Description : A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g/cm3. -Maths 9th
Last Answer : We have, the radius of a metallic sphere (r) = 4.9 cm ∴ Volume of the sphere = 4 / 3 πr3 = 4 / 3 × 22 / 7 × 4.9 × 4.9 × 4.9 = 493.005 cm3 ∵ Density of the metal used = 7.8 g/cm3 Hence, the mass of the shot - put = 493.005 × 7.8 = 3845.44
Description : A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th
Last Answer : Since 1 cm3 of water weighs 1 g. ∴ Volume of cyclinder vessel = 154 cm3 πr2h = 154 h = 154 × 7 / 22 × 3.5 × 3.5 h ;= 4 cm Hence, the depth of water is 4 cm.
Description : How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th
Last Answer : No.of balls = Volume of share / Volume of each ball = 4 / 3π × 8 × 8 × 8 / 4 / 3π × 2 × 2 × 2 = 64
Description : A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. -Maths 9th
Last Answer : NEED ANSWER
Last Answer : According to question find the radius of the sphere
Description : Find the radius of a sphere whose surface area is 154 cm square. -Maths 9th
Last Answer : Let 'r' be the radius of sphere Surface area of sphere = 4 πr2 ⇒ 154 = 4 πr2 ⇒ 154 = 4 x 22/7 x r2 ⇒ r 2 = 154 x 7/4 x 22 = 49/4 ⇒ r = 7/2 cm = 3.5 cm
Description : The surface area of a sphere of radius 5 cm -Maths 9th
Last Answer : Radius of the sphere (r1) = 5 cm Radius of the base of cone (r2) = 4 cm Let r сm be the height of the cone. Surface area of sphere = 4 πr2 ⇒ 4 π(5)2 = 100 π cm2 Curved surface area of cone = πrl = 4 πl ... ∴ Volume of cone = 1/3 πr2h = 1/3 x 22/7 x 42 x 3 = 352/7 cm3 = 50.29 cm3 (Approximately)
Description : In a sphere of radius 2 cm a cone of height 3 cm is inscribed. What is the ratio of volumes of the cone and sphere ? -Maths 9th
Last Answer : answer:
Description : The volume of the metal of a cylindrical pipe is 748 cm^3. The length of the pipe is 14 cm and its external radius is 9 cm -Maths 9th
Description : Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm x 8 cm x 8 cm. -Maths 9th
Last Answer : Volume of rectangular box=lbh=16(64)=1024cm3 Volume of sphere=34πr3=33.5238cm3 16 sphere=16(33.5238)=536.3808 Volume of liquid=1024−536.3808=488cm3
Last Answer : According to question find the volume of this liquid.
Description : A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th
Last Answer : ∵ PM = MQ = 1/2 = PQ = 45 cm and OP = 7.5 cm In right angled ΔOMP, using phthagoras theorem OM2 = OP2 - PM2 ⇒OM2 = 7.52 - 4.52 ⇒OM2 = 56.25 - 20.25 ⇒OM2 = 36 ∴ OM = √36 = 6 cm
Description : A metallic sheet is of rectangular shape with dimensions 48 cm x 36 cm. From each of its corners, a square of 8 cm is cut-off and an open box is made of the remaining sheet. Find the volume of the box. -Maths 9th
Last Answer : When squares of 8 cm is cutt-off from rectangulare sheet then, Length of box (l) = (98 - 8 - 8) = 32 cm Breadth of box (b) = (36 - 8 - 8) = 20 cm Height of box (h) = 8cm ∴ Volume of box = lbh = 32 x 20 x 8 = 5120 cm3
Description : A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder -Maths 9th
Last Answer : Surface area of sphere = 4πr2, where r is the radius of sphere (ii) Height of cylinder, h = r+r =2r Radius of cylinder = r CSA of cylinder formula = 2πrh = 2πr(2r) (using value of h) = 4πr2 (iii) Ratio ... sphere)/CSA of Cylinder) = 4r2/4r2 = 1/1 Ratio of the areas obtained in (i) and (ii) is 1:1.
