Given that,
Diameter of cylinder = 10 cm
So, the radius of the cylinder (r) = $\dfrac{10}{2}$ cm = 5 cm
∴ Length of wire in completely one round = 2πr = 3.14×5 cm = 31.4 cm
It is given that diameter of wire = 3 mm = $\dfrac{3}{10}$ cm
∴ The thickness of the cylinder covered in one round = $\dfrac{3}{10}$ m
Hence, the number of turns (rounds) of the wire to cover 12 cm will be
=$\dfrac{12}{\dfrac{3}{10}}$ =12x$\dfrac{10}{3}$ =40Now, the length of wire required to cover the whole surface = length of wire required to complete 40 rounds
40 x 31.4 cm = 1256 cm
Radius of the wire = $\dfrac{0.3}{2}$ = 0.15 cm
The volume of wire = Area of the cross-section of wire × Length of wire
= π$(0.15)^{2}$×1257.14
= 88.898 cm$^{3}$
We know,
Mass = Volume × Density
= 88.898×8.88
= 789.41 gm
Draw the diagram as follows:
Let us consider the ABA
Here,
AS = 3 cm, AC = 4 cm
So, Hypotenuse BC = 5 cm
We have got 2 cones on the same base AA’ where the radius = DA or DA’
Now, $\dfrac{AD}{CA}$ = $\dfrac{AB}{CB}$
By putting the value of CA, AB and CB, we get,
AD = $\dfrac{2}{5}$ cm
We also know,
$\dfrac{DB}{AB}$ = $\dfrac{AB}{CB}$
So, DB = $\dfrac{9}{5}$ cm
As, CD = BC-DB,
CD = $\dfrac{16}{5}$ cm
Now, the volume of the double cone will be
= $[\dfrac{1}{3}π \times (\dfrac{12}{5})^{2}\dfrac{9}{5}+\dfrac{1}{3}π \times (\dfrac{12}{5})^{2} \times \dfrac{16}{5}]cm^{3} $Solving this, we get
V = 30.14 cm$^{3} $
The surface area of the double cone will be
=$(π \times \dfrac{12}{5} \times 3)+(π \times \dfrac{12}{5} \times 4)cm^{2}$ =$π \times \dfrac{12}{5}[3+4]cm^{2}$= 52.75 $cm^{2}$
Given that the dimension of the cistern = 150 × 120 × 110
So, volume = 1980000 cm$^{3}$
Volume to be filled in cistern = 1980000 – 129600
= 1850400 cm$^{3}$
Now, let the number of bricks placed to be “n”
So, the volume of n bricks will be = n×22.5×7.5×6.5
Now, as each brick absorbs one-seventeenth of its volume, the volume will be
= $\dfrac{n}{(17)}$×(22.5×7.5×6.5)
For the condition given in the question,
The volume of n bricks has to be equal to the volume absorbed by n bricks + the volume to be filled in the cistern
Or, n×22.5×7.5×6.5 = 1850400+$\dfrac{n}{(17)}$ ×(22.5×7.5×6.5)
Solving this, we get
n = 1792.41
From the question, it is clear that
Total volume of 3 rivers = 3×[(Surface area of a river)×Depth]
Given,
Surface area of a river = [1072×($\dfrac{75}{1000}$)] km
And,
Depth = ($\dfrac{3}{1000}$) km
Now, volume of 3 rivers = 3×[1072×($\dfrac{75}{1000}$)]×($\dfrac{3}{1000}$)
= 0.7236 km$^{3}$
Now, the volume of rainfall = total surface area × total height of rain
=7280x$\dfrac{10}{100 \times 1000}km^{3}$= 0.7280 km$^{3}$
For the total rainfall to be approximately equivalent to the addition to the normal water of three rivers, the volume of rainfall has to be equal to the volume of 3 rivers.
But, 0.7280 km3 = 0.7236 km$^{3}$
So, the question statement is true.
Given,
Diameter of the upper circular end of the frustum part = 18 cm
So, radius ($r_$) = 9 cm
Now, the radius of the lower circular end of the frustum ($r_2$) will be equal to the radius of the circular end of the cylinder
So, $r_2$ = $\dfrac{8}{2}$ = 4 cm
Now, height ($h_1$) of the frustum section = 22 – 10 = 12 cm
And,
Height ($h_2$) of cylindrical section = 10 cm (given)
Now, the slant height will be-
$l$=$\sqrt{(r_1-r_2)^{2}+h_1^{2}}$Or,$l$ = 13 cm
Area of tin sheet required = CSA of frustum part + CSA of the cylindrical part
= $π(r_1+r_2)l+2πr_2h_2$
Solving this, we get
Area of tin sheet required = 782 $\dfrac{4}{7}$ cm$^{2}$
Consider the diagram
Let ABC be a cone. From the cone, the frustum DECB is cut by a plane parallel to its base. Here, $r_1$ and $r_2 $ are the radii of the frustum ends of the cone, and h is the frustum height.
