Measurement of Volume and Surface Area Questions in English

(A) Total floor area required for $5$ persons $= 16 \times 5 = 80\, m^2$.
Total volume of air required for $5$ persons $= 100 \times 5 = 500\, m^3$.
The volume of a cone is given by the formula $V = \frac{1}{3} \times A \times h$,where $A$ is the base area and $h$ is the height.
For the smallest tent,we set the base area $A = 80\, m^2$ and the volume $V = 500\, m^3$.
Substituting these values into the formula: $500 = \frac{1}{3} \times 80 \times h$.
Solving for $h$: $h = \frac{500 \times 3}{80} = \frac{1500}{80} = \frac{150}{8} = 18.75\, m$.

(B) Let $H = 30 \, cm$ be the height of the larger cone and $h$ be the height of the smaller cone.
Let $V$ be the volume of the larger cone and $v$ be the volume of the smaller cone.
Since the smaller cone is cut by a plane parallel to the base,the smaller cone and the larger cone are similar.
Therefore,the ratio of their volumes is equal to the cube of the ratio of their heights:
$\frac{v}{V} = \left( \frac{h}{H} \right)^3$
Given $\frac{v}{V} = \frac{1}{27}$,we have:
$\frac{1}{27} = \left( \frac{h}{30} \right)^3$
Taking the cube root on both sides:
$\frac{1}{3} = \frac{h}{30}$
$h = \frac{30}{3} = 10 \, cm$
The section is made at a height above the base equal to the difference between the height of the larger cone and the height of the smaller cone:
Height above base $= H - h = 30 - 10 = 20 \, cm$.

(C) The surface area of a sphere is given by the formula: $A = 4 \pi r^{2}$.
Given that $A = 346.5 \, cm^{2}$ and $\pi = \frac{22}{7}$.
Substituting the values into the formula:
$4 \times \frac{22}{7} \times r^{2} = 346.5$
$\Rightarrow r^{2} = \frac{346.5 \times 7}{4 \times 22}$
$\Rightarrow r^{2} = \frac{2425.5}{88}$
$\Rightarrow r^{2} = 27.5625$
Taking the square root on both sides:
$r = \sqrt{27.5625} = 5.25 \, cm$.
Thus,the radius of the sphere is $5.25 \, cm$.

(A) The volume of a pyramid is given by the formula: $V = \frac{1}{3} \times A \times h$,where $V$ is the volume,$A$ is the area of the base,and $h$ is the height.
Given: $V = 500\, m^{3}$ and $A = 30\, m^{2}$.
Substituting the values into the formula:
$500 = \frac{1}{3} \times 30 \times h$
$500 = 10 \times h$
$h = \frac{500}{10} = 50\, m$.
Therefore,the height of the pyramid is $50\, m$.

(A) The base is a right-angled triangle with sides $a = 5 \, cm$ and $b = 12 \, cm$.
The hypotenuse $c = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, cm$.
The perimeter of the base $P = 5 + 12 + 13 = 30 \, cm$.
The lateral surface area of the prism $= P \times \text{height} = 30 \times 10 = 300 \, cm^2$.
The area of the triangular base $A_{base} = \frac{1}{2} \times 5 \times 12 = 30 \, cm^2$.
The total surface area of the prism $= \text{Lateral surface area} + 2 \times A_{base} = 300 + 2(30) = 300 + 60 = 360 \, cm^2$.

(A) Let the side of the equilateral triangle base be $a$ and the height of the prism be $h$.
Lateral surface area of the prism $= 3 \times a \times h = 120 \, cm^2$.
$\therefore a \times h = \frac{120}{3} = 40 \, cm^2 \dots (1)$
Volume of the prism $=$ Area of base $\times$ height $= \frac{\sqrt{3}}{4} a^2 \times h = 40 \sqrt{3} \, cm^3$.
$\Rightarrow a^2 \times h = \frac{40 \sqrt{3} \times 4}{\sqrt{3}} = 160 \, cm^3 \dots (2)$
Dividing equation $(2)$ by $(1)$:
$\frac{a^2 \times h}{a \times h} = \frac{160}{40}$
$a = 4 \, cm$.
Thus,the side of the base is $4 \, cm$.

(D) The diameter of the lead ball is $4\, cm$,so its radius $r = 2\, cm$.
Volume of lead $= \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (2)^3 = \frac{4}{3} \pi (8) = \frac{32}{3} \pi\, cm^3$.
Let the thickness of the gold layer be $x\, cm$.
The total radius of the ball (lead + gold) becomes $(2 + x)\, cm$.
The volume of the gold layer is the difference between the total volume and the volume of the lead ball:
Volume of gold $= \frac{4}{3} \pi (2 + x)^3 - \frac{4}{3} \pi (2)^3$.
According to the problem,the volume of gold equals the volume of lead:
$\frac{4}{3} \pi ((2 + x)^3 - 2^3) = \frac{4}{3} \pi (2^3)$.
$(2 + x)^3 - 8 = 8$.
$(2 + x)^3 = 16$.
Taking the cube root on both sides:
$2 + x = \sqrt[3]{16} = \sqrt[3]{8 \times 2} = 2 \times \sqrt[3]{2}$.
Given $\sqrt[3]{2} = 1.259$,we have:
$2 + x = 2 \times 1.259 = 2.518$.
$x = 2.518 - 2 = 0.518\, cm$.

