Ganit Prakash - Class-X - Solid Objects
Let us work out 19 Mensuration
📘 Exercise 19 Solutions (Q1 - Q16)
Solid ObjectsExplanation:
Base diameter of the pillar \(= 20\text{ cm}\)
Radius of the base \( (r) = \frac{20}{2} = 10\text{ cm} \)
Height of the cylindrical portion \( (h_1) = 2.8\text{ m} = 280\text{ cm} \)
Height of the conical portion \( (h_2) = 42\text{ cm} \)
1. Volume of the cylindrical portion:
\[\begin{array}{l} V_1 = \pi r^2 h_1 \\ = \frac{22}{7} \times (10)^2 \times 280 \\ = \frac{22}{7} \times 100 \times 280 \\ = 22 \times 100 \times 40 \\ = 88000\text{ cm}^3 \end{array}\]2. Volume of the conical portion:
\[\begin{array}{l} V_2 = \frac{1}{3}\pi r^2 h_2 \\ = \frac{1}{3} \times \frac{22}{7} \times (10)^2 \times 42 \\ = \frac{22}{21} \times 100 \times 42 \\ = 22 \times 100 \times 2 \\ = 4400\text{ cm}^3 \end{array}\]3. Total Volume and Weight:
\[\begin{array}{l} \text{Total Volume } (V) = V_1 + V_2 \\ = 88000 + 4400 \\ = 92400\text{ cm}^3 \end{array}\]Weight of \( 1\text{ cm}^3 \) iron = \( 7.5\text{ gm} \)
\[\begin{array}{l} \text{Total Weight} = 92400 \times 7.5\text{ gm} \\ = 693000\text{ gm} \\ = \frac{693000}{1000}\text{ kg} \\ = 693\text{ kg} \end{array}\]Therefore, the weight of the iron pillar is \( 693\text{ kg} \).
Explanation:
For the cone:
Height \( (h) = 20\text{ cm} \)
Slant height \( (l) = 25\text{ cm} \)
Radius of the cone \( (r) = \sqrt{l^2 - h^2} \)
\[\begin{array}{l} r = \sqrt{25^2 - 20^2} \\ r = \sqrt{625 - 400} \\ r = \sqrt{225} = 15\text{ cm} \end{array}\]Volume of the cone \( (V_c) = \frac{1}{3}\pi r^2 h \)
\[\begin{array}{l} V_c = \frac{1}{3} \times \pi \times (15)^2 \times 20 \\ = \frac{1}{3} \times \pi \times 225 \times 20 \\ = 75 \times \pi \times 20 \\ = 1500\pi\text{ cm}^3 \end{array}\]For the cylinder:
Height \( (H) = 15\text{ cm} \)
Let the radius be \( R \).
Volume of the cylinder \( (V_{cyl}) = \pi R^2 H = \pi R^2 \times 15 \)
According to the problem, Volume of cylinder = Volume of cone:
\[\begin{array}{l} \pi R^2 \times 15 = 1500\pi \\ \Rightarrow 15R^2 = 1500 \quad \text{[Cancelling } \pi \text{]} \\ \Rightarrow R^2 = \frac{1500}{15} \\ \Rightarrow R^2 = 100 \\ \Rightarrow R = 10\text{ cm} \end{array}\]Base diameter of the cylinder = \( 2R = 2 \times 10 = 20\text{ cm} \).
Therefore, the base diameter of the cylinder is \( 20\text{ cm} \).
Explanation:
1. Volume of the 60 conical pieces:
Base diameter of one conical piece \(= 6\text{ cm} \Rightarrow\) Radius \( (r) = 3\text{ cm} \)
Height of one conical piece \( (h) = 4\text{ cm} \)
Volume of one conical piece \(= \frac{1}{3}\pi r^2 h\)
\[\begin{array}{l} = \frac{1}{3} \times \pi \times (3)^2 \times 4 \\ = \frac{1}{3} \times \pi \times 9 \times 4 \\ = 12\pi\text{ cm}^3 \end{array}\]Total volume of 60 pieces \(= 60 \times 12\pi = 720\pi\text{ cm}^3\)
2. Increased height in the cylindrical can:
Diameter of the cylindrical can \(= 24\text{ cm} \Rightarrow\) Radius \( (R) = 12\text{ cm} \)
Let the increased height of the water level be \( H \text{ cm} \).
