
\( r_c^3 = \frac{256}{3 \times 2} \)
\( r_c^3 = \frac{128}{3} \)
\( r_c^3 = 42.667 \)

\( \implies \) \( r_c = \sqrt[3]{42.667} \)
\( r_c \approx 3.49 \text{ cm} \)
Rounding this to one decimal place, the radius of the base of the cylinder is approximately 3.5 cm.
In simple words: We find how much space the sphere takes up. Then we make the cylinder take up the same amount of space, using the rule that its height is twice its radius. From this, we calculate the cylinder's radius and round it to one decimal place.

🎯 Exam Tip: Remember to express the cylinder's height in terms of its radius when its height equals its diameter. Always equate the volumes correctly and perform cubic root calculations carefully. Rounding to the specified decimal place is crucial for the final answer.