Volume of a Right Circular Cone Questions in English

(B) False.
For the largest possible right circular cone cut out of a cube of edge $r$,the height of the cone $(h)$ must be equal to the edge of the cube,so $h = r \, cm$.
The diameter of the base of the cone must be equal to the edge of the cube,so the radius of the base $(R)$ is $\frac{r}{2} \, cm$.
The volume of a cone is given by the formula $V = \frac{1}{3} \pi R^2 h$.
Substituting the values: $V = \frac{1}{3} \pi \left(\frac{r}{2}\right)^2 \cdot r = \frac{1}{3} \pi \left(\frac{r^2}{4}\right) \cdot r = \frac{1}{12} \pi r^3 \, cm^3$.
Since $\frac{1}{12} \pi r^3 \neq \frac{1}{6} \pi r^3$,the given statement is False.

(FALSE) Let the original radius of the cone be $r$ and the original height be $h$.
The volume of the original cone is $V_1 = \frac{1}{3} \pi r^{2} h$.
Now,the new radius $r' = \frac{r}{2}$ and the new height $h' = 2h$.
The new volume $V_2 = \frac{1}{3} \pi (r')^{2} h' = \frac{1}{3} \pi \left(\frac{r}{2}\right)^{2} (2h)$.
Simplifying this,$V_2 = \frac{1}{3} \pi \left(\frac{r^2}{4}\right) (2h) = \frac{1}{2} \left(\frac{1}{3} \pi r^{2} h\right) = \frac{1}{2} V_1$.
Since the new volume is half of the original volume,the statement is False.

(A) The diameter of the semi-circular sheet is $28 \, cm$,so its radius $R = 14 \, cm$. When this sheet is bent to form an open conical cup,the radius of the sheet becomes the slant height $(l)$ of the cone.
Therefore,$l = 14 \, cm$.
The circumference of the semi-circular arc becomes the circumference of the base of the cone. The length of the semi-circular arc is $\pi R = \pi \times 14$.
Let $r$ be the radius of the base of the cone. Then,$2 \pi r = 14 \pi$,which gives $r = 7 \, cm$.
Now,we find the height $(h)$ of the cone using the relation $l^2 = r^2 + h^2$:
$h = \sqrt{l^2 - r^2} = \sqrt{14^2 - 7^2} = \sqrt{196 - 49} = \sqrt{147} \approx 12.124 \, cm$.
The capacity (volume) of the conical cup is $V = \frac{1}{3} \pi r^2 h$.
$V = \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 12.124 \approx 622.38 \, cm^3$. Given the options,the closest value is $622.16 \, cm^3$.

(B) Given: Height $(h) = 21 \, cm$,Slant height $(l) = 28 \, cm$.
Using the relation $l^2 = r^2 + h^2$,we find the radius $(r)$:
$r = \sqrt{l^2 - h^2} = \sqrt{28^2 - 21^2} = \sqrt{784 - 441} = \sqrt{343} = 7\sqrt{7} \, cm$.
Now,the volume of the cone is given by $V = \frac{1}{3} \pi r^2 h$.
$V = \frac{1}{3} \times \frac{22}{7} \times (7\sqrt{7})^2 \times 21$.
$V = \frac{1}{3} \times \frac{22}{7} \times 49 \times 7 \times 21$.
$V = 22 \times 49 \times 7 = 7546 \, cm^3$.

(D) The formula for the volume of a right circular cone is $V = \frac{1}{3} \pi r^2 h$.
Given:
Radius $(r) = 6 \, cm$
Height $(h) = 7 \, cm$
$\pi = \frac{22}{7}$
Substituting the values into the formula:
$V = \frac{1}{3} \times \frac{22}{7} \times (6)^2 \times 7$
$V = \frac{1}{3} \times \frac{22}{7} \times 36 \times 7$
$V = \frac{1}{3} \times 22 \times 36$
$V = 22 \times 12$
$V = 264 \, cm^3$
Thus,the volume of the cone is $264 \, cm^3$.

(A) Given:
Radius of the cone $(r) = 3.5 \, cm = \frac{35}{10} \, cm = \frac{7}{2} \, cm$.
Height of the cone $(h) = 12 \, cm$.
The formula for the volume of a right circular cone is $V = \frac{1}{3} \pi r^2 h$.
Substituting the values:
$V = \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times 12$
$V = \frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 12$
$V = \frac{1}{3} \times 22 \times \frac{1}{2} \times 7 \times 6$
$V = 11 \times 14 = 154 \, cm^3$.
Thus,the volume of the cone is $154 \, cm^3$.

