Description : The radius of a spherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th
Last Answer : Surface area of a spherical balloon whose radius is 6 cm. = 4π × 6 × 6 cm2 Surface area of a spherical balloon whose radius is 12 cm. = 4π × 12 × 12 cm2 ∴ Ration of surface areas = 4π × 6 × 6 / 4π × 12 × 12 = 1 / 4 = 1 : 4
Description : The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th
Last Answer : NEED ANSWER
Last Answer : Find the ratios of the surface areas of the balloon in the two cases.
Description : The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. -Maths 9th
Last Answer : Let r1 and r2 be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. So r1 = 7cm r2 = 14 cm Now, Required ratio = (initial surface area)/(Surface area after pumping air into ... = (7/14)2 = (1/2)2 = ¼ Therefore, the ratio between the surface areas is 1:4.
Description : The radius of a spherical balloon increases -Maths 9th
Last Answer : Radius of the spherical balloon = r1 = 7 cm Surface area s1 of the balloon = 4 πr12 = 4 x 22/7 x 72 = 616 cm2 Radius of the spherical balloon when air is pumped into it = r2 = 14 cm Surface area S2 of the balloon = 4 πr22 = 4 x ... S 1/S 2 = 4 x 22/7 x 7 2/4 x 22/7 x14 2 = 1/4 ∴ S1 : S2 = 1 : 4
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 : 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 : A shopkeeper has one spherical laddoo of radius 5 cm. With the same amount of material, how many laddoos of radius 2.5 cm can be made ? -Maths 9th
Last Answer : Radius of a big spherical laddoo = 5 cm ∴ Volume of the big spherical laddoo = 4 / 3 π 5 5 5 cm3 Radius of a small spherical laddoo = 2.5 = 5 / 2 cm ∴ Volume of the small spherical laddoo = 4 / ... = 2 2 2 = 8 Hence , with the same amount of big laddoo, 8 small laddoos can be made.
Description : A shopkeeper has one spherical laddoo of radius 5 cm. With the same amount of material, how many laddoos of radius 2.5 cm can be made? -Maths 9th
Last Answer : Solution of the question
Description : The volume of a spherical balloon is increasing at a rate of ` 25 cm^(3)//sec`. Find the rate of increase of its curved surface when the radius of bal
Last Answer : The volume of a spherical balloon is increasing at a rate of ` 25 cm^(3)//sec`. Find the ... its curved surface when the radius of balloon is 5 cm.
Description : A balloon which always remains spherical, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius o
Last Answer : A balloon which always remains spherical, is being inflated by pumping in 900 cubic centimetres of gas ... is increasing when the radius is 15 cm.
Description : What is the force required (in Newtons) to hold a spherical balloon stationary in water at a depth of H from the air-water interface? The balloon is of radius 0.1 m and is filled with air. (A) 4πg/3 (B) 0.01 πgH/4 (C) 0.01 πgH/8 (D) 0.04 πgH/3
Last Answer : (A) 4πg/3
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 : 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 : If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th
Last Answer : According to question the radius of the circle passing through the points A, B and C .
Description : Find the length of a chord which is at a distance of 12 cm from the centre of a circle of radius 13 cm. -Maths 9th
Last Answer : Let AB be a chord of circle with centre O and radius 13cm. Draw OM perpendicular AB and join OA. In the right triangle OMA, we have OA2 = OM2 + AM2 ⇒ 132 = 122 + AM2 ⇒ AM2 = 169 - 144 ... . As the perpendicular from the centre of a chord bisects the chord.Therefore, AB = 2AM = 2 x 5 = 10cm.
Description : Find the amount of water displaced by a solid spherical ball of diameter 4.2 cm, when it is completely immersed in water. -Maths 9th
Last Answer : The amount of water displaced by a solid spherical ball when it is completely immersed in water is equal to its volume. Volume of a sphere of radius r is 34πr3 As the diameter of the ball is 4.2 cm, its radius r=2.1 cm Hence, volume of water displaced =34πr3=34×722×2.1×2.1×2.1=38.808 cm3
Last Answer : Solution of this question
Description : A spherical iron shell with external diameter 21 cm weighs 22775 x 5/21 grams. Find the thickness of the shell if the metal weighs 10 gms per cu cm. -Maths 9th
Last Answer : answer:
Description : Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the A distance between AB and CD is 6 cm, find the radius of the circle. -Maths 9th
Last Answer : Join OA and OC. Let the radius of the circle be r cm and O be the centre Draw OP⊥AB and OQ⊥CD. We know, OQ⊥CD, OP⊥AB and AB∥CD. Therefore, points P,O and Q are collinear. So, PQ=6 cm. Let OP=x. Then, ... r2=52+(2.5)2=25+6.25=31.25 ⇒r2=31.25⇒r=5.6 Hence, the radius of the circle is 5.6 cm
Description : If the lengths of a sides of a triangle are in the ratio 4 : 5 : 6 and the in-radius of the triangle is 3 cm, -Maths 9th
Last Answer : (b) 7.5 cm.Area of a triangle = \(rac{1}{2}\)x base x height= In-radius x semi-perimeter of the Δ \(\big[ ext{Using r =}rac{\Delta}{s}\big]\)Let the sides of triangle be 4x, 5x and 6x respectively. Given: In-radius = 3 ... x \(rac{15x}{2}\) = \(rac{6x}{2}\) x h ⇒ h = \(rac{45}{6}\) = 7.5 cm.
Description : The surface of a spharical balloon is increase of its volume when its radius is 6 cm.
Last Answer : The surface of a spharical balloon is increase of its volume when its radius is 6 cm.
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
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 radius of the sphere
Last Answer : According to question find the volume of this liquid.
Description : A circle has radius √2 cm. It is divided into two segments by a chord of length 2cm.Prove that the angle subtended by the chord at a point in major segment is 45 degree . -Maths 9th
Last Answer : Given radius =2 cm Therefore AO=2 cm Let OD be the perpendicular from O on AB And AB =2cm Therefore AD=1cm (perpendicular from the centre bisects the chord) Now in triangle AOD, AO=2 cm ... by a chord at the centre is double of the angle made by the chord at any poin on the circumference)
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 : 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 : Find the total surface area of a hemisphere of radius 10 cm -Maths 9th
Last Answer : Radius of hemisphere, r = 10cm Formula: Total surface area of hemisphere = 3πr2 = 3×3.14×102 = 942 The total surface area of given hemisphere is 942 cm2.
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 : 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 : 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 : 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 : 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. -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
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 : 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