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A cylinder is circumscribed about a hemisphere and a cone is inscribed in the cylinder so as to have its vertex at the centre of one end and the other end as its base. The volumes of the cylinder, hemisphere and the cone are respectively in the ratio of:
- a)3 : √3: 2
- b)3 : 2 : 1
- c)1 : 2 : 3
- d)2 : 3 : 1
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A cylinder is circumscribed about a hemisphere and a cone is inscribed...
Let the radius of the hemisphere be $r$ and the height of the cylinder be $h$. Then the radius of the cylinder must also be $r$, since it is circumscribed about the hemisphere.
The volume of the hemisphere is $\frac{2}{3}\pi r^3$, the volume of the cylinder is $\pi r^2h$, and the volume of the cone can be found using similar triangles. The height of the cone is $r$, and the slant height can be found using the Pythagorean theorem: $\sqrt{r^2+h^2}$. Therefore, the radius of the base of the cone is $\frac{r}{h}\sqrt{r^2+h^2}$, and the volume of the cone is $\frac{1}{3}\pi r^2h\left(\frac{r}{h}\sqrt{r^2+h^2}\right)^2=\frac{1}{3}\pi r^3\sqrt{r^2+h^2}$.
The ratio of the volumes is then:
$$\frac{\frac{2}{3}\pi r^3+\pi r^2h+\frac{1}{3}\pi r^3\sqrt{r^2+h^2}}{\frac{2}{3}\pi r^3:\pi r^2h:\frac{1}{3}\pi r^3\sqrt{r^2+h^2}}$$
Simplifying, we get:
$$\frac{2r^3+3r^2h+r^3\sqrt{r^2+h^2}}{2r^3:3r^2h:\frac{1}{3}r^3\sqrt{r^2+h^2}}=\frac{6+9\frac{h}{r}+\sqrt{r^2+h^2}}{6:9:\frac{1}{3}\sqrt{r^2+h^2}}$$
Since the ratio is given to us, we can set up an equation:
$$\frac{6+9\frac{h}{r}+\sqrt{r^2+h^2}}{6:9:\frac{1}{3}\sqrt{r^2+h^2}}=\frac{3}{1}:\frac{4}{3}:\frac{1}{3}$$
Simplifying, we get:
$$\frac{6+9\frac{h}{r}+\sqrt{r^2+h^2}}{2\sqrt{r^2+h^2}}=\frac{9}{2}$$
Solving for $\frac{h}{r}$, we get:
$$\frac{h}{r}=\frac{4}{3}$$
Therefore, the volumes are in the ratio of:
$$\frac{\frac{2}{3}\pi r^3+\pi r^2h+\frac{1}{3}\pi r^3\sqrt{r^2+h^2}}{\frac{2}{3}\pi r^3:\pi r^2h:\frac{1}{3}\pi r^3\sqrt{r^2+h^2}}=\frac{6+9\frac{h}{r}+\sqrt{r^2+h^2}}{6:9:\frac{1}{3}\sqrt{
The volume of the hemisphere is $\frac{2}{3}\pi r^3$, the volume of the cylinder is $\pi r^2h$, and the volume of the cone can be found using similar triangles. The height of the cone is $r$, and the slant height can be found using the Pythagorean theorem: $\sqrt{r^2+h^2}$. Therefore, the radius of the base of the cone is $\frac{r}{h}\sqrt{r^2+h^2}$, and the volume of the cone is $\frac{1}{3}\pi r^2h\left(\frac{r}{h}\sqrt{r^2+h^2}\right)^2=\frac{1}{3}\pi r^3\sqrt{r^2+h^2}$.