(B) Given: radius $r = 7 \, cm$ and slant height $l = 25 \, cm$.
First,we find the vertical height $h$ using the relation $l^2 = r^2 + h^2$:
$h = \sqrt{l^2 - r^2} = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24 \, cm$.
The volume $V$ of a conical vessel is given by the formula $V = \frac{1}{3} \pi r^2 h$.
Substituting the values:
$V = \frac{1}{3} \times \frac{22}{7} \times (7)^2 \times 24 \, cm^3$
$V = \frac{1}{3} \times \frac{22}{7} \times 49 \times 24 \, cm^3$
$V = 22 \times 7 \times 8 \, cm^3 = 1232 \, cm^3$.
Since $1000 \, cm^3 = 1 \, l$,the capacity in litres is:
$V = \frac{1232}{1000} \, l = 1.232 \, l$.
Thus,the required capacity of the conical vessel is $1.232 \, l$.

(C) Given: Height $(h) = 12 \,cm$ and slant height $(l) = 13 \,cm$.
First,we find the radius $(r)$ using the relation $l^2 = r^2 + h^2$.
$r = \sqrt{l^2 - h^2} = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5 \,cm$.
Now,the volume $(V)$ of the conical vessel is given by the formula $V = \frac{1}{3} \pi r^2 h$.
$V = \frac{1}{3} \times \frac{22}{7} \times (5)^2 \times 12 \,cm^3$.
$V = \frac{1}{3} \times \frac{22}{7} \times 25 \times 12 = 22 \times 25 \times \frac{4}{7} = \frac{2200}{7} \,cm^3$.
Since $1000 \,cm^3 = 1 \,l$,the capacity in litres is $\frac{2200}{7 \times 1000} \,l = \frac{22}{70} \,l = \frac{11}{35} \,l$.
Thus,the capacity of the conical vessel is $\frac{11}{35} \,l$.

(D) Given,height of the cone $(h) = 15 \, cm$.
Volume of the cone $(V) = 1570 \, cm^3$.
Let the radius of the base be $r \, cm$.
The formula for the volume of a cone is $V = \frac{1}{3} \pi r^2 h$.
Substituting the given values: $1570 = \frac{1}{3} \times 3.14 \times r^2 \times 15$.
Simplifying the equation: $1570 = 3.14 \times r^2 \times 5$.
$1570 = 15.7 \times r^2$.
$r^2 = \frac{1570}{15.7} = 100$.
$r = \sqrt{100} = 10 \, cm$.
Therefore,the radius of the base is $10 \, cm$.

(A) Volume of the cone $= 48 \pi \, cm^3$.
Height of the cone $(h) = 9 \, cm$.
Let $r$ be the radius of the base.
The formula for the volume of a cone is $V = \frac{1}{3} \pi r^2 h$.
Substituting the given values: $\frac{1}{3} \pi r^2 (9) = 48 \pi$.
$3 \pi r^2 = 48 \pi$.
$r^2 = \frac{48 \pi}{3 \pi} = 16$.
$r = \sqrt{16} = 4 \, cm$.
The diameter of the base is $d = 2r = 2 \times 4 = 8 \, cm$.

(B) Given, diameter of the conical pit $= 3.5 \, m$.
Radius $(r) = \frac{3.5}{2} \, m = 1.75 \, m$.
Depth $(h) = 12 \, m$.
The volume (capacity) of a cone is given by the formula $V = \frac{1}{3} \pi r^2 h$.
Substituting the values: $V = \frac{1}{3} \times \frac{22}{7} \times (1.75)^2 \times 12 \, m^3$.
$V = \frac{1}{3} \times \frac{22}{7} \times \frac{35}{20} \times \frac{35}{20} \times 12 \, m^3$.
$V = \frac{22}{7} \times \frac{7}{4} \times \frac{7}{4} \times 4 \, m^3 = 38.5 \, m^3$.
We know that $1 \, m^3 = 1000 \, litres = 1 \, kilolitre (kl)$.
Therefore, the capacity of the pit is $38.5 \, kl$.