(D) Let the radius of the larger sphere be $R$.
Volume of the larger sphere $= \frac{4}{3} \pi R^3$.
Volume of one cone with base radius $R$ and height $R$ is $V_{cone} = \frac{1}{3} \pi R^2 h = \frac{1}{3} \pi R^3$.
Let the radius of the smaller sphere formed by melting one cone be $r$.
Volume of the smaller sphere $= \frac{4}{3} \pi r^3$.
Equating the volumes: $\frac{4}{3} \pi r^3 = \frac{1}{3} \pi R^3$.
$4r^3 = R^3 \Rightarrow r^3 = \frac{R^3}{4} \Rightarrow r = \frac{R}{4^{1/3}} = \frac{R}{2^{2/3}}$.
The ratio of the surface area of the smaller sphere to the larger sphere is $\frac{4 \pi r^2}{4 \pi R^2} = \frac{r^2}{R^2}$.
Substituting $r$: $\frac{(R / 2^{2/3})^2}{R^2} = \frac{R^2 / 2^{4/3}}{R^2} = \frac{1}{2^{4/3}}$.
Thus,the ratio is $1 : 2^{4/3}$.

(A) The total volume of the ice cream is the sum of the volume of the hemisphere and the volume of the cone.
Given,radius of the hemisphere $r = 7 \, cm$.
Since the radius of the hemispherical part equals the height of the cone,the height of the cone $h = 7 \, cm$.
Volume of hemisphere $= \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 7^3 = \frac{2}{3} \times 22 \times 49 = \frac{2156}{3} \, cm^3$.
Volume of cone $= \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 7 = \frac{1}{3} \times 22 \times 49 = \frac{1078}{3} \, cm^3$.
Total volume $= \frac{2156}{3} + \frac{1078}{3} = \frac{3234}{3} = 1078 \, cm^3$.

(C) Let the original side of the cube be $s$. The original volume is $V_1 = s^3$.
If the side is increased by $10 \%$,the new side $s' = s + 0.1s = 1.1s$.
The new volume is $V_2 = (1.1s)^3 = 1.331s^3$.
The increase in volume is $V_2 - V_1 = 1.331s^3 - s^3 = 0.331s^3$.
The percentage increase in volume is $\frac{0.331s^3}{s^3} \times 100 \% = 33.1 \%$.

(D) Area of the rectangular plot $LMNO = 36 \times 28 = 1008 \, m^{2}$.
Area of the paths = Area of road parallel to length + Area of road parallel to breadth - Area of the common intersection square $PQRS$.
Area of paths = $(36 \times 5) + (28 \times 5) - (5 \times 5) = 180 + 140 - 25 = 295 \, m^{2}$.
Area of the rectangular plot excluding the area covered by the roads = $1008 - 295 = 713 \, m^{2}$.
Total cost of gravelling the plot = $713 \times 3.60 = ₹ 2566.80$.

(A) Given: Height of the cone $h = 12\, cm$ and radius of the cone $R = 6\, cm$.
Let the radius of the hemisphere (which is the same as the radius of the sphere) be $r\, cm$.
The volume of a hemisphere is given by $V_h = \frac{2}{3} \pi r^3$.
The volume of a cone is given by $V_c = \frac{1}{3} \pi R^2 h$.
According to the problem,$8$ bowlfuls of the hemisphere fill the conical vessel:
$8 \times V_h = V_c$
$8 \times \left( \frac{2}{3} \pi r^3 \right) = \frac{1}{3} \pi R^2 h$
Substitute the values:
$8 \times \frac{2}{3} \pi r^3 = \frac{1}{3} \pi (6)^2 (12)$
Divide both sides by $\frac{1}{3} \pi$:
$16 r^3 = 36 \times 12$
$16 r^3 = 432$
$r^3 = \frac{432}{16} = 27$
$r = \sqrt[3]{27} = 3\, cm$.
Thus,the radius of the sphere is $3\, cm$.

(A) In right $\Delta DAB$,by the Pythagorean theorem,$BD = \sqrt{AD^2 + AB^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\, cm$.
Area of $\Delta DAB = \frac{1}{2} \times AB \times AD = \frac{1}{2} \times 9 \times 12 = 54\, cm^2$.
For $\Delta BCD$,the sides are $BC = 14\, cm$,$CD = 13\, cm$,and $BD = 15\, cm$. The semi-perimeter $S = \frac{14 + 13 + 15}{2} = \frac{42}{2} = 21\, cm$.
Using Heron's formula,Area of $\Delta BCD = \sqrt{S(S-a)(S-b)(S-c)} = \sqrt{21(21-14)(21-13)(21-15)} = \sqrt{21 \times 7 \times 8 \times 6} = \sqrt{7056} = 84\, cm^2$.
Area of quadrilateral $ABCD = \text{Area of } \Delta DAB + \text{Area of } \Delta BCD = 54 + 84 = 138\, cm^2$.
Volume of the prism = Base Area $\times$ Height.
$2070 = 138 \times h \implies h = \frac{2070}{138} = 15\, cm$.
Lateral Surface Area $(LSA)$ = Perimeter of base $\times$ Height.
Perimeter = $AB + BC + CD + DA = 9 + 14 + 13 + 12 = 48\, cm$.
$LSA$ = $48 \times 15 = 720\, cm^2$.

