NCERT Solutions For Class 10 Maths Chapter 13

NCERT Solutions Class 10 Maths Surface Areas and Volumes

Ncert Solutions For Class 10 Maths Chapter 13 PDF Download

NCERT solutions for class 10 Maths chapter 13 Surface Areas and Volumes is an important chapter for board examination. Surface area and volume questions hold more marks in the board examination, 1 or 2 questions of 5 marks each is asked in the board examination. To score good marks in mathematics and in class 10th board examination it is recommended to solve NCERT questions provided at the end of each chapter.

NCERT Solution for class 10 maths chapter 13 is provided here so that students can have have a look at these questions whenever they are facing any difficulties while solving the NCERT questions.

NCERT Solutions Class 10 Maths Chapter 13 Exercises

Short answer questions:

  1. Three metallic solid cubes, whose edges are 3 cm, 4cm, and 5cm, are melted and formed into a single cube. Find the edges of the cube so formed.

Sol.  Here , edge of there metallic cubes are 3 cm, 4cm, and 5cm.

Volume of single cube = \(3^{3}+4^{3}+5^{3}\)

= 27 +64 +125

= 216 \(cm^{3}\)

\(\ Therefore, \, edge \, of \,single \,cube= \sqrt[3]{216}=6\)

 

  1. The rainwater from a roof of dimensions 22 m * 20m drains into a cylindrical vessel having diameter of base 2 m and height 3.5 m. If the rainwater collected from the roof just fill the cylindrical vessel, then find the rainfall in cm.

Sol. Let the required rainfall in cm be x

therefore according to the statement of the question, we have

Volume of rain = volume of cylinder

\(\ \frac{x}{100}\times 22\times 20 =\frac{22}{7}\times 1\times 1\times 3.5\)

 

x=\(\ \frac{22}{7}\times \frac{1\times 1\times 3.5\times 100}{22\times 20}\)

=2.5m

 

  1. A cone having radius 8 cm and height 12 cm is divide into two part by a plane through the mid-point of axis parallel to its base. Find the ratio of the volumes of two parts.

Sol.   Here,   \(\triangle ADE \sim \triangle ABC \,\, by \, AA \,\, similarity \,\, rule \frac{r}{4}=\frac{6}{12}\)

∴\( \frac{radius\, of \,smaller\, cone}{radius of larger cone}=\frac{Height of smaller cone}{Height of larger cone}\)

⇒ r=2cm

now,\(\frac{volume of upper part (cone)}{volume of lower part (frustum of cone)}\)

1

=\(\frac{\frac{1}{3}\pi r^{2}h}{\frac{1}{3}\pi h(r^{2}+R^{2}+R)}\)

=\(\frac{2\times 2\times 6}{6(4+16+8)}\)

=\(\frac{4}{28}\)

=\(\frac{1}{7}\)

So that required ratio is 1:7.

 

  1. Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing some water. Find the number of marble that should be dropped into the beaker so that the water level rises by 5.6 cm.

Sol. Let us assume that the marbles are spherical in shape,

Hence,Volume of each marble

=\(\frac{4}{3}\times \pi \times 0.7\times 0.7\times 0.7\)

Let n be the number of marbles that are required to raise the water  level  upto 5.6cm of the cylindrical beaker

\(\ therefore\) n times volume of each marble = Volume of cylindrical beaker of height 5.6cm.

\(n\times \frac{4}{3}\times\pi \times 0.7\times 0.7\times 0.7\) =\(\pi\times \frac{7}{2}\times \frac{7}{2}\times 5.6\)

n=\( \frac{7}{2}\times \frac{7}{2}\times 5.6\times \frac{3}{4}\times \frac{1}{0.7}\times                           \frac{1}{0.7}\times \frac{1}{0.7}\)

n= 150.

 

  1. Find the number of metallic circular disc with 1.5 cm base diameter and of height 0.2 cm. to be melted to from a right circular cylinder of diameter 4.5 cm and height 10 cm.

Sol.         Volume of each circular disc

\(= \pi \times \frac{1.5}{2}\times \frac{1.5}{2}\times 0.2 \,cm^{3}\)

Volume of right circular cylinder

=\( \pi \times \frac{4.5}{2}\times \frac{4.5}{2}\times 10\, cm^{3}\)

Let the number  of metallic circular disc be n

\(∴\,    n \times Volume \, of \, each \, circular \, disc = Volume of cylinder\)

 

\(∴\,  n\times \pi \times \frac{1.5}{2}\times \frac{1.5}{2}\times 0.2 = \pi \times \frac{4.5}{2}\times \frac{4.5}{2}\times 10\)

 

n=\( \frac{4.5}{2}\times \frac{4.5}{2}\times 10\times \frac{2}{1.5}\times \frac{2}{1.5}\times \frac{1}{0.2}\)

=450

 

  1. A heap of rice is in the form of a cone of diameter 9 m and height 3.5 m. find the volume of the rice. How much canvas cloth is required to just cover the heap?

Sol. Radius of conical heap of rice (r) = \(\frac{9}{2}= 4.5m\)

Height of conical heap of rice (h) = 3.5m

Therefore, Volume of the rice =\(\frac{1}{3}\times \pi \times r^{2}\times h\)

=\( \frac{1}{3}\times \frac{22}{7}\times 4.5\times 4.5\times 3.5\)

=74.25 \(m^{3}\)

 

Therefore the Slant height of the heap =\(\sqrt{(4.5)^{2}+(3.5)^{2}}\)

=\( \sqrt{20.25+12.25}\)

=\( \sqrt{32.5}= 5.7m\)

Area of the canves cloth required  =\(\frac{22}{7}\times 4.5\times 5.7\)

=\(80.6m^{3}\)

 

  1. Two cones with same base radius 8 cm and height 15 cm are joined together along their bases. Find the surface area of the shape so formed.

Sol.

Slant height of cone  = \(\sqrt{8^{2}+15^{2}}\)

=\( \sqrt{64+225}\)

=\( \sqrt{289}\)

=17cm

2

Total surface area of the shape so formed

= \(2\times  Curved\, surface\, of\, the\, original\, cone\, used\)

=\(2\times \frac{22}{7}\times 8\times 17\)

=\(854.86cm^{2}\)

 

  1. An ice-cream cone full of ice cream having radius 5cm and height 10cm as shown in the figure. Calculate the volume of ice-cream, provided that its \( \frac{1}{6}\) part is left unfilled with ice-cream.

