Class 10th, Surface and Volume, 13.5 (Solution 2)

Class 10th, Surface Areas and Volumes

 Exercise: 13.5 (Solution), Q5 to Q7 Also with some Important Questions

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Q5. An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see Fig.).

Solution:

Given,

Diameter of upper circular end of frustum part = 18 cm

So, radius (r1) = 9 cm

Now, the radius of the lower circular end of frustum (r2) will be equal to the radius of the circular end of the cylinder

So, r2 = 8/2= 4 cm

Now, height (h1) of the frustum section = 22 – 10 = 12 cm

And,

Height (h2) of cylindrical section = 10 cm (given)

Now, the slant height will be-

Or, l = 13 cm

Area of tin sheet required = CSA of frustum part + CSA of cylindrical part

= π(r1 + r2)l + 2πr2h2

Solving this we get,

Area of tin sheet required = 782 × (4/7) cm2

Q6. Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Solution:

Consider the diagram

Let ABC be a cone. From the cone the frustum DECB is cut by a plane parallel to its base. Here, r1 and r2 are the radii of the frustum ends of the cone and h be the frustum height.

Now, consider the ΔABG and ΔADF,

Here, DF||BG

So, ΔABG ~ ΔADF

Now, by rearranging we get,

The total surface area of frustum will be equal to the total CSA ot frustum + the area ol upper circular end + area of the lower circular end

= π(r1 + r2)l + πr22 + πr12

∴ Surface area of frustum = π[r1 + r2)l + r12 + r22]

Q7. Derive the formula for the volume of the frustum of a cone.

Solution:

Consider the same diagram as the previous question.

Now, approach the question in the same way as the previous one and prove that

ΔABG ~ ΔADF

Again,

Now, rearrange them in terms of h and h1

The total volume of frustum of the cone will be = Volume of cone ABC – Volume of cone ADE

= (⅓)πr12h1 – (⅓)πr22(h1 – h)

= (π/3)[r12h1 – r22(h1 – h)]

Now, solving this we get,

∴ Volume of frustum of the cone = (⅓)πh(r12 + r22 + r1r2) 

Some Important Question which are asked in Examination papers. Which are given below.

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Q1:-A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass?

Solution:-

Big cylinder (H) = 220 cmRadius of the base (R) = 24/12 = 12 cm

So, the volume of the big cylinder = πR2H

= π(12)2 × 220 cm3

= =99565.8 cm3

Now, the height of smaller cylinder (h) = 60 cm

Radius of the base (r) = 8 cm

So, the volume of the smaller cylinder = πr2h

= π(8)2 × 60 cm3

= 12068.5 cm3

∴ Volume of iron = Volume of the big cylinder + Volume of the small cylinder

99565.8 + 12068.5

=111634.5 cm3

We know,

Mass = Density x volume

So, mass of the pole = 8×111634.5

= 893 Kg (approx.)

Q2:-A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder?

Solution:-

It is given that radius of the sphere (R) = 4.2 cm

Also, Radius of cylinder (r) = 6 cm

Now, let height of cylinder = h

It is given that the sphere is melted into a cylinder.

So, Volume of Sphere = Volume of Cylinder

∴ (4/3)× π×R3 = π× r2 × h.

=> h = 2.74 cm

Q3:-A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform?

Solution:-

It is given that the shape of the well is in the shape of a cylinder with a diameter of 7 m

So, radius = 7/2 m

Also, Depth (h) = 20 m

Volume of the earth dug out will be equal to the volume of the cylinder

∴ Volume of Cylinder = π × r2 × h

= 22 × 7 × 5 m3

Let the height of the platform = H

Volume of soil from well (cylinder) = Volume of soil used to make such platform

π × r2 × h = Area of platform × Height of the platform

We know that the dimension of the platform is = 22 × 14

So, Area of platform = 22 × 14 m2

∴ π × r2 × h = 22 × 14 × H

Or, H = 2.5 m

Q4:-How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cn × 10 cm × 3.5 cm?

Solution:-

It is known that the coins are cylindrical in shape.

So, height (h1) of the cylinder = 2 mm = 0.2 cm

Radius (r) of circular end of coins =1.75/2 = 0.875 cm

Now, the number of coins to be melted to form the required cuboids be “n”

So, Volume of n coins = Volume of cuboids

n × π × r2 × h1 = l × b × h

n × π × (0.875)2× 0.2 = 5.5 × 10 × 3.5

Or, n = 400

Q5:-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. Find the surface area of the frustum?

Solution:-

Slant height (l) = 4 cm

Circumference of upper circular end of the frustum = 18 cm

∴ 2πr1 = 18

Or, r1 = 9/π

Similarly, circumference of lower end of the frustum = 6 cm

∴ 2πr2 = 6

Or, r2 = 6/π

Now, CSA of frustum = π (r1 + r2) × l

= π (9/π + 6/π) × 4

= 12 × 4 = 48 cm2

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