Volume of the displaced water = Volume of the 60 conical pieces
\[\begin{array}{l} \pi R^2 H = 720\pi \\ \Rightarrow \pi \times (12)^2 \times H = 720\pi \\ \Rightarrow 144H = 720 \quad \text{[Cancelling } \pi \text{]} \\ \Rightarrow H = \frac{720}{144} \\ \Rightarrow H = 5\text{ cm} \end{array}\]Therefore, the increased height of the water level is \( 5\text{ cm} \).
Explanation:
Let the common base radius be \( r \) and the common height be \( h \).
Curved Surface Area (CSA) of the cone \(= \pi r l = \pi r \sqrt{r^2 + h^2}\)
Curved Surface Area (CSA) of the cylinder \(= 2\pi r h\)
According to the problem, the ratio of their CSAs is \( 5:8 \):
\[\begin{array}{l} \frac{\pi r \sqrt{r^2 + h^2}}{2\pi r h} = \frac{5}{8} \end{array}\]Cancel out \( \pi r \) from the numerator and denominator:
\[\begin{array}{l} \frac{\sqrt{r^2 + h^2}}{2h} = \frac{5}{8} \\ \Rightarrow \frac{\sqrt{r^2 + h^2}}{h} = \frac{10}{8} \\ \Rightarrow \frac{\sqrt{r^2 + h^2}}{h} = \frac{5}{4} \end{array}\]Squaring both sides:
\[\begin{array}{l} \frac{r^2 + h^2}{h^2} = \left(\frac{5}{4}\right)^2 \\ \Rightarrow \frac{r^2}{h^2} + \frac{h^2}{h^2} = \frac{25}{16} \\ \Rightarrow \frac{r^2}{h^2} + 1 = \frac{25}{16} \\ \Rightarrow \frac{r^2}{h^2} = \frac{25}{16} - 1 \\ \Rightarrow \frac{r^2}{h^2} = \frac{9}{16} \end{array}\]Taking the square root of both sides:
\[\begin{array}{l} \frac{r}{h} = \frac{3}{4} \end{array}\]Therefore, the ratio of their base radii and heights is \( 3:4 \).
Explanation:
1. Volume of the large iron sphere:
Radius \( (R) = 8\text{ cm} \)
\[\begin{array}{l} V_1 = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (8)^3\text{ cm}^3 \end{array}\]2. Volume of one small marble ball:
Diameter \(= 1\text{ cm} \Rightarrow\) Radius \( (r) = \frac{1}{2}\text{ cm} = 0.5\text{ cm} \)
\[\begin{array}{l} V_2 = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \left(\frac{1}{2}\right)^3\text{ cm}^3 \end{array}\]3. Number of marble balls:
Number of balls \( (n) \) = Total volume / Volume of one small ball
\[\begin{array}{l} n = \frac{\frac{4}{3}\pi (8)^3}{\frac{4}{3}\pi \left(\frac{1}{2}\right)^3} \end{array}\]Cancel out \( \frac{4}{3}\pi \):
\[\begin{array}{l} n = \frac{8^3}{\left(\frac{1}{2}\right)^3} \\ n = \frac{512}{\frac{1}{8}} \\ n = 512 \times 8 \\ n = 4096 \end{array}\]Therefore, \( 4096 \) marble balls can be obtained.
Explanation:
1. Volume of the iron rod (Cylinder):
Radius \( (R) = 32 \) cm
Length/Height \( (H) = 35 \) cm
\[\begin{array}{l} \text{Volume of rod } (V_1) = \pi R^2 H \\ = \pi \times (32)^2 \times 35 \text{ cm}^3 \end{array}\]2. Volume of one solid cone:
Radius \( (r) = 8 \) cm
Height \( (h) = 28 \) cm
\[\begin{array}{l} \text{Volume of cone } (V_2) = \frac{1}{3} \pi r^2 h \\ = \frac{1}{3} \times \pi \times (8)^2 \times 28 \text{ cm}^3 \end{array}\]3. Number of solid cones:
\[\begin{array}{l} \text{Number of cones} = \frac{V_1}{V_2} \\ = \frac{\pi \times 32 \times 32 \times 35}{\frac{1}{3} \times \pi \times 8 \times 8 \times 28} \end{array}\]Cancel \( \pi \) and simplify:
\[\begin{array}{l} = \frac{32 \times 32 \times 35 \times 3}{8 \times 8 \times 28} \\ = \frac{(8 \times 4) \times (8 \times 4) \times 35 \times 3}{8 \times 8 \times 28} \\ = \frac{4 \times 4 \times 35 \times 3}{28} \\ = \frac{16 \times 35 \times 3}{28} \\ = \frac{16 \times 5 \times 3}{4} \\ = 4 \times 5 \times 3 \\ = 60 \end{array}\]Therefore, 60 solid cones can be made.
Explanation:
To waste the minimum quantity of wood, the cone must be inscribed perfectly within the cube. This means the base of the cone will touch the sides of the cube's face, and its height will be equal to the cube's edge.
Edge of the wooden cube = 4.2 dcm
Maximum base diameter of the cone = Edge of the cube = 4.2 dcm
Radius of the cone \( (r) = \frac{4.2}{2} = 2.1 \) dcm
Maximum height of the cone \( (h) = \text{Edge of the cube} = 4.2 \) dcm
Now, calculating the volume of the cone:
\[\begin{array}{l} \text{Volume} = \frac{1}{3} \pi r^2 h \\ = \frac{1}{3} \times \frac{22}{7} \times (2.1)^2 \times 4.2 \\ = \frac{1}{3} \times \frac{22}{7} \times 4.41 \times 4.2 \\ = \frac{22}{21} \times 18.522 \\ = 22 \times 0.882 \\ = 19.404 \text{ dcm}^3 \end{array}\]Therefore, the volume of the solid cone is 19.404 cubic dcm.
Explanation:
Let the common radius of the sphere and the cylinder be \( r \).
Let the height of the cylinder be \( h \).
Volume of the solid sphere \( = \frac{4}{3} \pi r^3 \)
Volume of the solid right circular cylinder \( = \pi r^2 h \)
According to the condition, their volumes are equal:
\[\begin{array}{l} \frac{4}{3} \pi r^3 = \pi r^2 h \end{array}\]Cancel \( \pi \) and \( r^2 \) from both sides:
\[\begin{array}{l} \frac{4}{3} r = h \\ \Rightarrow \frac{r}{h} = \frac{3}{4} \end{array}\]Therefore, the ratio of the radius and the height of the cylinder is 3 : 4.
Explanation:
1. Quantity of metal in each sphere (Volume):
Diameter of the sphere = 2.1 dcm
Radius \( (r) = \frac{2.1}{2} = 1.05 \) dcm
So, each sphere contains 4.851 dcm³ of metal.
2. Number of solid spheres:
Volume of the rectangular parallelepiped (cuboid) \( = \text{length} \times \text{breadth} \times \text{thickness} \)
\[\begin{array}{l} V_{\text{cuboid}} = 6.6 \times 4.2 \times 1.4 \\ = 38.808 \text{ dcm}^3 \end{array}\]Now, divide the total volume by the volume of one sphere:
\[\begin{array}{l} \text{Number of spheres} = \frac{V_{\text{cuboid}}}{\text{Volume of one sphere}} \\ = \frac{38.808}{4.851} \\ = 8 \end{array}\]Therefore, 8 solid spheres can be made.
Explanation:
Radius of the solid gold sphere \( (R) = 4.2 \) cm
\[\begin{array}{l} \text{Volume of the sphere} = \frac{4}{3} \pi R^3 \\ = \frac{4}{3} \pi (4.2)^3 \text{ cm}^3 \end{array}\]Radius of the right circular rod (cylinder) \( (r) = 2.8 \) cm
Let the length (height) of the rod be \( h \).
Since the sphere is recasted into the rod, their volumes are equal:
\[\begin{array}{l} \pi (2.8)^2 h = \frac{4}{3} \pi (4.2)^3 \end{array}\]Cancel \( \pi \) and solve for \( h \):
\[\begin{array}{l} (2.8 \times 2.8) \times h = \frac{4}{3} \times (4.2 \times 4.2 \times 4.2) \\ h = \frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 2.8 \times 2.8} \\ h = \frac{4}{3} \times \left(\frac{4.2}{2.8}\right) \times \left(\frac{4.2}{2.8}\right) \times 4.2 \\ h = \frac{4}{3} \times \frac{3}{2} \times \frac{3}{2} \times 4.2 \quad \left[\text{since } \frac{4.2}{2.8} = \frac{3}{2}\right] \\ h = 1 \times \frac{3}{2} \times 4.2 \\ h = 1.5 \times 4.2 \\ h = 6.3 \text{ cm} \end{array}\]Therefore, the length of the right circular rod is 6.3 cm.