3

Sol.  Radius of the cone (r)  = 5cm

Total height of ice-cream cone = 10cm

Height of cone = 10 – 5=5cm

Volume of the ice-cream = \(\frac{1}{3}\pi r^{2}h +\frac{2}{3}\pi r^{3}\)

=\( \frac{1}{3}\pi r^{2}(h+2r)\)

=\( \frac{1}{3}\times \frac{22}{7} \times 25(5+10)\)

=\( \frac{1}{3}\times \frac{22}{7} \times 25\times 15\)

=\( 392.86cm^{3}\)

Volume of ice-cream = \((1-\frac{1}{6})of 392.86\)

=\( \frac{5}{6}\times 392.86\)

=\(327.38cm^{3}\)

 

  1. How many spherical lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66cm, 42cm, and 21cm?

Sol.  Let n be the number of spherical lead shots.

\(\, n\times volume\, of\, each\, spherical\, lead\, shot\, =\, volume\, of\, the\, rectangular\, lead\, piece\)

 

\(n=\frac{4}{3}\times \frac{22}{7}\times \frac{4.2}{2}\times \frac{4.2}{2}\times \frac{4.2}{2}\)=\(66\times 42\times 21\)

 

=\(\frac{\times 66\times 42\times 21\times 3\times 7\times 8}{4\times 22\times 4.2\times 4.2\times 4.2}\)

=1500

Hence, the required number of spherical lead shots are 1500.

 

  1. A well 24m long, 0.4m thick and 6m high is constructed with the bricks each of dimensions \(25cm \times 16cm \times 10cm\). IF the motor occupies \(\frac{1}{10}\)  th of the well, then find the number of bricks used in constructing the well.

Sol.       Volume of the wall = \(24m \times 0.4m\times 6m\)

=\(2400\times 40\times 600cm^{3}\)

=\(57600000 cm^{3}\)

Volume of the wall occupied by motor = \(\frac{1}{10}\times 5700000\)

=\(5760000cm^{3}\)

Volume of the wall constructed with bricks = \( 57600000- 5760000cm^{3}\)

=51840000

Let n be the number of bricks required

\(∴\, n\times 25\times 16\times 10\) =51840000

\( \Rightarrow n = \frac{51840000}{25\times 16\times 10}\)=\( 12960\)

 

  1. A solid metallic hemisphere of radius 8 cm is melted and re-casted into a right circular cone of base radius 6 cm. Determine the height of the cone.

Sol.   Let h be the height of the cone.

Volume of the cone  = Volume of hemisphere

\(\frac{1}{3}\pi r^{2}h =\frac{2}{3}\pi r^{3}\)

 

\( \Rightarrow 6\times 6\times h = 2\times 8\times 8\times 8\)

h=\( \frac{2\times 8\times 8\times 8}{6\times 6}\)

h= 28.44

Hence, the height of cone is 28.44cm

 

  1. How many cubic centimeters  of iron is required to construct an open box whose external dimension are 36 cm, 25cm  and 16.5 cm provided the thickness of the iron  weights 7.5g, Find the weight of the box.

Sol.  The Volume of the external box =  \(36\times25\times16.5\)

=14850

Volume of the internal box=\(33\times22\times15\)

=\(10890cm^{3}\)

Therefore length =36-1.5-1.5 =33cm

breadth  =  25-1.5-1.5 =22cm

Height =16.5-1.5 =15cm

Volume of  iron used =14850-10890

=3960 cubic cm

 

Weight of the box = \(3960 \times 7.5\)

=29700g

=29.7kg

 

  1. Water flows at the rate of 10m/ minute through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the base is 40cm and depth 24 cm?

Sol. Radius of cylindrical pipe =\(\frac{5}{2}\)mm

=\( \frac{5}{2}\)cm

Length per min = 10m

=1000cm

\(\ therefore\, Volume\, of\, water\, flowing\, per\, minute\,  =\, \pi r^{2}h\)

 

=\(\frac{22}{7}\times \frac{5}{20}\times \frac{5}{20}\times 1000\)

Here, Radius of the conical vessel = 20cm

Depth of the conical vessel =24cm

Now Volume of the conical vessel=\(\frac{1}{3}\times \frac{22}{7}\times 20\times 20\times 24\)

Therefore Time required to the conical vessel

=\( \frac{\frac{1}{3}\times \frac{22}{7}\times 20\times 20\times 24}{\frac{22}{7}\times     \frac{5}{20}\times \frac{5}{20}\times 1000 }\)

=51minutes 12 second

  1. A factory manufactures 120000 pencils daily. All pencils are cylindrical in shape each of length 25cm , circumference of base as 1.5cm. Determine the cost of coloring the curved surface of the pencils manufactured in one day at the rate of 0.05 per \( dm^{3}\).

Sol.  Given that, Circumference of base of pencil =1.5cm

2\(\pi r\)=1.5cm

\( r=\frac{1.5}{2\pi }cm\)

So, Curved surface area of each pencil =\(2\pi \times \frac{1.5}{2\pi }\times 25\)

=\(37.5cm^{2}\)

Total Curved surface area of 120000 pencils

\(=0.375\times 120000=45000 dm^{2}\)

Total cost of coloring the C.S.A is 120000 pencils

=\(0.05 \times 45000=2250\)

 

  1. water is following at the rate of 15 km/h through a pipe diameter 14cm, a cuboidal pond which is 50 m long and 44m wide. IN what time will the level of water in pond rise by 21cm?

Sol. Hence, Volume of the pound =\(\times 50m\times 44m\times \frac{21}{100}m\)

=\(462m^{3}\)

Radius of pipe (r) =\( \frac{14}{2}\times \frac{1}{100}\)

=0.07m

Length of the pipe per hour= 15km

=15000m

Therefore Water collected per hour =\(\frac{22}{7}\times 0.07\times0.07 \times 15000\)

=\(231m^{3}\)}

 

Time required to rise the water level by 21cm

\(=\frac{462}{231}\)

=2hours

 

  1. A milk container of height 16cm is made of metal sheet in the from of a frustum of a cone with radii of its lower and upper ends as 8cm and 200cm 22 Rs. per liter which the container canhold.

