Mix Examples - Surface Areas and Volumes Questions in English

(D) The volume $V$ of a cubical tank is given by the formula $V = \text{length} \times \text{breadth} \times \text{height}$.
Given dimensions are $2 \,m \times 2 \,m \times 2 \,m = 8 \,m^{3}$.
To convert $m^{3}$ to $cm^{3}$, we know that $1 \,m = 100 \,cm$, so $1 \,m^{3} = (100 \,cm)^{3} = 1,000,000 \,cm^{3} = 10^{6} \,cm^{3}$.
Thus, $8 \,m^{3} = 8 \times 10^{6} \,cm^{3}$. (Option $A$ is correct).
We also know that $1 \,m^{3} = 1000 \text{ litres}$.
Therefore, $8 \,m^{3} = 8000 \text{ litres}$. (Option $C$ is correct).
Since $1000 \text{ litres} = 1 \text{ kilolitre}$, $8000 \text{ litres} = 8 \text{ kilolitres}$. (Option $B$ is correct).
Since all options $A, B,$ and $C$ are correct, the correct choice is $D$.

(C) For the given cone,the radius $r = \frac{\text{diameter}}{2} = \frac{10}{2} = 5 \, cm$.
The curved surface area $(CSA)$ of a cone is given by the formula $CSA = \pi r l$,where $l$ is the slant height.
Given that $CSA = 85 \pi \, cm^{2}$,we have:
$85 \pi = \pi \times 5 \times l$
Dividing both sides by $5 \pi$:
$l = \frac{85 \pi}{5 \pi} = 17 \, cm$.
Therefore,the slant height of the cone is $17 \, cm$.

(D) We know that $1\,m = 100\,cm = 10^2\,cm$.
To find the value of $1\,m^3$ in $cm^3$,we cube both sides of the equation:
$1\,m^3 = (10^2\,cm)^3$
$1\,m^3 = (10^2)^3\,cm^3$
Using the power rule $(a^m)^n = a^{m \times n}$,we get:
$1\,m^3 = 10^{2 \times 3}\,cm^3 = 10^6\,cm^3$.
Therefore,$1\,m^3 = 10^6\,cm^3$.

(C) cylinder with cone-shaped lids at both ends consists of the curved surface area of the cylinder and the curved surface areas of the two cones.
$1$. The curved surface area of the cylinder is $2\pi rh$,where $r$ is the radius and $h$ is the height of the cylinder.
$2$. The curved surface area of one cone is $\pi rl$,where $l$ is the slant height of the cone.
$3$. Since there are two such cones,their combined curved surface area is $2\pi rl$.
$4$. The total surface area is the sum of these areas: $2\pi rh + 2\pi rl = 2\pi r(h+l)$.

(B) The diameter of the hemisphere is $d = 1 \, cm$.
Therefore,the radius $r$ is $r = \frac{d}{2} = \frac{1}{2} \, cm$.
The formula for the volume of a hemisphere is $V = \frac{2}{3} \pi r^3$.
Substituting the value of $r$ into the formula:
$V = \frac{2}{3} \pi \left( \frac{1}{2} \right)^3$
$V = \frac{2}{3} \pi \left( \frac{1}{8} \right)$
$V = \frac{2 \pi}{24} = \frac{\pi}{12} \, cm^3$.

(B) To match the given parts,we identify the formulas for surface areas:
$1$. The Curved Surface Area $(CSA)$ of a cylinder is $2\pi rh$. Thus,$1-b$.
$2$. The Curved Surface Area $(CSA)$ of a sphere is $4\pi r^2$. Thus,$2-c$.
$3$. The Curved Surface Area $(CSA)$ of a cone is $\pi rl$. Thus,$3-d$.
$4$. The Total Surface Area $(TSA)$ of a hemisphere is $3\pi r^2$. Thus,$4-a$.
Therefore,the correct matching is $(1-b), (2-c), (3-d), (4-a)$.

(D) To match the given parts,we identify the standard volume formulas:
$1$. The volume of a cuboid with length $l$,breadth $b$,and height $h$ is given by $V = lbh$. Thus,$1$ matches with $d$.
$2$. The volume of a cone with radius $r$ and height $h$ is given by $V = \frac{1}{3} \pi r^{2} h$. Thus,$2$ matches with $a$.
$3$. The volume of a cylinder with radius $r$ and height $h$ is given by $V = \pi r^{2} h$. Thus,$3$ matches with $b$.
$4$. The volume of a sphere with radius $r$ is given by $V = \frac{4}{3} \pi r^{3}$. Thus,$4$ matches with $c$.
Therefore,the correct matching is $(1-d), (2-a), (3-b), (4-c)$.