A solid consists of a circular cylinder with an exact fitting rig… - Vidyadip

MCQ

A solid consists of a circular cylinder with an exact fitting right circular cone placed at the top. The height of the cone is $h.$ If the total volume of the solid is $3$ times the volume of the cone, then the height of the circular is:

A
$2\text{h}$
✓
$\frac{2\text{h}}{3}$
C
$\frac{3\text{h}}{2}$
D
$4\text{h}$

✓

Answer

Correct option: B.

$\frac{2\text{h}}{3}$

Let $r$ be the radius of the base of solid.
Clearly,
The volume of solid $= 3 \times$ volume of cone
Vol. of cone $+$ Vol. of cylinder $= 3$ Volume of cone
Vol. of cylinder $= 2$ Vol. of cone
$\pi\text{r}^2\text{x}=2\times\frac{1}{3}\pi\text{r}^2\text{h}$
$\text{x}=\frac{2}{3}\text{h}$
Thus,
The height of cylinder $=\frac{2}{3}\text{h}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Surface Areas And Volumes→Surface Areas And Volumes — all question types→Maths STD 10 question papers→Maharashtra Board STD 10 question papers→Maharashtra Board question paper generator→

Similar questions

If $\angle A =30^{\circ}$ then $\tan 2 A =$ ?

If $\alpha,\ \beta,\ \gamma$ are the zeros of the polynomial $x^3 - 6x^2 - x + 30$, then the value of $(\alpha\beta+\beta\gamma+\gamma\alpha)$ is:

Out of given triplets, which is a Pythagoras triplet ?

If $\sin \theta=\frac{4}{5}$ and $\cos \theta=\frac{3}{5}$, then $\tan \theta=$

If three coins are tossed simultaneously, then the probability of getting at least two heads, is:

Four alternative answers for each of the following questions are given. Choose the correct alternative.
Points A, B, C are on a circle, such that m(arc AB) = m(arc BC) = 120°. Nopoint, except point B, is common to the arcs. Which is the type of ∆ ABC?
A. Equilateral triangle
B. Scalene triangle
C. Right angled triangle
D. Isosceles triangle

If mode of a series exceeds its mean by $12$, then mode exceeds the median by:

It is given that the difference between the zeroes of $4x^2 - 8kx + 9$ is $4$ and $k > 0$. Then, $k = ?$

In triangles $\text{ABC}$ and $\text{DEF}, \angle\text{A}=\angle\text{E}=40^\circ, \text{AB : ED = AC : EF}$ and $\angle\text{F}=65^\circ,$ then $\angle\text{B}=$

$\triangle\text{ABC}\sim\triangle\text{DEF}.$ If $BC = 3\ cm, EF = 4\ cm$ and $\text{ar}(\triangle\text{ABC})=54\text{cm}^2,$ then $\text{ar}(\triangle\text{DEF}):$