Maharashtra Board Class 10 Maths Chapter 7 Mensuration Set 7.1 Solutions

Question 1. Find the volume of a cone if the radius of its base is 1.5 cm and its perpendicular height is 5 cm.
Answer:
Given: For the cone,
radius (r) = 1.5 cm,
perpendicular height (h) = 5 cm
To find: Volume of the cone.
Solution:
Volume of cone = \( \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \times \frac{22}{7} \times (1.5)^2 \times 5 \)
\( = \frac{1}{3} \times \frac{22}{7} \times 1.5 \times 1.5 \times 5 \)
\( = \frac{22}{7} \times 0.5 \times 1.5 \times 5 \)
\( = 11.785 \) cm\(^3\)
\( \approx 11.79 \) cm\(^3\)
\( \therefore \) The volume of the cone is 11.79 cm\(^3\).
In simple words: To find the volume of the cone, we use the formula \( \frac{1}{3} \pi r^2 h \). By substituting the given radius (r = 1.5 cm) and height (h = 5 cm) into the formula, we calculate the volume to be approximately 11.79 cubic centimeters.

🎯 Exam Tip: Remember the formula for the volume of a cone \( (\frac{1}{3} \pi r^2 h) \) and ensure accurate substitution and calculation for full marks.

Question 2. Find the volume of a sphere of diameter 6 cm. [\( \pi = 3.14 \)]
Answer:
Given: For the sphere, diameter (d) = 6 cm
To find: Volume of the sphere.
Solution:
Radius (r) \( = \frac{d}{2} = \frac{6}{2} = 3 \) cm
Volume of sphere \( = \frac{4}{3} \pi r^3 \)
\( = \frac{4}{3} \times 3.14 \times (3)^3 \)
\( = 4 \times 3.14 \times 3 \times 3 \)
\( = 113.04 \) cm\(^3\)
\( \therefore \) The volume of the sphere is 113.04 cm\(^3\).
In simple words: To calculate the volume of a sphere, we first determine the radius from the given diameter. Then, we apply the volume formula for a sphere, \( \frac{4}{3} \pi r^3 \), substituting the radius and the value of pi to get the final volume.

🎯 Exam Tip: Convert diameter to radius correctly before applying the volume of sphere formula. Pay attention to the given value of \( \pi \) if specified.

Question 3. Find the total surface area of a cylinder if the radius of its base is 5 cm and height is 40 cm. [\( \pi = 3.14 \)]
Answer:
Given: For the cylinder,
radius (r) = 5 cm,
height (h) = 40 cm
To find: Total surface area of the cylinder.
Solution:
Total surface area of cylinder \( = 2 \pi r (r + h) \)
\( = 2 \times 3.14 \times 5 (5 + 40) \)
\( = 2 \times 3.14 \times 5 \times 45 \)
\( = 1413 \) cm\(^2\)
The total surface area of the cylinder is 1413 cm\(^2\).
In simple words: The total surface area of a cylinder is found using the formula \( 2 \pi r (r + h) \). We substitute the given radius and height into this formula and perform the calculation to get the surface area in square centimeters.

🎯 Exam Tip: Correctly identify the formula for the total surface area of a cylinder. Be careful with order of operations and calculations involving \( \pi \).

Question 4. Find the surface area of a sphere of radius 7 cm.
Answer:
Given: For the sphere, radius (r) = 7 cm
To find: Surface area of the sphere.
Solution:
Surface area of sphere \( = 4 \pi r^2 \)
\( = 4 \times \frac{22}{7} \times (7)^2 \)
\( = 88 \times 7 \)
\( = 616 \) cm\(^2\)
\( \therefore \) The surface area of the sphere is 616 cm\(^2\).
In simple words: To find the surface area of a sphere, we use the formula \( 4 \pi r^2 \). By plugging in the given radius (r = 7 cm) and simplifying with the value of \( \pi \), we calculate the total surface area.

🎯 Exam Tip: Remember the formula for the surface area of a sphere. Ensure correct substitution of the radius and perform multiplication accurately.

