# Ncert Solutions For Class 10 Maths Ex 13.5

## Ncert Solutions For Class 10 Maths Chapter 13 Ex 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 = AreaWidth$\frac{Area}{Width}$

= 120π0.3=400π$\frac{120\pi }{0.3}=400\pi$

= 1256cm

Similarly, volume of wire can be calculated as follows:

V = πr2h$\pi r^{2}h$

= 3.14×0.152×1256=88.7364cm3$3.14\times 0.15^ {2}\times 1256=88.7364cm^ {3}$

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

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

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;

AC2=AB2+BC2$AC^{2}=AB^{2}+BC^{2}$

Or, AC2=32+42=9+16=25$AC^{2}=3^{2}+4^{2}=9+16=25$

Or, AC=5CM

In triangle ABC and triangle BDC;

ABC=BDC$\angle ABC=\angle BDC$ BAC=DBC$\angle BAC=\angle DBC$

Hence, ABCBDC$\bigtriangleup ABC\sim \bigtriangleup BDC$

So, now we get following equations:

ABAC=BDBC$\frac{AB}{AC}=\frac{BD}{BC}$

Or, 35=BD4$\frac{3}{5}=\frac{BD}{4}$

thereforeBD=3×45=2.4$\ therefore BD=\frac{3\times 4}{5}=2.4$

In triangle BDC;

DC2=BC2BD2$DC^{2}=BC^{2}-BD^{2}$

Or, DC2=422.42=165.76=10.24$DC^{2}=4^{2}-2.4^{2}=16-5.76=10.24$

thereforeDC=3.2$\ therefore DC=3.2$
= 13×3.14×2.42×1.8=10.85184cm3$\frac{1}{3}\times 3.14\times 2.4^{2}\times 1.8=10.85184cm^{3}$
Upper Cone: r = 2.4 cm, l = 3 cm, h = 1.8 cm

We know that, volume of cone is

= 13πr2h$\frac{1}{3}\pi r^{2}h$

Now, Curved surface area of cone

=πrl$\pi rl$

= 3.14×2.4×3=22.608cm3$3.14\times 2.4\times 3=22.608cm^{3}$

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

As we know that, volume of cone is

= 13πr2h$\frac{1}{3}\pi r^{2}h$

= 13×3.14×2.42×3.2=19.2916cm3$\frac{1}{3}\times 3.14\times 2.4^{2}\times 3.2=19.2916cm^{3}$

Now, Curved surface area of cone is

=πrl$\pi rl$

= 3.14×2.4×4=30.144cm2$3.14\times 2.4\times 4=30.144cm^{2}$

Total volume = 19.29216 + 10.85184=30.144cm3$^{3}$

thereforeTotalsurfacearea=30.144+22.608=52.752cm2$\ therefore\, Total\, surface\, area\, =\, 30.144 + 22.608 =52.752cm^{2}$

3: A cistern measuring 150 cm x 120 cm x 110 cm has 129600 cm3$cm^{3}$ 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 cm3$cm^{^{3}}$

Vacant space = Volume of cistern – Volume of water

= 1980000 – 129600 = 1850400 cm3$cm^{^{3}}$

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

= 22.5×7.5×6.5×1617$22.5\times 7.5\times 6.5\times \frac{16}{17}$

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

= 1850400×1722.5×7.5×6.5×16$\frac{1850400\times 17}{22.5\times 7.5\times 6.5\times 16}$

= 3145680017550=1792$\frac{31456800}{17550}=1792$

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$\ therefore$ Total curved surface area is

= 169π+80π=249+3.14$169\pi +80\pi=249+3.14$

= 781.86cm2$781.86cm^{2}$