Answer: Option C. -> 30%
Volume of the sphere :
$$ = \left[ {\frac{4}{3}\pi {{\left( {10} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3}$$
Surface area of the sphere :
$$ = \left[ {4\pi {{\left( {10} \right)}^2}} \right]{\text{ c}}{{\text{m}}^2}$$
∴ Required percentage :
$$\eqalign{
& = \left[ {\frac{{4\pi {{\left( {10} \right)}^2}}}{{\frac{4}{3}\pi {{\left( {10} \right)}^3}}}} \times 100 \right]\% \cr
& = 30\% \cr} $$

Answer: Option B. -> 288 cm3
Let the length of base be 3a cm and breadth be 2a cm
Total surface area of prism :
= [Perimeter of base × height] + [2 × Area of base]
= [2 (3a + 2a) × 12 + 2 × 3a × 2a] sq.cm
= (120a + 12a2) sq.cm
According to the question,
120a + 12a2 = 288
⇒ a2 + 10a = 24
⇒ a2 + 10a - 24 = 0
⇒ a2 + 12a - 2a - 24 = 0
⇒ a (a + 12) - 2 (a + 12) = 0
⇒ (a - 2)(a + 12) = 0
⇒ a = 2 because a $$ \ne $$ -12
∴ Volume of prism :
= Area of base × Height
= (3a × 2a × 12)cu.cm
= 72a2 cu.cm
= (72 × 2 × 2)cu.cm
= 288 cu.cm

Answer: Option D. -> 20 cm
Let the height of the cone be h cm
Then,
$$\eqalign{
& \frac{4}{3}\pi \times {\left( 5 \right)^3} = \frac{1}{3}\pi \times {\left( 5 \right)^2} \times h \cr
& \Rightarrow h = 20\,cm \cr} $$

Answer: Option D. -> 84000
Volume of each lead shot :
$$\eqalign{
& = \left[ {\frac{4}{3}\pi \times {{\left( {\frac{{0.3}}{2}} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3} \cr
& = \left( {\frac{4}{3} \times \frac{{22}}{7} \times \frac{{27}}{{8000}}} \right){\text{ c}}{{\text{m}}^3} \cr
& = \frac{{99}}{{7000}}{\text{ c}}{{\text{m}}^3} \cr} $$
∴ Number of lead shots :
$$\eqalign{
& = \left( {9 \times 11 \times 12 \times \frac{{7000}}{{99}}} \right) \cr
& = 84000 \cr} $$

Answer: Option C. -> 2.4 cm
Let the radius of the cone be R cm
Then,
$$\eqalign{
& \frac{1}{3}\pi \times {R^2} \times 75 = \frac{2}{3}\pi \times 6 \times 6 \times 6 \cr
& \Rightarrow {R^2} = \left( {\frac{{2 \times 6 \times 6 \times 6}}{{75}}} \right) \cr
& \Rightarrow {R^2} = \frac{{144}}{{25}} \cr
& \Rightarrow {R^2} = \frac{{{{\left( {12} \right)}^2}}}{{{{\left( 5 \right)}^2}}} \cr
& \Rightarrow R = \frac{{12}}{5} \cr
& \Rightarrow R = 2.4\,cm \cr} $$

Answer: Option C. -> 60π m2
$$\eqalign{
& l = 10\,m,h = 8\,m \cr
& So, \cr
& r = \sqrt {{l^2} + {h^2}} \cr
& \,\,\,\, = \sqrt {{{\left( {10} \right)}^2} - {{\left( 8 \right)}^2}} \cr
& \,\,\,\, = 6\,m \cr} $$
∴ Curved surface area :
$$\eqalign{
& = \pi rl \cr
& = \left( {\pi \times 6 \times 10} \right){m^2} \cr
& = 60\pi \,{m^2} \cr} $$

Answer: Option D. -> 81 cm
$$\eqalign{
& \frac{{\frac{4}{3}\pi {r^3}}}{{4\pi {r^2}}} = 27 \cr
& \Rightarrow r = 81\,cm \cr} $$

Answer: Option C. -> 251.2 cm2
Let the radius of the cone be r cm
Then,
$$\eqalign{
& \pi \times {\left( 8 \right)^2} \times 2 = \frac{1}{3} \times \pi \times {r^2} \times 6 \cr
& \Rightarrow r = 8 \cr} $$
Slant height,
$$\eqalign{
& l = \sqrt {{r^2} + {h^2}} \cr
& \,\,\, = \sqrt {{8^2} + {6^2}} \cr
& \,\,\, = \sqrt {100} \cr
& \,\,\, = 10\,cm \cr} $$
Curved surface area of cone :
$$\eqalign{
& = \pi rl \cr
& = \left( {3.14 \times 8 \times 10} \right){\text{ c}}{{\text{m}}^2} \cr
& = 251.2{\text{ c}}{{\text{m}}^2} \cr} $$

Answer: Option D. -> $$\sqrt 6 :\sqrt \pi $$
$$\eqalign{
& 4\pi {R^2} = 6{a^2} \cr
& \Rightarrow \frac{{{R^2}}}{{{a^2}}} = \frac{3}{{2\pi }} \cr
& \Rightarrow \frac{R}{a} = \frac{{\sqrt 3 }}{{\sqrt {2\pi } }} \cr} $$
$$\eqalign{
& \therefore \frac{{{\text{Volume of spere}}}}{{{\text{Volume of cube}}}} \cr
& = \frac{{\frac{4}{3}\pi {R^3}}}{{{a^3}}} \cr
& = \frac{4}{3}\pi {\left( {\frac{R}{a}} \right)^3} \cr
& = \frac{4}{3}\pi \frac{{3\sqrt 3 }}{{2\pi \sqrt {2\pi } }} \cr
& = \frac{{2\sqrt 3 }}{{\sqrt {2\pi } }} \cr
& = \frac{{\sqrt {12} }}{{\sqrt {2\pi } }} \cr
& = \frac{{\sqrt 6 }}{{\sqrt \pi }} \cr
& \text{or, }\sqrt 6 :\sqrt \pi \cr} $$

Answer: Option C. -> 24
Volume of sphere :
$$\eqalign{
& = \left( {\frac{4}{3}\pi \times {6^3}} \right){\text{c}}{{\text{m}}^{\text{3}}} \cr
& = \left( {288\pi } \right){\text{c}}{{\text{m}}^{\text{3}}} \cr} $$
Volume of each cone :
$$\eqalign{
& = \left( {\frac{1}{3}\pi \times {3^2} \times 4} \right){\text{c}}{{\text{m}}^{\text{3}}} \cr
& = \left( {12\pi } \right){\text{c}}{{\text{m}}^{\text{3}}} \cr} $$
∴ Number of cone :
$$\eqalign{
& = \frac{{288\pi }}{{12\pi }} \cr
& = 24 \cr} $$