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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 $15$ cm $\times 10$ cm $\times 3.5$ cm. The radius of each conical depression is $0.5$ cm and depth is $1.4$ cm. The radius and height of each cylindrical metal extension are $0.5$ cm and $4.9$ 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 $\sqrt{2.21} = 1.48$)
OR
A solid is in the shape of a cylinder with a solid hemisphere attached to one of its ends as shown in the figure.
Let the height and diameter of the cylinder be same (say H). Prove that the volume of the solid shown in the figure is same as the volume of a cone of height H and diameter '2H'. Also find the total surface area of the solid in terms of H.
The dimensions of the cuboidal block of wood are $15$ cm $\times 10$ cm $\times 3.5$ cm. The radius of each conical depression is $0.5$ cm and depth is $1.4$ cm. The radius and height of each cylindrical metal extension are $0.5$ cm and $4.9$ 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 $\sqrt{2.21} = 1.48$)
OR
A solid is in the shape of a cylinder with a solid hemisphere attached to one of its ends as shown in the figure.
Let the height and diameter of the cylinder be same (say H). Prove that the volume of the solid shown in the figure is same as the volume of a cone of height H and diameter '2H'. Also find the total surface area of the solid in terms of H.
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(a) (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)
$= \frac{7853}{15}$ cm$^3$ or $523.53$ 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)
$= \frac{12408}{700}$ cm$^2$ or $17.72$ cm$^2$ (1/2 Mark)
OR
(b) Radius of cylinder/hemisphere $= R = \frac{H}{2}$ (1/2 Mark)
Volume of the solid = Volume of cylinder + Volume of hemisphere
$= \pi R^2 H + \frac{2}{3} \pi R^3$ (1 Mark)
$= \pi (\frac{H}{2})^2 H + \frac{2}{3} \pi (\frac{H}{2})^3$
$= \frac{\pi H^3}{3}$ (1/2 Mark)
Volume of cone of height H and diameter $2H = \frac{1}{3} \pi \times H^2 \times H$
$= \frac{\pi H^3}{3}$ (1/2 Mark)
T.S.A. of the solid = C.S.A. of cylinder + Area of circle + C.S.A. of hemisphere
$= 2\pi \times \frac{H}{2} \times H + \pi \frac{H^2}{4} + 2\pi \frac{H^2}{4}$ (1/2 Mark)
$= \frac{7\pi H^2}{4}$ (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)
$= \frac{7853}{15}$ cm$^3$ or $523.53$ 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)
$= \frac{12408}{700}$ cm$^2$ or $17.72$ cm$^2$ (1/2 Mark)
OR
(b) Radius of cylinder/hemisphere $= R = \frac{H}{2}$ (1/2 Mark)
Volume of the solid = Volume of cylinder + Volume of hemisphere
$= \pi R^2 H + \frac{2}{3} \pi R^3$ (1 Mark)
$= \pi (\frac{H}{2})^2 H + \frac{2}{3} \pi (\frac{H}{2})^3$
$= \frac{\pi H^3}{3}$ (1/2 Mark)
Volume of cone of height H and diameter $2H = \frac{1}{3} \pi \times H^2 \times H$
$= \frac{\pi H^3}{3}$ (1/2 Mark)
T.S.A. of the solid = C.S.A. of cylinder + Area of circle + C.S.A. of hemisphere
$= 2\pi \times \frac{H}{2} \times H + \pi \frac{H^2}{4} + 2\pi \frac{H^2}{4}$ (1/2 Mark)
$= \frac{7\pi H^2}{4}$ (1/2 Mark)