Question 5. The dimensions of a cuboid are 44 cm, 21 cm, 12 cm. It is melted and a cone of height 24 cm is made. Find the radius of its base.
Answer:
Given: For the cuboid,
length (l) = 44 cm, breadth (b) = 21 cm,
height (h) = 12 cm
For the cone, height (H) = 24 cm
To find: Radius of base of the cone (r).
Solution:
Volume of cuboid = \( l \times b \times h \)
\( = 44 \times 21 \times 12 \) cm\(^3\)
Volume of cone \( = \frac{1}{3} \pi r^2 H \)
\( = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 24 \) cm\(^3\)
Since the cuboid is melted to form a cone,
\( \therefore \) volume of cuboid = volume of cone

\( \therefore 44 \times 21 \times 12 = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 24 \)

\( \therefore r^2 = \frac{44 \times 21 \times 12 \times 3 \times 7}{22 \times 24} \)
\( \therefore r^2 = 21 \times 21 \)
\( \therefore r = 21 \) cm ... [Taking square root of both sides]
\( \therefore \) The radius of the base of the cone is 21 cm.
In simple words: When a cuboid is melted and reshaped into a cone, their volumes remain equal. We calculate the cuboid's volume and equate it to the cone's volume formula, then solve for the unknown radius of the cone's base.

🎯 Exam Tip: The principle of conservation of volume is crucial here. Ensure you correctly set up the equation equating the volumes of the cuboid and the cone and solve for 'r' carefully.

Question 6. Observe the measures of pots in the given figures. How many jugs of water can the cylindrical pot hold?

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो बर्तनों को दर्शाता है: एक शंकु के आकार का पानी का जग और एक बेलनाकार पानी का बर्तन। शंकु के जग में आधार की त्रिज्या 3.5 सेमी और ऊंचाई 10 सेमी है, जबकि बेलनाकार बर्तन में आधार की त्रिज्या 7 सेमी और ऊंचाई 10 सेमी है। यह चित्रण दोनों बर्तनों के सापेक्ष आकार और आयामों को दर्शाता है।
Answer:
Given: For the conical water jug,
radius (r) = 3.5 cm, height (h) = 10 cm
For the cylindrical water pot,
radius (R) = 7 cm, height (H) = 10 cm
To find: Number of jugs of water the cylindrical pot can hold.
Solution:
Volume of conical jug \( = \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \times \pi \times 3.5^2 \times 10 \)
\( = \frac{1}{3} \times 3.5^2 \times 10 \pi \) cm\(^3\)
Volume of cylindrical pot \( = \pi R^2 H \)
\( = \pi \times 7^2 \times 10 \)
\( = 49 \times 10 \pi \) cm\(^3\)
Number of jugs \( = \frac{\text{Volume of cylindrical pot}}{\text{Volume of conical jug}} \)

\( = \frac{49 \times 10 \pi}{\frac{1}{3} \times 3.5^2 \times 10 \pi} \)

\( = \frac{49 \times 3}{3.5 \times 3.5} \)

\( = \frac{49 \times 3 \times 100}{35 \times 35} \)
\( = 12 \)
\( \therefore \) The cylindrical pot can hold 12 jugs of water.
In simple words: To find how many conical jugs a cylindrical pot can hold, we first calculate the volume of both the conical jug and the cylindrical pot. Then, we divide the volume of the cylindrical pot by the volume of the conical jug to get the number of jugs.

🎯 Exam Tip: Calculate the volumes of both shapes accurately using their respective formulas. Ensure to use consistent units and simplify fractions carefully when finding the ratio.

