Surface area of sphere is 154 cm². Let the radius of sphere = r cm surface area of sphere = 4πr² ⇒ 154 = 4 × 22/7 × r² ⇒ r² = 154 × 7/22 × 1/4 ⇒ r² = 77/4 ⇒ r = √77/4 = 7/2 Hence, the radius of sphere is 7/2 cm.
Surface area of sphere is 154 cm². Let the radius of sphere = r cm
surface area of sphere = 4πr²
⇒ 154 = 4 × 22/7 × r² ⇒ r² = 154 × 7/22 × 1/4 ⇒ r² = 77/4 ⇒ r = √77/4 = 7/2
Hence, the radius of sphere is 7/2 cm.
Let, the radius of earth = R, Therefore, the diameter of earth = 2 R According to question, diameter of moon = 1/4(2R), So, the radius of moon = (1/4(2R)/2) = 1/4 R Now, Surface area of moon/surface of earth = (4π(1/4R)²)/4π(R)² = (1/16R²)/R² = 1/16 Hence, the ratio of surface areas of moon to earthRead more
Let, the radius of earth = R, Therefore, the diameter of earth = 2 R
According to question, diameter of moon = 1/4(2R), So, the radius of moon
= (1/4(2R)/2) = 1/4 R
Now,
Surface area of moon/surface of earth = (4π(1/4R)²)/4π(R)² = (1/16R²)/R² = 1/16
Hence, the ratio of surface areas of moon to earth is 1: 16.
Internal radius of hemispherical bowl r = 5 cm and thickness = 0.25 cm Therefore, the outer radius of hemispherical bowl = R = 5 + 0.25 = 5.25 cm Outer curved surface area of hemispherical bowl = 2πR² = 2 × 22/7 × 5.25 × 5.25 = 2 × 22 × 0.75 × 5.25 = 173.25 cm² Hence, the outer curved surface area oRead more
Internal radius of hemispherical bowl r = 5 cm and thickness = 0.25 cm
Therefore, the outer radius of hemispherical bowl = R = 5 + 0.25 = 5.25 cm
Outer curved surface area of hemispherical bowl = 2πR²
= 2 × 22/7 × 5.25 × 5.25 = 2 × 22 × 0.75 × 5.25 = 173.25 cm²
Hence, the outer curved surface area of hemispherical bowl is 173.25 cm².
Length of matchbox l = 4 cm, breadth b = 2.5 cm and heigth h = 1.5 cm Volume of matchbox = lbh = 4 × 2.5 × 1.5 = 15 cm³ Therefore, volume of 12 machbox = 12 × 15 = 180 cm³ Hence, the volume of a packet containing 12 boxes is 180 cm³.
Length of matchbox l = 4 cm, breadth b = 2.5 cm and heigth h = 1.5 cm
Volume of matchbox = lbh = 4 × 2.5 × 1.5 = 15 cm³
Therefore, volume of 12 machbox = 12 × 15 = 180 cm³
Hence, the volume of a packet containing 12 boxes is 180 cm³.
Here, a = 3 and d = 8 - 3 = 5. Let, nth term of the A.P. is 78. Therefore, a_n = 78 ⇒ a + (n -1)d = 78 ⇒ 3 + (n -1)(5) = 78 ⇒ (n - 1)(5) = 75 ⇒ n - 1 = 15 ⇒ n = 16 Hence, 16th term of the A.P .: 3, 8, 13, 18, ... is 78.
Here, a = 3 and d = 8 – 3 = 5.
Let, nth term of the A.P. is 78.
Therefore, a_n = 78
⇒ a + (n -1)d = 78
⇒ 3 + (n -1)(5) = 78
⇒ (n – 1)(5) = 75
⇒ n – 1 = 15
⇒ n = 16
Hence, 16th term of the A.P .: 3, 8, 13, 18, … is 78.
Find the radius of a sphere whose surface area is 154 cm².