Description : Find the surface area of a sphere of radius: (i) 10.5cm (ii) 5.6cm (iii) 14cm -Maths 9th
Last Answer : Formula: Surface area of sphere (SA) = 4πr2 (i) Radius of sphere, r = 10.5 cm SA = 4 (22/7) 10.52 = 1386 Surface area of sphere is 1386 cm2 (ii) Radius of sphere, r = 5.6cm Using formula, SA = 4 (22 ... 75 cm Surface area of sphere = 4πr2 = 4 (22/7) 1.752 = 38.5 Surface area of a sphere is 38.5 cm2
Description : If the radius of a sphere is 2r, then its volume will be -Maths 9th
Last Answer : As, r=2r Volume of sphere = 4/3π(2r)^3 =32/3πr^3
Last Answer : (d) Given, radius of a sphere = 2r Volume of a sphere =4/3 π(Radius)3 = 4/3 π(2r)3 = 4/3 π 8r3 = (32 πr3)/3 cu units Hence the volume of a sphere is (32 πr3)/3 cu units.
Description : Find the ratio of surface area and volume of the sphere of unit radius. -Maths 9th
Last Answer : Required ratio = 4πr2 / 4/3.πr3 = 3 x 4 x π x (1)2 / 4 x π x (1)3 = 3/1 (Since, r = 1) i.e., 3 : 1
Description : The radius of sphere is 2r, then find its volume. -Maths 9th
Last Answer : Volume of the sphere = 4/3.π.(2r)3 = 32/3πr3
Description : If the radius of a sphere is doubled... -Maths 9th
Last Answer : Surface area of sphere = 4πr2 When radius is doubled then new surface area = 4π(2r)2 = 4π x 4r2 = 4(4πr2 ) = 4 x original surface area. ∴ Surface area becomes 4 tim es.
Description : In Fig., a right circular cylinder just encloses a sphere of radius r. Find -Maths 9th
Last Answer : (i) Surface areas S1 of the sphere = 4 πr2 (ii) We have Radius of the cylinder = r Height of the cylinder = h = 2r ∴ Curved surface area S2 of the cylinder ... 2 πrh = 2 πr x 2r = 4 πr2 (iii) S1/S2 = 4 πr2/4 πr2 = 1/1 ∴ S1 : S2 = 1 : 1
Description : A cylinder, a cone and a sphere are of the same radius -Maths 9th
Last Answer : Let r be the common radius of a cylinder, cone and a sphere. Then, height of the cylinder = Height of the cone = Height of the sphere = 2r Let 'I' be the slant height of the cone. Then l = root under( √r2 + h2) = root under( ... , S1 : S2 :S3 = 4 πr2 : √5 πr2 : 4 πr2 ∴ S1 : S2 : S3 = 4 : √5 : 4
Description : A sphere is cut into two equal halves and both the halves are painted from all the sides. The radius of the sphere is r unit and the -Maths 9th
Description : A square has its side equal to the radius of the sphere. The square revolves round a side to generate a surface of total area S. -Maths 9th
Description : A sphere, a cylinder and a cone respectively are of the same radius and same height. Find the ratio of their curved surfaces. -Maths 9th
Description : A sphere and a right circular cone of same radius have equal volumes. By what percentage does the height of the cone exceed its diameter ? -Maths 9th
Description : A spherical metal of radius 10 cm is melted and made into 1000 smaller spheres of equal sizes. In this process the surface area of the -Maths 9th
Last Answer : Option (C) is correct. Solution: Let the radius of the small spheres be r' cm. Volume of metal remains the same in both cases. So, vol of the spherical metal of radius 10 cm = total ... Total Surface area of 1000 smaller spheres: 1000*4π12 = 4000π Hence, the surface area increased by 10 times.
Description : How much ice-cream can be put into a cone with base radius 3.5 cm and height 12 cm ? -Maths 9th
Last Answer : Here, radius (r) = 3.5 cm and height (h) = 12 cm ∴ Amount of ice cream = 1 / 3 πr2h = 1 / 3 × 22 / 7 × 3.5 × 3.5 × 12 = 154 cm3
Description : The radius and height of a right circular cylinder are 42 cm & 63 cm respectively. Find its volume. a) 237564 cm3 b) 349272 cm3 c) 379252 cm3 d) 453213 cm3
Last Answer : Answer: B
Description : The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume. -Maths 9th
Last Answer : This answer was deleted by our moderators...