Explanation:
Diameter of the silver sphere = 6 dcm
Radius of the sphere \( (R) = \frac{6}{2} = 3 \) dcm
Length (height) of the right circular rod \( (h) = 1 \) dcm (from the original Bengali text).
Let the base radius of the rod be \( r \).
Since the sphere is recasted into the rod, their volumes must be equal:
\[\begin{array}{l} \pi r^2 = 36\pi \\ \Rightarrow r^2 = 36 \quad \text{[Cancelling } \pi \text{]} \\ \Rightarrow r = 6 \text{ dcm} \end{array}\]The diameter of the rod is \( 2r = 2 \times 6 = 12 \) dcm.
Therefore, the length of the diameter of the rod is 12 dcm.
Explanation:
First, ensure all units are consistent. Let's convert decimeters (dcm) to centimeters (cm).
Radius of the right circular rod \( (R) = 3.2 \) dcm = \( 3.2 \times 10 = 32 \) cm.
Let the length (height) of the rod be \( h \) cm.
Radius of one solid sphere \( (r) = 8 \) cm.
\[\begin{array}{l} \text{Volume of one sphere} = \frac{4}{3} \pi r^3 \\ = \frac{4}{3} \pi (8)^3 \text{ cm}^3 \end{array}\]The total volume of the rod is equal to the volume of the 21 spheres made from it:
\[\begin{array}{l} \pi (32)^2 h = 21 \times \left( \frac{4}{3} \pi (8)^3 \right) \end{array}\]Cancel \( \pi \) and simplify:
\[\begin{array}{l} 1024 \times h = 21 \times \frac{4}{3} \times 512 \\ 1024 \times h = 7 \times 4 \times 512 \\ 1024 \times h = 28 \times 512 \\ h = \frac{28 \times 512}{1024} \\ h = \frac{28}{2} \\ h = 14 \text{ cm} \end{array}\]Therefore, the length of the right circular rod is 14 cm (or 1.4 dcm).
Explanation:
1. Volume of the 100 iron spheres:
Diameter of one sphere = 21 cm = 2.1 dcm
Radius of one sphere \( (r) = \frac{2.1}{2} = 1.05 \) dcm
Total volume of 100 spheres:
\[\begin{array}{l} = 100 \times \left( \frac{4}{3} \times \frac{22}{7} \times \frac{9261}{8000} \right) \\ = 100 \times \frac{88}{21} \times \frac{9261}{8000} \\ = \frac{88}{21} \times \frac{9261}{80} \\ = \frac{11}{21} \times \frac{9261}{10} \\ = \frac{11 \times 441}{10} \\ = 485.1 \text{ cubic dcm} \end{array}\]2. Rise of water level in the tank:
Base area of the tank = Length × Breadth = \( 21 \times 11 = 231 \) sq. dcm.
Let the rise in water level be \( h \) dcm.
Volume of the displaced water = Base Area × Rise in height
\[\begin{array}{l} 231 \times h = 485.1 \\ \Rightarrow h = \frac{485.1}{231} \\ \Rightarrow h = 2.1 \text{ dcm} \end{array}\]Therefore, the rise of the water level is 2.1 dcm.
Explanation:
Since they have the same base diameter, they all have the same base radius. Let the common radius be \( r \).
The height of a hemisphere is equal to its radius (\( r \)).
Since all three solids have the same height, the common height \( h \) must be equal to \( r \). So, \( h = r \).
1. Volume of the solid cone (\( V_1 \)):
\[\begin{array}{l} V_1 = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (r) = \frac{1}{3}\pi r^3 \end{array}\]2. Volume of the solid hemisphere (\( V_2 \)):
\[\begin{array}{l} V_2 = \frac{2}{3}\pi r^3 \end{array}\]3. Volume of the solid cylinder (\( V_3 \)):
\[\begin{array}{l} V_3 = \pi r^2 h = \pi r^2 (r) = \pi r^3 \end{array}\]Ratio of their volumes:
\[\begin{array}{l} V_1 : V_2 : V_3 = \frac{1}{3}\pi r^3 : \frac{2}{3}\pi r^3 : \pi r^3 \end{array}\]Multiplying the entire ratio by 3 and cancelling \( \pi r^3 \):
\[\begin{array}{l} = 1 : 2 : 3 \end{array}\]Therefore, the ratio of their volumes is 1 : 2 : 3.
