1300 cm²

28 cm

3.5 cm

For a brick
l = 22.5 cm, b = 10 cm,
h = 7.5 cm
\(\therefore \)  Total surface area of a brick 
= 2 (lb + bh + hl)
= 2 (22.5 \(\times\) 10 + 10 \(\times\) 7.5 + 7.5 \(\times\) 22.5)
= 2 (225 + 75 + 168.75)
= 2(468.75) = 937.5 cm² = .09375 m²
\(\therefore \)  Number of bricks that can be painted out 
\(=\frac { 9.375 }{ .09375 } =100.\)

(i) For herbarium 
l = 30 cm, b = 25 cm,
h = 25 cm
\(\therefore \)  Area of the glass = 2 (lb + bh + hl)
= 2[(30)(25) + (25)(25) + (25)(30)]
= 2[750 + 625 + 750] = 4250 cm².
(ii) The tape needed for all the 12 edges
= 4 (l + b + h)
= 4(30 + 25 + 25) = 320 cm.

(i)   2r = 3.5 m
\(\Rightarrow\)  \(r=\frac { 3.5 }{ 2 } m\)
\(\Rightarrow\)   r = 1.75 m
h = 10 m
\(\therefore \) Inner curved surface area of the circular well = \(2\pi rh\)
\(=2\times \frac { 22 }{ 7 } \times 1.75\times 10=110{ m }^{ 2 }.\)
(ii) Cost of plastering the curved surface at the rate of Rs 40 per m² = Rs 110 \(\times\) 40 = Rs 4400.   

\(\because \) Diameter of the base   = 10.5 cm
\(\therefore \) Radius of the base (r) \(=\frac { 10.5 }{ 2 } cm\)
= 5.25 cm
Slant height (l) = 10 cm
\(\therefore \) Curved surface area of the cone = \(\pi rl\)
 \(=\frac { 22 }{ 7 } \times 5.25\times 10=165{ cm }^{ 2 }.\)

Let the dimensions of the box be 2k, 3k and 4k.
\(\therefore \)  Total surface area = 2(lb + bh + hl)
= 2(2k.3k + 3k.4k + 4k.2k)
= 52k²m²
Cost of covering at the rate of Rs 8 per m²
= 52k² \(\times\) 8 = Rs 416k²
Cost of covering at the rate of Rs 9.50 per m²
= 52k² \(\times\) 9.50 = Rs 494k²
Difference between the costs
= Rs 494k² - Rs 416k² = Rs 78k²
According to the question,
78k² = 1248
\(\Rightarrow\)  k² = 16 \(\Rightarrow\) k = 4
Hence, the dimensions of the box are 8 m, 12 m and 16 m.

Diameter = 50 cm
Radius (r) = \(\frac { 50 }{ 2 } cm=25cm\)
\( =\frac { 25 }{ 100 } m=\frac { 1 }{ 4 } m\)
Height (h) = 4 m
\(\therefore \)  Curved surface area of 1 pillar = \(2\pi rh\)
\(=2\times \frac { 22 }{ 7 } \times \frac { 1 }{ 4 } \times 4=\frac { 44 }{ 7 } { m }^{ 2 }\)
\(\therefore \) Curved surface area of 10 pillars
\(=\frac { 44 }{ 7 } \times 10{ m }^{ 2 }=\frac { 440 }{ 7 } { m }^{ 2 }\)
\(\therefore \)   Cost of painting at the rate of Rs 14 per square metre = \(\frac { 440 }{ 7 } \times 14=\)  Rs 880.

(a) Volume of well in cuboidal shape V₁= l \(\times\)b \(\times\) h₁
Given, l=1.5 m, b = 1.5 m and h₁=7m
\(\therefore\) V₁=1.5 \(\times\) 1.5 \(\times\) 7 
= 15.75 m³
\(\because\) Rate of digging well = Rs 75/m³   (given)
\(\because\) Cost of digging a well of cuboidal shape
= 15.75 \(\times\) 75
= Rs 1181.25
Volume of well in cylindrical shape V₂ = \(\pi\)r²h₂
Given, r = 75 cm = 0.75 m, h₂= 7m
\(\therefore \ { V }_{ 2 }=\frac { 22 }{ 7 } \times 0.7\times 0.75\times 7\)
= 12.375 m³
\(\therefore\) Cost of digging a well of cylindrical shape = 12.375 \(\times\) 75
= Rs 928.125
(b) Economical approach.

Length of the rod \(=\sqrt { { \left( 10 \right) }^{ 2 }+{ \left( 10 \right) }^{ 2 }+{ \left( 5 \right) }^{ 2 } } \)

Number of cubes = \(\frac { { \left( 20 \right) }^{ 3 } }{ { \left( 5 \right) }^{ 3 } } =64\)

Required number \(=\frac { 60\times 30\times 30 }{ 15\times 6\times 4 } =150\)

Height of the cone(h) = 24 cm
Let r cm be the radius of the base and l cm an can be the slant height of the cone, then
\(l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
\(=\sqrt { { r }^{ 2 }+{ 24 }^{ 2 } } \)
\(=\sqrt { { r }^{ 2 }+576 } \)
Now, curved surface area=\(\pi\)rl
\(\Rightarrow \ \frac { 22 }{ 7 } \times r\times \sqrt { { r }^{ 2 }+576 } =550\)
\(\Rightarrow \ r\sqrt { { r }^{ 2 }+576 } =550\times \frac { 7 }{ 22 } \)
\(\Rightarrow \ r\sqrt { { r }^{ 2 }+576 } =175\)
Squaring both the sides we get
r² (r² + 576) = 30625
(r²)² + 576r²-30625 = 0
Let r² = x
x² + 576x - 30625 = 0
\(\Rightarrow\) x² + 625x - 49x - 30625 = 0
\(\Rightarrow\) x(x + 625)-  49(x + 625) = 0
\(\Rightarrow\) (x + 625)(x - 49) = 0
\(\Rightarrow\) x + 625=0 or x - 49 = 0
\(\Rightarrow\) x = 625 or x = 49
not possible x = 49
\(\therefore\) r²= 49
\(\Rightarrow\) r = 7 cm
Volume & the cone = \(\frac { 1 }{ 3 } \pi { r }^{ 2 }h\)
\(=\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times { 7 }^{ 2 }+24\)
= 1232 cm³.

Let r and R be the radii of the smaller and larger spheres respectively, we have
\(r=\frac { 5 }{ 2 } cm\)
Volume of the smaller sphere \(=\frac { 4 }{ 3 } \pi { r }^{ 3 }=\frac { 4 }{ 3 } \pi { \left( \frac { 5 }{ 2 } \right) }^{ 3 }\)
\(=\frac { 4 }{ 3 } \times \pi \times \frac { 125 }{ 8 } { cm }^{ 3 }\)
Density of metal\(=\frac { mass }{ Valume } \)
\(=\frac { 740 }{ \frac { 4 }{ 3 } \pi \times \frac { 125 }{ 8 } } g\quad { cm }^{ 3 }\)      ...........(i)
Volume of larger sphere = \(\frac { 4 }{ 3 } \pi { R }^{ 3 }\)
Density of metal=\(\frac { mass }{ Volume } =\frac { 5920 }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } \)    ......(ii)
From (i) and (ii), we have
\(\frac { 740 }{ \frac { 4 }{ 3 } \pi \times \frac { 125 }{ 8 } } =\frac { 5920 }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } \)
\(\Rightarrow \quad { R }^{ 3 }=\frac { 5920\times 125 }{ 740\times 8 } \)
= 125
\(\Rightarrow\) R = 5 cm.