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

1 Answer

Answer :

Convert all to metres: 5 km = 5000 m 14 cm = 0.14 m 7 cm = 0.07 m Find the radius: Radius = Diameter ÷ 2 Radius = 0.14 ÷ 2 = 0.07 m Find the amount of water that flowed out in an hour: Volume = πr²h Volume = π (0.07)² (5000) = 77 m³ per hour Find the amount of water needed in the tank: Volume = Length x Breadth x Height Volume = 50 x 44 x 0.07 = 154 m³ Find the number of hours needed: Number of hours = 154 ÷ 77 = 2 hours It takes 2 hours to fill up the tank to rise by 7 cm

Related questions

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

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.

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.

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

Description : A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4cm. -Maths 9th

Last Answer : Let r1 and r2 Inner and outer radii of cylindrical pipe r1 = 4/2 cm = 2 cm r2 = 4.4/2 cm = 2.2 cm Height of cylindrical pipe, h = length of cylindrical pipe = 77 cm (i) curved surface ... CSA of roller = (500 31680) cm2 = 15840000 cm2 = 1584 m2. Therefore, area of playground is 1584 m2. Answer!

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

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.

Description : A design is made on a rectangular tile of dimensions 50 cm x 17 cm as shown in figure. -Maths 9th

Last Answer : According to question find the the total area of the design and the remaining area of the tiles.

Description : A design is made on a rectangular tile of dimensions 50 cm x 17 cm as shown in figure. -Maths 9th

Last Answer : According to question find the the total area of the design and the remaining area of the tiles.

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:

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.

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.

Description : 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

Last Answer : Height of cylinder, h = 14cm Let the diameter of the cylinder be d Curved surface area of cylinder = 88 cm2 We know that, formula to find Curved surface area of cylinder is 2πrh. So 2πrh =88 cm2 (r is the ... 88 cm2 2r = 2 cm d =2 cm Therefore, the diameter of the base of the cylinder is 2 cm.

Description : Nidhi has to find the area of a sphere whose diameter was 14 cm. -Maths 9th

Last Answer : Area is two-dimensional while 4 πr represents a length.

Description : A semi-circle of diameter 14 cm has three chords of equal length connecting the two end points of the diameter so as -Maths 9th

Last Answer : (c)\(\bigg(rac{147 imes\sqrt3}{4}\bigg)\) cm2Take the trapezoid ABCD in the semi-circle with centre O such that AD = DC = CB. Now, complete the circle and draw an identical trapezoid in the other semicircle also. Then, ADCBEF ... {1}{2}\)x \(rac{3\sqrt3}{2}\) x 49 = \(rac{147 imes\sqrt3}{4}\) cm2.

Description : In a rectangular field of dimension 60 m x 50 m, -Maths 9th

Last Answer : Area of rectangular field = length x breadth = 60 x 50 = 3,000 m2 Now, a = 50 m, b = 45 m, c = 35 m s = (a + b + c)/2 = (50 + 45 + 35)/2 = 130/2 = 65 m By Heron's formula ... m2 ( approximately ) Hence, the remaining area = Area of rectangle - Area of triangle = 3,000 - 764.85 = 2,235.15 m2

Description : A rectangular piece of paper is 22 cm long and 10 cm wide. -Maths 9th

Last Answer : Since rectangular piece of paper of rolled along its length. ∴ 2πr = 22 r = 22 × 7 / 2 × 22 = 3.5 cm Height of cyclinder (h) = 10 cm ∴ Volume of cyclinder = πr2h = 22 / 7 × 3.5 × 3.5 × 10 = 385 cm3.

Description : A rectangular piece of paper is 22 cm long and 10 cm wide. -Maths 9th

Last Answer : Since rectangular piece of paper of rolled along its length. ∴ 2πr = 22 r = 22 × 7 / 2 × 22 = 3.5 cm Height of cyclinder (h) = 10 cm ∴ Volume of cyclinder = πr2h = 22 / 7 × 3.5 × 3.5 × 10 = 385 cm3.

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

Last Answer : This answer was deleted by our moderators...

Description : The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs.10 per m2 is Rs.15000, find the height of the hall. -Maths 9th

Last Answer : Let length, breadth, and height of the rectangular hall be l, b, and h respectively. Area of four walls = 2lh+2bh = 2(l+b)h Perimeter of the floor of hall = 2(l+b) = 250 m Area of four walls = 2( ... of paining the walls is Rs. 15000. 15000 = 2500h Or h = 6 Therefore, the height of the hall is 6 m.

