1.5 m/s

780 m

Let R = 3 cm, r = 1 cm and h= 7-1 6 cm
∴ Slant height = \(\sqrt { { h }^{ 2 }+(R-r)^{ 2 } } \)
  = \(\sqrt { { 6 }^{ 2 }+(3-1)^{ 2 } } \)
= \(\sqrt { 36+4 } =\sqrt { 40 } \)
=6.32 cm
Now, external surface area of the shuttlecock =C.S.A. of the frustum of cone + C.S.A. of base hemisphere 
= \(\pi l(R+r)+2\pi { r }^{ 2 }\)
= \(\frac { 22 }{ 7 } \times 6.32\times (3+1)+2\times \frac { 22 }{ 7 } \times 1\times 1\)
=\(\frac { 22 }{ 7 } \times 6.32\times 4+\frac { 44 }{ 7 } \)   = 79.45+6.29
= 85.74 cm²
Hence, the required external surface area of the  shuttlecock is 85.74 cm² 

Let internal radius of the pipe = x cm
Speed of water = 2.52 km/h = 2520 m/h
\(\therefore \) Volume of water that flows in half an hour = \(\frac { 1 }{ 2 } \pi r^{ h }h\)
\(=\frac { 1 }{ 2 } \pi \times \frac { x }{ 100 } \times \frac { x }{ 100 } \times 2520=\frac { 126\pi x^{ 2 } }{ 1000 } m^{ 3 }\)
Volume of water cylindrical tank
= \(\pi \times \frac { 40 }{ 100 } \times \frac { 40 }{ 100 } \times 3.15\quad m^{ 3 }\)
\(\Rightarrow \frac { 126\pi x^{ 2 } }{ 1000 } =\pi \times \frac { 40 }{ 100 } \times \frac { 40 }{ 100 } \times 3.15\)
\(x^{ 2 }=\frac { 40 }{ 100 } \times \frac { 40 }{ 100 } \times 3.15\times \frac { 1000 }{ 126 } \)
\(x^{ 2 }=4\Rightarrow x=2cm\)
\(\therefore \) Internal diameter = 4cm

Let breadth of the room be x m 
\(\Rightarrow \) Length of the room is
1\(\frac { 1}{2 } \) x  = \(\frac {3 }{2 } \)x m and height be h m
Area of the floor
= \(\frac { Cost \ of \ carpeting }{ cost \ of \ carpeting \ 1 \ m^{ 2 } } \)
\(=\frac { 175.50 }{ 3.25 } =54m^{ 2 }\)
\(\Rightarrow x\times \frac { 3 }{ 2 } x=54\)
\(\Rightarrow x^{ 2 }=\frac { 54\times 2 }{ 3 } \Rightarrow x=6m\)
\(\therefore \) Length of the room = \(\frac { 3 }{ 2 } \times 6=9m\)
and breadth of the room= 6m
Area to be papered = area of 4 walls - area of door and windows 
= [2 x h(9+6)-8] m²
= (30h-8) m² ...(i)
Also, area to be papered
\(=\frac { total \ cost \ of \ papering \ }{ cost \ of \ papering \ 1 \ m^{ 2 } } =\frac { 240.80 }{ 1.40 } m^{ 2 }\)
\(\frac { 2408 }{ 14 } =172m^{ 2 }\) ....(ii)
From (i) and (ii)
30h-8 = 172
\(\Rightarrow \) 30h-8 = 172
\(\Rightarrow \) h= 6m

Diameter of pencil = 7 mm
\(\therefore \) Radius = 3.5 mm = 0.35 cm
Length of pencil = 10 cm
Volume = \(\pi r^{ h }\)
= \(\pi \times 0.35\times \times 0.35\times 10\)
= 1.225 \(\pi cm^{ 3 }\)
Diameter of lead = 1 mm
\(\therefore \) Radius = \(\frac { 1 }{ 2 } mm\quad =0.05\quad cm\)
Length = 10 cm
Volume = \(\pi \times 0.05\times 0.05\times 10\)
= 0.025 \(\pi cm^{ 3 }\)
\(\therefore \)Volume of the wood = Volume of pencil - volume of lead
= 1.225 \(\pi -0.025\pi =1.2\pi cm^{ 3 }\)
Specific gravity of wood = 0.7 g/cm³
\(\therefore \) Weight of wood = 1.2 \(\pi \times 0.7=0.84\pi g\)
Specific gravity of lead (graphite ) = 2.1 g/cm³
\(\therefore \) Weight of lead = 2.1X \(0.025\pi \)
= \(0.0525\pi g\)
Total weight =  \(0.0525\pi g+0.84\pi g\)
= 2.805 g.

R₁ = 7m ; R² =13cm
Slant height of cone L = 12m
Slant height of frustum, I = \(\sqrt { (8)^{ 2 }+6^{ 2 } } =10m\)
= \(\sqrt { 8^{ 2 }+6^{ 2 } } =10m\)
Area of canvas required = curved surface area of frustum + curved surface area of cone 
= \(\pi (R_{ 1 }-R_{ 2 })l+\pi R_{ 1 }L\)
\(=\pi (13+7)\times 10+\pi \times 7\times 12\)
\(=200\pi +84\pi =284\pi m^{ 2 }\)

Volume of hollow sphere = volume of cone
 \(\frac{4}{3}\pi (R^3 - r^3)=\frac{1}{3}\pi r^2h\)
\(\Rightarrow \ \frac{4}{3}(5^3 - r^3)=\frac{1}{3}\times(7)^2 \times8\)
\(\Rightarrow \ \frac{4}{3}(125 - r^3)=\frac{1}{3}\times49 \times8\)
\(​​\Rightarrow \ 500 -4r^3 = 49\times 8\)
\(\Rightarrow \ -4r^3 = 392 - 500\)
\(\Rightarrow \ -4r^3 = -108\)
\(\Rightarrow \ r^3 =27\)
\(\Rightarrow \ r =3\) cm
Diameter, d = 6 cm

Apparent capacity Of glass =  \(\pi r^{ 2 }h\)
= 3.14 \(\times \left( \frac { 5 }{ 2 } \right) ^{ 2 }\times 10cm^{ 3 }=196.25cm^{ 3 }\)
Actual capacity of glass = apparent capacity — volume of hemispherical part 
= 196.25 cm³ - \(\frac { 2 }{ 3 } \times 3.14\times \left( \frac { 5 }{ 2 } \right) ^{ 3 }=196.25cm^{ 3 }=32.70\quad cm^{ 3 }=163.55cm^{ 2 }\)

Height of cylinder = 20 cm
radius of cylinder = 3.5 cm = radius of each hemisphere
Total surface area of the article = 2X C.S.A of a hemisphere
= \(2\times 2\pi r^{ 2 }+2\pi rh\rightleftharpoons 2\pi r(2r+h)\)
= \(2\times \frac { 22 }{ 7 } \times 3.5\left[ 2\times 3.5+20 \right] \)
= 44 x 0.5[7+20] = 44 x 0.5 x 27 cm² = 594.0 cm²