Explanation:
1. Volume of the lead in the hollow sphere:
External radius \( (R) = 6 \) cm
Thickness = 1 cm
Internal radius \( (r) = 6 - 1 = 5 \) cm
2. Length of the right circular rod:
Radius of the rod \( (r_{rod}) = 2 \) cm
Let the length (height) of the rod be \( h \) cm.
Since the sphere is melted to form the rod, their volumes are equal:
\[\begin{array}{l} 4\pi h = \frac{364\pi}{3} \\ \Rightarrow h = \frac{364}{4 \times 3} \quad \text{[Cancelling } \pi \text{]} \\ \Rightarrow h = \frac{91}{3} \\ \Rightarrow h = 30\frac{1}{3} \text{ cm} \end{array}\]Therefore, the length of the rod is \( 30\frac{1}{3} \) cm.
[Hints: The circumcircle inscribed in a rectangular figure, the length of the diameter of the circle is equal to the length of the side of the square.]
Explanation:
First, convert all units to meters since the answer is required in m³.
Length of the log \( (h) = 2 \) m
Side of the square cross-section \( (a) = 14 \) dcm = 1.4 m
1. Initial volume of the rectangular log (Cuboid):
\[\begin{array}{l} V_{original} = \text{Base Area} \times \text{Length} \\ = a^2 \times h \\ = (1.4)^2 \times 2 \\ = 1.96 \times 2 = 3.92 \text{ m}^3 \end{array}\]2. Volume of the right circular log (Cylinder) with minimum waste:
To waste the minimum amount of wood, the largest possible cylinder must be cut out. The diameter of this cylinder will be equal to the side of the square cross-section (as per the hint).
Diameter of the cylinder = 1.4 m
Radius of the cylinder \( (r) = \frac{1.4}{2} = 0.7 \) m
This is the amount of wood that remains.
3. Amount of wood wasted:
\[\begin{array}{l} \text{Wasted Wood} = V_{original} - V_{cylinder} \\ = 3.92 - 3.08 \\ = 0.84 \text{ m}^3 \end{array}\]Therefore, the amount of wood remained is 3.08 m³ and the amount of wood wasted is 0.84 m³.
📘 17. V.S.A.
(A) M.C.Q.Explanation:
Volume of the solid sphere with radius \( r \) = \( \frac{4}{3}\pi r^3 \)
Let the base radius of the new cone be \( R \). Assuming its height \( h = r \):
Volume of the cone = \( \frac{1}{3}\pi R^2 h = \frac{1}{3}\pi R^2 r \)
Since the sphere is melted to form the cone, their volumes are equal:
\[\begin{array}{l} \frac{1}{3}\pi R^2 r = \frac{4}{3}\pi r^3 \\ \Rightarrow R^2 = 4r^2 \quad \text{[Cancelling } \frac{1}{3}\pi r \text{ from both sides]} \\ \Rightarrow R = \sqrt{4r^2} \\ \Rightarrow R = 2r \end{array}\]Correct Option: (a) \( 2r \) unit
Explanation:
Let the common radius of the cone and the cylinder be \( r \).
Let the height of the cone be \( h \).
Volume of the solid cone = \( \frac{1}{3}\pi r^2 h \)
Volume of the solid cylinder = \( \pi r^2 H \), where \( H = 5 \text{ cm} \).
Since the cone is melted to form the cylinder, their volumes are equal:
\[\begin{array}{l} \frac{1}{3}\pi r^2 h = \pi r^2 \times 5 \\ \Rightarrow \frac{1}{3}h = 5 \quad \text{[Cancelling } \pi r^2 \text{ from both sides]} \\ \Rightarrow h = 15 \text{ cm} \end{array}\]Correct Option: (b) 15 cm.
Explanation:
The largest sphere that can be placed inside a cylinder will have its diameter constrained by the dimensions of the cylinder.
The diameter of the cylinder is \( 2r \). The height of the cylinder is \( 2r \).
Since the height and diameter are equal, a sphere of diameter \( 2r \) will fit perfectly inside, touching the sides, the top, and the bottom.
Therefore, the maximum diameter of the sphere is \( 2r \) units.
Correct Option: (b) \( 2r \) unit
Explanation:
The largest solid cone that can be carved out of a solid hemisphere of radius \( r \) will have its base as the base of the hemisphere and its apex at the topmost point of the hemisphere's curved surface.
