To do:We have to derive the formula for the volume of the frustum of a cone.Solution:Let $ABC$ be a cone.From the cone the frustum $DECB$ is cut by a plane parallel to its base.$r_1$ and $r_2$ be the radii of the frustum ends of the cone and $h$ be the height of the frustum.In $\triangle ABG$ and $\triangle ADF$$DF \| BG$Therefore, $\triangle ABG \sim \triangle ADF$This implies, $\frac{D F}{B G}=\frac{A F}{A G}=\frac{A D}{A B}$$\frac{r_{2}}{r_{1}}=\frac{h_{1}-h}{h_{1}}=\frac{l_{1}-l}{l_{1}}$$\frac{r_{2}}{r_{1}}=1-\frac{h}{h_{1}}=1-\frac{l}{l_{1}}$$1-\frac{h}{\mathrm{h}_{1}}=\frac{r_{2}}{r_{1}}$$\frac{h}{\mathrm{h}_{1}}=1-\frac{r_{2}}{r_{1}}$$\frac{h}{\mathrm{h}_{1}}=\frac{r_1-r_{2}}{r_{1}}$$\frac{h_1}{h}=\frac{r_1}{r_1-r_2}$$h_1=\frac{r_1h}{r_1-r_2}$Volume of frustum of the cone $=$ Volume of cone $ABC -$ Volume of cone $ADE$$=\frac{1}{3}\pi r_1^2h_1 -\frac{1}{3}\pi r_2^2(h_1 - h)$$= \frac{\pi}{3}[r_1^2h_1-r_2^2(h_1 - h)]$$=\frac{\pi}{3}[r_{1}^{2}(\frac{h r_{1}}{r_{1}-r_{2}})-r_{2}^{2}(\frac{h r_{1}}{r_{1}-r_{2}}-h)]$$=\frac{\pi}{3}[(\frac{h r_{1}^{3}}{r_{1}-r_{2}})-r_{2}^{2}(\frac{h ... Read More

To do:We have to derive the formula for the curved surface area and total surface area of the frustum of the cone.Solution:Let $ABC$ be a cone. From the cone the frustum $DECB$ is cut by a plane parallel to its base.$r_1$ and $r_2$ be the radii of the frustum ends of the cone and $h$ be the height of the frustum.In $\triangle ABG$ and $\triangle ADF$$DF \| BG$Therefore, $\triangle ABG \sim \triangle ADF$This implies, $\frac{D F}{B G}=\frac{A F}{A G}=\frac{A D}{A B}$$\frac{r_{2}}{r_{1}}=\frac{h_{1}-h}{h_{1}}=\frac{l_{1}-l}{l_{1}}$$\frac{r_{2}}{r_{1}}=1-\frac{h}{h_{1}}=1-\frac{l}{l_{1}}$$1-\frac{l}{\mathrm{l}_{1}}=\frac{r_{2}}{r_{1}}$$\frac{l}{l_{1}}=1-\frac{r_{2}}{r_{1}}$$\frac{l}{l_{1}}=\frac{r_{1}-r_{2}}{r_{1}}$$\Rightarrow l_{1}=\frac{r_{1} l}{r_{1}-r_{2}}$..............(i)Curved surface area of frustum $D E C B=$ Curved surface area of cone $A B C$-Curved surface ... Read More

Given:In one fortnight of a given month, there was a rainfall of \( 10 \mathrm{~cm} \) in a river valley. The area of the valley is \( 7280 \mathrm{~km}^{2} \).To do:We have to show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each \( 1072 \mathrm{~km} \) long, \( 75 \mathrm{~m} \) wide and \( 3 \mathrm{~m} \) deep.Solution:Height of the rain $=10\ cm$$=\frac{10}{100}\ m$$=\frac{10}{100\times1000}\ km$Total rainfall in the river valley $=$ area of the valley $\times$ height of the rain$=7280\times\frac{10}{100\times1000}$$=0.7280\ km^3$Volume of normal water in each river $=1072\times\frac{75}{1000}\times\frac{3}{1000}$$= 0.2412\ km^3$Volume ... Read More

Given:A cistern, internally measuring \( 150 \mathrm{~cm} \times 120 \mathrm{~cm} \times 110 \mathrm{~cm} \), has \( 129600 \mathrm{~cm}^{3} \) of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water.Dimensions of each is \( 22.5 \mathrm{~cm} \times 7.5 \mathrm{~cm} \times 6.5 \mathrm{~cm} \).To do:We have to find the number of bricks that can be put in without overflowing the water.Solution:The dimensions of the cistern are \( 150 \mathrm{~cm} \times 120 \mathrm{~cm} \times 110 \mathrm{~cm} \)This implies, Volume of the cistern $= 1980000\ ... Read More

Given:A right triangle, whose sides are \( 3 \mathrm{~cm} \) and \( 4 \mathrm{~cm} \) (other than hypotenuse) is made to revolve about its hypotenuse. To do:We have to find the volume and surface area of the double cone so formed.Solution:Let $AB=3\ cm$ and $BC=4\ cm$ be the two sides of the right triangle $ABC$.This implies, using Pythagoras theorem, $AC^2=AB^2+BC^2$$AC^2=3^2+4^2$$=9+16$$=25$$\Rightarrow AC=\sqrt{25}$$=5\ cm$When the triangle $ABC$ is revolved about the hypotenuse $AC$, the following double cone is formed.In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{BDC}$, $\angle \mathrm{ABC}=\angle \mathrm{CDB}=90^{\circ}$           ($\mathrm{BD} \perp \mathrm{AC})$)$\angle \mathrm{BCA}=\angle \mathrm{BCD}$           (Common)Therefore, by AA similarity, ... Read More

Given:A copper wire, \( 3 \mathrm{~mm} \) in diameter, is wound about a cylinder whose length is \( 12 \mathrm{~cm} \), and diameter \( 10 \mathrm{~cm} \), so as to cover the curved surface of the cylinder.The density of copper is \( 8.88 \mathrm{~g} \mathrm{per} \mathrm{cm}^{3} \).To do:We have to find the length and mass of the wire.Solution:Diameter of the cylinder $(d)= 10\ cm$This implies, Radius of the cylinder $(r) = \frac{10}{2}\ cm$$= 5\ cm$Length of the wire one complete round $= 2 \pi r$$= 2\times3.14\times5\ cm$$= 31.4\ cm$The diameter of the wire $= 3\ mm$$= \frac{3}{10}\ cm$This implies, Radius ... Read More

To do:We have to formulate the given problems as a pair of equations, and hence find their solutions. Solution:(i) Let the speed of the current be $x\ km/hr$ and the speed of her rowing in still water be $y\ km/hr$.Upstream speed $=y−x\ km/hr$Downstream speed $=y+x\ km/hr$$Time=\frac{speed}{distance}$Ritu can row downstream 20 km in 2 hours.Time taken $=\frac{20}{y+x}$$2=​\frac{20}{y+x}$$2(y+x)=20$$y+x=10$.....(i)Ritu can row upstream 4 km in 2 hours.Time taken $=\frac{4}{y-x}$$2=​\frac{4}{y-x}$$2(y-x)=4$$y-x=2$.....(ii)Adding equations (i) and (ii), we get, $y+x+y-x=10+2$$2y=12$$y=\frac{12}{2}$ $y=6\ km/hr$This implies, $6-x=2$$x=6-2$$x=4\ km/hr$Therefore, The speed of her rowing in still water is 6 km/hr and the speed of the current is 4 km/hr. (ii) Let the number of ... