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A wall mounted lamp, made of fabric, is shown below. Lamp has cuboidal shape, open from top and bottom. A spherical bulb of diameter $7$ cm is latched with a very thin rod. (Ignore the rod while making calculations.)
Dimensions of the cuboid are $24$ cm $\times 12$ cm $\times 17$ cm.
(i) Find the surface area of the bulb.
(ii) What could be the maximum diameter of the bulb if at least $1$ cm space is left from each side?
(iii) (a) Find the area of the fabric used if there is a fold of $2$ cm on top and bottom edges.
OR
(iii) (b) Find the space available inside the lamp.
Dimensions of the cuboid are $24$ cm $\times 12$ cm $\times 17$ cm.
(i) Find the surface area of the bulb.
(ii) What could be the maximum diameter of the bulb if at least $1$ cm space is left from each side?
(iii) (a) Find the area of the fabric used if there is a fold of $2$ cm on top and bottom edges.
OR
(iii) (b) Find the space available inside the lamp.
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Sol. (i) Surface area of the bulb = $4 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = 154$ cm$^2$ (I Mark)
(ii) Maximum diameter of the bulb = Minimum side length $- 2$ cm
$= 12 - 2 = 10$ cm (I Mark)
(iii) (a) With $2$ cm extra cloth for top and bottom edges,
new dimensions are $24$ cm $\times 12$ cm $\times 21$ cm (I Mark)
Area of fabric used = $2 \times 21 \times (24 + 12) = 1512$ cm$^2$ (II Mark)
OR
(iii) (b) Space available = $24 \times 12 \times 17 - \frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}$ (I Mark)
$= 4896 - \frac{539}{3}$ (II Mark)
$= \frac{14149}{3}$ cm$^3$ or $4716.3$ cm$^3$ (approx.)
(ii) Maximum diameter of the bulb = Minimum side length $- 2$ cm
$= 12 - 2 = 10$ cm (I Mark)
(iii) (a) With $2$ cm extra cloth for top and bottom edges,
new dimensions are $24$ cm $\times 12$ cm $\times 21$ cm (I Mark)
Area of fabric used = $2 \times 21 \times (24 + 12) = 1512$ cm$^2$ (II Mark)
OR
(iii) (b) Space available = $24 \times 12 \times 17 - \frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}$ (I Mark)
$= 4896 - \frac{539}{3}$ (II Mark)
$= \frac{14149}{3}$ cm$^3$ or $4716.3$ cm$^3$ (approx.)