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A ball bearing has a radius of 0.7 cm. Find its surface area.
A ball bearing is spherical in shape with a given radius r = 0.7 cm. The surface area of a sphere is given by the formula SA = 4πr². Substituting the radius and using the approximation π ≈ 22/7, we get SA = 4 × (22/7) × (0.7)² = 4 × (22/7) × 0.49 = 6.16 cm². For more NCERT Solutions of ClassRead more
A ball bearing is spherical in shape with a given radius r = 0.7 cm. The surface area of a sphere is given by the formula SA = 4πr². Substituting the radius and using the approximation π ≈ 22/7, we get SA = 4 × (22/7) × (0.7)² = 4 × (22/7) × 0.49 = 6.16 cm².
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-14/
See lessTwo solid spheres made of the same metal have weights 5920 g and 740 g. Determine the radius of the larger sphere, if the diameter of the smaller one is 5 cm.
Because both solid spheres are made of the same metal, their weights are directly proportional to their volumes. The ratio of their volumes is V₁ / V₂ = 5920 / 740 = 8. Since volume is proportional to r³, the ratio of their radii is R / r = ∛8 = 2. The smaller sphere's radius is 5 / 2 = 2.5 cm, so tRead more
Because both solid spheres are made of the same metal, their weights are directly proportional to their volumes. The ratio of their volumes is V₁ / V₂ = 5920 / 740 = 8. Since volume is proportional to r³, the ratio of their radii is R / r = ∛8 = 2. The smaller sphere’s radius is 5 / 2 = 2.5 cm, so the larger sphere’s radius is 2 × 2.5 = 5 cm.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-14/
See lessThe diameter of the moon is approximately one fourth the diameter of the earth. Given that the moon and earth are both roughly spherical, find the ratio of their surface areas.
Let the diameter of the Earth be D and the diameter of the Moon be D/4, meaning their radii are in the ratio r_moon / r_earth = 1 / 4. The surface area of a sphere is given by SA = 4πr². Therefore, the ratio of their surface areas is SA_moon / SA_earth = (r_moon / r_earth)² = (1/4)² = 1 / 16, whichRead more
Let the diameter of the Earth be D and the diameter of the Moon be D/4, meaning their radii are in the ratio r_moon / r_earth = 1 / 4. The surface area of a sphere is given by SA = 4πr². Therefore, the ratio of their surface areas is SA_moon / SA_earth = (r_moon / r_earth)² = (1/4)² = 1 / 16, which is 1 : 16.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-14/
See lessIs there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
Yes, such a temperature exists. Setting F = C = x in the conversion formula gives x = (9/5)x + 32. Rearranging the terms gives x − (9/5)x = 32, which simplifies to −(4/5)x = 32. Multiplying both sides by −5/4 results in x = −40. Therefore, −40°C is numerically equal to −40°F. For more NCERT SRead more
Yes, such a temperature exists. Setting F = C = x in the conversion formula gives x = (9/5)x + 32. Rearranging the terms gives x − (9/5)x = 32, which simplifies to −(4/5)x = 32. Multiplying both sides by −5/4 results in x = −40. Therefore, −40°C is numerically equal to −40°F.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-13/
See lessIf the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?
When C = 0°C, substituting into the linear equation gives F = (9/5)(0) + 32 = 32°F. When F = 0°F, the relation becomes 0 = (9/5)C + 32, which leads to (9/5)C = −32. Multiplying by 5 and dividing by 9 gives C = −160/9°C or approximately −17.78°C. For more NCERT Solutions of Class 9 Maths GanitRead more
When C = 0°C, substituting into the linear equation gives F = (9/5)(0) + 32 = 32°F. When F = 0°F, the relation becomes 0 = (9/5)C + 32, which leads to (9/5)C = −32. Multiplying by 5 and dividing by 9 gives C = −160/9°C or approximately −17.78°C.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-13/
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