Maths M.C.Q (1 Marks)

Given, diameter of cylinder $=$ Diameter of hemisphere $= 0.5cm$

$[$since, both hemispheres are attach with cylinder$]$
$\therefore$ Radius of cylinder $(r) =$ radius of hemisphere $(\text{r})=\frac{0.5}{2}=0.25\text{cm}$
$[\because\text{diameter}=2\times\text{radius}]$
and total length of capsule $= 2\ cm$
$\therefore$ Length of cylindrical part of capsule,
$h =$ Length of capsule $-$ Radius of both hemispheres
$= 2 - (0.25 + 0.25) = 1.5\ cm$
Now, capacity of capsule $=$ Volume of cylindrical part $+ 2 \times$ Volume of hemisphere
$=\pi\text{r}^2\text{h}+2\times\frac{2}{3}\pi\text{r}^3$
$\big[\because$ volume of cylinder $=\pi\times(\text{radius})^2 \times$ height and volume of hemispere $=\frac{2}{3}\pi(\text{radiud})^3\big]$
$=\frac{22}{7}\big[(0.25)^2\times1.5+\frac{4}{3}\times(0.25)^3\big]$
$=\frac{22}{7}[0.09375+0.0208]$
$=\frac{22}{7}\times0.11455=0.36\text{cm}^3$
Hence, the capacity of capsule is $0.36 \ cm^3$​​​​​​​

Volume of the wall $=270 \times 300 \times 350=28350000 \ cm^3$
$[\because$ volume of chboid $=$ lenth $\times $ breadth $\times $ height$]$
Since, $\frac{1}{8}$ space of wall is covered by mortar.
So, remainig space of wall $=$ Volume of wall $-$ Volume of mortar
$=28350000-28350000\times\frac{1}{8}$
$=28350000-3543750=24806250\text{cm}^3$
Now, volume of one birck $= 22.5 \times 1125 \times 875 = 2214.844\ cm^3$
$[\because$ volume of chboid $=$ lenth $\times $ breadth $\times $ height$]$
$\therefore$ Required number of bricks $=\frac{24806250}{2214.844}=11200\ (\text{approx})$
Hence, the number of used to construct the wall is $11200.$

Because curved surface area of a hemisphere is $2 w^2$ and here, we join two solid hemispheres along their bases of radius $r$, from which we get a solid sphere.
Hence, the curved surface area of new solid $=2\pi\text{r}^2+2\pi\text{r}^2=4\pi\text{r}^2$

Given, internal diameter of spherical shell $= 4\ cm$
and exiernal diameter of shell $= 8\ cm$
$\therefore$ Internal radius of spherical shell $\text{r}_1=\frac{4}{2}\text{cm}=2\text{cm}$
$[\because\text{diameter}=2\times\text{radius}]$
and external radius of shell, $\text{r}_2=\frac{8}{2}=4\text{cm}$
$[\because\text{diameter}=2\times\text{radius}]$
Now, volume of the spherical shell
$=\frac{4}{3}\pi\big[\text{r}^3_2-\text{r}^3_1\big]$
$\left[\because\right.$ volume of the spherical shell $\left.=\frac{4}{3} \pi\left\{(\text { extermal radius })^3-(\text { internal radius })^3\right\}\right]$
$\frac{4}{3}\pi(4^3-2^3)$
$\frac{4}{3}\pi(64-8)$
$=\frac{224}{3}\pi\text{ \ cm}^3$
Let height of cone $= h \ cm$
Diameter of the base of cone $= 8\ cm$
$\therefore$ Radius of the base of cone $=\frac{8}{2}=4\text{cm}$
$[\because\text{diameter}=2\times\text{radius}]$
According to the question,
Volume of cone $=$ Volume of spherical shell
$\Rightarrow\ \ \frac{1}{3}\pi(4)^2\text{ h}=\frac{224}{3}\pi$
$\Rightarrow\ \text{h}=\frac{224}{16}=14\text{cm}$
$\Big[\because\text{volume of cone}=\frac{1}{3}\times\pi\times(\text{radius})^2\times(\text{height})\Big]$
Hence, the height of the cone is $14\ cm.$

Given, edge of the cude $= 22\ cm$
$\therefore$ Volume of the cude $=(22)^3=10648 \ cm^2$
Also, given diameter of marble $= 0.5\ cm$
$\therefore$ Radius of a marble, $\text{r}=\frac{0.5}{2}=0.25\text{cm}$
$[\because\text{diameter}=2\times\text{radius}]$
Volume of one marble $=\frac{4}{3}\pi\text{r}^3=\frac{4}{3}\times\frac{22}{7}\times(0.25)^3$
$\Big[\because\text{volume of sphere}=\frac{4}{3}\times\pi\times(\text{radius})^3\Big]$
$=\frac{1.375}{21}=0.0655\text{cm}^3$
Filled space of cude $=$ Volume of the cube $\frac{1}{8} \times$ Volume of cube
$=10648-10648\times\frac{1}{8}$
$=10648\times\frac{7}{8}=9317\text{cm}^3$
$\therefore\ \text{Required number of marbles}=\frac{\text{Total space filled by marbles in a cube}}{\text{Volume of one marble}}$
$=\frac{9317}{0.0655}=142244\ (\text{approx})$
Hence, the number of marbles that the cube can accomodate is $142244.$