Mukta had to make a model of a cylindrical -Maths 9th

Mukta had to make a model of a cylindrical -Maths 9th
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Radius of the base of the cylindrical kaleidoscope = r = 3.5 cm  Height of kaleidoscope = h = 25 cm  Chart paper required = curved surface area of kaleidoscope = 2 πrh = 2 x 22/7 x 3.5 x 25 = 550 cm2
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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 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 1 m and base diameter 140 cm from a metal sheet. -Maths 9th

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

Last 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

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.

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.

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 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 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 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 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 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 : Volume of milk in 1 glass =πr2h =π×(3.5)2×12=461.58cm2​for 1600 students milk needed is =1600×461.58=738258litre​

A cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m2. -Maths 9th

Description : A cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m2. -Maths 9th

Last Answer : NEED ANSWER

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 cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m2. -Maths 9th

Description : A cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m2. -Maths 9th

Last Answer : Solution of the question

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

The circumference of the base of a cylindrical -Maths 9th

Description : The circumference of the base of a cylindrical -Maths 9th

Last Answer : Height of the cylindrical vessel (h) = 25 cm Let r cm be the radius of the base of the cylindrical vessel. Circumference of the base = 2 πr = 132 cm ⇒ 2 x 22/7 x r = 132 ⇒ r = 132 x 7/2 x 22 = 21 ... 21 x 21 x 25 = 34650 cm3 ∴ Volume of water which vessel can hold = 34650 L/ 1000 L = 34.65 L

A cylindrical tube opened at both the ends -Maths 9th

Description : A cylindrical tube opened at both the ends -Maths 9th

Last Answer : Outer radius of the cylindrical tube (R) = 16/2 cm = 8 cm Inner radius of the cylindrical tube (r) = (8 - 2) cm = 6 cm Length of the tube (h) = 100 cm Volume of iron used in making the cylindrical tube = πR2h - πr2h = πh(R2 - r2) = 22/7 x100(82 – 62) = 22/7 x 100 x 14 x 2 = 8800 cm3

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 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 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

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

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 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

Last Answer : answer:

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

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

Last Answer : answer:

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

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=34​πr3 No. of balls =34​πr34πr3​=

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

Last Answer : answer:

Three identical balls fit snugly into a cylindrical can. The radius of the spheres is equal to the radius of the can and the balls just touch the -Maths 9th

Description : Three identical balls fit snugly into a cylindrical can. The radius of the spheres is equal to the radius of the can and the balls just touch the -Maths 9th

Last Answer : hope its clear

How to make Working model on the area of trapezium -Maths 9th

Description : How to make Working model on the area of trapezium -Maths 9th

Last Answer : NEED ANSWER

How to make Working model on the area of trapezium -Maths 9th

Description : How to make Working model on the area of trapezium -Maths 9th

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working model on area of trapezium -Maths 9th

Description : working model on area of trapezium -Maths 9th

Last Answer : NEED ANSWER

a working model on area of trapezium -Maths 9th

Description : a working model on area of trapezium -Maths 9th

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working model on area of trapezium -Maths 9th

Description : working model on area of trapezium -Maths 9th

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a working model on area of trapezium -Maths 9th

Description : a working model on area of trapezium -Maths 9th

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On planet A , each alien has 3 eyes and on planet B , each alien has 4 eyes. A group of aliens from planets A and B landed on earth. They had 30 eyes in all. Assuming number of aliens from planets A and B respectively , express the situation in the form of linear equation in two variables. -Maths 9th

Description : On planet A , each alien has 3 eyes and on planet B , each alien has 4 eyes. A group of aliens from planets A and B landed on earth. They had 30 eyes in all. Assuming number of ... planets A and B respectively , express the situation in the form of linear equation in two variables. -Maths 9th

Last Answer : answer:

How had India kept her relationship with the world in ancient times? -Maths 9th

Description : How had India kept her relationship with the world in ancient times? -Maths 9th

Last Answer : INDIA's contact with the world have continued through the ages but her relationships through the land route are much older than her maritime contacts.The various passes across the mountains in the ... Panchtantra, the Indian numerals and the decimal system thus could reach many parts of the world.

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.

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

In a school marks obtained by 80 students are given in the table. Draw a histogram. Also, make frequency polygon. -Maths 9th

Description : In a school marks obtained by 80 students are given in the table. Draw a histogram. Also, make frequency polygon. -Maths 9th

Last Answer : Here, class size = 135 - 305 = 10 ∴ Lower limit of first class interval is 305 - 10 / 2 = 300 Upper limit of first class interval is 305 + 10 / 2 = 310 Thus, first class interval is 300 - 310 Required histogram and frequency polygon is given on the graph paper .

The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Description : The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Last Answer : Solution of this question

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

In a school marks obtained by 80 students are given in the table. Draw a histogram. Also, make frequency polygon. -Maths 9th

Description : In a school marks obtained by 80 students are given in the table. Draw a histogram. Also, make frequency polygon. -Maths 9th

Last Answer : Here, class size = 135 - 305 = 10 ∴ Lower limit of first class interval is 305 - 10 / 2 = 300 Upper limit of first class interval is 305 + 10 / 2 = 310 Thus, first class interval is 300 - 310 Required histogram and frequency polygon is given on the graph paper .

The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Description : The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Last Answer : Solution of this question

How much paper of each shade is needed to make a kite given in figure, in which ABCD is a square with diagonal 44 cm. -Maths 9th

Description : How much paper of each shade is needed to make a kite given in figure, in which ABCD is a square with diagonal 44 cm. -Maths 9th

Last Answer : NEED ANSWER

How much paper of each shade is needed to make a kite given in figure, in which ABCD is a square with diagonal 44 cm. -Maths 9th

Description : How much paper of each shade is needed to make a kite given in figure, in which ABCD is a square with diagonal 44 cm. -Maths 9th

Last Answer : According to question ABCD is a square with diagonal 44 cm.

If two intersecting chords of a circle make equal angles ...... -Maths 9th

Description : If two intersecting chords of a circle make equal angles ...... -Maths 9th

Last Answer : Given: AB and CD are two chords of a circle with centre O, intersecting at point E. PQ is a diameter through E, such that ∠AEQ = ∠DEQ. To prove: AB = CD Construction: Draw OL perpendicular AB and ... the circle. Since, chords of a circle which are equidistant from the centre are equal. ∴ AB = CD

For annual day, Sakshi and Nidhi were asked to make one rangoli -Maths 9th

Description : For annual day, Sakshi and Nidhi were asked to make one rangoli -Maths 9th

Last Answer : In △ABC and △PQR, BC = QR (Given) ⇒ 1/2BC = 1/2QR ⇒ BM = QN In triangles ABM and PQN, we have AB = PQ (Given) BM = QN (Proved above) AM = PN ... BC = QR (Given) ∴ ΔΑΒC ≅ ΔPOR (SAS congruence criterion) Participation, beauty, hardworking.

A metallic sheet is of rectangular shape with dimensions 28m × 36m. From each of its corners, a square is cut off so as to make an open box. -Maths 9th

Description : A metallic sheet is of rectangular shape with dimensions 28m × 36m. From each of its corners, a square is cut off so as to make an open box. -Maths 9th

Last Answer : R.E.F image Volume of box =l×b×h From the diagram l=48−2(8) ∵ Two square formed side =32m b=36−2(8) =20m Also h=8m from question ∴ Volume =32×20×8 =5120m3