The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th
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Find the ratios of the surface areas of the balloon in the two cases.
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The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

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

The radius of a spherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

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

The radius of a spherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

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

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

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.

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

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.

A hemispherical bowl is made of steel 0.5 cm thick. The inside radius of the bowl is 4 cm. What is the volume of steel used in making the bowl ? -Maths 9th

Description : A hemispherical bowl is made of steel 0.5 cm thick. The inside radius of the bowl is 4 cm. What is the volume of steel used in making the bowl ? -Maths 9th

Last Answer : answer:

Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside. -Maths 9th

Description : Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside. -Maths 9th

Last Answer : Here, radius of hemispherical dome (r) = 14 m Surface area of dome = 2πr2 = 2 × 22 / 7 × 14 × 14 = 1232 m2 Hence , total surface area to be whitewashed from outside is 1232 m2

Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside. -Maths 9th

Description : Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside. -Maths 9th

Last Answer : Here, radius of hemispherical dome (r) = 14 m Surface area of dome = 2πr2 = 2 × 22 / 7 × 14 × 14 = 1232 m2 Hence , total surface area to be whitewashed from outside is 1232 m2

The radius of a spherical balloon increases -Maths 9th

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

A hemispherical bowl is made of steel 0.25 cm thick. -Maths 9th

Description : A hemispherical bowl is made of steel 0.25 cm thick. -Maths 9th

Last Answer : Outer radius of hemispherical bowl = R = r + 0.25 = 5 + 0.25 = 5.25 cm Outer curved surface area of bowl = 2 πR2 = 2 x 22/7 x (5.25)2 = 2 x 22 x 5.25 x 0.75 = 173.25 cm2

A hemispherical bowl has its external diameter equal to 10 cm and its thickness is 1 cm. What is the whole surface area of the bowl ? -Maths 9th

Description : A hemispherical bowl has its external diameter equal to 10 cm and its thickness is 1 cm. What is the whole surface area of the bowl ? -Maths 9th

Last Answer : External radius of hemispherical bowl = 5 cm Internal radius of the bowl = (5 – 1) cm = 4 cm Surface area of external portion = 2π(5)2 = 50 p sq. cm Surface area of internal portion = 2π(4)2 = ... = 91π sq. cm = (91×227)(91×227) sq. cm = 13 × 22 sq. cm = 286 cm2

A hemispherical bowl is 176 cm round the brim. Supposing it to be half full, how many persons may be served from it in hemispherical -Maths 9th

Description : A hemispherical bowl is 176 cm round the brim. Supposing it to be half full, how many persons may be served from it in hemispherical -Maths 9th

Last Answer : Let the radius of the hemispherical bowl be r cm Then, 2πr=176⇒r=176×72×22⇒r=282πr=176⇒r=176×72×22⇒r=28 Volume of liquid in the bowl : =12×( ... : =Volume of liquid in the bowlVolume of 1 glass=(14×28×282×2×2)= 1372

How much ice-cream can be put into a cone with base radius 3.5 cm and height 12 cm ? -Maths 9th

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

How much ice-cream can be put into a cone with base radius 3.5 cm and height 12 cm ? -Maths 9th

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

A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2. -Maths 9th

Description : A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2. -Maths 9th

Last Answer : Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm Formula for Surface area of hemispherical bowl = 2πr2 = 2 (22/7) (5.25)2 = 173.25 Surface area of hemispherical ... of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm2 is Rs 27.72.

Two hemispherical domes are to be painted -Maths 9th

Description : Two hemispherical domes are to be painted -Maths 9th

Last Answer : Let radii of the bases of two domes be r and R. ∴ 2πr = 17.6 ⇒ 2 x 22/7 x r = 17.6 ⇒ r = 17.6 x 7/2 x 22 = 2.8 cm and 2 πR = 70.4 ⇒ 2 x 22/7 x R = 70.4 ⇒ R = 70.4 x 7/2 x 22 = ... .2 x 11.2 = 49.28 + 788.48 cm2 = 837.76 cm2 Cost of painting at the rate of ₹10 per cm2 = 837.76 x 10 = ₹8377.6

A hemispherical tank is made up of an iron -Maths 9th

Description : A hemispherical tank is made up of an iron -Maths 9th

Last Answer : Inner radius of the hemispherical tank (r) = 1 m Outer radius of the hemispherical tank (R) = 1 + 0.01 = 1.01 m Volume of iron used to make the hemispherical tank = 2/3 πR3 - 2/3πr3 = 2/3π(R3 - r3) = 2/3 x 22/7 [(1.01)3 - 13] = 44/21(1.0303 - 1) = 44/21 x 0.0303 = 0.06349 m3

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

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 .

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

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 .

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

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.

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

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

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

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.

The surface of a spharical balloon is increase of its volume when its radius is 6 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.

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

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

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

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

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 : 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

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

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=34​πr3=33.5238cm3 16 sphere=16(33.5238)=536.3808 Volume of liquid=1024−536.3808=488cm3

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 : 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 : According to question find the radius of the sphere

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

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 : According to question find the volume of this liquid.

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

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)

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

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

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

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.

Find the total surface area of a hemisphere of radius 10 cm -Maths 9th

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.

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

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.

Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base -Maths 9th

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.

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

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

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

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

How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

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

A cone is 8.4 cm high and the radius of its base is 2.1 cm. -Maths 9th

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

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

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.

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

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

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

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.

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

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

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

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

How many balls, each of radius 2 cm can be made from a solid sphere of lead of radius 8 cm ? -Maths 9th

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

A cone is 8.4 cm high and the radius of its base is 2.1 cm. -Maths 9th

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

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

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

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

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.

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

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.

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

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 : NEED ANSWER