5 Mark Question

Question 15 Marks

To make an open fish tank, a glass sheet of 2 mm gauge is used. The outer length, breadth and height of the tank are 60.4 cm, 40.4 cm and 40.2 cm respectively. How much maximum volume of water will be contained in it ?
Given: Thickness of the glass = 2 mm,
outer length of the tank = 60.4 cm,
outer breadth of the tank = 40.4 cm,
outer height of the tank = 40.2 cm
To find: Volume of water fish tank contains

Answer

i. Thickness oldie glass = 2 mm.
$=\frac{2}{10} cm$
= 0.2 cm
Outerlengthofthetank = 60.4 cm
∴ Inner length oldie tank (l) = Outer length – thickness oldie glass on both sides
= 60.4 – 0.2 – 0.2
= 60cm
Outer breadth oldie tank = 40.4 cm
∴ Inner breadth of the tank (b) = 40.4 – 0.2 – 0.2
= 40 cm
Outer height of the tank = 40.2 cm
∴Inner height of the tank (h) = 40.2 – 0.2
= 40 cm

ii. Maximum volume of water that can be contained in the tank = volume of the tank
= l x b x h
= 60 x 40 x 40
= 96000 cubic cm.
∴ The fishtank can contain maximum of 96000 cubic cm. water in it.

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Question 25 Marks

In a field, dry fodder for the cattle is heaped in a conical shape. The height of the cone is 2.1 m and diameter of base is 7.2 m. Find the volume of the heap of the fodder. If it is to be covered by polythene in rainy se&son then how much minimum polythene
sheet is needed?$\left(\pi=\frac{22}{7}\right.$ and $\left.\sqrt{17.37}=4.17\right]$
Given: Height of the heap (h) = 2.1 m.
diameter of the base (d) = 7.2 m
$\therefore$ Radius of the base $(r)=\frac{d}{2}=\frac{7.2}{2}=3.6 m$
To find: Volume of the heap of the fodder and polythene sheet required

Answer

$\text { i. Volume of the heap of fodder }=\frac{1}{3} \pi r^2 h$
$=\frac{1}{3} \times \frac{22}{7} \times(3.6)^2 \times 2.1$
$=\frac{1}{3} \times \frac{22}{7} \times 3.6 \times 3.6 \times 2.1$
$=1 \times 22 \times 1.2 \times 3.6 \times 0.3$
$=28.51 \text { cubic metre }$
ii. Now,$ l^2 = r^2 + h^2$
$= (3.6)^2 + (2.1)^2$
$= 12.96 + 4.41$
$\therefore l^2 =17.37$
$\therefore 1^2=\sqrt{17.37} \ldots$.[Taking square root on both sides]= 4.17 m
iii. Area of the polythene sheet needed to cover the heap of the fodder = Curved surface area of the conical heap
= πrl
$=\frac{22}{7} \times 3.6 \times 4.17$
= 47.18 sq.m
∴ The volume of the heap of the fodder is 28.51 cubic metre and a polythene sheet of 47.18 sq.m will be required to cover it.

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Question 35 Marks

Find the volume of a cone, if its total surface area is $7128 sq . cm$ and radius of base is 28
$cm .\left(\pi=\frac{22}{7}\right)$
Given: Radius $(r)=28 cm$,
Total surface area of cone $=7128$ sq.cm
To find: Volume of the cone

Answer

$\text { i. Total surface area of cone }=\pi r ( l + r )$
$\therefore 7128= y \times 28 \times(1+28)$
$\therefore 7128=22 \times 4 \times(1+28)$
$\therefore I +28=\frac{7128}{22 \times 4}$
$\therefore I +28=81$
$\therefore I =81-28$
$\therefore I =53 cm $
$\text { ii. Now, } l ^2= r ^2+ h ^2$
$\therefore 53^2=28^2+ h ^2$
$\therefore 2809=784+ h ^2$
$\therefore 2809-784= h ^2$
$\therefore h ^2=2025$
$\therefore h =\sqrt{2025} \ldots . . \text { [Taking square root on both sides] }$
$=45 cm $
$ \therefore \quad \text { Volume of cone } =\frac{1}{3} \pi r^2 h$
$=\frac{1}{3} \times \frac{22}{7} \times 28^2 \times 45$
$=\frac{1}{3} \times \frac{22}{7} \times 28 \times 28 \times 45$
= 22 x 4 x 28 x 15
= 36960 cubic.cm
∴ The volume of the cone is 36960 cubic.cm.

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