(D) Let the radius of the cylinder be $r$ and its height be $h$.
Let the radius of the sphere be $R$.
Given that the radius of the cylinder is equal to the diameter of the sphere,we have $r = 2R$,which implies $R = r/2$.
The volume of the cylinder is $V_c = \pi r^2 h$.
The volume of the sphere is $V_s = \frac{4}{3} \pi R^3$.
Since the volumes are equal,$\pi r^2 h = \frac{4}{3} \pi R^3$.
Substituting $R = r/2$ into the equation:
$\pi r^2 h = \frac{4}{3} \pi (r/2)^3$
$\pi r^2 h = \frac{4}{3} \pi (r^3 / 8)$
$r^2 h = \frac{4}{24} r^3$
$r^2 h = \frac{1}{6} r^3$
Dividing both sides by $r^2$ (assuming $r \neq 0$):
$h = \frac{1}{6} r$
Therefore,the ratio of the height to the radius of the cylinder is $h/r = 1/6$.

(C) When a rectangular sheet is rolled along its length to form a cylinder,the length of the sheet becomes the circumference of the base of the cylinder,and the breadth of the sheet becomes the height of the cylinder.
Given: Length of sheet $= 12 \, cm$,Breadth of sheet $= 5 \, cm$.
Circumference of the cylinder base $= 2 \pi R = 12 \, cm$.
$R = \frac{12}{2 \pi} = \frac{6}{\pi} \, cm$.
Height of the cylinder $(h) = 5 \, cm$.
Volume of the cylinder $= \pi R^2 h = \pi \left( \frac{6}{\pi} \right)^2 \times 5$.
Volume $= \pi \times \frac{36}{\pi^2} \times 5 = \frac{180}{\pi} \, cm^3$.

(A) The total surface area of a cone is given by the formula: $\text{Total Surface Area} = \pi r(r + l)$,where $r$ is the base radius and $l$ is the slant height.
Given: height $h = 24 \text{ cm}$,radius $r = 7 \text{ cm}$.
First,calculate the slant height $l$ using the Pythagorean theorem: $l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \text{ cm}$.
Now,substitute the values into the formula: $\text{Total Surface Area} = \frac{22}{7} \times 7 \times (7 + 25)$.
$= 22 \times 32 = 704 \text{ cm}^2$.

(D) Let the number of bricks be $n$.
Volume of the pile $= 20 \, m^3 = 20 \times 100 \times 100 \times 100 \, cm^3 = 20,000,000 \, cm^3$.
Volume of one brick $= \text{length} \times \text{breadth} \times \text{height} = 25 \, cm \times 12.5 \, cm \times 8 \, cm$.
$25 \times 12.5 = 312.5$.
$312.5 \times 8 = 2500 \, cm^3$.
Since there is no gap between the bricks, the total volume of the pile is equal to the total volume of $n$ bricks:
$n \times (\text{Volume of one brick}) = \text{Volume of the pile}$.
$n \times 2500 = 20,000,000$.
$n = \frac{20,000,000}{2500} = \frac{200,000}{25} = 8000$.
Therefore, the number of bricks is $8000$.

(C) The volume of a right circular cone is given by the formula $V = \frac{1}{3} \pi r^2 h$.
For the first cone:
Radius $r_1 = 1.6 \ cm$,Height $h_1 = 3.6 \ cm$.
Volume $V_1 = \frac{1}{3} \pi (1.6)^2 (3.6) = \pi \times 1.6 \times 1.6 \times 1.2 \ cm^3$.
When the cone is melted and recast,the volume remains constant. Let the height of the new cone be $H$.
For the second cone:
Radius $r_2 = 1.2 \ cm$,Height $H$.
Volume $V_2 = \frac{1}{3} \pi (1.2)^2 H$.
Equating the volumes $(V_1 = V_2)$:
$\frac{1}{3} \pi (1.2)^2 H = \pi (1.6)^2 (1.2)$
$H = \frac{1.6 \times 1.6 \times 1.2 \times 3}{1.2 \times 1.2}$
$H = \frac{2.56 \times 3}{1.2} = \frac{7.68}{1.2} = 6.4 \ cm$.

(D) Let the radius of the base of the cone be $r$ units.
The volume $v$ is given by $v = \frac{1}{3} \pi r^{2} h$.
The curved surface area $c$ is given by $c = \pi r \sqrt{h^{2} + r^{2}}$.
Now,substitute these into the expression $3 \pi v h^{3} - c^{2} h^{2} + 9 v^{2}$:
$3 \pi v h^{3} = 3 \pi (\frac{1}{3} \pi r^{2} h) h^{3} = \pi^{2} r^{2} h^{4}$.
$c^{2} h^{2} = (\pi r \sqrt{h^{2} + r^{2}})^{2} h^{2} = \pi^{2} r^{2} (h^{2} + r^{2}) h^{2} = \pi^{2} r^{2} h^{4} + \pi^{2} r^{4} h^{2}$.
$9 v^{2} = 9 (\frac{1}{3} \pi r^{2} h)^{2} = 9 (\frac{1}{9} \pi^{2} r^{4} h^{2}) = \pi^{2} r^{4} h^{2}$.
Substituting these back into the expression:
$(\pi^{2} r^{2} h^{4}) - (\pi^{2} r^{2} h^{4} + \pi^{2} r^{4} h^{2}) + (\pi^{2} r^{4} h^{2}) = \pi^{2} r^{2} h^{4} - \pi^{2} r^{2} h^{4} - \pi^{2} r^{4} h^{2} + \pi^{2} r^{4} h^{2} = 0$.