Sol. Height of the frustum of cone(h)= 16 cm

Radii of lower and upper end are

R=8cm and R = 20cm

 

Therefore Volume of the milk =\(\frac{1}{3}\pi h (R^{2}+ r^{2}+Rr)\)

=\( \frac{1}{3}\times \frac{22}{7}\times 16(20^{2}+ 8^{2}+20(8))\)

=\(\frac{22\times 16}{21}\times (400+64+160)\)

=\( \frac{22\times 16}{21}\times 624\)

=\(10459.43cm^{3}\)

=\(\frac{10459.43}{1000}litres\)

=\(10.459litres\)

Total cost of the milk =\(22\times 10.459Rs\)

=\(230.10Rs\)

  1. A rocket is in the from of a right circular cylinder close at the lower end surmounted by a cone with the same radius as that of the cylinder .The diameter and height of the cylinder are 6cm and 12 cm respectively. If the slant height of the conical portion is 5cm, find the total surface area and volume of the rocket. [use \(\pi =3.14\)]

Sol.         Height of the cone =\(\sqrt{5^{2}-3^{2}}\)

=\(\sqrt{25-9}  =\sqrt{16}=4\)

4

Volume of the rocket   =\(\pi r^{2}H+\frac{1}{3}\pi r^{2}h\)

=\(\pi r^{2}(H+\frac{1}{3}h)\)

=\(3.14\times 9(12+\frac{1}{3}\times 4)\)

=\(3.14\times 19\times\frac{40}{3}\)

=\(376.8cm^{3}\)

Total surface area =\(\pi r^{2}+2\pi rh+\pi rl\)

=\(\pi r(r+2h+l)\)

=\(3.14\times 3(3+24+5)\)

=\(3.14\times 3\times 32\)

=\(301.44cm^{2}\)

  1. A hemispherical bowl of internal radius 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped bottle

Sol. Volume of liquid in hemispherical bowl  \(=\frac{2}{3}\times \frac{22}{7}\times 9\times 9\times 9cm^{3}\)

Volume of each of cylindrical bottle = \(\frac{22}{7}\times 1.5\times 1.5\times 4cm^{3}\)

Let n be the number of cylindrical shaped bottles.

\(\ therefore, \, n\times \frac{22}{7}\times 1.5\times 1.5\times 4\)

\(n=\frac{2\times 9\times 9\times 9}{3\times 1.5\times 1.5\times 4}\)

=\(54\)

 

Hence, the required number of bottles needed to empty the blow are 54.

  1. A pen stand made of wood is in the shape of a cuboid with four conical depression and cubical depression to hold the cuboid are 10cm, 5cm and 4cm.The radius of each of the conical depression is 0.5 cm. find the volume of the wood in the enter stand.

Sol. Required volume of wood in stand = Volume of cuboid – 4 (Volume of one conical depression ) – Volume of cubical depression

\(=10\times 5\times 4-\frac{1}{3}\times \frac{22}{7}\times0.5\times 0.5\times 2.1-3\times 3\times 3\)

=200-2.2-27

=170.8

 

Exercise 13.1

 

  1. Two cubes each having volume 64 \(cm^{3}\) are connected end to end. Calculate the surface area of the cuboid formed.

Ans- Volume of each cube=\(64cm^{3}\)

If “x” is the side of cube, \(x^{3}=64\)

\(\Rightarrow x=\sqrt[3]{64} \Rightarrow x=4cm.\)

 

\(\ Therefore surface area of a cube=6x^{2} =6(4)^{2} =6(16) =96cm^{2}\)

 

\(\ therefore\, surface\, area\, of\, resulting\, cuboid\)

 

\(=2(surface area of each cube)-(surface area of one side)\)

 

=\((2\times 96)-2x^{2} =192-2(4)^{2} =192-32 =160cm^{2}.\)

 

  1. A container which is in the shape of a hollow hemisphere is covered from the top by a hollow cylinder with diameter of 14 cm and the total height of the container is 13 cm. Calculate the inner surface area of the container.

Ans.-Radius = 7 cm

Height of cylindrical portion = 13 – 7 = 6 cm

Curved surface are of cylindrical portion can be calculated as follows:

 

\(=2\pi rh\)

\(=2\times 22\div 7\times 7\times 6\)

\(=264cm^{2}\)

 

Curved surface area of hemispherical portion can be calculated as follows:

 

\(=2\pi r^{2}\)

\(=2\times 22\div 7\times 7\times 7\)

\(=308cm^{2}\)

Total surface are = 308 + 264 = 572 sq cm

 

3) A toy in a shape of a cone having radius 3.5 cm, mounted on a hemisphere with same radius. The total height of the toy is 15.5 cm. Calculate the total surface area of the toy.

Ans.-Radius of cone = 3.5 cm, height of cone = 15.5 – 3.5 = 12 cm

 

Slant height of cone can be calculated as follows:

\(a=\sqrt{h^{2}+r^{2}}\)

\(a=\sqrt{12^{2}+3.5^{2}}\)

\(a=\sqrt{144+12.5}\)

\(a=\sqrt{156.25}=12.5cm\)

 

Curved surface area of cone can be calculated as follows:

 

\(=\pi ra\)

\(=22\div 7\times 3.5\times 12.5\)

\(137.5cm^{2}\)

 

Curved surface area of hemispherical portion can be calculated as follows:

 

\(=2\pi r^{2}\)

\(=2\times 22\div 7\times 3.5\times 3.5\)

\(77cm^{2}\)

 

Hence, total surface area = 137.5 + 77 = 214.5 sq cm.

 

  1. A cubical block with side 7 cm is surmounted by a hemisphere. Identify the greatest diameter the hemisphere which can be formed? Determine the surface area of the solid.

 

Ans.-The greatest diameter = side of the cube = 7 cm

Surface Area of Solid = Surface Area of Cube – Surface Area of Base of Hemisphere + Curved Surface Area of hemisphere

Surface Area of Cube = 6 x Side\(^{2}\)

= 6 x 7 x 7 = 294 sq cm

Surface Area of Base of Hemisphere

\(=\pi r^{2}\)

\(22\div 7\times 3.5^{2}\)\(=38.5cm^{2}\)

Curved Surface Area of Hemisphere = 2 x 38.5 = 77 sq cm

Total Surface Area = 294 – 38.5 + 77 = 332.5 sq cm

 

  1. A hemispherical dint is removed by cutting out from one side of a cubical wooden block in such a way that the diameter ‘d’ of the hemisphere becomes equal to the edge of the cube. Calculate the surface area of the remaining solid.

 

Ans.- This question can be solved like previous question. Here the curved surface of the hemisphere is a dint, unlike a projection in the previous question-:

Total Surface Area-:

\(6a^{2}-\pi \left ( a\div 2 \right )^{2}\pi+2\pi \left ( a\div 2 \right )^{2}\)

\(=6a^{2}+\pi\left ( \frac{a}{2} \right )^{2}\)

\(=\frac{1}{4}a^{2}\times \left ( \pi +24 \right )\)

 

  1. A medicine capsule having the shape of a cylinder consisting two hemispheres stuck to each of its ends. The length of the total capsule is 14 mm with diameter of 5 mm. Calculate its surface area.Ans.- Height of Cylinder = 14 – 5 = 9 cm, radius = 2.5 cm

Curved Surface Area of Cylinder
\(2\pi rh\)

\(=2\pi\times 2.5\times 9\)
\(=45\pi cm^{2}\)
Curved Surface Area of two Hemispheres

\(=4\pi r^{2}\)\(=4\pi\times2.5^{2}\)

\(=25\pi cm^{2}\)

Total Surface Area
\(=45\pi +25\pi\)
\(=70\pi =220cm^{2}\)

  1. A tent having the shape of a cylinder is covered by a conical top. The height of the cylindrical part is 2.1 m with a diameter of 4 m having a slant height of 2.8 m, calculate the area of the fabric used for making the tent. Also, calculate the cost of the fabric of the tent at the rate of Rs 500 per m2. (Note that the bottom of the tent can’t be covered with fabric).

