Measurement of Volume and Surface Area Questions in English

(B) Let the radius of the base of the hemisphere be $r$ units. Since the bases are equal,the radius of the cone is also $r$ units.
Given that the heights are equal,the height of the cone $h = r$ (as the height of a hemisphere is equal to its radius).
The slant height $l$ of the cone is given by $l = \sqrt{r^2 + h^2} = \sqrt{r^2 + r^2} = \sqrt{2r^2} = r\sqrt{2}$.
The curved surface area of the hemisphere is $2\pi r^2$.
The curved surface area of the cone is $\pi rl = \pi r(r\sqrt{2}) = \pi r^2\sqrt{2}$.
The ratio of the curved surface area of the hemisphere to the curved surface area of the cone is $2\pi r^2 : \pi r^2\sqrt{2} = 2 : \sqrt{2}$.

(C) Let the common height be $h$. For a hemisphere,the height is equal to its radius,so $h = r_3$. Given that the heights of the cone,cylinder,and hemisphere are equal,we have $h = h_{cone} = h_{cylinder} = r_{hemisphere}$.
Let the radii be $r_1 = 2x$,$r_2 = 3x$,and $r_3 = 1x = x$. Since the height of the hemisphere is its radius,$h = x$.
Volume of cone $V_1 = \frac{1}{3} \pi r_1^2 h = \frac{1}{3} \pi (2x)^2 (x) = \frac{4}{3} \pi x^3$.
Volume of cylinder $V_2 = \pi r_2^2 h = \pi (3x)^2 (x) = 9 \pi x^3$.
Volume of hemisphere $V_3 = \frac{2}{3} \pi r_3^3 = \frac{2}{3} \pi (x)^3 = \frac{2}{3} \pi x^3$.
The ratio $V_1 : V_2 : V_3 = \frac{4}{3} \pi x^3 : 9 \pi x^3 : \frac{2}{3} \pi x^3$.
Multiplying by $\frac{3}{\pi x^3}$,we get $4 : 27 : 2$.

(C) The area of a square base can be calculated using its diagonal $d$ as: $\text{Area} = \frac{1}{2} \times d^2$.
Given $d = 24 \sqrt{2} \text{ m}$,the area is $\frac{1}{2} \times (24 \sqrt{2})^2 = \frac{1}{2} \times 576 \times 2 = 576 \text{ m}^2$.
The volume $V$ of a pyramid is given by the formula: $V = \frac{1}{3} \times \text{Base Area} \times \text{height} (h)$.
Substituting the given values: $1728 = \frac{1}{3} \times 576 \times h$.
$1728 = 192 \times h$.
$h = \frac{1728}{192} = 9 \text{ m}$.

(B) Let the height of the original cone be $H = 9 \, cm$ and the radius of its base be $R = 3 \, cm$.
Let the radius of the upper circular surface of the frustum be $r$ and the height of the small cone cut from the top be $h'$.
By similar triangles,the ratio of the radius to the height is constant: $\frac{r}{h'} = \frac{R}{H} = \frac{3}{9} = \frac{1}{3}$,so $h' = 3r$.
The volume of the original cone is $V_{total} = \frac{1}{3} \pi R^2 H = \frac{1}{3} \times \frac{22}{7} \times 3^2 \times 9 = \frac{22 \times 27}{7} = \frac{594}{7} \, cm^3$.
The volume of the small cone at the top is $V_{small} = \frac{1}{3} \pi r^2 h' = \frac{1}{3} \pi r^2 (3r) = \pi r^3$.
The volume of the frustum is $V_{frustum} = V_{total} - V_{small} = 44 \, cm^3$.
Substituting the values: $\frac{594}{7} - \frac{22}{7} r^3 = 44$.
Multiply by $7$: $594 - 22r^3 = 308$.
$22r^3 = 594 - 308 = 286$.
$r^3 = \frac{286}{22} = 13$.
Therefore,$r = \sqrt[3]{13} \, cm$.

(A) Let the radii of the two cylinders be $r_1$ and $r_2$,and their heights be $h_1$ and $h_2$ respectively.
Given,the ratio of radii $r_1 : r_2 = 2 : 3$ and the ratio of heights $h_1 : h_2 = 5 : 4$.
The curved surface area $(CSA)$ of a cylinder is given by the formula $CSA = 2 \pi r h$.
The ratio of the curved surface areas of the two cylinders is $\frac{CSA_1}{CSA_2} = \frac{2 \pi r_1 h_1}{2 \pi r_2 h_2}$.
Simplifying this,we get $\frac{CSA_1}{CSA_2} = \frac{r_1}{r_2} \times \frac{h_1}{h_2}$.
Substituting the given ratios: $\frac{CSA_1}{CSA_2} = \frac{2}{3} \times \frac{5}{4} = \frac{10}{12} = \frac{5}{6}$.
Thus,the ratio of their curved surface areas is $5:6$.