(C) Given,volume of the cone $(V) = 9856 \, cm^3$.
Diameter of the base $= 28 \, cm$.
Radius of the base $(r) = \frac{28}{2} \, cm = 14 \, cm$.
Let the height of the cone be $h \, cm$.
The formula for the volume of a cone is $V = \frac{1}{3} \pi r^2 h$.
Substituting the values: $9856 = \frac{1}{3} \times \frac{22}{7} \times (14)^2 \times h$.
$9856 = \frac{1}{3} \times \frac{22}{7} \times 14 \times 14 \times h$.
$9856 = \frac{1}{3} \times 22 \times 2 \times 14 \times h$.
$9856 = \frac{616}{3} \times h$.
$h = \frac{9856 \times 3}{616}$.
$h = 16 \times 3 = 48 \, cm$.
Thus,the height of the cone is $48 \, cm$.

(D) Volume of the cone $(V) = 9856 \, cm^3$.
Diameter of the base $= 28 \, cm$.
Radius of the base $(r) = \frac{28}{2} = 14 \, cm$.
Formula for the volume of a cone is $V = \frac{1}{3} \pi r^2 h$.
$9856 = \frac{1}{3} \times \frac{22}{7} \times 14 \times 14 \times h$.
$9856 = \frac{1}{3} \times 22 \times 2 \times 14 \times h$.
$9856 = \frac{616}{3} \times h$.
$h = \frac{9856 \times 3}{616} = 16 \times 3 = 48 \, cm$.
Slant height $(\ell) = \sqrt{r^2 + h^2} = \sqrt{14^2 + 48^2} = \sqrt{196 + 2304} = \sqrt{2500} = 50 \, cm$.

(B) The sides of the right triangle are $5 \, cm, 12 \, cm$,and $13 \, cm$.
Since the right-angled triangle is revolved about the $12 \, cm$ side,the height of the resulting cone is $h = 12 \, cm$ and the radius of the base is $r = 5 \, cm$.
The volume of a cone is given by the formula $V = \frac{1}{3} \pi r^2 h$.
Substituting the values,we get:
$V = \frac{1}{3} \times \pi \times (5)^2 \times 12 \, cm^3$
$V = \frac{1}{3} \times \pi \times 25 \times 12 \, cm^3$
$V = \pi \times 25 \times 4 \, cm^3$
$V = 100 \pi \, cm^3$.
Thus,the volume of the solid obtained is $100 \pi \, cm^3$.

(C) $1$. When the triangle is revolved about the side $12 \, cm$:
Height $(h) = 12 \, cm$,Radius $(r) = 5 \, cm$.
Volume $V_1 = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \pi \times (5)^2 \times 12 = 100 \pi \, cm^3$.
$2$. When the triangle is revolved about the side $5 \, cm$:
Height $(h) = 5 \, cm$,Radius $(r) = 12 \, cm$.
Volume $V_2 = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \pi \times (12)^2 \times 5 = 240 \pi \, cm^3$.
$3$. Ratio of the volumes:
Ratio $= \frac{V_1}{V_2} = \frac{100 \pi}{240 \pi} = \frac{100}{240} = \frac{5}{12} = 5:12$.

(A) For a given cone:
Radius $(r) = 10 \, cm$ and slant height $(l) = 26 \, cm$.
Using the relation $l^2 = r^2 + h^2$,we find the height $(h)$:
$h^2 = l^2 - r^2$
$h^2 = (26)^2 - (10)^2$
$h^2 = 676 - 100 = 576$
$h = \sqrt{576} = 24 \, cm$.
The volume of a cone is given by the formula $V = \frac{1}{3} \pi r^2 h$.
$V = \frac{1}{3} \times 3.14 \times (10)^2 \times 24$
$V = 3.14 \times 100 \times 8$
$V = 314 \times 8 = 2512 \, cm^3$.

(D) Given: Radius of the base $(r)$ = $14 \, cm$,Slant height $(l)$ = $50 \, cm$.
First,we find the vertical height $(h)$ of the cone using the relation $r^2 + h^2 = l^2$.
$14^2 + h^2 = 50^2$
$196 + h^2 = 2500$
$h^2 = 2500 - 196 = 2304$
$h = \sqrt{2304} = 48 \, cm$.
The volume of a cone is given by the formula $V = \frac{1}{3} \pi r^2 h$.
$V = \frac{1}{3} \times \frac{22}{7} \times 14 \times 14 \times 48$
$V = \frac{1}{3} \times 22 \times 2 \times 14 \times 48$
$V = 22 \times 2 \times 14 \times 16$
$V = 9856 \, cm^3$.
Therefore,the volume of the cone is $9856 \, cm^3$.