Surface area of sphere is 154 cm². Let the radius of sphere = r cm surface area of sphere = 4πr² ⇒ 154 = 4 × 22/7 × r² ⇒ r² = 154 × 7/22 × 1/4 ⇒ r² = 77/4 ⇒ r = √77/4 = 7/2 Hence, the radius of sphere is 7/2 cm.
Surface area of sphere is 154 cm². Let the radius of sphere = r cm
See lesssurface area of sphere = 4πr²
⇒ 154 = 4 × 22/7 × r² ⇒ r² = 154 × 7/22 × 1/4 ⇒ r² = 77/4 ⇒ r = √77/4 = 7/2
Hence, the radius of sphere is 7/2 cm.
The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Let, the radius of earth = R, Therefore, the diameter of earth = 2 R According to question, diameter of moon = 1/4(2R), So, the radius of moon = (1/4(2R)/2) = 1/4 R Now, Surface area of moon/surface of earth = (4π(1/4R)²)/4π(R)² = (1/16R²)/R² = 1/16 Hence, the ratio of surface areas of moon to earthRead more
Let, the radius of earth = R, Therefore, the diameter of earth = 2 R
See lessAccording to question, diameter of moon = 1/4(2R), So, the radius of moon
= (1/4(2R)/2) = 1/4 R
Now,
Surface area of moon/surface of earth = (4π(1/4R)²)/4π(R)² = (1/16R²)/R² = 1/16
Hence, the ratio of surface areas of moon to earth is 1: 16.
A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Internal radius of hemispherical bowl r = 5 cm and thickness = 0.25 cm Therefore, the outer radius of hemispherical bowl = R = 5 + 0.25 = 5.25 cm Outer curved surface area of hemispherical bowl = 2πR² = 2 × 22/7 × 5.25 × 5.25 = 2 × 22 × 0.75 × 5.25 = 173.25 cm² Hence, the outer curved surface area oRead more
Internal radius of hemispherical bowl r = 5 cm and thickness = 0.25 cm
See lessTherefore, the outer radius of hemispherical bowl = R = 5 + 0.25 = 5.25 cm
Outer curved surface area of hemispherical bowl = 2πR²
= 2 × 22/7 × 5.25 × 5.25 = 2 × 22 × 0.75 × 5.25 = 173.25 cm²
Hence, the outer curved surface area of hemispherical bowl is 173.25 cm².
A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes?
Length of matchbox l = 4 cm, breadth b = 2.5 cm and heigth h = 1.5 cm Volume of matchbox = lbh = 4 × 2.5 × 1.5 = 15 cm³ Therefore, volume of 12 machbox = 12 × 15 = 180 cm³ Hence, the volume of a packet containing 12 boxes is 180 cm³.
Length of matchbox l = 4 cm, breadth b = 2.5 cm and heigth h = 1.5 cm
See lessVolume of matchbox = lbh = 4 × 2.5 × 1.5 = 15 cm³
Therefore, volume of 12 machbox = 12 × 15 = 180 cm³
Hence, the volume of a packet containing 12 boxes is 180 cm³.
Which term of the AP : 3, 8, 13, 18, . . . ,is 78?
Here, a = 3 and d = 8 - 3 = 5. Let, nth term of the A.P. is 78. Therefore, a_n = 78 ⇒ a + (n -1)d = 78 ⇒ 3 + (n -1)(5) = 78 ⇒ (n - 1)(5) = 75 ⇒ n - 1 = 15 ⇒ n = 16 Hence, 16th term of the A.P .: 3, 8, 13, 18, ... is 78.
Here, a = 3 and d = 8 – 3 = 5.
See lessLet, nth term of the A.P. is 78.
Therefore, a_n = 78
⇒ a + (n -1)d = 78
⇒ 3 + (n -1)(5) = 78
⇒ (n – 1)(5) = 75
⇒ n – 1 = 15
⇒ n = 16
Hence, 16th term of the A.P .: 3, 8, 13, 18, … is 78.