(i) Given that the radius of solid sphere is r and the radius of new sphere is r'. Volume of solid sphere = 4/3 πr³ Therefore, the volume of 27 solid spheres = 27 × 4/3 πr³ According to question : Volume of new sphere = volume of 27 solid spheres ⇒ 4/3π (r')³ = 27 × 4/3πr³ ⇒ (r')³ = 27 × r³ ⇒ r' = 3Read more
(i) Given that the radius of solid sphere is r and the radius of new sphere is r’.
Volume of solid sphere = 4/3 πr³
Therefore, the volume of 27 solid spheres = 27 × 4/3 πr³
According to question :
Volume of new sphere = volume of 27 solid spheres
⇒ 4/3π (r’)³ = 27 × 4/3πr³
⇒ (r’)³ = 27 × r³
⇒ r’ = 3 × r
Hence, the radius of new sphere is 3r.
(ii) Surface area of new sphere S’ = 4πr²
Surface area of new sphere S’ = 4π(r’)² = 4π(3r)² = 36πr²
Therefore,
s/s’ = 4πr²/36πr² = 1/9
Hence, the ratio of S and S’ is 1:9.
The internal radius of hemispherical tank r = 1 m and thickness 1 cm = 0.01 m Therefore, the outer radius R = 1 + 0.01 = 1.01 m Volume of hemispherical tank = 2/3π(R³ - r³) = 2/3 × 22/7 × [(1.01)³ -1³] = 2/3 × 22/2 × (1.030301 - 1) = 2/3 × 22/7 × 0.030301 = 0.06348 m³ Hence, the volume of the iron uRead more
The internal radius of hemispherical tank r = 1 m and thickness 1 cm = 0.01 m
Therefore, the outer radius R = 1 + 0.01 = 1.01 m
Volume of hemispherical tank = 2/3π(R³ – r³)
= 2/3 × 22/7 × [(1.01)³ -1³]
= 2/3 × 22/2 × (1.030301 – 1)
= 2/3 × 22/7 × 0.030301
= 0.06348 m³
Hence, the volume of the iron used to make the tank is 0.06348 m³.
(I) Radius of spherical ball r = 28/2 = 14 cm Volume of water displaced by spherical ball = 4/3πr³ = 4/3 × 22/7 × 14 × 14 × 14 × = 4/3 × 22 × 2 × 14 × 14 = 11498(2/3) cm³ Hence, the volume of water displaced by spherical ball is 11498(2/3) cm³. (II) Radius of spherical ball r = 0.21/2 = 0.105 m VoluRead more
(I) Radius of spherical ball r = 28/2 = 14 cm
Volume of water displaced by spherical ball = 4/3πr³
= 4/3 × 22/7 × 14 × 14 × 14 × = 4/3 × 22 × 2 × 14 × 14 = 11498(2/3) cm³
Hence, the volume of water displaced by spherical ball is 11498(2/3) cm³.
(II) Radius of spherical ball r = 0.21/2 = 0.105 m
Volume of water displaced by spherical ball = 4/3πr³
= 4/3 × 22/7 × 0.105 × 0.105 × 0.105 = 4 × 22 × 0.005 × 0.63 × 0.63 = 0.004861 m³
Hence, the volume of water displaced by spherical ball is 0.004861 m³.
Radius of metallic ball r = 4.2/2 = 2.1 cm Therefore, the volume of metallic ball = 4/3πr³ = 4/3 × 22/7 × 2.1 × 2.1 × 2.1 = 4 × 22 × 0.1 × 2.1 × 2.1 = 38.808 cm² Here, the mass of 1 cm³ = 8.9 × 38.808 = 345.39g(approx) Hence, the mass of the ball is 345. 39 gram.
Radius of metallic ball r = 4.2/2 = 2.1 cm
Therefore, the volume of metallic ball = 4/3πr³
= 4/3 × 22/7 × 2.1 × 2.1 × 2.1
= 4 × 22 × 0.1 × 2.1 × 2.1
= 38.808 cm²
Here, the mass of 1 cm³ = 8.9 × 38.808 = 345.39g(approx)
Hence, the mass of the ball is 345. 39 gram.
Let, the radius of Earth be R Therefore, the diameter of Earth = 2R According to question, diameter of moon = 1/4(2R) So, the radius of moon = 1/4(2R)/2 = (1/4)R Volume of moon/Volume of Earth = 4/3π(1/4R)³/4/3π(R)³ = 1/64R³/R³ = 1/64 ⇒ Volume of moon = 1/64 × volume of Earth Hence, the volume of moRead more
Let, the radius of Earth be R
Therefore, the diameter of Earth = 2R
According to question, diameter of moon = 1/4(2R)
So, the radius of moon = 1/4(2R)/2 = (1/4)R
Volume of moon/Volume of Earth = 4/3π(1/4R)³/4/3π(R)³ = 1/64R³/R³ = 1/64
⇒ Volume of moon = 1/64 × volume of Earth
Hence, the volume of moon is 1/64 the volume of Earth.
Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S′. Find the:
(i) Given that the radius of solid sphere is r and the radius of new sphere is r'. Volume of solid sphere = 4/3 πr³ Therefore, the volume of 27 solid spheres = 27 × 4/3 πr³ According to question : Volume of new sphere = volume of 27 solid spheres ⇒ 4/3π (r')³ = 27 × 4/3πr³ ⇒ (r')³ = 27 × r³ ⇒ r' = 3Read more
(i) Given that the radius of solid sphere is r and the radius of new sphere is r’.
Volume of solid sphere = 4/3 πr³
Therefore, the volume of 27 solid spheres = 27 × 4/3 πr³
According to question :
Volume of new sphere = volume of 27 solid spheres
⇒ 4/3π (r’)³ = 27 × 4/3πr³
⇒ (r’)³ = 27 × r³
⇒ r’ = 3 × r
Hence, the radius of new sphere is 3r.
(ii) Surface area of new sphere S’ = 4πr²
See lessSurface area of new sphere S’ = 4π(r’)² = 4π(3r)² = 36πr²
Therefore,
s/s’ = 4πr²/36πr² = 1/9
Hence, the ratio of S and S’ is 1:9.
A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
The internal radius of hemispherical tank r = 1 m and thickness 1 cm = 0.01 m Therefore, the outer radius R = 1 + 0.01 = 1.01 m Volume of hemispherical tank = 2/3π(R³ - r³) = 2/3 × 22/7 × [(1.01)³ -1³] = 2/3 × 22/2 × (1.030301 - 1) = 2/3 × 22/7 × 0.030301 = 0.06348 m³ Hence, the volume of the iron uRead more
The internal radius of hemispherical tank r = 1 m and thickness 1 cm = 0.01 m
See lessTherefore, the outer radius R = 1 + 0.01 = 1.01 m
Volume of hemispherical tank = 2/3π(R³ – r³)
= 2/3 × 22/7 × [(1.01)³ -1³]
= 2/3 × 22/2 × (1.030301 – 1)
= 2/3 × 22/7 × 0.030301
= 0.06348 m³
Hence, the volume of the iron used to make the tank is 0.06348 m³.
Find the amount of water displaced by a solid spherical ball of diameter (i) 28 cm (ii) 0.21 m
(I) Radius of spherical ball r = 28/2 = 14 cm Volume of water displaced by spherical ball = 4/3πr³ = 4/3 × 22/7 × 14 × 14 × 14 × = 4/3 × 22 × 2 × 14 × 14 = 11498(2/3) cm³ Hence, the volume of water displaced by spherical ball is 11498(2/3) cm³. (II) Radius of spherical ball r = 0.21/2 = 0.105 m VoluRead more
(I) Radius of spherical ball r = 28/2 = 14 cm
Volume of water displaced by spherical ball = 4/3πr³
= 4/3 × 22/7 × 14 × 14 × 14 × = 4/3 × 22 × 2 × 14 × 14 = 11498(2/3) cm³
Hence, the volume of water displaced by spherical ball is 11498(2/3) cm³.
(II) Radius of spherical ball r = 0.21/2 = 0.105 m
See lessVolume of water displaced by spherical ball = 4/3πr³
= 4/3 × 22/7 × 0.105 × 0.105 × 0.105 = 4 × 22 × 0.005 × 0.63 × 0.63 = 0.004861 m³
Hence, the volume of water displaced by spherical ball is 0.004861 m³.
The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm³?
Radius of metallic ball r = 4.2/2 = 2.1 cm Therefore, the volume of metallic ball = 4/3πr³ = 4/3 × 22/7 × 2.1 × 2.1 × 2.1 = 4 × 22 × 0.1 × 2.1 × 2.1 = 38.808 cm² Here, the mass of 1 cm³ = 8.9 × 38.808 = 345.39g(approx) Hence, the mass of the ball is 345. 39 gram.
Radius of metallic ball r = 4.2/2 = 2.1 cm
See lessTherefore, the volume of metallic ball = 4/3πr³
= 4/3 × 22/7 × 2.1 × 2.1 × 2.1
= 4 × 22 × 0.1 × 2.1 × 2.1
= 38.808 cm²
Here, the mass of 1 cm³ = 8.9 × 38.808 = 345.39g(approx)
Hence, the mass of the ball is 345. 39 gram.
The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
Let, the radius of Earth be R Therefore, the diameter of Earth = 2R According to question, diameter of moon = 1/4(2R) So, the radius of moon = 1/4(2R)/2 = (1/4)R Volume of moon/Volume of Earth = 4/3π(1/4R)³/4/3π(R)³ = 1/64R³/R³ = 1/64 ⇒ Volume of moon = 1/64 × volume of Earth Hence, the volume of moRead more
Let, the radius of Earth be R
See lessTherefore, the diameter of Earth = 2R
According to question, diameter of moon = 1/4(2R)
So, the radius of moon = 1/4(2R)/2 = (1/4)R
Volume of moon/Volume of Earth = 4/3π(1/4R)³/4/3π(R)³ = 1/64R³/R³ = 1/64
⇒ Volume of moon = 1/64 × volume of Earth
Hence, the volume of moon is 1/64 the volume of Earth.