Question 7. A cylinder and a cone have equal bases. The height of the cylinder is 3 cm and the area of its base is 100 cm\(^2\). The cone is placed up on the cylinder. Volume of the solid figure so formed is 500 cm\(^3\). Find the total height of the figure

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक संयुक्त ठोस आकृति को दर्शाता है जो एक बेलन के ऊपर रखे गए एक शंकु से बनी है। बेलन का निचला भाग और शंकु का आधार समान त्रिज्या साझा करते हैं। बेलन की ऊंचाई 'h' और शंकु की ऊंचाई 'H' के रूप में दिखाई गई है, जो इस समग्र आकृति की कुल ऊंचाई निर्धारित करते हैं।
Answer:
Given: For the cylindrical part,
height (h) = 3 cm,
area of the base (\( \pi r^2 \)) = 100 cm\(^2\)
Volume of the entire figure = 500 cm\(^3\)
To find: Total height of the figure.
Solution:
A cylinder and a cone have equal bases.
\( \therefore \) they have equal radii.
radius of cylinder = radius of cone = r
Area of base = 100 cm\(^2\)
\( \therefore \pi r^2 = 100 \) ...(i)
Let the height of the conical part be H.
Volume of the entire figure
= Volume of the cylinder + Volume of cone

\( \therefore 500 = \pi r^2 h + \frac{1}{3} \pi r^2 H \)

\( \therefore 500 = \pi r^2 (h + \frac{H}{3}) \)

\( \therefore 500 = 100 (3 + \frac{H}{3}) \) ...[ From (i)]

\( \therefore 3 + \frac{H}{3} = \frac{500}{100} \)

\( \therefore 3 + \frac{H}{3} = 5 \)

\( \therefore \frac{H}{3} = 5 - 3 \)

\( \therefore \frac{H}{3} = 2 \)

\( \therefore H = 6 \) cm
\( \therefore \) Height of conical part (H) = 6 cm
Total height of the figure = h + H
= 3 + 6
= 9 cm
\( \therefore \) The total height of the figure is 9 cm.
In simple words: The total height of the figure is the sum of the height of the cylindrical part and the height of the conical part.

🎯 Exam Tip: The total height is simply the sum of individual heights when figures are stacked. Double-check your arithmetic for the final sum.

Question 8. In the given figure, a toy made from a hemisphere, a cylinder and a cone is shown. Find the total area of the toy.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक खिलौने को दर्शाता है जो एक अर्धगोले, एक बेलन और एक शंकु को जोड़कर बनाया गया है। शंकु शीर्ष पर है, बेलन बीच में है, और अर्धगोला आधार पर है। शंकु की ऊंचाई 4 सेमी, बेलन की ऊंचाई 40 सेमी है, और सभी तीनों भागों की त्रिज्या 3 सेमी है। यह चित्रण खिलौने की समग्र संरचना और उसके आयामों को दर्शाता है।
Answer:
Given: For the conical Part,
height (h) = 4 cm, radius (r) = 3 cm
For the cylindrical part,
height (H) = 40 cm, radius (r) = 3 cm
For the hemispherical part,
radius (r) = 3 cm
To find: Total area of the toy.
Solution:
Slant height of cone (l) \( = \sqrt{h^2 + r^2} \)
\( = \sqrt{4^2 + 3^2} \)
\( = \sqrt{16 + 9} \)
\( = \sqrt{25} = 5 \) cm
Curved surface area of cone \( = \pi r l \)
\( = \pi \times 3 \times 5 \)
\( = 15 \pi \) cm\(^2\)
Curved surface area of cylinder \( = 2 \pi r H \)
\( = 2 \times \pi \times 3 \times 40 \)
\( = 240 \pi \) cm\(^2\)
Curved surface area of hemisphere \( = 2 \pi r^2 \)
\( = 2 \times \pi \times 3^2 \)
\( = 18 \pi \) cm\(^2\)
Total area of the toy
= Curved surface area of cone + Curved surface area of cylinder + Curved surface area of hemisphere
\( = 15 \pi + 240 \pi + 18 \pi \)
\( = 273 \pi \) cm\(^2\)
\( \therefore \) The total area of the toy is 273\( \pi \) cm\(^2\).
In simple words: The total surface area of this composite toy is the sum of the curved surface areas of its three individual parts: the cone, the cylinder, and the hemisphere. First, calculate the slant height of the cone, then find the curved surface area for each component, and add them up.

🎯 Exam Tip: For composite solids, the total surface area is usually the sum of the *exposed* surface areas of its parts. Don't include areas where parts join. Use the correct formulas for each shape's curved surface area.