Radius of the cone's base = Radius of the hemisphere = \( r \)
Height of the cone = Radius of the hemisphere = \( r \)
Volume of the cone = \( \frac{1}{3}\pi \times (\text{radius})^2 \times (\text{height}) \)
\[\begin{array}{l} V = \frac{1}{3}\pi (r)^2 (r) \\ V = \frac{\pi r^3}{3} \text{ cubic units} \end{array}\]Correct Option: (d) \( \frac{\pi r^3}{3} \text{ unit}^3 \)
Explanation:
When the largest possible sphere is cut out from a solid cube, the sphere touches all six inner faces of the cube.
The diameter of this inscribed sphere will be exactly equal to the distance between two opposite faces of the cube, which is the edge length of the cube.
Therefore, the diameter of the sphere = Edge of the cube = \( x \) units.
Correct Option: (a) \( x \) unit
📘 17. V.S.A.
(B) True or FalseIf two solid hemispheres of same type whose base radii are \( r \) unit each and if they are connected along base, then the total surface area of the connected solid is \( 6\pi r^2 \) sq. unit.
Explanation:
When two solid hemispheres of the same radius \( r \) are connected along their flat circular bases, they form a complete solid sphere.
The total surface area of a complete sphere of radius \( r \) is \( 4\pi r^2 \).
The flat circular bases (each of area \( \pi r^2 \)) are hidden inside the joined solid and do not contribute to the outer surface area.
Therefore, the total surface area is \( 4\pi r^2 \), not \( 6\pi r^2 \).
Answer: False
The base radius of a solid right circular cone is \( r \) unit, height is \( h \) unit and slant height is \( l \) uint. The base of the cone is joined along a base of a right circular cylinder. If the base radii and heights of the cylinder and cone are same, then the total surface area of the connected solid is \( (\pi r l + 2\pi r h + 2\pi r^2) \) sq. unit.
Explanation:
The connected solid is composed of a cone sitting on top of a cylinder. Its total exposed surface area consists of three parts:
- 1. Curved surface area of the cone = \( \pi r l \)
- 2. Curved surface area of the cylinder = \( 2\pi r h \)
- 3. The bottom flat circular base of the cylinder = \( \pi r^2 \)
The top base of the cylinder and the base of the cone are joined together internally, so they are not part of the external surface area.
Thus, the correct total surface area = \( \pi r l + 2\pi r h + \pi r^2 \).
The given expression incorrectly includes \( 2\pi r^2 \) (which would mean two flat bases are exposed).
Answer: False
📘 17. V.S.A.
(C) Fill in the blanksExplanation:
When two hemispheres are attached to both flat ends (plane surfaces) of a cylinder, the resulting shape is like a capsule.
The entire outer surface consists solely of the curved parts of the three individual solids. The flat circular bases are all hidden internally at the joints.
Therefore, Total Surface Area = (Curved surface area of the top hemisphere) + (Curved surface area of the middle cylinder) + (Curved surface area of the bottom hemisphere).
Answer: cylinder
Explanation:
A standard sharpened pencil (which is "cut at one face" by a sharpener) consists of two main geometric shapes. The main unsharpened body is a cylinder, and the sharpened tip forms a cone.
Answer: cylinder
Explanation:
According to the principle of conservation of mass/volume in solid geometry, when one solid object is melted down and completely recast into a new shape (without any wastage), the total amount of material remains unchanged. Therefore, their volumes are identical.
Answer: are equal
📘 18. Short answer type question (S.A.)
Solid ObjectsExplanation:
Let the common radius of the cone and the cylinder be \( r \).
Height of the cone \( (h_1) = 15 \) cm.
Let the height of the solid cylinder be \( h_2 \) cm.
Since the cylinder is made by melting the cone, their volumes must be equal:
\[\begin{array}{l} \text{Volume of Cylinder} = \text{Volume of Cone} \\ \pi r^2 h_2 = \frac{1}{3}\pi r^2 h_1 \end{array}\]Canceling \( \pi r^2 \) from both sides:
\[\begin{array}{l} h_2 = \frac{1}{3} h_1 \\ \Rightarrow h_2 = \frac{1}{3} \times 15 \\ \Rightarrow h_2 = 5 \text{ cm} \end{array}\]Answer: The height of the solid cylinder is 5 cm.
