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A pen stand is made up of wood and is in the shape of a cuboid with four conical depressions and cylindrical extensions to hold pens. The depressions are hollowed out from the wooden cuboidal stand and cylindrical extensions are made up of metal attached separately as shown in the figure given below :
The dimensions of the cuboidal block of wood are $15 \text{ cm} \times 10 \text{ cm} \times 3.5 \text{ cm}$. The radius of each conical depression is $0.5 \text{ cm}$ and depth is $1.4 \text{ cm}$. The radius and height of each cylindrical metal extension are $0.5 \text{ cm}$ and $4.9 \text{ cm}$ respectively.
(i) Find the volume of wood in the entire stand.
(ii) Find the inner surface area of a conical depression along with that of the cylindrical extension. (Use $\sqrt{221} = 14.8$)
The dimensions of the cuboidal block of wood are $15 \text{ cm} \times 10 \text{ cm} \times 3.5 \text{ cm}$. The radius of each conical depression is $0.5 \text{ cm}$ and depth is $1.4 \text{ cm}$. The radius and height of each cylindrical metal extension are $0.5 \text{ cm}$ and $4.9 \text{ cm}$ respectively.
(i) Find the volume of wood in the entire stand.
(ii) Find the inner surface area of a conical depression along with that of the cylindrical extension. (Use $\sqrt{221} = 14.8$)
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(i) Volume of wood = volume of cuboid - $4 \times$ volume of cone
$= 15 \times 10 \times 3.5 - 4 \times (\frac{1}{3} \times \frac{22}{7} \times 0.5 \times 0.5 \times 1.4)$ (2 Marks)
$= \frac{7853}{15} \text{ cm}^3$ or $523.53 \text{ cm}^3$ (1/2 Mark)
(ii) $l = \sqrt{(1.4)^2 + (0.5)^2} = \sqrt{2.21} = 1.48$ (1 Mark)
Inner curved surface area = C.S.A. of cone + C.S.A. of cylinder
$= (\frac{22}{7} \times 0.5 \times 1.48 + 2 \times \frac{22}{7} \times 0.5 \times 4.9)$ (1 Mark)
$= \frac{12408}{700} \text{ cm}^2$ or $17.72 \text{ cm}^2$ (1/2 Mark)
$= 15 \times 10 \times 3.5 - 4 \times (\frac{1}{3} \times \frac{22}{7} \times 0.5 \times 0.5 \times 1.4)$ (2 Marks)
$= \frac{7853}{15} \text{ cm}^3$ or $523.53 \text{ cm}^3$ (1/2 Mark)
(ii) $l = \sqrt{(1.4)^2 + (0.5)^2} = \sqrt{2.21} = 1.48$ (1 Mark)
Inner curved surface area = C.S.A. of cone + C.S.A. of cylinder
$= (\frac{22}{7} \times 0.5 \times 1.48 + 2 \times \frac{22}{7} \times 0.5 \times 4.9)$ (1 Mark)
$= \frac{12408}{700} \text{ cm}^2$ or $17.72 \text{ cm}^2$ (1/2 Mark)