Ans.-  Radius of cylinder = 2 m, height = 2.1 m and slant height of conical top = 2.8 m

Curved Surface Area of cylindrical portion

\(=2\pi rh\)

\(=2\pi \times 2\times 2.5\)

\(=8.4\pi m^{2}\)

Curved Surface Area of conical portion

\(=\pi rl\)

\(=\pi\times 2\times 2.8=5.6\pi m^{2}\)

Total CSA

\(=8.4\pi +5.6\pi\)

\(=14\times\frac{22}{7}=44m^{2}\)

Cost of fabric = Rate x Surface Area

= 500 x 44 = Rs. 22000

 

  1. From a solid cylinder having height and diameter of 2.4 cm and 1.4 cm respectively, a conical cavity with same height and same diameter is hollowed out. Calculate the total surface area of the remaining solid to the closest \(cm^{2}\).

Ans.- Radius = 0.7 cm and height = 2.4 cm

Total Surface Area of Structure = Curved Surface Area of Cylinder + Area of top of cylinder + Curved Surface Area of Cone

Curved Surface Area of Cylinder

= \(2\pi rh\)

\(=2\pi\times 0.7\times 2.4=3.36\pi cm^{2}\)

Area of top

\(=\pi r^{2}\)

\(=\pi \times 0.7^{2}\)

\(=0.49\pi cm^{2}\)

Slant height of cone can be calculated as follows:

\(=l=\sqrt{r^{2}+h^{2}}\)

\(=\sqrt{2.4^{2}+0.7^{2}}\)

\(=\sqrt{5.76+0.49}\)

\(=\sqrt{6.25}=2.5cm\)

Curved Surface Area of Cone

\(=\pi rl\)

\(=\pi\times 0.7\times 2.5\)

\(=1.75\pi cm^{2}\)

Hence, remaining surface area of structure

\(=3.36\pi +0.49\pi +1.75\pi\)

\(=5.6\pi =17.6cm^{2}\)

\(=18cm^{2} \left ( approx \right )\)

 

  1. A wooden article has been made by removing out a hemisphere from each end of a solid cylinder, as given in figure. If the height and radius of base of the cylinder is 10 cm and 3.5 cm respectively, then calculate the complete surface area of the article

Ans.- Radius = 3.5 cm, height = 10 cm

Total Surface Area of Structure = CSA of Cylinder + CSA of two hemispheres

Curved Surface Area of Cylinder

\(=2\pi rh\)

\(=2\pi \times 3.5\times 10\)

\(=70\pi cm^{2}\)

Surface Area of Sphere

\(=4\pi r^{2}\)

\(=4\pi\times 3.5^{2}\)

\(=49\pi\)

Total Surface Area

\(=70\pi +49\pi=119\pi\)

\(=119\times \frac{22}{7}=374cm^{2}\)

 

 

Exercise 13.2

1: A solid cone placed on a hemisphere with both their radii whose value is equal to 1 cm and the height of the cone is equal to its radius. Calculate the volume of the solid in terms of π.

Ans.-  radius = 1 cm, height = 1 cm

Volume of hemisphere

\(=\frac{2}{3}\pi r^{3}\)

\(=\frac{2}{3}\pi\times 1^{3}\)

\(=\frac{2}{3}\pi cm^{2}\)

Volume of cone

\(=\frac{1}{3}\pi r^{2}h\)

\(=\frac{1}{3}\pi\times 1^{2}\times 1=\frac{1}{3}\pi cm^{3}\)

Total volume

\(=\frac{2}{3}\pi+\frac{1}{3}\pi =\pi cm^{3}\)

 

2: Harish, who is student, he was asked to make a model of shape similar to a cylinder which contains two cones connected at its two ends by using a metal sheet. The diameter and the height of the model are 3 cm and 12 cm respectively. If both the cone has a height of 2 cm, calculate the volume of air present in the model that Harish made. (Assume that both the inner and outer dimension of the model is almost same).

Ans.- Height of cylinder = 12 – 4 = 8 cm, radius = 1.5 cm, height of cone = 2 cm

Volume of cylinder

\(=\pi r^{2}h\)

\(=\pi \times 1.5^{2}\times 8=18\pi cm^{3}\)

Volume of cone

\(=\frac{1}{3}\pi r^{2}h\)

\(=\frac{1}{3}\pi\times 1.5^{2}\times 2\)

\(=1.5\pi cm^{3}\)

Total volume

\(=1.5\pi +1.5\pi +18\pi\)

\(=21\pi =66cm^{3}\)

 

3: A sweet, which contains sugar syrup up to about 30% of its volume. Calculate approximately how much syrup will be available in 45 sweets, each shaped like a cylinder having two hemispherical ends with length 5 cm and 2.8 cm diameter.

Ans.-  Length of cylinder = 5 – 2.8 = 2.2 cm, radius = 1.4 cm

Volume of cylinder

\(=\pi r^{2}h\)

\(=\pi \times 1.4^{2}\times 2.2\)

\(=4.312\pi cm^{3}\)

Volume of two hemispheres

=\(\frac{4}{3}\pi r^{3}\)

\(=\frac{4}{3}\pi\times 1.4^{3}\)

\(=\frac{10.976}{3}\pi cm^{3}\)

Total volume

\(=4.312\pi +\frac{10.976}{3}\pi\)

Volume of syrup = 30% of total volume

\(=\pi \left ( 4.312+\frac{10.976}{3} \right )\times \frac{30}{100}\)

\(=\frac{23.912}{3}\times \frac{30}{100}\times \frac{22}{7}=7.515cm^{3}\)

Volume of syrup in 45 sweets = 45 x 7.515 = 338.184 \(cm^{3}\)

 

4: A flower pot that is made of wood having the shape of a cuboid with four conical depressions to hold flowers. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Calculate the volume of wood in the entire pot.

Ans.- Dimensions of cuboid is 15 cm x 10 cm x 3.5 cm, radius of cone is 0.5 cm, depth of cone is 1.4 cm

As we know Volume of cuboid = length x width x height

So, \(15\times 10\times 3.5=525cm^{3}\)

Now volume of cone \(=\frac{1}{3}\pi r^{2}h\)

\(=\frac{1}{3}\times \frac{22}{7}\times 0.5^{2}\times 1.4=\frac{11}{30}cm^{3}\)

\(\ therefore\) Hence volume of wood =Volume of cuboid – 6 x volume of cone

\(=525-6\times \frac{11}{30}\)

\(=525-\frac{11}{5}=522.8cm^{3}\)

 

5: A container is in the shape of an inverted cone. The height and the radius at the top (which is open) of the container are 8 cm and 5 cm respectively. The container is filled with water up to the upper edge. When lead shots, each of which is a sphere of radius 0.5 cm are dropped inside the vessel, quarter quantity of the water flows out. Calculate the number of lead shots dropped in the vessel.