Now, consider the ΔABG and ΔADF,
Here, DF||BG
So, ΔABG ~ ΔADF
$\dfrac{DF}{BG}$=$\dfrac{AF}{AG}$=$\dfrac{AD}{AB}$
$\dfrac{r_2}{r_1}$=$\dfrac{h_1-h}{h_1}$=$\dfrac{l_1-l}{l_1}$
$\dfrac{r_2}{r_1}$=$1-\dfrac{h}{h_1}$=$1-\dfrac{l}{l_1}$
$\dfrac{1-l}{l_1}$=$\dfrac{r_2}{r_1}$
$\dfrac{l}{l_1}$=$1-\dfrac{r_2}{r_1}$=$\dfrac{r_1-r_2}{r_1}$
Now, by rearranging, we get
${l_1}$=$\dfrac{r_1l}{r_1-r_2}$
CSA of frustum DECB=CSA of cone ABC-CSA cone ADE
=$πr_1l_1-πr_2(l_1-l)$
=$πr_1(\dfrac{lr_1}{r_1-r_2})-πr_2[dfrac{r_1l}{r_1-r_2}-l]$
=$\dfrac{πr_1^{2}l}{r_1-r_2}-πr_2(\dfrac{r_1l-r_1l+r_2l}{r_1-r_2})$
=$\dfrac{πr_1^{2}l}{r_1-r_2}-\dfrac{πr_2^{2}l}{r_1-r_2}$
=$πl[\dfrac{r_1^{2}-r_2^{2}}{r_1-r_2}]$
CSA of frustum =$π(r_1+r_2)l$
The total surface area of the frustum will be equal to the total CSA of the frustum + the area of the upper circular end + the area of the lower circular end
= $π(r_1+r_2)l+πr_2^{2}+πr_1^{2}$
∴ Surface area of frustum = $π[(r_1+r_2)l+r_1^{2}+r_2^{2}]$
Consider the same diagram as the previous question.
Now, approach the question in the same way as the previous one and prove that
ΔABG ~ ΔADF
Again,
$\dfrac{DF}{BG}$=$\dfrac{AF}{AG}$=$\dfrac{AD}{AB}$
$\dfrac{r_2}{r_1}$=$\dfrac{h_1-h}{h_1}$=$\dfrac{l_1-l}{l_1}$
Now, rearrange them in terms of h and $h_1$
$\dfrac{r_2}{r_1}$=$1-\dfrac{h}{h_1}$=$1-\dfrac{l}{l_1}$
$1-\dfrac{h}{h_1}$=$\dfrac{r_2}{r_1}$
$\dfrac{h}{h_1}$=$1-\dfrac{r_2}{r_1}$=$\dfrac{r_1-r_2}{r_1}$
$\dfrac{h_1}{h}$=$\dfrac{r_1}{r_1-r_2}$
$h_1$=$\dfrac{r_1h}{r_1-r_2}$
The total volume of the frustum of the cone will be = Volume of cone ABC – Volume of cone ADE
= ($\dfrac{1}{3})πr_1^{2}h_1 -(\dfrac{1}{3})πr_2^{2}(h_1 – h)$
= $(\dfrac{π}{3})[r_1^{2}h_1-r_2^{2}(h_1 – h)]$
=$\dfrac{π}{3}[r_1^{2}(\dfrac{hr_1}{r_1-r_2})-r_2^{2}(\dfrac{hr_1}{r_1-r_2}-h)]$
=$\dfrac{π}{3}[(\dfrac{hr_1^{3}}{r_1-r_2})-r_2^{2}(\dfrac{hr_1-hr_1+hr_2}{r_1-r_2})]$
=$\dfrac{π}{3}[\dfrac{hr_1^{3}}{r_1-r_2}-\dfrac{hr_2^{3}}{r_1-r_2}]$
=$\dfrac{π}{3}h[\dfrac{r_1^{3}-r_2^{3}}{r_1-r_2}]$
=$\dfrac{π}{3}h[\dfrac{(r_1-r_2)(r_1^{2}+r_2^{2}+r_1r_2)}{r_1-r_2}]$
Now, solving this, we get
∴ The volume of frustum of the cone = $(\dfrac{1}{3})πh(r12+r22+r1r2)$