(D) Given: Volume of cone $V = 1232 \, m^3$,Base area $A = 154 \, m^2$,Width of canvas $w = 2 \, m$.
Step $1$: Find the radius $r$.
$A = \pi r^2 = 154$
$\frac{22}{7} \times r^2 = 154 \Rightarrow r^2 = \frac{154 \times 7}{22} = 49 \Rightarrow r = 7 \, m$.
Step $2$: Find the height $h$.
$V = \frac{1}{3} \pi r^2 h = 1232$
$\frac{1}{3} \times 154 \times h = 1232 \Rightarrow h = \frac{1232 \times 3}{154} = 8 \times 3 = 24 \, m$.
Step $3$: Find the slant height $l$.
$l = \sqrt{h^2 + r^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \, m$.
Step $4$: Find the area of the canvas (Curved Surface Area).
$CSA = \pi r l = \frac{22}{7} \times 7 \times 25 = 550 \, m^2$.
Step $5$: Find the length of the canvas.
$Length = \frac{Area}{Width} = \frac{550}{2} = 275 \, m$.

(D) Let the height of the conical glass be $h \text{ cm}$.
Since the height is equal to the diameter of the rim,the radius of the rim $r = \frac{h}{2} \text{ cm}$.
The volume of the conical glass is given by $V_{\text{glass}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (\frac{h}{2})^2 h = \frac{\pi h^3}{12}$.
The diameter of a spherical water drop is $\frac{1}{10} \text{ cm}$,so its radius $r_d = \frac{1}{20} \text{ cm}$.
The volume of one drop is $V_{\text{drop}} = \frac{4}{3} \pi r_d^3 = \frac{4}{3} \pi (\frac{1}{20})^3 = \frac{4 \pi}{3 \times 8000} = \frac{\pi}{6000} \text{ cm}^3$.
Given that $32,000$ drops fill the glass,$V_{\text{glass}} = 32000 \times V_{\text{drop}}$.
$\frac{\pi h^3}{12} = 32000 \times \frac{\pi}{6000}$.
$\frac{h^3}{12} = \frac{32}{6} = \frac{16}{3}$.
$h^3 = \frac{16}{3} \times 12 = 16 \times 4 = 64$.
$h = \sqrt[3]{64} = 4 \text{ cm}$.

(B) Volume of the rectangular block $= 11 \times 10 \times 5 = 550 \text{ m}^3$.
Since $1 \text{ m} = 10 \text{ dm}$,$1 \text{ m}^3 = 1000 \text{ dm}^3$.
Volume of the block $= 550 \times 1000 = 550,000 \text{ dm}^3$.
Diameter of each spherical bullet $= 5 \text{ dm}$,so radius $r = 2.5 \text{ dm} = \frac{5}{2} \text{ dm}$.
Volume of one spherical bullet $= \frac{4}{3} \pi r^3 = \frac{4}{3} \times \pi \times (\frac{5}{2})^3 = \frac{4}{3} \times \pi \times \frac{125}{8} = \frac{125\pi}{6} \text{ dm}^3$.
Given $\pi > 3$,the volume of one bullet is $V_b = \frac{125\pi}{6} > \frac{125 \times 3}{6} = 62.5 \text{ dm}^3$.
The number of bullets $n = \frac{\text{Total Volume}}{\text{Volume of one bullet}} = \frac{550,000}{V_b} < \frac{550,000}{62.5} = 8800$.
Since the volume of each bullet is strictly greater than $62.5 \text{ dm}^3$,the total number of bullets must be less than $8800$.

(A) The volume of the rectangular block is given by $V = \text{length} \times \text{breadth} \times \text{height} = 21 \times 77 \times 24 \, cm^3$.
Let the radius of the sphere be $r \, cm$.
Since the block is melted into a sphere, the volume of the sphere must be equal to the volume of the rectangular block.
Volume of sphere $= \frac{4}{3} \pi r^3$.
Equating the volumes: $\frac{4}{3} \times \frac{22}{7} \times r^3 = 21 \times 77 \times 24$.
Solving for $r^3$: $r^3 = \frac{21 \times 77 \times 24 \times 3 \times 7}{4 \times 22}$.
Simplifying the expression: $r^3 = \frac{21 \times 77 \times 24 \times 21}{88} = \frac{21 \times 77 \times 6 \times 21}{22} = 21 \times 7 \times 3 \times 21 = 21 \times 21 \times 21 = 21^3$.
Therefore, $r = 21 \, cm$.

(A) Let the radius of the cone be $r\, cm$.
Given height $h = 24\, cm$ and volume $V = 1232\, cm^{3}$.
The formula for the volume of a cone is $V = \frac{1}{3} \pi r^{2} h$.
Substituting the values: $\frac{1}{3} \times \frac{22}{7} \times r^{2} \times 24 = 1232$.
$r^{2} = \frac{1232 \times 3 \times 7}{22 \times 24} = 49$.
Therefore,$r = \sqrt{49} = 7\, cm$.
The slant height $l$ is given by $l = \sqrt{h^{2} + r^{2}} = \sqrt{24^{2} + 7^{2}} = \sqrt{576 + 49} = \sqrt{625} = 25\, cm$.
The curved surface area of the cone is $\pi r l = \frac{22}{7} \times 7 \times 25 = 22 \times 25 = 550\, cm^{2}$.

(A) Let the original edge of the cube be $a$.
Original surface area $S_1 = 6a^2$.
New edge length $a' = a + 0.50a = 1.5a$.
New surface area $S_2 = 6(1.5a)^2 = 6(2.25a^2) = 13.5a^2$.
Increase in surface area $= S_2 - S_1 = 13.5a^2 - 6a^2 = 7.5a^2$.
Percentage increase $= \left(\frac{7.5a^2}{6a^2}\right) \times 100 = 1.25 \times 100 = 125 \%$.
Alternatively,using the successive percentage formula for area change: $\text{Percentage change} = (x + y + \frac{xy}{100}) \%$,where $x = y = 50$.
Percentage increase $= (50 + 50 + \frac{50 \times 50}{100}) \% = (100 + 25) \% = 125 \%$.

(C) Let the radii of the two spheres be $r_1$ and $r_2$ respectively.
The surface area of a sphere is given by $S = 4 \pi r^2$.
Given the ratio of surface areas: $\frac{4 \pi r_1^2}{4 \pi r_2^2} = \frac{4}{9}$.
This simplifies to $\frac{r_1^2}{r_2^2} = \frac{4}{9}$,which implies $\frac{r_1}{r_2} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.
The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
The ratio of their volumes is $\frac{V_1}{V_2} = \frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3} = \left(\frac{r_1}{r_2}\right)^3$.
Substituting the ratio of radii: $\left(\frac{2}{3}\right)^3 = \frac{8}{27}$.
Thus,the ratio of their volumes is $8:27$.

(D) Let the original edge of the cube be $a$. The original surface area is $S_1 = 6a^2$.
If the edge is increased by $50 \%$,the new edge $a' = a + 0.5a = 1.5a$.
The new surface area is $S_2 = 6(1.5a)^2 = 6(2.25a^2) = 13.5a^2$.
The increase in surface area is $S_2 - S_1 = 13.5a^2 - 6a^2 = 7.5a^2$.
The percentage increase is $\frac{7.5a^2}{6a^2} \times 100 = 1.25 \times 100 = 125 \%$.
Alternatively,using the successive percentage change formula for area (two dimensions): $x + y + \frac{xy}{100} = 50 + 50 + \frac{50 \times 50}{100} = 100 + 25 = 125 \%$.

(D) The volume of the sphere is given by $V = \frac{4}{3} \pi r^3$,where $r$ is the radius of the sphere.
Given diameter $= 18 \, cm$,so radius $r = 9 \, cm$.
Volume $= \frac{4}{3} \times \pi \times (9)^3 = \frac{4}{3} \times \pi \times 729 = 972 \pi \, cm^3$.
The wire is a cylinder with length $h = 108 \, m = 10800 \, cm$.
Let the radius of the wire be $R \, cm$.
The volume of the wire is $\pi R^2 h = \pi R^2 \times 10800$.
Since the volume remains constant,$\pi R^2 \times 10800 = 972 \pi$.
$R^2 = \frac{972}{10800} = 0.09$.
$R = \sqrt{0.09} = 0.3 \, cm$.
The diameter of the wire is $2R = 2 \times 0.3 = 0.6 \, cm$.

(B) The diameter of the semicircular sheet is $28 \, cm$,so its radius $R = 14 \, cm$.
When this sheet is bent into a cone,the radius of the semicircle becomes the slant height $l$ of the cone,so $l = 14 \, cm$.
The circumference of the base of the cone is equal to the arc length of the semicircle.
Circumference of base $= \pi R = \frac{22}{7} \times 14 = 44 \, cm$.
Let $r$ be the radius of the base of the cone. Then $2 \pi r = 44$.
$2 \times \frac{22}{7} \times r = 44 \implies r = 7 \, cm$.
The height $h$ of the cone is given by $h = \sqrt{l^2 - r^2} = \sqrt{14^2 - 7^2} = \sqrt{196 - 49} = \sqrt{147} = 7\sqrt{3} \, cm$.
Using $\sqrt{3} \approx 1.732$,$h \approx 7 \times 1.732 = 12.124 \, cm$.
The capacity (volume) of the cup is $V = \frac{1}{3} \pi r^2 h$.
$V = \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 7\sqrt{3} = \frac{1078\sqrt{3}}{3} \approx 622.36 \, cm^3$.

(A) The surface area $S$ of a sphere with radius $r$ is given by $S = 4 \pi r^{2}$.
The volume $V$ of a sphere with radius $r$ is given by $V = \frac{4}{3} \pi r^{3}$.
We need to calculate the value of $\frac{S^{3}}{V^{2}}$.
First,calculate $S^{3} = (4 \pi r^{2})^{3} = 64 \pi^{3} r^{6}$.
Next,calculate $V^{2} = (\frac{4}{3} \pi r^{3})^{2} = \frac{16}{9} \pi^{2} r^{6}$.
Now,divide $S^{3}$ by $V^{2}$:
$\frac{S^{3}}{V^{2}} = \frac{64 \pi^{3} r^{6}}{\frac{16}{9} \pi^{2} r^{6}} = \frac{64 \pi^{3} r^{6} \times 9}{16 \pi^{2} r^{6}} = \frac{64 \times 9}{16} \times \frac{\pi^{3}}{\pi^{2}} = 4 \times 9 \times \pi = 36 \pi$.
Thus,the value is $36 \pi$.

(C) The edge of the ice cube is $a = 14 \, cm$.
To form the largest cylinder from this cube,the diameter of the cylinder must be equal to the edge of the cube,and the height of the cylinder must also be equal to the edge of the cube.
Therefore,the radius $r = \frac{a}{2} = \frac{14}{2} = 7 \, cm$.
The height $h = a = 14 \, cm$.
The volume of the cylinder $V = \pi r^2 h$.
Substituting the values: $V = \frac{22}{7} \times 7 \times 7 \times 14$.
$V = 22 \times 7 \times 14 = 154 \times 14 = 2156 \, cm^3$.

(D) Let the initial side of the cube be $x$. The initial volume $V_1 = x^3$.
If the side is increased by $100 \%$,the new side becomes $x + 1.00x = 2x$.
The new volume $V_2 = (2x)^3 = 8x^3$.
The increase in volume is $V_2 - V_1 = 8x^3 - x^3 = 7x^3$.
The percentage increase in volume is $\frac{V_2 - V_1}{V_1} \times 100 = \frac{7x^3}{x^3} \times 100 = 700 \%$.

(D) The volume of the solid sphere is given by $V = \frac{4}{3} \pi r^3$. Given $r = 1 \, cm$,the volume is $V = \frac{4}{3} \pi (1)^3 = \frac{4}{3} \pi \, cm^3$.
The wire is in the shape of a cylinder. The volume of a cylinder is $V = \pi r_w^2 h$,where $r_w$ is the radius of the wire and $h$ is the length of the wire.
Given $h = 100 \, cm$,the volume of the wire is $V = \pi r_w^2 (100) = 100 \pi r_w^2 \, cm^3$.
Since the sphere is melted to form the wire,their volumes must be equal:
$100 \pi r_w^2 = \frac{4}{3} \pi$
Dividing both sides by $\pi$:
$100 r_w^2 = \frac{4}{3}$
$r_w^2 = \frac{4}{300} = \frac{1}{75}$
$r_w = \sqrt{\frac{1}{75}} = \frac{1}{\sqrt{25 \times 3}} = \frac{1}{5 \sqrt{3}}$
Using $\sqrt{3} = 1.732$:
$r_w = \frac{1}{5 \times 1.732} = \frac{1}{8.66} \approx 0.115 \, cm$.
Rounding to two decimal places,the radius is approximately $0.11 \, cm$.

(C) Volume of the earth taken out $= (7.5 \times 6 \times 0.8) \, m^3 = 36 \, m^3$.
Area of the remaining field $= (18 \times 15) - (7.5 \times 6) \, m^2 = 270 - 45 = 225 \, m^2$.
Let the rise in the level of the field be $h$ meters.
Since the volume of the earth taken out is spread over the remaining area,we have:
$225 \times h = 36$.
$h = \frac{36}{225} \, m = 0.16 \, m$.
Converting to centimeters: $0.16 \times 100 = 16 \, cm$.

(C) Area of the base $= \frac{\sqrt{3}}{4} \times 4^2 = 4 \sqrt{3} \text{ cm}^2$.
The distance from the centroid of the equilateral triangle to the midpoint of any side is $r = \frac{1}{3} \times \text{median} = \frac{1}{3} \times \frac{\sqrt{3}}{2} \times 4 = \frac{2 \sqrt{3}}{3} \text{ cm}$.
Let the height of the pyramid be $h$ and the slant height be $l$. Given $h = \frac{l}{2}$,so $l = 2h$.
In the right-angled triangle formed by the height,the distance $r$,and the slant height $l$,we have $l^2 = h^2 + r^2$.
Substituting $l = 2h$,we get $(2h)^2 = h^2 + (\frac{2 \sqrt{3}}{3})^2$.
$4h^2 - h^2 = \frac{4 \times 3}{9} \Rightarrow 3h^2 = \frac{4}{3} \Rightarrow h^2 = \frac{4}{9} \Rightarrow h = \frac{2}{3} \text{ cm}$.
Volume $= \frac{1}{3} \times \text{Area of base} \times h = \frac{1}{3} \times 4 \sqrt{3} \times \frac{2}{3} = \frac{8 \sqrt{3}}{9} \text{ cm}^3$.

(A) First,convert the speed of water from $km/h$ to $m/h$: $15\, km/h = 15000\, m/h$.
Next,convert the pipe cross-section dimensions from $dm$ to $m$: $2\, dm = 0.2\, m$ and $1.5\, dm = 0.15\, m$.
The volume of water flowing through the pipe in one hour is: $V_{pipe} = 0.2\, m \times 0.15\, m \times 15000\, m/h = 450\, m^3/h$.
The required volume of water in the tank is: $V_{tank} = 150\, m \times 100\, m \times 3\, m = 45000\, m^3$.
The time required to fill the tank to a depth of $3\, m$ is: $\text{Time} = \frac{V_{tank}}{V_{pipe}} = \frac{45000\, m^3}{450\, m^3/h} = 100\, hours$.

(C) The height of the cylindrical part is $h_1 = 3\, m$.
The total height of the tent is $H = 13.5\, m$.
The height of the conical part is $h_2 = H - h_1 = 13.5 - 3 = 10.5\, m$.
The radius of the base is $r = 14\, m$.
Slant height of the cone $(l) = \sqrt{r^2 + h_2^2} = \sqrt{14^2 + 10.5^2} = \sqrt{196 + 110.25} = \sqrt{306.25} = 17.5\, m$.
Curved surface area of the cone $= \pi r l = \frac{22}{7} \times 14 \times 17.5 = 22 \times 2 \times 17.5 = 770\, m^2$.
Curved surface area of the cylinder $= 2 \pi r h_1 = 2 \times \frac{22}{7} \times 14 \times 3 = 2 \times 22 \times 2 \times 3 = 264\, m^2$.
Total surface area to be painted $= 770 + 264 = 1034\, m^2$.
Total cost of painting $= 1034 \times 2 = ₹ 2068$.