The ratio of the volumes is then:
$$\frac{\frac{2}{3}\pi r^3+\pi r^2h+\frac{1}{3}\pi r^3\sqrt{r^2+h^2}}{\frac{2}{3}\pi r^3:\pi r^2h:\frac{1}{3}\pi r^3\sqrt{r^2+h^2}}$$
Simplifying, we get:
$$\frac{2r^3+3r^2h+r^3\sqrt{r^2+h^2}}{2r^3:3r^2h:\frac{1}{3}r^3\sqrt{r^2+h^2}}=\frac{6+9\frac{h}{r}+\sqrt{r^2+h^2}}{6:9:\frac{1}{3}\sqrt{r^2+h^2}}$$
Since the ratio is given to us, we can set up an equation:
$$\frac{6+9\frac{h}{r}+\sqrt{r^2+h^2}}{6:9:\frac{1}{3}\sqrt{r^2+h^2}}=\frac{3}{1}:\frac{4}{3}:\frac{1}{3}$$
Simplifying, we get:
$$\frac{6+9\frac{h}{r}+\sqrt{r^2+h^2}}{2\sqrt{r^2+h^2}}=\frac{9}{2}$$
Solving for $\frac{h}{r}$, we get:
$$\frac{h}{r}=\frac{4}{3}$$
Therefore, the volumes are in the ratio of:
$$\frac{\frac{2}{3}\pi r^3+\pi r^2h+\frac{1}{3}\pi r^3\sqrt{r^2+h^2}}{\frac{2}{3}\pi r^3:\pi r^2h:\frac{1}{3}\pi r^3\sqrt{r^2+h^2}}=\frac{6+9\frac{h}{r}+\sqrt{r^2+h^2}}{6:9:\frac{1}{3}\sqrt{
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A cylinder is circumscribed about a hemisphere and a cone is inscribed...
3:2:1
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A cylinder is circumscribed about a hemisphere and a cone is inscribed in the cylinder so as to have its vertex at the centre of one end and the other end as its base. The volumes of the cylinder, hemisphere and the cone are respectively in the ratio of:a)3 : √3: 2b)3 : 2 : 1c)1 : 2 : 3d)2 : 3 : 1Correct answer is option 'B'. Can you explain this answer?
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A cylinder is circumscribed about a hemisphere and a cone is inscribed in the cylinder so as to have its vertex at the centre of one end and the other end as its base. The volumes of the cylinder, hemisphere and the cone are respectively in the ratio of:a)3 : √3: 2b)3 : 2 : 1c)1 : 2 : 3d)2 : 3 : 1Correct answer is option 'B'. Can you explain this answer? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about A cylinder is circumscribed about a hemisphere and a cone is inscribed in the cylinder so as to have its vertex at the centre of one end and the other end as its base. The volumes of the cylinder, hemisphere and the cone are respectively in the ratio of:a)3 : √3: 2b)3 : 2 : 1c)1 : 2 : 3d)2 : 3 : 1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cylinder is circumscribed about a hemisphere and a cone is inscribed in the cylinder so as to have its vertex at the centre of one end and the other end as its base. The volumes of the cylinder, hemisphere and the cone are respectively in the ratio of:a)3 : √3: 2b)3 : 2 : 1c)1 : 2 : 3d)2 : 3 : 1Correct answer is option 'B'. Can you explain this answer?.
A cylinder is circumscribed about a hemisphere and a cone is inscribed in the cylinder so as to have its vertex at the centre of one end and the other end as its base. The volumes of the cylinder, hemisphere and the cone are respectively in the ratio of:a)3 : √3: 2b)3 : 2 : 1c)1 : 2 : 3d)2 : 3 : 1Correct answer is option 'B'. Can you explain this answer? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about A cylinder is circumscribed about a hemisphere and a cone is inscribed in the cylinder so as to have its vertex at the centre of one end and the other end as its base. The volumes of the cylinder, hemisphere and the cone are respectively in the ratio of:a)3 : √3: 2b)3 : 2 : 1c)1 : 2 : 3d)2 : 3 : 1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cylinder is circumscribed about a hemisphere and a cone is inscribed in the cylinder so as to have its vertex at the centre of one end and the other end as its base. The volumes of the cylinder, hemisphere and the cone are respectively in the ratio of:a)3 : √3: 2b)3 : 2 : 1c)1 : 2 : 3d)2 : 3 : 1Correct answer is option 'B'. Can you explain this answer?.
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