Description : A passenger train 165 m long which is running at a speed of 30km/hr. In what time will it pass a man who is running at a speed of 3 km/hr in the opposite direction. In which the train is moving. A)14 sec B) 17sec C)16sec D)18sec

Last Answer : ANSWER: D Explanation:  Speed of the train = 30 km/hr  Speed of man = 3 km/hr  Length of the train = 165m If direction is given in opposite direction then we add both speed.  That is (30 + 3 )= 33 km/hr ... can easily get the distance of m/s  So it cover 165 m  i.e., 165 * 6 /55 = 18 sec.

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

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.

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

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

Description : A train 125 m long passes a man, running at 5 kmph in the same direction in which the train is going, in 10 seconds. The speed of the train is: A.35 km/hr B.50 km/hr C.48 km/hr D.55 km/hr E.None of these

Last Answer : Answer – B (50 km/hr) Explanation – Let speed be S Relative speed = total lengths/time [ in this case man length is neglible compare to train so neglected] (S – 5) x 5/18 = 125 /10 S- 5 = (125/10)x(18/5) S-5 = 45 S=50 km/hr

Description : A metallic sheet is of rectangular shape with dimensions 48 cm x 36 cm. From each of its corners, a square of 8 cm is cut-off and an open box is made of the remaining sheet. Find the volume of the box. -Maths 9th

Last Answer : When squares of 8 cm is cutt-off from rectangulare sheet then, Length of box (l) = (98 - 8 - 8) = 32 cm Breadth of box (b) = (36 - 8 - 8) = 20 cm Height of box (h) = 8cm ∴ Volume of box = lbh = 32 x 20 x 8 = 5120 cm3

Description : A metallic sheet is of rectangular shape with dimensions 48 cm x 36 cm. From each of its corners, a square of 8 cm is cut-off and an open box is made of the remaining sheet. Find the volume of the box. -Maths 9th

Last Answer : When squares of 8 cm is cutt-off from rectangulare sheet then, Length of box (l) = (98 - 8 - 8) = 32 cm Breadth of box (b) = (36 - 8 - 8) = 20 cm Height of box (h) = 8cm ∴ Volume of box = lbh = 32 x 20 x 8 = 5120 cm3

Description : A rectangular paper 11 cm by 8 cm can be exactly wrapped to cover the curved surface of a cylinder of height 8 cm . -Maths 9th

Last Answer : answer:

Description : The volume of a certain rectangular solid is 8 cm^3. Its total surface area is 32 cm^2 and its three dimensions are in geometric progression. -Maths 9th

Last Answer : (b) 32 Let the edges of the solid be a, ar, ar2. Then, Volume = a x ar x ar2 = a3r3 = (ar)3. Given (ar)3 = 8 ⇒ ar = 2 Also, surface area = 2(a x ar + ar x ar2 + a × ar2) = 2(a2r + ... Given, 2ar (a + ar + ar2) = 32 ⇒ 4(a + ar + ar2) = 32 ; Sum of lengths of all edges = 32.

Description : Senthil started bike at 5 a.m to reach a temple. After going temple, his bike went out of order. Consequently he rested for 45 mins and came back to his house walking all the way. Senthil reached home at 8 a.m. if bike ... /hr, then on bike he covered a distance of. a) 3.12 b) 2.76 c) 4.13 d) 2.14

Last Answer : D Time taken = 2 hr 15 mins = 2 ¼ hrs Time = 9/4 hrs Let the required distance be x km Then x/20 + x/1 = 9/4 (X+20x) / 20 = 9/4 21x / 20 = 9/4 84x = 180 X= 2.14 km Therefore the required diatance is 2.14 km

Description : A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. -Maths 9th

Last Answer : Let ABCD is a rectangular plot having a measurement of 40 m long and 15 m front. ∴ Length of inner-rectangle, EF = 40 - 3 - 3 = 34 m and breadth of inner-rectangle, FG =15 - 2 - 2 = ... [∴ area of a rectangle = length x breadth] Hence, the largest area where the house can be constructed in 374 m2

Description : A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. -Maths 9th

Last Answer : Let ABCD is a rectangular plot having a measurement of 40 m long and 15 m front. ∴ Length of inner-rectangle, EF = 40 - 3 - 3 = 34 m and breadth of inner-rectangle, FG =15 - 2 - 2 = ... [∴ area of a rectangle = length x breadth] Hence, the largest area where the house can be constructed in 374 m2.