Question 9. In the given figure, a cylindrical wrapper of flat tablets is shown. The radius of a tablet is 7 mm and its thickness is 5 mm. How many such tablets are wrapped in the wrapper?

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक बेलनाकार रैपर को दर्शाता है जिसमें गोलियाँ पैक की गई हैं। रैपर की ऊंचाई 10 सेमी और व्यास 14 मिमी है। यह चित्रण गोलियों के आकार और रैपर के अंदर उनकी व्यवस्था को दर्शाता है, जिससे यह समझने में मदद मिलती है कि कितनी गोलियां रैपर के भीतर समा सकती हैं।
Answer:
Given: For the cylindrical tablets,
radius (r) = 7 mm,
thickness = height(h) = 5 mm
For the cylindrical wrapper,
diameter (D) = 14 mm, height (H) = 10 cm
To find: Number of tablets that can be wrapped.
Solution:
Radius of wrapper (R) \( = \frac{\text{Diameter}}{2} = \frac{14}{2} = 7 \) mm
Height of wrapper (H) = 10 cm
\( = 10 \times 10 \) mm
\( = 100 \) mm
Volume of a cylindrical wrapper \( = \pi R^2 H \)
\( = \pi (7)^2 \times 100 \)
\( = 4900 \pi \) mm\(^3\)
Volume of a cylindrical tablet \( = \pi r^2 h \)
\( = \pi (7)^2 \times 5 \)
\( = 245 \pi \) mm\(^3\)
No. of tablets that can be wrapped \( = \frac{\text{volume of a cylindrical wrapper}}{\text{volume of a cylindrical tablet}} \)

\( = \frac{4900 \pi}{245 \pi} \)
\( = 20 \).
\( \therefore \) 20 tablets can be wrapped in the wrapper.
In simple words: To find the number of tablets that fit in the wrapper, we calculate the volume of both the wrapper and a single tablet. Ensure all units are consistent (convert cm to mm). Then, divide the total volume of the wrapper by the volume of one tablet.

🎯 Exam Tip: Always check and convert units to be consistent before starting calculations. In this case, convert cm to mm for the wrapper's height to match the tablet's dimensions.

Question 10. The given figure shows a toy. Its lower part is a hemisphere and the upper part is a cone. Find the volume and the surface area of the toy from the measures shown in the figure.
(\( \pi = 3.14 \))

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक खिलौने को दर्शाता है जो एक शंकु और एक अर्धगोले के संयोजन से बना है। शंकु ऊपरी भाग है जिसकी ऊंचाई 4 सेमी और त्रिज्या 3 सेमी है। अर्धगोला निचला भाग है जिसकी त्रिज्या 3 सेमी है। यह चित्रण खिलौने की समग्र ज्यामितीय संरचना और उसके आयामों को दर्शाता है।
Answer:
Given: For the conical part,
height (h) = 4 cm, radius (r) = 3 cm
For the hemispherical part,
radius (r) = 3 cm
To find: Volume and surface area of the toy.
Solution:
Now, volume of the toy
= Volume of cone + volume of hemisphere
\( = \frac{1}{3} \pi r^2 h + \frac{2}{3} \pi r^3 \)
\( = 12 \pi + 18 \pi \)
\( = 30 \pi \)
\( = 30 \times 3.14 \)
\( = 94.20 \) cm\(^3\)
Also, surface area of the toy
= Curved surface area of cone + Curved surface area of hemisphere
\( = 15 \pi + 18 \pi \)
\( = 33 \pi \)
\( = 33 \times 3.14 \)
\( = 103.62 \) cm\(^2\)
\( \therefore \) The volume and surface area of the toy are 94.20 cm\(^3\) and 103.62 cm\(^2\) respectively.
In simple words: To find the total volume and surface area of this toy, calculate the volume and curved surface area of the cone and the hemisphere separately. Then, sum these individual values to get the total volume and total curved surface area of the combined figure.

🎯 Exam Tip: Remember to calculate both volume and surface area. For surface area, only consider the exposed surfaces, not the joining base. Ensure to use the given \( \pi \) value if specified and calculate cone's slant height if needed.

