Surface Areas and Volumes Test - 12

Given: Side of the cube $\left(a\right) = 3 m$

$\therefore$ Surface Area of Cube $=6a^2=6\times 3\times 3=54\mathrm{sq}. \mathrm{cm}$ 

Now, Cost of painting the cubical box of  $1 \mathrm{sq}. \mathrm{m}=\mathrm{Rs}. 2$ 

$\therefore$ Cost of painting the cubical box of $54\mathrm{sq}·\mathrm{m}=54\times 2=\mathrm{Rs}.108$

Hence, the correct option is (B).

Given: Surface Area $= 100\pi$ sq. cm

$\Rightarrow 4\pi r^2=100\pi$

$\Rightarrow r^{2 }= 25$

$\Rightarrow r = 5 cm$

Hence, the correct option is (A).

Since, the total surface area of cylinder of radius, $\mathrm{r}$ and height, $\mathrm{h}=2\mathrm{\pi rh}+2{\mathrm{\pi r}}^2$

When one cylinder is placed over the other cylinder of same height and radius, then height of the new cylinder $=2\mathrm{h}$ and radius of the new cylinder $=\mathrm{r}$

$\therefore$ Total surface area the new cylinder $=2\mathrm{\pi r}\left(2\mathrm{h}\right)+2{\mathrm{\pi r}}^2=4\pi rh+2{\mathrm{\pi r}}^2$

Hence, the correct option is (A).

Given: Diameter of hemisphere $= 2 cm$, then Radius $= \frac{2}{2}= 1 cm$

We know that, total surface area of the hemisphere $=2\pi r^2+{\mathrm{\pi r}}^2=3{\mathrm{\pi r}}^2$

$\therefore 3\pi r^2=3\pi {\left(1\right)}^2=3\pi {\mathrm{cm}}^2$

Hence, the correct option is (A).

We have,

The radius of the two cones is equal.

The ratio of the slant height,

$l_1:l_2=4 : 3$

We know that the curved surface area of the cone $=\mathrm{\pi rl}$

So, the required ratio $=\frac{\mathrm{\pi r}{l}_1}{\mathrm{\pi r}{l}_2}$

$=\frac{l_1}{l_2}$

$=\frac{4}{3}=4 : 3$

Hence, the correct option is (D).

Let $r$ be the radius of the cylinder. 

Given: Circumference of cylinder $=44\mathrm{cm}$

$\Rightarrow 2\pi r=44$

$\Rightarrow 2\times \frac{22}{7}\times r=44$

$\Rightarrow r=7\mathrm{cm}$

Hence, the correct option is (A).

Given : two identical solid cubes of side '$x$'. If two cubes are joined end to end, then

Length of resulting cuboid $=x+x=2x$

Breadth of resulting cuboid $=x$

And Height of the resulting cuboid $=x$

$\therefore$ TSA of cuboid $=2(lb+bh+hl)$

$=2(2x\times x+x\times x+x\times 2x)$

$=2\left(5x^2\right)=10x^2$

Hence, the correct option is (B).

Here, Radius of the sphere $= 2 cm$

$\therefore$ CSA of sphere $= 4\pi r^2 = 4\pi {\left(2\right)}^2$

$= 16\pi c{m}^2$

Hence, the correct option is (C).

Here $r=\frac{h}{2}$

$\therefore$ TSA of a solid cylinder $=2\pi r(r+h)$

$=2\pi \times \frac{h}{2}\left(\frac{h}{2}+h\right)$

$=\pi h\left(\frac{3h}{2}\right)$

$=\frac{3}{2}\pi h^2$ sq. units

Hence, the correct option is (B).

Given: Volume of sphere $=\frac{9}{16}\pi c{m}^3$

$\Rightarrow \frac{4}{3}\pi r^3=\frac{9}{16}\pi$

$\Rightarrow r^3=\frac{9}{16}\times \frac{3}{4}$

$\Rightarrow r^3=\frac{3\times 3}{4\times 4}\times \frac{3}{4}$

$\Rightarrow r=\frac{3}{4}\mathrm{cm}$

Hence, the correct option is (D).

Given: TSA of solid hemisphere $= 12\pi$ sq. cm

We know that, TSA of solid hemisphere is $3{\mathrm{\pi r}}^2$

$\Rightarrow 3\pi r^2 = 12\pi$

$\Rightarrow r^2 = 4$

$\Rightarrow r = 2 cm$

$\therefore$ CSA of hemisphere $= 2\pi r^{2 }= 2\pi {\left(2\right)}^2 = 8\pi$ sq. cm

Hence, the correct option is (C).

If two solid hemispheres of same base radius '$x$' cm are joined together along their bases, then the CSA of the new solid formed is $4\pi r^2=4\pi x^2 c{m}^2$

Because curved surface area of a hemisphere is $2\pi r^2 c{m}^2$

Hence, the correct option is (D).

Given: $r=1\mathrm{cm}$ and $h=1\mathrm{cm}$

A curved surface area of cylinder $= 2\pi rh$

$=2\pi \times 1\times 1$

$=2\pi {\mathrm{cm}}^2$

Hence, the correct option is (D).

Since, CSA of a right circular cylinder $= 2\pi rh$

Therefore, according to question,

CSA of a right circular cylinder $= 2\pi xz$ sq. units

Hence, the correct option is (D).

Let the radius of a sphere be $\mathrm{r} \mathrm{cm}$

and the radius of the right circular cylinder is $\mathrm{R} \mathrm{cm}$.

Given, Height $=12 cm$ and $R=\frac{Diameter}{2}=\frac{12}{2}=6 cm$

Then according to question,

$4{\mathrm{\pi r}}^2=2\mathrm{\pi Rh}$

$\Rightarrow 2{\mathrm{r}}^2=6\times 12$

$\Rightarrow {\mathrm{r}}^2=36$

$\Rightarrow \mathrm{r}=6$

$= 6 \mathrm{cm}$

Hence, the correct option is (C).