Description : Nidhi has to find the area of a sphere whose diameter was 14 cm. -Maths 9th
Last Answer : Area is two-dimensional while 4 πr represents a length.
Description : Find the volume of a sphere whose surface area is 154 cm sq. -Maths 9th
Last Answer : Let r cm be the radius of sphere. Surface area of the sphere = 4 πr2 ⇒ 154 = 4 πr2 ⇒ 4 x 22/7 x r2 = 154 r 2 = 154 x 7/4 x 22 = 72/22 ⇒ r = 7/2 Volume of sphere = 4/3 πr3 = 4/3 x 22/7 x 7/2 x 7/2 x 7/2 cm3 = 539/3 cm3 = 179.2/3 cm3
Description : A cube of side 5 cm contain a sphere -Maths 9th
Last Answer : Each side of the cube (a) = 5 cm Diameter of the sphere (2r) = 5 cm . ∴ Radius of the sphere (r) = 5/2 cm Volume of the cube = a3 = 53 cm3 = 125 cm3 Volume of the sphere = 4/3 πr3 = 4/3 x ... /2 x 5/2 = 65.476 cm3 Volume of gap between cube and sphere = 125.000 cm3 - 65.476 cm3 = 59.524 cm3
Description : A hollow METALLIC sphere has an inner radius A and an outer radius B. If charge is placed on the inner surface, that is at radius A, where is the charge located after it has come to rest?
Last Answer : ANSWER: ON THE SPHERE'S OUTER SURFACE or OUTSIDE RADIUS B
Description : Calculate the edge of the cube if its volume is 1331 cm3 -Maths 9th
Last Answer : Let the edge of the cube be a Volume of a cube = a×a×a 1331= a×a×a 11 =a ( as 1331 is the cube of 11 Therefore , Edge of a cube is 11 cm
Description : The diameter of a metallic ball is 4.2 cm. -Maths 9th
Last Answer : Radius of the metallic ball = 4.2/2 cm = 2.1 cm Volume of the ball = 4/3 πr3 = 4/3 x 22/7 x 2.1 x 2.1 x 2.1 = 38.808 cm3 Mass of the ball = density x volume = 8.9 g/cm3 x 38.808 cm3 = 345.39 g
Description : A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. -Maths 9th
Last Answer : Radius of conical cap, r = 7 cm Height of conical cap, h = 24cm Slant height, l2 = (r2+h2) = (72+242) = (49+576) = (625) Or l = 25 cm CSA of 1 conical cap = πrl = (22/7)×7×24 = 550 CSA of 10 caps = (10×550) cm2 = 5500 cm2 Therefore, the area of the sheet required to make 10 such caps is 5500 cm2.
Description : A cone of height 7 cm and base radius 1 cm is carved from a cuboidal block of wood 10 cm × 5 cm × 2 cm -Maths 9th
Last Answer : 92239223% Volume of cone = 1313πr2h = 13×227×1×7=22313×227×1×7=223 cu. cm Volume of cubical block = (10 × 5 × 2) cm3 = 100 cm3 ∴ Wastage of wood = (100−227)100×100(100−227)100×100 = 27832783% = 92239223%
Description : The radius of a sphere decreases from 10 cm to `9.9` cm. Find (i) approximate decrease in its volume. (ii) approximate decrease in its surface.
Last Answer : The radius of a sphere decreases from 10 cm to `9.9` cm. Find (i) approximate decrease in its volume. (ii) approximate decrease in its surface.
Description : Mass of steel sphere having density 7850 kg m-3 and radius 0.15 m is A. 112 kg B. 290 kg C. 110.9 kg D. 300 kg
Last Answer : 110.9 kg
Description : The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th
Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm
Description : A cone is 8.4 cm high and the radius of its base is 2.1 cm. -Maths 9th
Last Answer : Volume of cone = Volume of sphere 1 / 3π(2.1)2 × 8.4 = 4 / 3 πr3 ⇒ r3 = (2.1)2 × 8.4 / 4 = (2.1)3 ⇒ r = 2.1 cm ∴ Radius of the sphere = 2.1 cm