Explanation:
Let the common radius of the cone and the sphere be \( r \).
Let the height of the cone be \( h \).
The problem states that their volumes are equal:
\[\begin{array}{l} \text{Volume of Cone} = \text{Volume of Sphere} \\ \frac{1}{3}\pi r^2 h = \frac{4}{3}\pi r^3 \end{array}\]Canceling \( \frac{1}{3}\pi r^2 \) from both sides:
\[\begin{array}{l} h = 4r \end{array}\]We need to find the ratio of the diameter of the sphere to the height of the cone.
Diameter of the sphere = \( 2r \)
\[\begin{array}{l} \text{Ratio} = \frac{\text{Diameter of sphere}}{\text{Height of cone}} = \frac{2r}{h} \end{array}\]Substituting \( h = 4r \):
\[\begin{array}{l} \text{Ratio} = \frac{2r}{4r} = \frac{1}{2} \end{array}\]Answer: The ratio of the diameter of the sphere and the height of the cone is 1 : 2.
Explanation:
Since they all have equal diameters, they must all have equal radii. Let the common radius be \( r \).
For a sphere, the height is its diameter. So, the common height \( h = 2r \).
1. Volume of the Cylinder:
\[\begin{array}{l} V_{\text{cylinder}} = \pi r^2 h = \pi r^2 (2r) = 2\pi r^3 \end{array}\]2. Volume of the Cone:
\[\begin{array}{l} V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (2r) = \frac{2}{3}\pi r^3 \end{array}\]3. Volume of the Sphere:
\[\begin{array}{l} V_{\text{sphere}} = \frac{4}{3}\pi r^3 \end{array}\]Ratio of their volumes:
\[\begin{array}{l} V_{\text{cylinder}} : V_{\text{cone}} : V_{\text{sphere}} = 2\pi r^3 : \frac{2}{3}\pi r^3 : \frac{4}{3}\pi r^3 \end{array}\]Divide by \( \pi r^3 \):
\[\begin{array}{l} = 2 : \frac{2}{3} : \frac{4}{3} \end{array}\]Multiply by 3 to clear the fractions:
\[\begin{array}{l} = 6 : 2 : 4 \end{array}\]Divide by 2 to get the simplest ratio:
\[\begin{array}{l} = 3 : 1 : 2 \end{array}\]Answer: The ratio of their volumes is 3 : 1 : 2.
Explanation:
The "surface area of the parts" refers to their curved surface areas (since the flat base connects the two parts internally).
Let the common radius be \( r \), slant height of the cone be \( l \), and height of the cone be \( h \).
Curved Surface Area of the Hemisphere = \( 2\pi r^2 \)
Curved Surface Area of the Cone = \( \pi r l \)
According to the problem:
\[\begin{array}{l} \pi r l = 2\pi r^2 \end{array}\]Canceling \( \pi r \) from both sides:
\[\begin{array}{l} l = 2r \end{array}\]We know the relationship for a cone: \( l^2 = r^2 + h^2 \).
Substitute \( l = 2r \):
\[\begin{array}{l} (2r)^2 = r^2 + h^2 \\ \Rightarrow 4r^2 = r^2 + h^2 \\ \Rightarrow 4r^2 - r^2 = h^2 \\ \Rightarrow 3r^2 = h^2 \\ \Rightarrow \frac{r^2}{h^2} = \frac{1}{3} \\ \Rightarrow \frac{r}{h} = \frac{1}{\sqrt{3}} \end{array}\]Answer: The ratio of the radius and height of the cone is \( 1 : \sqrt{3} \).
Explanation:
Let the common radius of the cone and the sphere be \( r \).
Let the height of the cone be \( h \).
Volume of the Cone = \( \frac{1}{3}\pi r^2 h \)
Volume of the Sphere = \( \frac{4}{3}\pi r^3 \)
According to the problem, Volume of Sphere = 2 × Volume of Cone:
\[\begin{array}{l} \frac{4}{3}\pi r^3 = 2 \times \left( \frac{1}{3}\pi r^2 h \right) \end{array}\]Canceling \( \frac{1}{3}\pi r^2 \) from both sides:
\[\begin{array}{l} 4r = 2h \\ \Rightarrow 2r = h \\ \Rightarrow \frac{h}{r} = \frac{2}{1} \end{array}\]Answer: The ratio of the height and base radius of the cone is 2 : 1.
Hi Please, do not Spam in Comments.