Ans.- Given, radius of cone is 5 cm, height of cone is 8 cm, radius of sphere is 0.5 cm

So volume of cone=\(\frac{1}{3}\pi r^{2}h\)

\(\frac{1}{3}\pi\times 5^{2}\times 8\)

\(\frac{200}{3}\pi cm^{3}\)

Similarly volume of lead shot=\(\frac{4}{3}\pi r^{3}\)

\(\frac{4}{3}\pi \times 0.5^{3}\)

\(\frac{1}{6}\pi cm^{3}\)

Now number of lead shots will be

=\(\frac{200}{3}\pi \times \frac{1}{4}\div \frac{1}{6}\pi\)

=\(\frac{50\pi }{3}\times \frac{6}{\pi }=100\)

 

6: A solid metal pole consisting of a cylinder of height and base diameter as 220 cm and 24 cm respectively is surmounted by another cylinder of height and radius as 60 cm and 8 cm respectively. Calculate the mass of the pole, given that 1 cm3 of iron has approximately 8g mass.

Ans.- Here radius of bigger cylinder = 12 cm, height of bigger cylinder = 220 cm

Similarly, radius of smaller cylinder = 8 cm, height of smaller cylinder = 60 cm

As we know volume of bigger cylinder=\(\pi r^{2}h\)

=\(\pi\times 12^{2}\times 220\)

=\(31680\pi cm^{3}\)

Now volume of small cylinder=\(\pi r^{2}h\)

=\(\pi \times 8^{2}\times 60\)

=\(3840\pi cm^{3}\)

Therefore total volume=\(31680\pi +3840\pi\)

=\(35520\pi cm^{3}\)

Hence, Mass= Density. Volume

=\(8\times 35520\pi =892262.4gm\)= \(892.3kg\)

 

7: A solid containing of a right circular cone of height and radius are 120 cm and 60 cm respectively standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of liquid such that it touches the bottom. Calculate the volume of liquid left in the cylinder, if the radius and height of the cylinder are 60 cm and 180 cm respectively.

Ans.- Here, Radius of cone = 60 cm, height of cone = 120 cm

Radius of hemisphere = 60 cm

Radius of cylinder = 60 cm, height of cylinder = 180 cm

So as we know that volume of cone is =\(\frac{1}{3}\pi r^{2}h\)

=\(\frac{1}{3}\pi \times 60^{2}\times 120\)

=\(144000\pi cm^{3}\)

Volume of hemisphere

=\(\frac{4}{3}\pi r^{3}\)

\(\frac{2}{3}\pi \times 60^{3}=144000cm^{3}\)

Similarly, volume of solid=\(\left ( 144000+144000 \right )\pi\)

=\(288000\pi cm^{3}\)

Now volume of cylinder=\(\pi r^{2}h\)

=\(\pi \times 60^{2}\times 180=648000\pi cm^{3}\)

Hence the total volume of liquid left in the cylinder is

=\(\left ( 648000-288000 \right )\pi =360000\pi\)

=\(1130400cm^{3}\)

 

 

8: A spherical glass vessel having a cylindrical neck of height 8 cm and diameter 2 cm; the radius of the spherical part is 4.25 cm. By determining the amount of water it holds, a child calculates its volume to be 345 cm3. Identify whether he is correct, taking the above as the inside measurements, and π = 3.14.

Ans.- Here, Radius of cylinder = 1 cm, height of cylinder = 8 cm, radius of sphere = 4.25 cm

As we know that volume of cylinder=\(\pi r^{2}h\)

=\(\pi \times 1^{2}\times 8\)

=\(8\pi cm^{3}\)

Similarly volume of sphere=\(\frac{4}{3}\pi r^{3}\)

=\(\frac{4}{3}\pi \times \left ( \frac{8.5}{2} \right )^{3}\)

=\(\frac{614125}{6000}\pi cm^{3}\)

\(\ therefore\),the total volume will be

=\(\left (\frac{614125}{6000}+8 \right )\pi\)

=\(\left (\frac{614125+48000}{6000} \right )\pi =346.51cm^{3}\)

 

Exercise 13.3

1: A metallic sphere having radius of 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Calculate the height of the cylinder.

Ans.- Here, Radius of sphere is 4.2 cm, radius of cylinder is 6 cm

As we know that volume of sphere

=\(\frac{4}{3}\pi r^{3}\)

=\(\frac{4}{3}\pi\times 4.2^{3}\)

Similarly volume of cylinder

=\(\pi r^{2}h=\pi \times 6^{2}\times h\)

Since volume of cylinder = Volume of sphere

Hence, height of cylinder.

 

2: Metallic spheres having radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Calculate the radius of the resulting sphere.

Ans.- Here, Radii of spheres = 6 cm, 8 cm, 10 cm

As we know that volume of sphere=\(\frac{4}{3}\pi r^{3}\)

So total volume of three spheres is

=\(\frac{4}{3}\pi\left ( 6^{3}+8^{3}+10^{3} \right )\)

=\(\frac{4}{3}\pi\left (616+512+1000 \right )\)

=\(\frac{4}{3}\pi\times 1728\)

Hence radius of biggest sphere is

=\(\sqrt[3]{1728}=12cm\).

 

3: A 20 m deep well having diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Calculate the height of the platform.

Ans.- Here, Radius of well = 3.5 m, depth of well = 20 m

So the dimension of rectangular platform =\(22m\times 14m\)

Now the volume of earth dug out

=\(\pi r^{2}h\)

=\(\pi\times 3.5^{2}\times 20=770m^{3}\)

Area of top of platform = Area of Rectangle – Area of Circle

(as the circular portion of mouth of well is open)

=\(22\times 14-\pi \times 3.5^{2}\)

=\(308-38.5=269.5m^{2}\)

So as we know, height=volume/area

=\(\frac{770}{269.5}=2.85m\)

 

4: A well is dug with diameter 3 m and a depth of 14 m. The earth taken out of it has been spread in a same quantity surrounding the well forming a shape of circular ring of width 4 m to form an embankment. Calculate the height of the embankment.

Ans.- Here, Radius of well = 1.5 m, depth of well = 14 m, width of embankment = 4 m

So the radius circular embankment=4+1.5=5.5m

Now the volume of earth dug out,

=\(\pi r^{2}h\)

=\(\pi \times 1.5^{2}\times 14=31.5\pi m^{3}\)

Area of top of platform = (Area of bigger circle – Area of smaller circle)

So as we know, height=volume/area

=\(\frac{31.5\pi }{28\pi }=1.125m\)

 

5: A vessel shaped like a right circular cylinder with diameter 12 cm and height 15 cm is full of ice cream. The ice cream needs to be filled into cones having height 12 cm and diameter 6 cm, forming a hemispherical shape on the top. Calculate the number of such cones which can be filled with ice cream.

