Description : If in a cylinder, radius is doubled and height is halved, then find its curved surface area. -Maths 9th
Last Answer : Let r and h be radius and height of the cyclinder, then C.S.A. = 2πrh Now, radius is doubled and height is halved. ∴ New radius = 2r and new height = h / 2 New C.S.A. = 2π × 2r × h / 2 = 2πrh .
Description : A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th
Last Answer : since curved surface of half of the spherical ball = 56.57 cm2 ∴ 2πr2 = 56.57 ⇒ r2 = 56.57 / 2 × 3.14 = 9 ⇒ r = 3 cm Now, volume of spherical ball = 4 / 3 πr3 = 4 / 3 × 3.14 × 3 × 3 × 3 = 113.04 cm3
Description : In a cylinder, radius is doubled and height is halved, then curved surface area will be -Maths 9th
Last Answer : The curved surface area will remain same. So, there is no change in the curved surface area of cylinder . Hence the curved surface area will remain same.
Description : If three cylinders of radius r and height h are placed vertically such that the curved surface of each cylinder touches the curved surfaces -Maths 9th
Last Answer : hr2 (3-√−π2)(3−π2) The bases of the three cylinders when placed as given are as shown in the figure : Let the radius of the base of each cylinder = r cm. We are required to find the volume of air. ... ∠C = 60º) = 3 x 60o360o πr2=πr2260o360o πr2=πr22 ∴ Required volume = (3-√r2−π2r2)h=(3-√−π2)r2h.
Description : If the ratio of curved surface area and total surface area of a cylinder is 1 : 3, then find the volume of cylinder when the height is 2 cm. -Maths 9th
Last Answer : Let the radius and height of the cylinder be r and h, respectively . Given that, Curved surface area / Total surface area = 1/3 ⇒ 2πrh / 2πr(h + r) = 1/3 ⇒ 3h = h + r ⇒ r = 2h = 4cm ∴ volume of cylinder πr2h = π × (4)2 × 2 = 32π 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 : 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
Last Answer : answer:
Description : The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7 ) -Maths 9th
Last Answer : Height of cylinder, h = 14cm Let the diameter of the cylinder be d Curved surface area of cylinder = 88 cm2 We know that, formula to find Curved surface area of cylinder is 2πrh. So 2πrh =88 cm2 (r is the ... 88 cm2 2r = 2 cm d =2 cm Therefore, the diameter of the base of the cylinder is 2 cm.
Description : Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base -Maths 9th
Last Answer : Slant height of cone, l = 14 cm Let the radius of the cone be r. (i) We know, CSA of cone = πrl Given: Curved surface area of a cone is 308 cm2 (308 ) = (22/7) r 14 308 = 44 r r = 308 ... Total surface area of cone = 308+(22/7) 72 = 308+154 Therefore, the total surface area of the cone is 462 cm2.
Description : The base of a right triangular prism is an equilateral triangle. If the height is halved and each side of the base is doubled, find the ratio of the -Maths 9th
Last Answer : 1 : 2 Let each side of the base of the original prism be a units and the height of the prism be h units. Then Required ratio = Vol. of original prismVol. of new prismVol. of original ... )2×h3√4×(2a)2×h234×(a)2×h34×(2a)2×h2 = 2a2h4a2h2a2h4a2h = 1 : 2.
Description : if seve times the curved surface area (A) of a cylinder is equal to 44 times the proudct of base radius (r) and height (h) then what si the formula wi
Last Answer : if seve times the curved surface area (A) of a cylinder is equal to 44 times the proudct of base ... (h) then what si the formula with subject A?
Description : A rectangular paper 11 cm by 8 cm can be exactly wrapped to cover the curved surface of a cylinder of height 8 cm . -Maths 9th
Description : the curved surface area of a cylinder is 154 cm. the total surface area of the cylinder is three times its curved surface area. find the volume of the cylinder. -Maths 9th
Last Answer : T.S.A = 3*154 = 462 cm² C.S.A = 154 cm² C.S.A = 2πrh T.S.A = 2πr(r+h) Now, In T.S.A = 2πrr + 2πrh 462 = 2πrr + 2πrh 462 = 2*22/7*r*r + 154 462 - 154 = 2*22/7*r*r 308*7/2*22 = r*r 49 = r*r R = 7 cm ... 7*h 154/44 = h 7/2 =h H = 3.5 cm or 7/2 cm Now volume = πrrh = 22/7 * 7* 7 *7/2 = 11*49 = 539 cm³
Description : The curved surface area of a cylinder is 154 cm2. -Maths 9th
Last Answer : Since curved surface area of a cyclinder = 154 cm2 [given] Total surface area of cyclinder = 3 curved surface area 2πrh + 2πr2 = 3 154 154 + 2πr2 = 462 2πr2 = 462 - 154 = 308 r2 = 308 7 / 2 22 = 49 ... 154 / 44 = 3.5 cm ∴ Volume of cyclinder = πr2h = 22 / 7 7 7 3.5 = 539 cm3
Description : The area of the curved surface and the area of the base of a right circular cylinder are a square cm and b square cm respectively -Maths 9th
Description : The curved surface of a cylinder is developed into a square whose diagonal is 2√2 cm. The area of the base of the cylinder (in cm^2) is -Maths 9th
Description : If the radius of a cylinder is doubled -Maths 9th
Last Answer : Volume of cylinder = πr2h New volume = πR2H = π.(2r)2.(h/2) = 2( πr2h) ∴ New volume = 2 x original volume.
Description : If S denotes the area of the curved surface of a right circular cone of height h end semi-vertical angle a, then S equals -Maths 9th
Description : Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area -Maths 9th
Last Answer : Radius of the base of cone = diameter/ 2 = (10.5/2)cm = 5.25cm Slant height of cone, say l = 10 cm CSA of cone is = πrl = (22/7)×5.25×10 = 165
Description : If radius of a batch basket centrifuge is halved & the r.p.m. is doubled, then the (A) Linear speed of the basket is doubled (B) Linear speed of the basket is halved (C) Centrifugal force is doubled (D) Capacity of centrifuge is increased
Last Answer : (C) Centrifugal force is doubled
Description : A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. -Maths 9th
Last Answer : Inner radius of hemispherical bowl = 5cm Thickness of the bowl = 0.25 cm Outer radius of hemispherical bowl = (5+0.25) cm = 5.25 cm Formula for outer CSA of hemispherical bowl = 2πr2, where r is radius of ... 22/7) (5.25)2 = 173.25 Therefore, the outer curved surface area of the bowl is 173.25 cm2.
Description : A cylindrical rod of iron whose height is eight times its radius is melted and cast into spherical balls each of half the radius of the cylinder. -Maths 9th
Last Answer : Let radius of iron rod = r ∴∴ Height = 8r ∴∴ Volume of iron rod =π×(r)2×8r⇒8πr3=π×(r)2×8r⇒8πr3 ⇒⇒ Radius of spherical ball =r2=r2 Volume of spherical ball =43π(r2)3=43π(r2)3 Let n balls are casted ∴n×43π(r38)=8πr3∴n×43π(r38)=8πr3 ⇒n6=8⇒n=48
Description : A cylindrical rod of iron whose radius is one-fourth of its height is melted and cast into spherical balls of the same radius as that of the cylinder. -Maths 9th
Last Answer : Let radius of cylindrical rod =r ⇒ height =4r Volume of cylindrical rod =πr2h =πr2(4r) =4πr3 Volume of spherical balls of radius r=34πr3 No. of balls =34πr34πr3=
Description : A solid right circular cylinder of radius 8 cm and height 2 cm is melted and cast into a right circular cone of height 3 times that of the cylinder. -Maths 9th
Last Answer : Height of cone = 3 times height of cylinder = 3 3 = 9 cm Volume of cylinder = volume of cone r2 = 8 8 r = 8 cm l2 = h2 + r2 = (9)2 + (8)2 l = = 12 cm C.S.A (cone) = = 301.71 cm2
Description : The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2 -Maths 9th
Last Answer : Slant height of conical tomb, l = 25m Base radius, r = diameter/2 = 14/2 m = 7m CSA of conical tomb = πrl = (22/7)×7×25 = 550 CSA of conical tomb= 550m2 Cost of white-washing 550 m2 area, which is Rs (210×550)/100 = Rs. 1155 Therefore, cost will be Rs. 1155 while white-washing tomb.
Description : The magnitude of the volume of a closed right circular cylinder of unit height divided by the magnitude of the total surface area of the -Maths 9th
Description : A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m2 . -Maths 9th
Last Answer : Diameter of the pillar = 50 cm ∴ Radius (r) = 502m = 25 m = 14m and height (h) = 3.5m Curved surface area of a pillar = 2πrh ∴ Curved surface area to be painted = 112m2 ∴ Cost of painting of 1 m2 pillar = Rs. 12.50 ∴ Cost of painting of 112 m2 pillar = Rs. ( 112 x 12.50 ) = Rs. 68.75.
Description : A cone of height 24 cm has a curved surface -Maths 9th
Last Answer : Height of the cone (h) = 24 cm Let r сm be the radius of the base and l cm be the slant height of the cone. Then, l = root under (√r2+ h2 ) = root under (√r2 + 242) = root under (√r2 + 576) Now, Curved surface ... ⇒ r = 7 cm ∴ Volume of the cone = 1/3πr2h = 1/3 x 22/7 x 72 x 24 = 1232 cm3
Description : The total surface area of a cone whose radius is r/2 and slant height 2l is -Maths 9th
Last Answer : Total surface area of cone = πr(r+l) Given, radius = r/2 and slant height = 2l Therefore, new total surface area of cone = πr/2(r/2+2l) = π(r/4^2+rl) = πr(l+r/4)
Last Answer : Radius (r)=r/2 & slant height=2l TSA (S)=PIE R (l+r) =22/7×r/2(2l+r/2) =11/7×r(2l+r/2)
Description : For minimum curved surface area and given volume, the ration of the height and radius of base of a cone is :
Last Answer : For minimum curved surface area and given volume, the ration of the height and radius of base of a cone is : A. ` ... : 1` C. `1 : 2` D. None of these
Description : The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. -Maths 9th
Last Answer : Inner radius of circular well, r = 3.5/2m = 1.75m Depth of circular well, say h = 10m (i) Inner curved surface area = 2πrh = (2 (22/7 ) 1.75 10) = 110 Therefore, the inner curved surface ... area = Rs (110 40) = Rs.4400 Therefore, the cost of plastering the curved surface of the well is Rs. 4400.
Description : Find (i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5m high. -Maths 9th
Last Answer : Height of cylindrical tank, h = 4.5m Radius of the circular end , r = (4.2/2)m = 2.1m (i) the lateral or curved surface area of cylindrical tank is 2πrh = 2 (22/7) 2.1 4.5 m2 = (44 0.3 ... ) = 87.12 m2 This implies, S = 95.04 m2 Therefore, 95.04m2 steel was used in actual while making such a tank.
Description : The curved surface area of a right circular -Maths 9th
Last Answer : Curved surface area of cylinder = 2 πrh ⇒ 88 = 2 x 22/7 x r x 14 ⇒ r = 88 x 7/2 x 22 x 14 = 1 ∴ Diameter of the base of cylinder = 2r = 2 x 1 = 2 cm
Description : Curved surface area of a cone is -Maths 9th
Last Answer : Slant height of the cone (l) = 14 cm Curved surface area of the cone = 308 cm2 Let 'r' be the radius of the base of cone (i) Curved surface area of cone = πrl ∴ 22/7 x r x 14 = 308 ⇒ r = 308 x 7/22 x 14 ... ii) Total surface area of cone = πr(r + l) = 22/7 x 7(7 + 14) = 22/7 x 7 x 21 = 462 cm2
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 : If the curved surface area of a closed cylinder is `2pirh` ad base area is `pir^(2)`, then total surface area of that cylinder A is ______
Last Answer : If the curved surface area of a closed cylinder is `2pirh` ad base area is `pir^(2)`, then total surface area of that cylinder A is ______
Description : If the base of right rectangular prism remains constant and the measures of the lateral edges are halved, then its volume will be reduced by : -Maths 9th
Description : The radius of a right circular cone is 3 cm and its height is 4 cm. The curved surface of the cone will be (1) 12 sq. cm (2) 15 sq. cm (3) 18 sq. cm (4) 21 sq. cm
Last Answer : (2) 15 sq. cm
Description : Frame the formula: The lateral surface area of a cylinder S is equal to twice the product of `pi`, radius (r) and height (h) of the cylinder.
Last Answer : Frame the formula: The lateral surface area of a cylinder S is equal to twice the product of `pi`, radius (r) ... `S=pirh` C. `2S=pirh` D. `S=2pirh`
Description : If the length of a simple pendulum is halved then its period of oscillation is - (1) doubled (2) halved (3) increased by a factor √ 2 (4) decreased by a factor √ 2
Last Answer : (4) decreased by a factor √ 2
Description : If we double the temperature of an ideal gas, then it's average kinetic energy will be A. halved B. triple the original C. fourth times of original D. doubled
Last Answer : doubled
Description : Two substances are in equilibrium in a reversible chemical reaction. If the concentration of each substance is doubled, then the value of the equilibrium constant will be (A) Same (B) Doubled (C) Halved (D) One fourth of its original value
Last Answer : (A) Same
Description : If the length of a simple pendulum is halved then its period of oscillation is (1) doubled (2) halved (3) increased by a factor root 2 (4) decreased by a factor root 2
Last Answer : decreased by a factor root 2
Description : If the length of a wire is halved and the cross-sectional area is doubled, the resistance will be _____________. A. quartered B. unchanged C. doubled D. quadrupled
Last Answer : Answer: A
Description : A wire of uniform area of cross-section A length L and resistance R is cut into two parts. Resistivity of each part a. remains the same b. is doubled c. is halved d. becomes zero
Last Answer : a. remains the same