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

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

1 Answer

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
by

Related questions

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.

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

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.

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

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

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

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

Last Answer : 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

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

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

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

Which of the listed precautions should be taken when cleaning the internals of a motor with compressed air? A. Open the machine on both ends so as to allow the air and dust to escape. B. Be certain that the circuit breaker is opened and tagged on the feeder panel. C. Be certain that the air is clean and as dry as possible. D. All of the above.

Description : Which of the listed precautions should be taken when cleaning the internals of a motor with compressed air? A. Open the machine on both ends so as to allow the air and dust to escape. B. Be certain that the ... panel. C. Be certain that the air is clean and as dry as possible. D. All of the above.

Last Answer : Answer: D

Two tube A and B can fill a cistern in 48 mins and 64 mins. If both the tubes are opened simultaneously, after how much time B should be closed so that the cistern is full in 36 minutes? A) 20 mins B) 16mins C) 22mins D) 14mins

Description : Two tube A and B can fill a cistern in 48 mins and 64 mins. If both the tubes are opened simultaneously, after how much time B should be closed so that the cistern is full in 36 minutes? A) 20 mins B) 16mins C) 22mins D) 14mins

Last Answer : B Let B be closed after ‘x’ mins. Then part filled by (A+B) in x mins + part filled by A in (36 – x) mins = 1 Therefore x(1/48 + 1/64) + (36 – x)* 1/48 = 1 7x/192 + 36 – x/48 = 1 7x + 4(36-x)/192 = 1 7x+144 – 4x = 192 3x = 48 =>x = 16 Here B must be closed after 16 mins.

Which of these is a tiny, cylindrical, multi-celled carbon tube used for minimizing instruments used in space applications?

Description : Which of these is a tiny, cylindrical, multi-celled carbon tube used for minimizing instruments used in space applications?

Last Answer : nanotube

A circuit breaker differs from a fuse in that a circuit breaker _____________. A. melts and must be replaced B. is enclosed in a tube of insulating material with metal ferrules at each end C. gives no visual indication of having opened the circuit D. trips to break the circuit and may be reset

Description : A circuit breaker differs from a fuse in that a circuit breaker _____________. A. melts and must be replaced B. is enclosed in a tube of insulating material with metal ferrules at each end C. gives no visual indication of having opened the circuit D. trips to break the circuit and may be reset

Last Answer : Answer: D

A tube can fill a tank completely in 18 hours. After half the tank is filled , one more similar tube is opened. What is the total time taken to fill the tank completely ? A) 14hrs 20min B) 13hrs 30min C) 13hrs 10min D) 14hrs 30min

Description : A tube can fill a tank completely in 18 hours. After half the tank is filled , one more similar tube is opened. What is the total time taken to fill the tank completely ? A) 14hrs 20min B) 13hrs 30min C) 13hrs 10min D) 14hrs 30min

Last Answer : C A tube can fill the half tank in 9hrs Now another similar tube opened 1/18+1/18 = 2/18 = 1/9 Remaining half tank filled in 4.5hrs Total time = 9+4.5 =13.5 = 13hrs 30min

A tube can fill a cistern in 18hrs.After half the cistern is filled, three more similar tubes are opened. What is the total time taken to fill the cistern completely ? A) 9hrs 52min B) 10hrs 15 min C) 9hrs 45 min D) 10hrs 30min

Description : A tube can fill a cistern in 18hrs.After half the cistern is filled, three more similar tubes are opened. What is the total time taken to fill the cistern completely ? A) 9hrs 52min B) 10hrs 15 min C) 9hrs 45 min D) 10hrs 30min

Last Answer : Time taken to fill half of the cistern = 18/2 = 9 hrs After half the cistern is filled , three more similar tubes are opened. We know that one tube can in 18 hrs can fill = V litres(i.e. fully fill a cistern) ... V/2 will be 2 hrs 15 min Total time = 9 hrs + 2 hrs + 15 min = 11 hrs 15 min

Surgical removal or cutting and ligation of the ends of oviduct or small part of the fallopian tube is removed or tied up is known as

Description : Surgical removal or cutting and ligation of the ends of oviduct or small part of the fallopian tube is removed or tied up is known as

Last Answer : Surgical removal or cutting and ligation of the ends of oviduct or small part of the fallopian ... tubectomy B. oviductomy C. castration D. vasectomy

During the start-up of the circuit shown in the illustration, it is noted that the ends of component 'C' remain lighted but the tube does not illuminate. The cause of this problem is ____________. EL-0081 A. component 'A' is open B. component 'D' is closed C. component 'C' is the wrong wattage D. component 'B' contacts are stuck closed

Description : During the start-up of the circuit shown in the illustration, it is noted that the ends of component 'C' remain lighted but the tube does not illuminate. The cause of this problem is ____________. EL- ... ' is closed C. component 'C' is the wrong wattage D. component 'B' contacts are stuck closed

Last Answer : Answer: D

What is indicated by gradual blackening at the ends of component 'C' shown in the illustration? EL-0081 A. The unit is in danger of exploding. B. The tube is nearing the end of its useful life. C. The circuit voltage is too high. D. The circuit current is too high.

Description : What is indicated by gradual blackening at the ends of component 'C' shown in the illustration? EL-0081 A. The unit is in danger of exploding. B. The tube is nearing the end of its useful life. C. The circuit voltage is too high. D. The circuit current is too high.

Last Answer : Answer: B

A hollow circular tube with a 2.3-cm I.D. and 2.5-cm O.D. is rigidly supported at its ends. A 2.5 kN-m torque is applied at the center of this tube. What is the maximum shear stress acting on this tube? (a) 2.87Mpa (b) 4.27 Mpa (c) 6.92 Mpa (d) 10.2 Mpa

Description : A hollow circular tube with a 2.3-cm I.D. and 2.5-cm O.D. is rigidly supported at its ends. A 2.5 kN-m torque is applied at the center of this tube. What is the maximum shear stress acting on this tube? (a) 2.87Mpa (b) 4.27 Mpa (c) 6.92 Mpa (d) 10.2 Mpa

Last Answer : (a) 2.87Mpa

If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Description : If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Last Answer : Let p(x) = ax3 + x2 - 2x + b Since (x+1) and (x-1) are the factors of p(x), ∴ p(-1) = 0 and p(1) = 0 ∴ p(-1) = a(-1)3 + (-1)2 - 2 (-1) + b = 0 ⇒ - a + 1 + 2 + b = 0 ⇒ a - b = 3 ---- (i) ... 0 ⇒ a + 1 - 2 + b = 0 ⇒ a + b = 1 ----- (ii) solving equations (i) and (ii) we get a = 2 and b = -1

If two opposite sides of a cyclic quadrilateral are parallel , then prove that - (a) remaining two sides are equal (b) both the diagonals are equal -Maths 9th

Description : If two opposite sides of a cyclic quadrilateral are parallel , then prove that - (a) remaining two sides are equal (b) both the diagonals are equal -Maths 9th

Last Answer : Let ABCD be quadrilateral with ab||cd Join be. In triangle abd and CBD, Angle abd=angle cdb(alternate angles) Anglecbd=angle adb(alternate angles) Bd=bd(common) Abd=~CBD by asa test Ad=BC by cpct Since ad ... c(from 1) Ad =bc(proved above) Triangle adc=~bcd by sas test Ac=bd by cpct Hence proved

If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Description : If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Last Answer : Show that p = r.

A point both of whose coordinates are negative will lie in -Maths 9th

Description : A point both of whose coordinates are negative will lie in -Maths 9th

Last Answer : (c) A point both of whose coordinates are negative will lie in III quadrant because, in III quadrant x-coordinate and y-coordinate both are negative.

If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation. -Maths 9th

Description : If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation. -Maths 9th

Last Answer : (b) By property, if we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation remains the same i.e., the solution of the linear equation is remains unchanged.

If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Description : If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Last Answer : Let p(x) = ax3 + x2 - 2x + b Since (x+1) and (x-1) are the factors of p(x), ∴ p(-1) = 0 and p(1) = 0 ∴ p(-1) = a(-1)3 + (-1)2 - 2 (-1) + b = 0 ⇒ - a + 1 + 2 + b = 0 ⇒ a - b = 3 ---- (i) ... 0 ⇒ a + 1 - 2 + b = 0 ⇒ a + b = 1 ----- (ii) solving equations (i) and (ii) we get a = 2 and b = -1

If two opposite sides of a cyclic quadrilateral are parallel , then prove that - (a) remaining two sides are equal (b) both the diagonals are equal -Maths 9th

Description : If two opposite sides of a cyclic quadrilateral are parallel , then prove that - (a) remaining two sides are equal (b) both the diagonals are equal -Maths 9th

Last Answer : Let ABCD be quadrilateral with ab||cd Join be. In triangle abd and CBD, Angle abd=angle cdb(alternate angles) Anglecbd=angle adb(alternate angles) Bd=bd(common) Abd=~CBD by asa test Ad=BC by cpct Since ad ... c(from 1) Ad =bc(proved above) Triangle adc=~bcd by sas test Ac=bd by cpct Hence proved

If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Description : If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Last Answer : Show that p = r.

A point both of whose coordinates are negative will lie in -Maths 9th

Description : A point both of whose coordinates are negative will lie in -Maths 9th

Last Answer : (c) A point both of whose coordinates are negative will lie in III quadrant because, in III quadrant x-coordinate and y-coordinate both are negative.

If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation. -Maths 9th

Description : If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation. -Maths 9th

Last Answer : (b) By property, if we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation remains the same i.e., the solution of the linear equation is remains unchanged.