Question 11. Find the surface area and the volume of a beach ball shown in the figure.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक गोलाकार बीच बॉल को दर्शाता है। बॉल का व्यास 42 सेमी है, जो इसकी समग्र माप को इंगित करता है। यह चित्रण बॉल के आकार को स्पष्ट रूप से प्रस्तुत करता है, जिससे छात्र इसके आयामों को आसानी से समझ सकें।
Answer:
Given: For the spherical ball,
diameter (d) = 42 cm
To find: Surface area and volume of the beach ball.
Solution:
Radius (r) \( = \frac{d}{2} = \frac{42}{2} = 21 \) cm
Surface area of sphere \( = 4 \pi r^2 \)
\( = 4 \times 3.14 \times (21)^2 \)
\( = 4 \times 3.14 \times 21 \times 21 \)
\( = 5538.96 \) cm\(^2\)
Volume of sphere \( = \frac{4}{3} \pi r^3 \)
\( = \frac{4}{3} \times 3.14 \times (21)^3 \)
\( = 4 \times 3.14 \times 7 \times 21 \times 21 \)
\( = 38772.72 \) cm\(^3\)
\( \therefore \) The surface area and the volume of the beach ball are 5538.96 cm\(^2\) and 38772.72 cm\(^3\) respectively.
In simple words: For a spherical beach ball, first calculate the radius from the given diameter. Then, use the formulas for the surface area \( (4 \pi r^2) \) and volume \( (\frac{4}{3} \pi r^3) \) of a sphere, substituting the radius to find both values.

🎯 Exam Tip: Remember to calculate both surface area and volume. Ensure you use the correct formulas for a sphere and convert diameter to radius accurately for calculations.

Question 12. As shown in the figure, a cylindrical glass contains water. A metal sphere of diameter 2 cm is immersed in it. Find the volume of the water.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक बेलनाकार गिलास को दर्शाता है जिसमें पानी भरा है, और उसके अंदर एक धातु का गोला डूबा हुआ है। गिलास की ऊंचाई 30 सेमी और व्यास 14 सेमी है। धातु के गोले का व्यास 2 सेमी है। यह चित्रण पानी में डूबे हुए गोले की स्थिति और संबंधित आयामों को स्पष्ट करता है, जिससे पानी के आयतन की गणना की जा सके।
Answer:
Given: For the metal sphere,
diameter (d) = 2 cm
For the cylindrical glass, diameter (D) = 14 cm,
height of water in the glass (H) = 30 cm
To find: Volume of water in the glass.
Solution:
Let the radii of the sphere and glass be r and R respectively.
Radius of sphere (r) \( = \frac{d}{2} = \frac{2}{2} = 1 \) cm
Radius of glass (R) \( = \frac{D}{2} = \frac{14}{2} = 7 \) cm
Now, Volume of sphere \( = \frac{4}{3} \pi r^3 \)
\( = \frac{4}{3} \pi (1)^3 \)
\( = \frac{4}{3} \pi \) cm\(^3\)
Volume of water with sphere in it \( = \pi R^2 H \)
\( = \pi \times (7)^2 \times 30 \)
\( = 1470 \pi \) cm\(^3\)
Volume of water in the glass
= Volume of water with sphere in it – Volume of sphere

\( = 1470 \pi - \frac{4}{3} \pi \)

\( = \frac{4410 \pi - 4 \pi}{3} \)

\( = \frac{4406 \pi}{3} \)
\( = 1468.67 \pi \) cm\(^3\)
(OR)
\( = \frac{4406}{3} \times \frac{22}{7} \)

\( = \frac{96932}{21} \)
\( = 4615.80 \) cm\(^3\)
\( \therefore \) The volume of the water in the glass is 1468.67 \( \pi \) cm\(^3\) (i.e. 4615.80 cm\(^3\)).
In simple words: When a sphere is immersed in water within a cylindrical glass, the volume of the water displaced is equal to the volume of the sphere. To find the actual volume of water, calculate the total volume of water and the sphere, then subtract the sphere's volume.