(C) Let the original diameter be $D$. The radius is $r = D/2$.
The surface area of a sphere is $A = 4 \pi r^2 = 4 \pi (D/2)^2 = \pi D^2$.
If the diameter is decreased by $25 \%$,the new diameter $D' = D - 0.25D = 0.75D$.
The new surface area $A' = \pi (D')^2 = \pi (0.75D)^2 = 0.5625 \pi D^2$.
The decrease in surface area is $A - A' = \pi D^2 - 0.5625 \pi D^2 = 0.4375 \pi D^2$.
The percentage decrease is $\frac{0.4375 \pi D^2}{\pi D^2} \times 100 \% = 43.75 \%$.

(A) Let the radius be $r = 10\, cm$ and the height be $h = 4\, cm$.
Let the value added to either the radius or the height be $x\, cm$.
Case $1$: When $x$ is added to the radius,the new radius is $(10+x)\, cm$ and the height remains $4\, cm$.
The new volume $V_1 = \pi(10+x)^2 \times 4$.
Case $2$: When $x$ is added to the height,the radius remains $10\, cm$ and the new height is $(4+x)\, cm$.
The new volume $V_2 = \pi(10)^2 \times (4+x)$.
According to the problem,the increase in volume is the same,which implies $V_1 = V_2$ (since the original volume is subtracted from both sides,the net increase is equal if the final volumes are equal).
$\pi(10+x)^2 \times 4 = \pi(10)^2(4+x)$
$(10+x)^2 \times 4 = 100(4+x)$
$4(100 + 20x + x^2) = 400 + 100x$
$400 + 80x + 4x^2 = 400 + 100x$
$4x^2 - 20x = 0$
$4x(x - 5) = 0$
Since $x$ cannot be $0$,we have $x = 5\, cm$.

(C) Let the radius and height of the cone be $r$ and $h$ respectively.
The volume of the cone is given by $V_{cone} = \frac{1}{3} \pi r^2 h = 27\, \pi\, cm^3$.
The volume of the cylinder with the same radius $r$ and height $h$ is $V_{cylinder} = \pi r^2 h$.
The empty space inside the cylinder after placing the cone is the difference between the volume of the cylinder and the volume of the cone.
Volume of empty space $= V_{cylinder} - V_{cone} = \pi r^2 h - \frac{1}{3} \pi r^2 h = \frac{2}{3} \pi r^2 h$.
Since $\frac{1}{3} \pi r^2 h = 27\, \pi$,we have $\pi r^2 h = 3 \times 27\, \pi = 81\, \pi$.
Therefore,the volume of water needed $= 81\, \pi - 27\, \pi = 54\, \pi\, cm^3$.

(B) Area of land $= 1 \, Km^2 = (1000 \, m) \times (1000 \, m) = 1,000,000 \, m^2$.
Height of rainfall $= 2 \, cm = 0.02 \, m$.
Total volume of rain $= \text{Area} \times \text{Height} = 1,000,000 \, m^2 \times 0.02 \, m = 20,000 \, m^3$.
Volume of water collected in the pool $= 50 \% \text{ of } 20,000 \, m^3 = 0.5 \times 20,000 = 10,000 \, m^3$.
Base area of the pool $= 100 \, m \times 10 \, m = 1,000 \, m^2$.
Increase in water level $= \frac{\text{Volume collected}}{\text{Base area}} = \frac{10,000 \, m^3}{1,000 \, m^2} = 10 \, m$.

(C) Let the radius of the base of the cylinder be $r = 3.5 \, cm$.
Since the sphere fits exactly into the can,the radius of the sphere is also $r = 3.5 \, cm$,and the height of the water level after placing the sphere is $H = 2r = 7 \, cm$.
The volume of the water in the can is equal to the total volume of the cylinder up to height $H$ minus the volume of the sphere.
Volume of water $= \pi r^2 (2r) - \frac{4}{3} \pi r^3 = 2 \pi r^3 - \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3$.
Let the initial depth of the water be $h$. The volume of this water is $\pi r^2 h$.
Equating the volumes: $\pi r^2 h = \frac{2}{3} \pi r^3$.
Solving for $h$: $h = \frac{2}{3} r = \frac{2}{3} \times 3.5 = \frac{7}{3} \, cm$.

(A) Let the initial radius be $r$ and height be $h$. The initial lateral surface area is $A_1 = 2 \pi r h$.
The new height $h' = 6h$. The base area is $\pi r^2$. The new base area is $\frac{1}{9} \pi r^2 = \pi (r')^2$,which implies $(r')^2 = \frac{1}{9} r^2$,so $r' = \frac{1}{3} r$.
The new lateral surface area is $A_2 = 2 \pi r' h' = 2 \pi (\frac{1}{3} r) (6h) = 2 \pi r h \times (\frac{1}{3} \times 6) = 2 \pi r h \times 2$.
Thus,the lateral surface area increases by a factor of $2$.

(B) Let the radius of the cone be $r\, cm$.
Given,volume $V = \frac{1}{3} \pi r^2 h = 1232\, cm^3$ and height $h = 24\, cm$.
Substituting the values: $\frac{1}{3} \times \frac{22}{7} \times r^2 \times 24 = 1232$.
$r^2 = \frac{1232 \times 3 \times 7}{22 \times 24} = 49$.
Therefore,$r = \sqrt{49} = 7\, cm$.
Now,find the slant height $l = \sqrt{h^2 + r^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25\, cm$.
The curved surface area of the cone is given by $A = \pi rl$.
$A = \frac{22}{7} \times 7 \times 25 = 550\, cm^2$.

(C) Let the side of the square base be $x\, cm.$
The height of the prism is $h = 15\, cm.$
The total surface area of a prism is given by the formula: $\text{Total Surface Area} = \text{Lateral Surface Area} + 2 \times \text{Base Area}.$
Lateral Surface Area $= \text{Perimeter of base} \times h = (4x) \times 15 = 60x.$
Base Area $= x^2.$
Given,$\text{Total Surface Area} = 608\, cm^2.$
So,$60x + 2x^2 = 608.$
Dividing by $2,$ we get $x^2 + 30x - 304 = 0.$
Factoring the quadratic equation: $x^2 + 38x - 8x - 304 = 0 \Rightarrow x(x + 38) - 8(x + 38) = 0.$
$(x - 8)(x + 38) = 0.$
Since the side length cannot be negative,$x = 8\, cm.$
Volume of the prism $= \text{Base Area} \times h = x^2 \times h = 8^2 \times 15 = 64 \times 15 = 960\, cm^3.$

(A) The volume of a solid hemisphere is given by the formula $V = \frac{2}{3} \pi r^3$.
Given $V = 19404 \, cm^3$,we have $\frac{2}{3} \times \frac{22}{7} \times r^3 = 19404$.
Solving for $r^3$: $r^3 = \frac{19404 \times 3 \times 7}{2 \times 22} = 9261$.
Taking the cube root: $r = \sqrt[3]{9261} = 21 \, cm$.
The total surface area of a solid hemisphere is given by $TSA = 3 \pi r^2$.
$TSA = 3 \times \frac{22}{7} \times 21 \times 21 = 3 \times 22 \times 3 \times 21 = 4158 \, cm^2$.

(A) $1$. Radius of the pipe $(r)$ = $14/2 = 7 \text{ cm} = 0.07 \text{ m}$.
$2$. Speed of water = $5 \text{ km/h} = 5000 \text{ m/h}$.
$3$. Volume of water flowing through the pipe in $1 \text{ hour} = \pi r^2 h = (22/7) \times (0.07)^2 \times 5000 = (22/7) \times 0.0049 \times 5000 = 22 \times 0.0007 \times 5000 = 77 \text{ m}^3$.
$4$. Volume of water required in the tank = $\text{Length} \times \text{Width} \times \text{Height} = 50 \text{ m} \times 44 \text{ m} \times (7/100) \text{ m} = 50 \times 44 \times 0.07 = 154 \text{ m}^3$.
$5$. Time taken = $\text{Total volume required} / \text{Volume per hour} = 154 / 77 = 2 \text{ hours}$.

(D) Let the length,breadth,and height of the cuboid be $x, y,$ and $z$ cm,respectively.
Given the areas of three consecutive faces:
$xy = 12 \, cm^2$
$yz = 20 \, cm^2$
$zx = 15 \, cm^2$
Multiplying these three equations together:
$(xy) \cdot (yz) \cdot (zx) = 12 \cdot 20 \cdot 15$
$x^2 y^2 z^2 = 3600$
Taking the square root of both sides:
$xyz = \sqrt{3600}$
$xyz = 60$
Since the volume $V$ of a cuboid is given by $V = xyz$,the volume is $60 \, cm^3$.

(B) $1$. Calculate the volume of the cistern: The cistern is cylindrical with radius $r = 5 \, m$ and height $h = 2 \, m$. Volume $V = \pi r^2 h = \pi \times (5)^2 \times 2 = 50\pi \, m^3$.
$2$. Calculate the volume of water flowing through the pipe per hour: The pipe has radius $r_p = 10 \, cm = 0.1 \, m$. The speed of water is $3 \, km/h = 3000 \, m/h$. The volume of water flowing per hour is $V_p = \pi r_p^2 \times \text{speed} = \pi \times (0.1)^2 \times 3000 = \pi \times 0.01 \times 3000 = 30\pi \, m^3/h$.
$3$. Calculate the time required: Time = $\frac{\text{Volume of cistern}}{\text{Volume of water per hour}} = \frac{50\pi}{30\pi} = \frac{5}{3} \, hours$.
$4$. Convert to hours and minutes: $\frac{5}{3} \, hours = 1 \, hour + \frac{2}{3} \times 60 \, minutes = 1 \, hour \, 40 \, minutes$.

(B) The volume of the water displaced in the cylindrical beaker is equal to the total volume of the submerged marbles.
Radius of the beaker $R = \frac{7}{2} = 3.5\, cm$.
Height of the water rise $h = 5.6\, cm$.
Volume of raised water $= \pi R^2 h = \frac{22}{7} \times 3.5 \times 3.5 \times 5.6 = 215.6\, cm^3$.
Radius of a marble $r = \frac{1.4}{2} = 0.7\, cm$.
Volume of one marble $= \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 0.7 \times 0.7 \times 0.7 = \frac{4}{3} \times 22 \times 0.1 \times 0.49 = \frac{4.312}{3}\, cm^3$.
Let the number of marbles be $n$.
$n \times (\text{Volume of one marble}) = \text{Volume of raised water}$
$n \times \frac{4.312}{3} = 215.6$
$n = \frac{215.6 \times 3}{4.312} = 50 \times 3 = 150$.
Therefore,$150$ marbles were dropped into the beaker.