(C) Let the height of the cylinder be $h \, cm$ and the radius of the base be $r \, cm$.
According to the question,the total surface area is $2 \pi r^2 + 2 \pi r h = 462 \, cm^2 \dots (1)$.
The curved surface area is given as $\frac{1}{3}$ of the total surface area:
Curved surface area $= 2 \pi r h = \frac{1}{3} \times 462 = 154 \, cm^2$.
Substituting this into equation $(1)$:
$2 \pi r^2 + 154 = 462$
$2 \pi r^2 = 462 - 154 = 308$
$2 \times \frac{22}{7} \times r^2 = 308$
$r^2 = \frac{308 \times 7}{44} = 7 \times 7 = 49$
$r = 7 \, cm$.
Now,using $2 \pi r h = 154$:
$2 \times \frac{22}{7} \times 7 \times h = 154$
$44 \times h = 154$
$h = \frac{154}{44} = 3.5 \, cm$ (or $\frac{7}{2} \, cm$).
The volume of the cylinder is $V = \pi r^2 h = \frac{22}{7} \times 7 \times 7 \times 3.5 = 22 \times 7 \times 3.5 = 539 \, cm^3$.

(D) Let the radius and the height be $r$ and $h$ respectively.
The curved surface area of a cylinder is $2 \pi rh$.
The curved surface area of a cone is $\pi rl$,where $l = \sqrt{h^2 + r^2}$.
Given the ratio of their curved surface areas is $8:5$:
$\frac{2 \pi rh}{\pi r \sqrt{h^2 + r^2}} = \frac{8}{5}$
$\frac{2h}{\sqrt{h^2 + r^2}} = \frac{8}{5}$
$\frac{h}{\sqrt{h^2 + r^2}} = \frac{4}{5}$
Squaring both sides:
$\frac{h^2}{h^2 + r^2} = \frac{16}{25}$
$25h^2 = 16h^2 + 16r^2$
$9h^2 = 16r^2$
$\frac{r^2}{h^2} = \frac{9}{16}$
$\frac{r}{h} = \frac{3}{4}$
Thus,the ratio of their radius and height is $3:4$.

(D) Let the radius of the base be $r$ units and the height of the cone be $h$ units.
The surface area of the hemispherical part is $2 \pi r^2$.
The curved surface area of the conical part is $\pi r l$,where $l = \sqrt{r^2 + h^2}$ is the slant height.
According to the problem,the surface areas of the two parts are equal:
$2 \pi r^2 = \pi r \sqrt{r^2 + h^2}$
Dividing both sides by $\pi r$ (assuming $r \neq 0$):
$2r = \sqrt{r^2 + h^2}$
Squaring both sides:
$4r^2 = r^2 + h^2$
$3r^2 = h^2$
Taking the square root of both sides:
$\sqrt{3}r = h$
Therefore,the ratio of the radius to the height is:
$\frac{r}{h} = \frac{1}{\sqrt{3}}$,which is $1: \sqrt{3}$.

(D) The area of the base of an equilateral triangle is given by the formula: $\text{Area} = \frac{\sqrt{3}}{4} \times \text{side}^2$.
Substituting the given side length of $6 \, cm$:
$\text{Area} = \frac{\sqrt{3}}{4} \times 6^2 = \frac{\sqrt{3}}{4} \times 36 = 9 \sqrt{3} \, cm^2$.
The volume of a right prism is calculated as: $\text{Volume} = \text{Area of base} \times \text{height}$.
Given the volume is $108 \sqrt{3} \, cm^3$,we have:
$108 \sqrt{3} = 9 \sqrt{3} \times h$.
Dividing both sides by $9 \sqrt{3}$:
$h = \frac{108 \sqrt{3}}{9 \sqrt{3}} = 12 \, cm$.

(A) Let the number of required coins be $x$.
The volume of a single coin (which is a cylinder) is given by $V = \pi r^2 h$.
For each coin,$r = 0.75\, cm$ and $h = 0.2\, cm$.
Total volume of $x$ coins $= x \times \pi \times (0.75)^2 \times 0.2$.
The volume of the target right circular cylinder is $V = \pi R^2 H$.
Here,$R = 3\, cm$ and $H = 8\, cm$.
Volume of cylinder $= \pi \times (3)^2 \times 8 = \pi \times 9 \times 8 = 72\pi\, cm^3$.
Equating the volumes:
$x \times \pi \times (0.75)^2 \times 0.2 = 72\pi$
$x \times (0.5625) \times 0.2 = 72$
$x \times 0.1125 = 72$
$x = \frac{72}{0.1125}$
$x = \frac{720000}{1125} = 640$.
Thus,the number of coins required is $640$.

(D) Let the radius of the original sphere be $r \, m$.
The new radius is $(r + 2) \, m$.
The surface area of a sphere is given by $A = 4 \pi r^2$.
According to the problem,the increase in surface area is $704 \, m^2$:
$4 \pi (r + 2)^2 - 4 \pi r^2 = 704$
Divide by $4 \pi$:
$(r + 2)^2 - r^2 = \frac{704}{4 \pi}$
$(r^2 + 4r + 4) - r^2 = \frac{176}{\pi}$
$4r + 4 = \frac{176}{22/7}$
$4(r + 1) = \frac{176 \times 7}{22}$
$4(r + 1) = 8 \times 7$
$4(r + 1) = 56$
$r + 1 = 14$
$r = 13 \, m$.

(C) right circular cylinder circumscribes a hemisphere such that their bases are common.
Let the radius of the hemisphere be $r$.
Since the cylinder circumscribes the hemisphere,the radius of the cylinder is also $r$,and the height of the cylinder is equal to the radius of the hemisphere,which is $r$.
Volume of the hemisphere $= \frac{2}{3} \pi r^3$.
Volume of the cylinder $= \pi r^2 h = \pi r^2 (r) = \pi r^3$.
The ratio of the volume of the hemisphere to the volume of the cylinder is $\frac{\frac{2}{3} \pi r^3}{\pi r^3} = \frac{2}{3}$.
Thus,the ratio is $2:3$.

(C) The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
Total volume of the three spherical balls $= \frac{4}{3} \pi (1^3 + 2^3 + 3^3) = \frac{4}{3} \pi (1 + 8 + 27) = \frac{4}{3} \pi (36) \, cm^3$.
Given that $25\%$ of the material is lost during the melting process.
Remaining volume $= \frac{4}{3} \pi (36) \times (1 - 0.25) = \frac{4}{3} \pi (36) \times 0.75 = \frac{4}{3} \pi (36) \times \frac{3}{4} = \frac{4}{3} \pi (27) \, cm^3$.
Let the radius of the new spherical ball be $R$.
Then,$\frac{4}{3} \pi R^3 = \frac{4}{3} \pi (27)$.
$R^3 = 27$.
$R = 3 \, cm$.

(B) The length of the diagonal of a cube is given by the formula $d = a\sqrt{3}$, where $a$ is the side length of the cube.
Given that the diagonal $d = 6 \text{ cm}$.
Therefore, $a\sqrt{3} = 6 \text{ cm}$.
Solving for $a$, we get $a = \frac{6}{\sqrt{3}} = \frac{6 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3} \text{ cm}$.
The volume $V$ of a cube is given by $V = a^3$.
Substituting the value of $a$, we get $V = (2\sqrt{3})^3 = 2^3 \times (\sqrt{3})^3 = 8 \times 3\sqrt{3} = 24\sqrt{3} \text{ cm}^3$.

(C) When a sphere of radius $r$ is divided into $4$ identical parts (by cutting it along two perpendicular planes passing through the center),each part is a spherical wedge (specifically,a spherical quadrant).
$1$. The curved surface area of the original sphere is $4 \pi r^{2}$. Since the sphere is divided into $4$ equal parts,the curved surface area of each part is $\frac{1}{4} \times 4 \pi r^{2} = \pi r^{2}$.
$2$. The total curved surface area of all $4$ parts is $4 \times \pi r^{2} = 4 \pi r^{2}$.
$3$. Each of the $4$ parts has two flat semi-circular faces created by the cuts. Since there are $4$ parts,there are $4 \times 2 = 8$ semi-circular faces in total.
$4$. The area of one semi-circular face is $\frac{1}{2} \pi r^{2}$.
$5$. The total area of these $8$ flat faces is $8 \times \frac{1}{2} \pi r^{2} = 4 \pi r^{2}$.
$6$. The total surface area of the $4$ parts is the sum of the total curved surface area and the total flat surface area:
Total Surface Area $= 4 \pi r^{2} + 4 \pi r^{2} = 8 \pi r^{2}$ square units.