🎯 Exam Tip: The volume of water remaining in the glass after immersion is the initial volume of water minus the volume of the sphere. Ensure correct calculation of radii and careful subtraction of volumes.

Question 1. The length, breadth and height of an oil can are 20 cm, 20 cm and 30 cm respectively as shown in the adjacent figure. How much oil will it contain? (1 litre = 1000 cm\(^3\)) (Textbook pg. no.141)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक घनाकार तेल के डिब्बे को दर्शाता है। डिब्बे की लंबाई 20 सेमी, चौड़ाई 20 सेमी और ऊंचाई 30 सेमी है। यह चित्रण डिब्बे के आयामों को स्पष्ट रूप से प्रस्तुत करता है, जिससे उसमें समा सकने वाले तेल की मात्रा की गणना की जा सके।
Answer:
Given: For the cuboidal can,
length (l) = 20 cm,
breadth (b) = 20 cm,
height (h) = 30 cm
To find: Oil that can be contained in the can.
Solution:
Volume of cuboid = \( l \times b \times h \)
\( = 20 \times 20 \times 30 \)
\( = 12000 \) cm\(^3\)
\( = \frac{12000}{1000} \) litres
\( = 12 \) litres
\( \therefore \) The oil can will contain 12 litres of oil.
In simple words: To find the capacity of the oil can, calculate its volume using the formula for a cuboid (\( l \times b \times h \)). Then, convert the volume from cubic centimeters to liters using the conversion factor (1 liter = 1000 cm\(^3\)).

🎯 Exam Tip: Remember the formula for the volume of a cuboid. Don't forget to convert the volume from cm\(^3\) to litres as requested, using the correct conversion factor.

Question 2. The adjoining figure shows the measures of a Joker's cap. How much cloth is needed to make such a cap? (Textbook pg. no. 141)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक जोकर की टोपी को दर्शाता है जो शंकु के आकार की है। टोपी की आधार त्रिज्या 10 सेमी और तिर्यक ऊंचाई (slant height) 21 सेमी है। यह चित्रण टोपी के आयामों को स्पष्ट रूप से प्रस्तुत करता है, जिससे उसे बनाने के लिए आवश्यक कपड़े की मात्रा की गणना की जा सके।
Answer:
Given: For the conical cap,
radius (r) = 10 cm,
slant height (l) = 21 cm
To find: Cloth required to make the cap.
Solution:
Cloth required to make the cap
= Curved surface area of the conical cap
\( = \pi r l = \frac{22}{7} \times 10 \times 21 \)
\( = 22 \times 10 \times 3 \)
\( = 660 \) cm\(^2\)
\( \therefore 660 \) cm\(^2\) of cloth will be required to make the cap.
In simple words: To determine the amount of cloth needed for a conical cap, we calculate its curved surface area using the formula \( \pi r l \). Substitute the given radius and slant height to find the total area of the material.

🎯 Exam Tip: For objects like a cap (open at the base), only the curved surface area is required, not the total surface area. Ensure you have the correct formula and accurately calculate the slant height if not directly provided.

Question 3. As shown in the adjacent figure, a sphere is placed in a cylinder. It touches the top, bottom and the curved surface of the cylinder. If radius of the base of the cylinder is 'r',
(i) what is the ratio of the radii of the sphere and the cylinder ?
(ii) what is the ratio of the curved surface area of the cylinder and the surface area of the sphere?
(iii) what is the ratio of the volumes of the cylinder and the sphere? (Textbook pg. no. 141)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक बेलन के अंदर एक गोले को दर्शाता है। गोला बेलन के शीर्ष, तल और वक्र सतह को छूता है। इसका अर्थ है कि गोले का व्यास बेलन की ऊंचाई के बराबर है, और गोले की त्रिज्या बेलन के आधार की त्रिज्या के बराबर है। यह ज्यामितीय संबंधों को स्पष्ट करता है।
Answer:
Solution:
Radius of base of cylinder = radius of sphere
\( \therefore \) Radius of sphere = r
Also, height of cylinder = diameter of sphere
\( \therefore h = d \)
\( \therefore h = 2r \) ...(i)
(i) Radius of sphere \( / \) Radius of cylinder \( = \frac{r}{r} = \frac{1}{1} \)
\( \therefore \) radius of sphere : radius of cylinder = 1 : 1.
(ii) Curved surface area of cylinder \( / \) Surface area of sphere \( = \frac{2 \pi r h}{4 \pi r^2} \)
\( = \frac{h}{2r} \)

