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A solid is in the form of a cone of vertical height $9 cm$ mounted on the top base of a right circular cylinder of height $40 cm$ . The radius of the base of the cone and that of the cylinder are both equal to $7 cm$ . Find the weight of the solid if $1 c m^{3}$ of the solid weighs $4 gm .$

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26.488 kg

25.488 kg

24.488 kg

23.488 kg

answer is A.

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Detailed Solution

Given a cone of vertical height

9 cm

mounted on the top base of a right circular cylinder of height

40 cm

and radius of the base of the cone and that of the cylinder are both equal to

7 cm

Now,
Volume of Cylinder of Radius "R"(7cm) and height "h"(40cm)

= π R^{2} h

Volume of a cone

= \frac{1}{3} π r^{2} h

, where

r

(7cm) is the radius of the base of the cone and

h

(9cm) is the height.
Volume of the solid

=

Volume of the cylindrical part

+

Volume of conical part
So, the Volume of the solid =

(\frac{22}{7} ​ \times 7 \times 7 \times 40) + (\frac{1}{3} ​ \times \frac{22}{7} \times 7^{2} \times 9)

\Rightarrow 6160 + 462

\Rightarrow 6622 c m^{3}

According to the question,

1 {cm}^{3}

4 gm

\Rightarrow 6622 {cm}^{3}

6622 \times 4

\Rightarrow 6622 {cm}^{3} = 26488 g

\Rightarrow 6622 {cm}^{3} = 26.488 kg

Therefore, the weight of the solid is

26.488 kg

.
Hence, the correct option is 1.

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