Ans.- Here, Radius of cylinder = 6 cm, height of cylinder = 15 cm

Similarly, Radius of cone = 3 cm, height of cone = 12 cm

Given, Radius of hemispherical top on ice cream = 3 cm

Now as we know that, volume of cylinder

= \(\pi r^{2}h\)

= \(\pi \times 6^{2}\times 15=540\pi cm^{3}\)

Volume of cone is

=\(\frac{1}{3}\times \pi \times 3^{2}\times 12\)

=\(36\pi cm^{3}\)

Now, volume of hemisphere is

=\(\frac{2}{3}\pi r^{3}\)

= \(\frac{2}{3}\times \pi \times 3^{3}=18\pi cm^{3}\)

So volume of ice-cream will be

= \(\left ( 36+18 \right )\pi =54\pi cm^{3}\)

Hence the number of ice creams = Volume of cylinder/Volume of ice cream

\(\frac{540\pi }{54\pi }=10\)

 

6: How many silver coins, 1.75 cm in diameter and of thickness 2 mm, can be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Ans.- Here, Radius of coin = 0.875 cm, height = 0.2 cm

Dimensions of cuboid = 5.5 cm x 10 cm x 3.5 cm

As we know that the volume of coin is

=\(\pi r^{2}h\)

= \(\pi \times 0.875^{2}\times 0.2\)= \(0.48125cm^{3}\)

Similarly, volume of cuboid is= \(5.5\times 10\times 3.5=192.5cm^{3}\)

Therefore Number of coins

= \(\frac{192.5}{0.48125}=400\)

 

7: A cylindrical bucket with 32 cm high and having a radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, calculate the radius and slant height of the heap.

Ans.- Given, Radius of cylinder = 18 cm, height = 32 cm

Height of cone = 24 cm

As we know that the volume of cylinder is

= \(\pi r^{2}h\)

= \(\pi\times 18^{2}\times 32\)

As, Volume of cone = Volume of cylinder

So, volume of cone is

= \(\frac{1}{3}\pi r^{2}\times 24\)

Hence, the radius of cone can be calculated as follows:

\(r^{2}=\frac{3\times \pi \times 18^{2}\times 32 }{\pi \times 24}\)

Or, \(r^{2}=18^{2}\times 2^{2}\)

Or, r = 36cm

So now the slant height of conical heap can be calculated as follows:

l = \(\sqrt{h^{2}+r^{2}}\)

= \(\sqrt{24^{2}+36^{2}}\)

= \(\sqrt{576+1296}=\sqrt{1872}\)

= \(36\sqrt{13}cm\)

 

8: Water in a canal with 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Ans.- Given, Depth = 1.5 m, width = 6 m, height of standing water = 0.08 m

In 30 minutes, length of water column = 5 km = 5000 m

Volume of water in 30 minutes = 1.5 x 6 x 5000 = 45000 cubic m

So as we know that,

Area = Volume/Height

= \(\frac{45000}{0.08}=562500m^{2}\)

 

9: A farmer connects a pipe with internal diameter of 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If the water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Ans.- Given,  Radius of pipe = 10 cm = 0.1 m, length = 3000 m/h

Radius of tank = 5 m, depth = 2 m

So the volume of water in 1 hr. through pipe is

= \(\pi r^{2}h\)

= \(\pi \times 0.1^{2}\times 3000\)= \(30\pi m^{3}\)

Similarly, the volume of tank is

= \(\pi r^{2}h\)

= \(\pi \times 5^{2}\times 2=50\pi m^{3}\)

Hence the time taken to fill the tank is = Volume of tank/Volume of water in 1 hr.

= \(\frac{50\pi }{30\pi }\)=1 hr. 40min

 

Exercise 13.4

1: A drinking glass with the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Calculate the capacity of the glass.

Ans.- Given, R = 2, r = 1 cm and h = 14 cm

We know that, volume of frustum is

\(=\frac{1}{3}\pi h\left ( R^{2} +r^{2}+Rr\right )\)

\(=\frac{1}{3}\pi \times 14\left ( 2^{2}+1^{2}+2 \right )\)

\(=\frac{1}{3}\times \frac{22}{7 }\times 14\times 7\)

\(=102\frac{2}{3}cm^{3}\)

 

2: The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Calculate the curved surface area of the frustum.

Ans.- Given, Slant height l = 4 cm, perimeters = 18 cm and 6 cm

Radii can be calculated as follows:

We know that, Radius= \(\frac{Perimeter}{2\pi }\)

= \(\frac{18}{2\pi }=\frac{9}{\pi }\)

And radius = \(\frac{6}{2\pi }=\frac{3}{\pi }\)

Therefore the curved surface area of frustum is

= \(\pi \left ( R+r \right )l\)

= \(\pi\left ( \frac{9}{\pi }+\frac{3}{\pi } \right )4\)= \(48cm^{2}\)

 

3: A fez, the cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, calculate the area of material used for making it.

Ans.- Given,  R = 10 cm, r = 4 cm, slant height = 15 cm

We know that, Curved surface area of frustum is

= \(\pi \left ( R+r \right )l\)

= \(\pi \left ( 10+4 \right )15\)

= \(\frac{22}{7}\times 14\times 15= 660\)

Now area of upper base is

= \(\pi r^{2}=\pi \times 4^{2}\)

= \(16\pi =50\frac{2}{7}\)

So the total surface area is

\(660+50\frac{2}{7}=710\frac{2}{7}cm^{2}\)

 

4: A vessel, opened from the top and made up of an aluminium sheet, is in the form of a frustum of a cone having height, radii of its lower and upper ends are 16cm, 8 cm and 20 cm, respectively. Calculate the cost of the milk which can completely fill the vessel, at the rate of Rs 20 per litre. Also calculate the cost of aluminium sheet used to make the vessel, if it costs Rs 8 per 100 cm2.

Ans.- Given,  Height of frustum = 16 cm, R = 20 cm, r = 8 cm

We know that, volume of frustum is

= \(\frac{1}{3}\pi h\left ( R^{2}+r^{2}+Rr \right )\)

= \(\frac{1}{3}\pi \times 16\left ( 20^{2}+8^{2}+160 \right )\)

= \(\frac{1}{3}\times \frac{22}{7}\times 16\left ( 400+64+160 \right )\)

= \(\frac{1}{3}\times \frac{22}{7}\times 16\times 604= 10449.92cm^{3}\)

Cost of milk @ Rs. 20 per 1000 cubic cm

10.44992 x 20 = Rs. 208.99

For calculating surface area, we need to find slant height which can be calculated as follows:

\(l= \sqrt{h^{2}+\left ( R- r \right )^{2}}\)

= \(\sqrt{16^{2}+\left ( 20- 8 \right )^{2}}\)

= \(\sqrt{256+144}=\sqrt{400}=20cm\)

Now the surface area of frustum is

= \(\pi \left ( R+r \right )l+\pi r^{2}\)

= \(\pi \left [ \left ( R+r \right ) l+r^{2}\right ]\)

= \(\pi \left [ \left ( 20+8 \right ) 20+8^{2}\right ]\)

= \(\pi \left ( 560+64 \right )\)

= \(\frac{22}{7}\times 624= 1959.36\)

Therefore Cost of metal sheet @ Rs. 8 per 100 sq cm = 19.5936 x 8 = Rs. 156.75

 

5: A metallic right circular cone of  20 cm high and whose vertical angle is \(60^{\circ}\) is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, calculate the length of the wire.

Ans.-  Here volume of frustum will be equal to the volume of wire and by using this relation we can calculate the length of the wire.

In the given figure; AO = 20 cm and hence height of frustum LO = 10 cm

In triangle AOC we have angle CAO = 30 (half of vertical angle of cone BAC)

Therefore;

\(tan 30^{\circ}= \frac{OC}{AO}\)

Or, \(\frac{1}{\sqrt{3}}=\frac{OC}{20}\)

Or, OC= \(\frac{20}{\sqrt{3}}\)

Using similarity cirteria in triangles AOC and ALM it can be shown that LM = 10/√3 (because LM bisects the cone through its height)

Similarly, LO = 10 cm

Volume of frustum can be calculated as follows:

V = \(\frac{1}{3}\pi \times 10\left ( \frac{400}{3}+\frac{100}{3} +\frac{200}{3}\right )\)

= \(\frac{7000}{9}\pi \, cub cm\)

Volume of cylinder is given as follows:

\(\pi r^{2}h\)

Or, \(\pi \times \left ( \frac{1}{32} \right )^{2}\times h\)

= \(\frac{7000}{9}\pi\)

Or, h= \(\frac{7000}{9}\times 1024= 796444.44cm\)

= 7964.4m

 

Exercise 13.5

1: A copper wire having a diameter of 3mm, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Calculate the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.

Ans.-Given, For copper wire: Diameter is = 3 mm = 0.3 cm

For cylinder; length h is = 12 cm, d = 10 cm

Density of copper is = 8.88 gm cm3

So now the curved surface area of cylinder can be calculated as follows:

 

We know that-: Length = \(\frac{Area}{Width}\)

= \(\frac{120\pi }{0.3}=400\pi\)

= 1256cm

Similarly, volume of wire can be calculated as follows:

V = \(\pi r^{2}h\)

= \( 3.14\times 0.15^ {2}\times 1256=88.7364cm^ {3} \)

Hence, mass can be calculated as-: Mass = Density*Volume

\(8.88\times 88.7364=788g(approx)\)

 

2: A right angled triangle, whose sides with 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Calculate the volume and surface area of the double cone so formed.

Ans.- In the given triangle ABC;

\(AC^{2}=AB^{2}+BC^{2}\)

Or, \(AC^{2}=3^{2}+4^{2}=9+16=25\)

Or, AC=5CM

In triangle ABC and triangle BDC;

\(\angle ABC=\angle BDC\)

\(\angle BAC=\angle DBC\)

Hence, \(\bigtriangleup ABC\sim \bigtriangleup BDC\)

So, now we get following equations:

\(\frac{AB}{AC}=\frac{BD}{BC}\)

Or, \(\frac{3}{5}=\frac{BD}{4}\)

\(\ therefore BD=\frac{3\times 4}{5}=2.4\)

In triangle BDC;

\(DC^{2}=BC^{2}-BD^{2}\)

Or, \(DC^{2}=4^{2}-2.4^{2}=16-5.76=10.24\)

\(\ therefore DC=3.2\)
= \(\frac{1}{3}\times 3.14\times 2.4^{2}\times 1.8=10.85184cm^{3}\)
Upper Cone: r = 2.4 cm, l = 3 cm, h = 1.8 cm

We know that, volume of cone is

= \(\frac{1}{3}\pi r^{2}h\)

Now, Curved surface area of cone

=\(\pi rl\)

= \(3.14\times 2.4\times 3=22.608cm^{3}\)

Similarly, for lower cone: r = 2.4 cm, l = 4 cm, h = 3.2 cm

As we know that, volume of cone is

= \(\frac{1}{3}\pi r^{2}h\)

= \(\frac{1}{3}\times 3.14\times 2.4^{2}\times 3.2=19.2916cm^{3}\)

Now, Curved surface area of cone is

=\(\pi rl\)

= \(3.14\times 2.4\times 4=30.144cm^{2}\)

Total volume = 19.29216 + 10.85184=30.144cm\(^{3}\)

\(\ therefore\, Total\, surface\, area\, =\, 30.144 + 22.608 =52.752cm^{2}\)

 

3: A cistern measuring 150 cm x 120 cm x 110 cm has 129600 \(cm^{3}\) water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one seventeenth of its own volume of water. Identify how many bricks can be put in without overflowing the water, each brick being 22.5 cm x 7.5 cm x 6.5 cm?

Ans.- Here volume of cistern is =  length x width x depth

= 150 x 120 x 110

= 1980000 \(cm^{^{3}}\)

Vacant space = Volume of cistern – Volume of water

= 1980000 – 129600 = 1850400 \( cm^{^{3}}\)

Volume of brick = length x width x height

= 22.5 x 7.5 x 6.5

Since the brick absorbs one seventeenth its volume hence the remaining volume will be equal to 16/17 the volume of brick

Remaining volume is

= \(22.5\times 7.5\times 6.5\times \frac{16}{17}\)

\(\ therefore\) Number of bricks = Remaining volume of cistern/remaining volume of brick

= \(\frac{1850400\times 17}{22.5\times 7.5\times 6.5\times 16}\)

 

= \(\frac{31456800}{17550}=1792\)

 

5: An funnel made of a metallic sheet consisting of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height and the diameter of the cylindrical portion are 22 cm and 8 cm respectively and diameter of the top of the funnel is 18 cm, calculate the area of the metallic sheet required to make the funnel.

Ans.- Curved surface area of cylinder

Slant height of frustum can be calculated as follows:

Curved surface area of frustum

\(\ therefore\) Total curved surface area is

= \(169\pi +80\pi=249+3.14\)

= \(781.86cm^{2}\)

 

Exercise 13.5

1. A matchbox measures 5cm × 1cm × 3.5cm. Determine the volume of a packet containing 14 such boxes?

Ans.- Given the dimension of the matchbox = 5cm × 1cm × 3.5cm

Let us assume, l = 5cm, b = 1cm, h = 3.5cm

As we know that, Volume of one matchbox = (l × b × h)

= \(\left ( 5\times 1\times 3.5 \right )cm^{3}=17.5cm^{3}\)

\(∴The\:volume\:of\:a\:packet\:containing\:14\:such\:boxes=\left ( 17.5\times 14 \right )cm^{3}=245cm^{3}\)

 

2.  A cuboidal water tank is 10 m long, 2 m wide and 7.5 m deep. How many litres of water can it hold? (\(1m^{3}\) = 1000 l)

Ans.- Dimensions of water tank = 10m × 2m × 7.5m

Let us assume, l = 10m, b = 2m, h = 7.5m

Therefore Volume of the tank = \(\left ( l\times b\times h \right )m^{3}\)

= \(\left ( 10\times 2\times 7.5 \right )m^{3}=150m^{3}\)

Hence, the tank can hold = 150 x 1000 litres = 150000 litres of water.

 

3. A cuboidal vessel is 25 m long and 12 m wide. Determine the height that must be made to hold 400 cubic metres of a liquid?

Ans.– Given, Length = 25 m , Breadth = 12 m and Volume = 400\(m^{3}\)

As we know, Volume of cuboid = Length x Breadth x Height

Therefore, Height = Volume of cuboid/(Length × Breadth)
= \(\frac{400}{25\times 12}m=1.33m\)

 

4.  Find the value of digging a cuboidal pit 15 m long, 5 m broad and 3 m deep at the rate of Rs.50 per \(m^{3}\).

Ans.- Here, length = 15m, breadth = 5m and height = 3m

As we know that, Volume of the pit = \(\left ( l\times b\times h \right )m^{3}\)

= \(\left (15\times 5\times 3 \right )m^{3}=225m^{3}\)

The rate of digging is = Rs.50 per \(m^{3}\)

∴ The total value of digging the pit = Rs.(225 x 50)

= Rs.11250

5.The capacity of a cuboidal tank is 2,10000 litres of water. Calculate the breadth, given the length is 3.5m and depth is 20m.

Ans.- Given, length = 3.5m, depth = 15m and volume = 30000 litres

As we know that, \(1m^{3}=1000\:litres\)

\(∴ 210000\:litres=\frac{210000}{1000}m^{3}\)= 210\(1m^{3}\)

Breadth = \(\frac{volume\:of\:cuboid}{length\times depth}\)

= \(\frac{210}{\left ( 3.5\times 20 \right )}m\)

= 3m

 

  1. A village, with a population of 6000, requires 200 litres of water per head per day. It has a tank measuring 30m × 25m × 8m. Justify the number of days that will take to empty the water tank ?

Ans.- Given, the dimension of the tank = 30m × 25m × 8m

So, l = 30m, b = 25m and h = 8m

As we know that, the total capacity of the tank = \(\left ( 30\times25\times 8 \right )m^{3}=6000m^{3}\)

Water required for a single person per day = 200 litres

The requirement of water for 6000 person in a single day = (6000 x 200) litres

= \(\frac{\left ( 6000\times 200 \right )}{1000}=1200m^{3}\)

Hence, the number of days the water will last = (the capacity of the tank /water required per day) = \(\left ( \frac{6000}{1200} \right )=5\)

∴ The water lasts for 5 days.

 

  1. A warehouse measures 50m × 35m × 25m. Calculate the maximum number of wooden box each measuring 2.5m × 1.5m × 1m that can be stored in the warehouse.

Ans.- Given the dimensions of the warehouse = 50m x 35m x 25m

As we know that, the volume of the warehouse will be = \(\left ( lbh \right ) m^{3}\)

= \(\left (50\times 35\times 25 \right ) m^{3}=43750m^{3}\)

Now, the dimension of box = 2.5m x 1.5m x 1m

Similarly, volume of 1 box = \(\left (2.5\times 1.5\times 1 \right ) m^{3}=3.75m^{3}\)

Hence, Number of box that can be stored =  volume of warehouse / volume of 1 box = \(\frac{43750}{3.75}=11666.666=11666\)

 

  1. A solid cuboid having side 20 cm is cut into 16 cubes of equal volume. Calculate the side of the new cuboid and also calculate the ratio between their surface areas.

Ans.- Here the edge of the cube = 20cm

 

So, Volume of the cuboid = \(\left ( edge \right )^{3}cm^{3}\)

 

= \(\left ( 20\times20\times 20 \right )cm^{3}=8000cm^{3}\)

 

Now, The number of smaller cube = 16

 

So, the volume of 1 small cube = \(\frac{8000}{16}cm^{3}=500cm^{3}\)

 

Let us assume the side of small cube as ‘p’

 

\(p^{3}=500\:\:\Rightarrow p=7.937\:\left ( approx \right )\)

 

Hence, the surface area the cube = \(7.937\left ( side \right )^{2}\)

 

Therefore, the ratio of their surface area

 

= (7.937 x 20 x 20)/(7.937 x 7.937 x 7.937)

 

= \(\frac{40}{1.585}\) = 40 :1.585

 

  1. A river 5 m deep and 60 m wide is flowing at a rate of 6 km per hour. Estimate the amount of water that will fall into the sea in a minute?

 

Ans.- Given, Depth (h) = 5m

Width (b) = 60m

 

So, the rate of flow of water (l) = 6km per hour= \(\left ( \frac{6000}{60} \right )m\:per\:minute=100m\:per\:minute\)

 

Therefore, the volume of water flowing into the sea in a minute = \(lbh\:m^{3}\)

 

\(\left ( 100\times 60\times 5 \right )m^{3}=30000m^{3}\)

5

Class 10th is very important phase of a students life, in class 10 students are introduced to many new topics which are very much essential for class 11. The marks score in class 10th helps a student to choose their preferred stream in class 11th. The Central Board of Secondary Education or CBSE is responsible for conducting the board examination of class 10th and class 12th. The NCERT textbooksare one of the best study materials for the students of class 10th, the questions provided at the end of each chapter are very much important from examination point of view, sometimes the questions in board exams are directly asked from the NCERT textbooks, so solving the NCERT questions will give an added advantage to the students who are preparing for their upcoming class 10th board examination.

Apart from solving NCERT questions students should also solve the previous year questions as well as the CBSE sample papers, solving the CBSE sample papers and previous year questions will help the students to know about the different types of questions asked in the examination. Knowing the pattern of the questions and the difficulty level of the questions will also help the students to prepare for the upcoming board examination.

Keep visiting BYJU’S for latest CBSE sample papers, previous year questions. At BYJU’S we also provide CBSE notes for classes 6th to 12th.

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