\( = \frac{2r}{2r} \) ...[From (i)]

\( = \frac{1}{1} \)
\( \therefore \) curved surface area of cylinder : surface area of sphere = 1:1.
(iii) Volume of cylinder \( / \) Volume of sphere \( = \frac{\pi r^2 h}{\frac{4}{3} \pi r^3} \)
\( = \frac{3h}{4r} \)

\( = \frac{3(2r)}{4r} \) ...[From (i)]

\( = \frac{3}{2} \)
\( \therefore \) volume of cylinder : volume of sphere = 3 : 2.
In simple words: When a sphere perfectly fits inside a cylinder, there are direct ratios between their radii, curved surface areas, and volumes. The key is to establish the relationship between the cylinder's height and the sphere's radius (h = 2r).

🎯 Exam Tip: For such "sphere in cylinder" problems, the crucial relation is that the height of the cylinder equals the diameter of the sphere (h = 2r), and their radii are equal. Remember to derive and use this relationship for all ratio calculations.

Question 4. Finding volume of a sphere using cylindrical beaker and water. (Textbook, pg. no. 142)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक बेलनाकार बीकर और एक गोलाकार गेंद को दर्शाता है जिसका उपयोग पानी के विस्थापन के माध्यम से गोले के आयतन को ज्ञात करने के लिए किया जाता है। बीकर के अंदर एक पट्टी लगी है जिसमें दो निशान हैं, और एक गेंद उसमें डूबी हुई है। यह प्रक्रियात्मक चित्रण पानी के स्तर में वृद्धि को समझने में मदद करता है।
Answer:
(i) Take a ball and a beaker of the same radius.
(ii) Cut a strip of paper of length equal to the diameter of the beaker.
(iii) Draw two lines on the strip dividing it into three equal parts.
(iv) Stick this strip on the beaker straight up from the bottom.
(v) Fill the water in the beaker upto the first mark of the strip from bottom.
(vi) Push the ball in the beaker so that it touches the bottom.
Observe how much water level rises.
You will notice that the water level has risen exactly upto the total height of the strip.
Try to obtain the formula for volume of sphere using the volume of the cylindrical beaker.
Solution:
Suppose volume of beaker up to height 2r is V.
V \( = \pi r^2 h \)
\( \therefore V = \pi r^2 (2r) \) ...[ h = 2r]
\( \therefore V = 2 \pi r^3 \)
The strip divides the height (2r) into three equal parts. So each mark is at \( \frac{2r}{3} \), \( \frac{4r}{3} \), and \( 2r \).
The water is filled up to the first mark from the bottom, so initial water height is \( \frac{2r}{3} \).
Volume of water initially \( = \pi r^2 (\frac{2r}{3}) = \frac{2}{3} \pi r^3 \).
When the ball is pushed in, the water level rises exactly up to the total height of the strip, which is \( 2r \).
The final volume (water + ball) is \( 2 \pi r^3 \).
Volume of the ball = (Final volume) - (Initial volume of water)
\( = 2 \pi r^3 - \frac{2}{3} \pi r^3 \)
\( = \frac{6 \pi r^3 - 2 \pi r^3}{3} \)
\( = \frac{4}{3} \pi r^3 \)
\( \therefore \) Volume of a sphere \( = \frac{4}{3} \pi r^3 \).
In simple words: This experiment demonstrates that when a sphere is immersed in a cylinder such that it perfectly touches the sides, top, and bottom, the volume of water displaced (which is the volume of the sphere) is exactly two-thirds of the volume of the cylinder. This leads to the formula for the volume of a sphere.

🎯 Exam Tip: This activity provides a visual and practical derivation of the sphere's volume formula. Understand the relationship between the volumes of the sphere and the circumscribing cylinder when the sphere touches all sides.