Description : A 250 m goods train takes 50 s to cross a person who is going in the same direction with the speed of 8 km/h. After crossing that person, the train can reach next station in 2 hr. How long that person takes to reach that station after being crossed by them? A) 2 ½ hr B) 5hr C) 3 1/2hr D) 1hr

Last Answer : ANSWER:A Explanation:  Speed of the person = 8 kmph  = (8*5/18)m/s  = 20/9 m/s  Then relative speed of train = (x-20/9)m/s  As train takes 50 sec to cross the person  Therefore 50 =250/(x-20/ ... 20,000 m  = 20 km Thus time taken by the person to cover the distance of 20km= 20/8 hr=2 ½ hr

Description : A man rows to a place 48km distant and back in 14 hours. He finds that he can row 4km with the stream in the same time as 3km against the stream. Find the rate of the stream. A.2 km/hr B.1 km/hr C.3 km/hr D.3.5km/hr

Last Answer : Answer- B Basic Formula: Speed of the stream = ½ (a-b) km / hr Speed = distance traveled / time taken Explanation: Suppose he moves 4km downstream in x hours Then, downstream a= 4 / x km/hr Speed upstream b = 3/ x km/hr ... = 1/2 a=8 km/hr ,b = 6 km/hr rate of stream = ½ (8 - 6 ) = 1 km/hr

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.

Description : A tank internally measuring 150 cm × 120 cm × 100 cm has 1281600 cm^3 of water in it. Porous bricks are placed in the water until the tank -Maths 9th

Last Answer : Volume of a brick = 20 cm 6 cm 4 cm = 480 cm3. Water absorbed by one brick = (110 480)(110 480) cm3 = 48 cm3. Let x bricks be placed in the water. Then, x bricks absorb 48x cm3 of water. ... tank ⇒ 1281600 + 480xx - 48xx = 150 x 120 x 100 ⇒ 432xx = 1800000 - 1281600 = 518400 ⇒ xx = 1200.

Description : A cuboidal water tank is 6 m long, -Maths 9th

Last Answer : Volume of cuboidal tank = 1 x b x h = 6 m x 5 m x 4.5 m = 135 m3 = 135 x 1000 L = 135000 L

Description : The outer and inner diameters of a circular pipe are 6 cm and 4 cm respectively. If its length is 10 cm, then what is the total surface -Maths 9th

Last Answer : answer:

Description : Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is: A.40 m B.55 m C.65 m D.50 m E.None of these

Last Answer : Answer- D (50) Explanation – Relative speed= total lengths/time (46-36) x 5/18 = [L + L ] / 36 10 x (5/18) x36 x (1/2)= L L=50 m

Description : Ravi walks to and fro to a Gym. He spends 30 minutes in gym. If he walks at speed of 20 km an hour, he returns to home at 8.00 a.m. If he walks at 30 km an hour, he returns to home at 7.30 a.m. How fast must he walk in order to return at 7.15 hours? a) 40 km/hr b) 30 km/hr c) 60 km/hr d) 50 km/hr

Last Answer :  A As per the question, let D be the total distance and ‘t’ is the time taken. So we have: D=20t 20t =30(t-0.5) 20t=30t-15 10t=15 t=3/2 D= 30 km Now, for the condition given we have: 30=S(t-3/4) 30=S(3/2-3/4) 30=S((6-3)/4) 30=S(3/4) S=40 km/hr

Description : The dimensions of a rectangular settling tank are: length 24 m, width 6 m and depth 3 m. If 2 hour detention period for tanks is recommended, the rate of flow of sewage per hour, is A. 204 cu.m B. 208 cu.m C. 212 cu.m D. 216 cu.m

Last Answer : ANS: D

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 34​πr3 As the diameter of the ball is 4.2 cm, its radius r=2.1 cm Hence, volume of water displaced =34​πr3=34​×722​×2.1×2.1×2.1=38.808 cm3

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 : Solution of this question

Description : Normal values of overflow rate for plain sedimentation tank using coagulants, is (A) 750 to 1000 litres/hr/m2 (B) 1000 to 1250 litres/hr/m2 (C) 1250 to 1500 litres/hr/m2 (D) 1500 litres/hr/m

Last Answer : (B) 1000 to 1250 litres/hr/m2

Description : Normal values of overflow rate for plain sedimentation tank, is  (A) 250 to 500 litres/hr/m2 (B) 500 to 750 litres/hr/m2 (C) 750 to 1000 litres/hr/m2 (D) 1000 to 1250 litres/hr/m

Last Answer : (B) 500 to 750 litres/hr/m

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

Last Answer : Say h = height of the frame of lampshade, looks like cylindrical shape r = radius Total height is h = (2.5+30+2.5) cm = 35cm and r = (20/2) cm = 10cm Use curved surface area formula to find the ... 2πrh = (2 (22/7) 10 35) cm2 = 2200 cm2 Hence, 2200 cm2 cloth is required for covering the lampshade.

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

Last Answer : Radius of the base of cone = diameter/ 2 = (10.5/2)cm = 5.25cm Slant height of cone, say l = 10 cm CSA of cone is = πrl = (22/7)×5.25×10 = 165

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

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: