It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. -Maths 9th

It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. -Maths 9th
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Answer :

Radius of the closed cylindrical tank = 140/2 cm  = 70 cm = 0.7 m Height of the closed cylindrical tank = 1 m  Area of metal sheet required = 2πr(r + h) = 2 x 22/7 x 0.7 (1 + 0.7) = 7.48 m2
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Find the area of the sheet required to make closed cylindrical vessel of height 1 m and diameter 140 cm. -Maths 9th

Description : Find the area of the sheet required to make closed cylindrical vessel of height 1 m and diameter 140 cm. -Maths 9th

Last Answer : Required sheet = T.S.A. of cyclinder = 2πr (h+r) = 2 × 22 / 7 × 70 / 100(1 + 70 / 100) = 2 × 22 × 0.1 × 1.7 = 7.48 m2

Find the area of the sheet required to make closed cylindrical vessel of height 1 m and diameter 140 cm. -Maths 9th

Description : Find the area of the sheet required to make closed cylindrical vessel of height 1 m and diameter 140 cm. -Maths 9th

Last Answer : Required sheet = T.S.A. of cyclinder = 2πr (h+r) = 2 × 22 / 7 × 70 / 100(1 + 70 / 100) = 2 × 22 × 0.1 × 1.7 = 7.48 m2

It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same? -Maths 9th

Description : It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same? -Maths 9th

Last Answer : Let h be the height and r be the radius of a cylindrical tank. Height of cylindrical tank, h = 1m Radius = half of diameter = (140/2) cm = 70cm = 0.7m Area of sheet required = Total surface are of tank = 2πr( ... [2 (22/7) 0.7(0.7+1)] = 7.48 Therefore, 7.48 square meters of the sheet are required.

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

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.

Two cans have the same height equal to 21 cm. One can is cylindrical, the diameter of whose base is 10 cm. -Maths 9th

Description : Two cans have the same height equal to 21 cm. One can is cylindrical, the diameter of whose base is 10 cm. -Maths 9th

Last Answer : (c) 450 cm3. Required difference in capacities = 227227 x (5)2 x 21~ (10)2 x 21 = (1650 ~ 2100) cm3 = 450 cm3

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

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.

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.

A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. -Maths 9th

Description : A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. -Maths 9th

Last Answer : Given: Radius of cone, r = diameter/2 = 40/2 cm = 20cm = 0.2 m Height of cone, h = 1m Slant height of cone is l, and l2 = (r2+h2) Using given values, l2 = (0.22+12) = (1.04) Or l ... (32.028 12) = Rs.384.336 = Rs.384.34 (approximately) Therefore, the cost of painting all these cones is Rs. 384.34.

A closed iron tank 12 m long 9 m wide and 4 m deep is to be made . Determine the cost of iron sheet used at the rate of rs 5 per meter , sheet being 2 m wide. -Maths 9th

Description : A closed iron tank 12 m long 9 m wide and 4 m deep is to be made . Determine the cost of iron sheet used at the rate of rs 5 per meter , sheet being 2 m wide. -Maths 9th

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In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. -Maths 9th

Description : In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. -Maths 9th

Last Answer : Height of cylindrical pipe = Length of cylindrical pipe = 28m Radius of circular end of pipe = diameter/ 2 = 5/2 cm = 2.5cm = 0.025m Now, CSA of cylindrical pipe = 2πrh, where r = radius and h = height of ... = 2 (22/7) 0.025 28 m2 = 4.4m2 The area of the radiating surface of the system is 4.4m2.

A cylindrical container of height 14 m and base 12 m contains oil. The oil is to be transferred to one cylindrical can, -Maths 9th

Description : A cylindrical container of height 14 m and base 12 m contains oil. The oil is to be transferred to one cylindrical can, -Maths 9th

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A semi-circular sheet of metal of diameter 28 cm -Maths 9th

Description : A semi-circular sheet of metal of diameter 28 cm -Maths 9th

Last Answer : When semi-circular sheet is bent to form an open conical cup, the radius of the sheet becomes slant height of the cup and the semi-circular part of the sheet becomes the circumference of the base of the cone. ∴ Slant height of the ... 3 x 22/7 x 7 x 7 x 7√3 = 1078/3.√3 = 359.3 x 1.732 = 622.31 cm3

A semicircular thin sheet of a metal of diameter 28 cm is bent and an open conical cup is made. What is the capacity of the cup ? -Maths 9th

Description : A semicircular thin sheet of a metal of diameter 28 cm is bent and an open conical cup is made. What is the capacity of the cup ? -Maths 9th

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he frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. -Maths 9th

Description : he frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. -Maths 9th

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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

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

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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

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Water is following at the rate of 5 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 50 m long -Maths 9th

Description : Water is following at the rate of 5 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 50 m long -Maths 9th

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Description : From a wooden cylindrical block, whose diameter is equal to its height, a sphere of maximum possible volume is carved out. -Maths 9th

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A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Description : A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Last Answer : Diameter d = 7 cm Radius r = 7 / 2 cm and h = 12 cm ∴ V = πr2h = 22 / 7 × 7 / 2 × 7 / 2 × 12 = 462 Total milk for 1600 students = 462 × 1600 = 739200 cm3 = 739200 / 1000 litres = 739.2 litres .

A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Description : A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Last Answer : Diameter d = 7 cm Radius r = 7 / 2 cm and h = 12 cm ∴ V = πr2h = 22 / 7 × 7 / 2 × 7 / 2 × 12 = 462 Total milk for 1600 students = 462 × 1600 = 739200 cm3 = 739200 / 1000 litres = 739.2 litres .

A school provides milk to the students daily in a cylindrical glasses of diameter 7 cm. -Maths 9th

Description : A school provides milk to the students daily in a cylindrical glasses of diameter 7 cm. -Maths 9th

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A school provides milk to the students daily in a cylindrical glasses of diameter 7 cm. -Maths 9th

Description : A school provides milk to the students daily in a cylindrical glasses of diameter 7 cm. -Maths 9th

Last Answer : According to question find the litres of milk is needed to serve 1600 students.

A person donates cylindrical bowls of diameter 7 cm -Maths 9th

Description : A person donates cylindrical bowls of diameter 7 cm -Maths 9th

Last Answer : Radius of cylindrical bowl = 7/2 cm = 3.5 cm Height of the bowl filled with soup (h) = 4 cm Volume of soup for 1 patient = πr2h = 22/7 x 3.5 x 3.5 x 4 = 154 cm3 ∴ Volume of soup for ... ( ∴ 1L = 1000 cm3) = 38.5 L The person is kind hearted, caring and contributing for the welfare of society.

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

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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.

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

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.

Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m -Maths 9th

Description : Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m -Maths 9th

Last Answer : Radius of cone, r = 24/2 m = 12m Slant height, l = 21 m Formula: Total Surface area of the cone = πr(l+r) Total Surface area of the cone = (22/7)×12×(21+12) m2 = 1244.57m2

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 : 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

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What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m -Maths 9th

Description : What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m -Maths 9th

Last Answer : Solution: Height of conical tent, h = 8m Radius of base of tent, r = 6m Slant height of tent, l2 = (r2+h2) l2 = (62+82) = (36+64) = (100) or l = 10 Again, CSA of conical tent = πrl = (3.14 6 ... .2) 3] = 188.4 L-0.2 = 62.8 L = 63 Therefore, the length of the required tarpaulin sheet will be 63 m.

How many metres of cloth 5 m wide will be required to make a conical tent, the radius of whose base is 7 m and whose height is 24 m? -Maths 9th

Description : How many metres of cloth 5 m wide will be required to make a conical tent, the radius of whose base is 7 m and whose height is 24 m? -Maths 9th

Last Answer : Given, radius (r) = 7 m and height (h) = 24m ∴ Slant height (l) = √h2 + r2 = √242 + 72 = √625 = 25 m ∴ Length of canvas required

How many metres of 5 m wide cloth will be required to make a conical tent, the radius of whose base is 3.5 m and height is 12 m ? -Maths 9th

Description : How many metres of 5 m wide cloth will be required to make a conical tent, the radius of whose base is 3.5 m and height is 12 m ? -Maths 9th

Last Answer : l = √h2 + r2 = √(3.5)2 + (12)2 = √12.25 + 144 = √156.25 = 12.5 m Curved surface area = πrl = 22 / 7 × 3.5 × 12.5 = 137.5 m2 Area of cloth = 137.5 m2 Length of cloth required = C.S.A. / Width l = 137.5 / 5 = 27.5 m

How many metres of cloth 5 m wide will be required to make a conical tent, the radius of whose base is 7 m and whose height is 24 m? -Maths 9th

Description : How many metres of cloth 5 m wide will be required to make a conical tent, the radius of whose base is 7 m and whose height is 24 m? -Maths 9th

Last Answer : Given, radius (r) = 7 m and height (h) = 24m ∴ Slant height (l) = √h2 + r2 = √242 + 72 = √625 = 25 m ∴ Length of canvas required

How many metres of 5 m wide cloth will be required to make a conical tent, the radius of whose base is 3.5 m and height is 12 m ? -Maths 9th

Description : How many metres of 5 m wide cloth will be required to make a conical tent, the radius of whose base is 3.5 m and height is 12 m ? -Maths 9th

Last Answer : l = √h2 + r2 = √(3.5)2 + (12)2 = √12.25 + 144 = √156.25 = 12.5 m Curved surface area = πrl = 22 / 7 × 3.5 × 12.5 = 137.5 m2 Area of cloth = 137.5 m2 Length of cloth required = C.S.A. / Width l = 137.5 / 5 = 27.5 m

How many square metres of canvas is required for a conical tent whose height is 3.5 m and the radius of the base is 12 m? -Maths 9th

Description : How many square metres of canvas is required for a conical tent whose height is 3.5 m and the radius of the base is 12 m? -Maths 9th

Last Answer : Slantheight5l = h2+r2 ​=(3.5)2+(12)2​ total canvas required = πrl =π×12×12.5 =471 sq m​

How many square metres of canvas is required for a conical tent whose height is 3.5 m and the radius of the base is 12 m? -Maths 9th

Description : How many square metres of canvas is required for a conical tent whose height is 3.5 m and the radius of the base is 12 m? -Maths 9th

Last Answer : Solution of this question

The slant height and base diameter.... -Maths 9th

Description : The slant height and base diameter.... -Maths 9th

Last Answer : Radius of the base of the conical tomb (r) = 14/2 m = 7 m Slant height of conical tomb (l) = 25 m Curved surface area of conical tomb = πrl = 22/7 x 7 x 25 = 550 m2 Cost of white-washing 1 m2 = ₹ 210/100 = ₹ 2.1 ∴ Cost of white-washing 550 m2 = ₹ 550 x 2.1 = ₹ 1155

The capacity of a closed cylindrical -Maths 9th

Description : The capacity of a closed cylindrical -Maths 9th

Last Answer : Height of the cylindrical vessel (h) = 1 m Capacity of the cylindrical vessel = 15.4 L = 15.4/1000 m 3 = 0.0154 m3 Let 'r' m be the radius of the base of the cylindrical vessel Volume of the cylindrical vessel = πr2h ⇒ πr2h = 0 ... 22/7 x 0.07(0.07 + 1) m2 = 1/7 x 44 x 0.07 x 1.07 m2 = 0.4708 m2

A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of Rs. 10 per m -Maths 9th

Description : A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of Rs. 10 per m -Maths 9th

Last Answer : Answer We have, r=0.7m, h=8m ∴ Total surface area = 2πr2+2πrh=2πr(r+h)=2×722​×0.7×8.7m2 Required cost = Rs. {2×722​×0.7×8.7×10}=Rs.382.80

Diameter of the base of a cone is 10.5 cm -Maths 9th

Description : Diameter of the base of a cone is 10.5 cm -Maths 9th

Last Answer : Radius of cone (r) = 10.5/2 cm Slant height of cone (l) = 10 cm Curved surface area of cone = πrl = 22/7 x 10.5/2 x 10 = 165 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

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Find the area of a triangle with base =20cm and height are 10 cm. -Maths 9th

Description : Find the area of a triangle with base =20cm and height are 10 cm. -Maths 9th

Last Answer : Area of a triangle = 1/2 × Base × Altitude ( height ) therefore., Area of a triangle = 1/2 × 20 cm× 10cm = 10cm ×10cm = 100 cm^2

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

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The base of a right prism is an equilateral triangle with a side 6 cm and its height is 18 cm. Find its volume, -Maths 9th

Description : The base of a right prism is an equilateral triangle with a side 6 cm and its height is 18 cm. Find its volume, -Maths 9th

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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

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%

The base in a right prism is an equilateral triangle of side 8 cm and the height of the prism is 10 cm. The volume of the prism is -Maths 9th

Description : The base in a right prism is an equilateral triangle of side 8 cm and the height of the prism is 10 cm. The volume of the prism is -Maths 9th

Last Answer : ⇒ Area of equilateral triangle =43 ( s i d e)2 =43 ( 8)2 =43 64 ... =3 3 2 . 5 5 4 cm3. =3 3 2 . 5 5 4 cc

A right triangular prism of height 18 cm and of base sides 5 cm, 12 cm and 13 cm is transformed into another right triangular prism on a base -Maths 9th

Description : A right triangular prism of height 18 cm and of base sides 5 cm, 12 cm and 13 cm is transformed into another right triangular prism on a base -Maths 9th

Last Answer : Vol. of △ ular prism = Area of △ ular base × height. ∴ Area of triangular base = area of triangle PQR By heron's formula. S=S(s−a)(s−b)(s−c)​where S=2a+b+c​∴Areaof△PQR= S=23+4+5​=6 S=6(6−3)(6−4)(6−5)​=3×2×3×2×1​=6cm2 ∴ vol. of Prism =6×10 =60cm3Answer.

Water flows in a tank 150 m × 100 m at the base, through a pipe whose cross-section is 2 dm by 1.5 dm, at a speed of 15 km per hour. -Maths 9th

Description : Water flows in a tank 150 m × 100 m at the base, through a pipe whose cross-section is 2 dm by 1.5 dm, at a speed of 15 km per hour. -Maths 9th

Last Answer : Volume of water discharged through the pipe = Volume increase of the tank First, consider the pipe. Volume discharge through pipe = length breadth speed time Let the time taken to fill the tank to 3 m depth be t. ... the tank V=150 100 3 cu. m ⇒ V=450 100 Therefore, 450t=450 100 ⇒ t=100 hours

The inner diameter of a cylindrical wooden -Maths 9th

Description : The inner diameter of a cylindrical wooden -Maths 9th

Last Answer : Inner radius of the cylindrical pipe (r) = 24/2 cm = 12 cm Outer radius of the cylindrical pipe (R) = 28/2 cm = 14 cm Length of the cylindrical pipe (h) = 35 cm Volume of the wood used in making the cylindrical pipe ... x 0.6) g = 3432 g = 3432/1000 kg = 3.432 kg Thus, the mass of pipe = 3.43 kg

A cylindrical vessel of diameter 12 cm contains `800picm^(3)` of water. A cylindrical glass piece of diameter `8.0cm` and height ``8.0 cm is placed in

Description : A cylindrical vessel of diameter 12 cm contains `800picm^(3)` of water. A cylindrical glass piece of diameter `8.0cm` and height ``8.0 cm is placed in

Last Answer : A cylindrical vessel of diameter 12 cm contains `800picm^(3)` of water. A cylindrical glass piece of diameter ... is 1.50 and that of water is 1.33.