(A) The formula for the volume of a cone is given by $V = \frac{1}{3} \pi r^2 h$,where $\pi r^2$ is the area of the base.
Given,the area of the base $(A)$ = $1386 \, cm^2$ and the height $(h)$ = $9 \, cm$.
Substituting these values into the formula:
$V = \frac{1}{3} \times A \times h$
$V = \frac{1}{3} \times 1386 \times 9$
$V = 1386 \times 3$
$V = 4158 \, cm^3$.
Therefore,the volume of the cone is $4158 \, cm^3$.

(B) Given: Radius $(r) = 7 \, cm$, Slant height $(l) = 25 \, cm$.
First, find the vertical height $(h)$ using the relation $l^2 = r^2 + h^2$.
$h^2 = l^2 - r^2 = 25^2 - 7^2 = 625 - 49 = 576$.
$h = \sqrt{576} = 24 \, cm$.
The volume $(V)$ of the cone is given by $V = \frac{1}{3} \pi r^2 h$.
$V = \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24 = 22 \times 7 \times 8 = 1232 \, cm^3$.
Since $1000 \, cm^3 = 1 \, litre$, the capacity is $\frac{1232}{1000} = 1.232 \, litres$.

(C) Given: Radius $(r)$ = $21 \,cm$,Slant height $(l)$ = $29 \,cm$.
First,find the height $(h)$ of the cone using the relation $l^2 = r^2 + h^2$.
$h^2 = l^2 - r^2 = 29^2 - 21^2 = 841 - 441 = 400$.
$h = \sqrt{400} = 20 \,cm$.
The volume $(V)$ of a cone is given by $V = \frac{1}{3} \pi r^2 h$.
$V = \frac{1}{3} \times \frac{22}{7} \times 21 \times 21 \times 20 = 22 \times 3 \times 7 \times 20 = 9240 \,cm^3$.
Since $1000 \,cm^3 = 1 \,litre$,the capacity is $9240 / 1000 = 9.240 \,litres$.

(A) Given: Radius $(r)$ = $35 \,cm$,Slant height $(l)$ = $91 \,cm$.
First,find the vertical height $(h)$ using the relation $l^2 = r^2 + h^2$.
$h = \sqrt{l^2 - r^2} = \sqrt{91^2 - 35^2} = \sqrt{(91 - 35)(91 + 35)} = \sqrt{56 \times 126} = \sqrt{7056} = 84 \,cm$.
The volume $(V)$ of the cone is given by $V = \frac{1}{3} \pi r^2 h$.
$V = \frac{1}{3} \times \frac{22}{7} \times 35 \times 35 \times 84 = 22 \times 35 \times 5 \times 84 / 3 = 22 \times 35 \times 5 \times 28 = 107800 \,cm^3$.
Since $1000 \,cm^3 = 1 \,litre$,the capacity is $107800 / 1000 = 107.8 \,litres$.

(N/A) When $\Delta ABC$ is revolved about side $AB$,the height of the cone $h = AB = 6.3 \, cm$ and the radius $r = BC = 8.4 \, cm$. The volume $V_1 = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (8.4)^2 \times 6.3 = 465.696 \, cm^3$. When $\Delta ABC$ is revolved about side $BC$,the height of the cone $h = BC = 8.4 \, cm$ and the radius $r = AB = 6.3 \, cm$. The volume $V_2 = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (6.3)^2 \times 8.4 = 349.272 \, cm^3$.

(C) The volume $V$ of a cone is given by the formula $V = \frac{1}{3} \pi r^2 h$,where $r$ is the radius and $h$ is the height.
Given: $V = 616 \, cm^3$ and $h = 12 \, cm$.
Using $\pi \approx \frac{22}{7}$,we have:
$616 = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 12$
$616 = 4 \times \frac{22}{7} \times r^2$
$616 = \frac{88}{7} \times r^2$
$r^2 = \frac{616 \times 7}{88}$
$r^2 = 7 \times 7 = 49$
$r = 7 \, cm$.
The diameter $d$ is $2r = 2 \times 7 = 14 \, cm$.

(D) The formula for the volume of a cone is $V = \frac{1}{3} \pi r^2 h$.
Given:
Radius $(r)$ = $7 \, cm$
Height $(h)$ = $30 \, cm$
Using $\pi = \frac{22}{7}$:
$V = \frac{1}{3} \times \frac{22}{7} \times (7)^2 \times 30$
$V = \frac{1}{3} \times \frac{22}{7} \times 49 \times 30$
$V = 22 \times 7 \times 10$
$V = 1540 \, cm^3$.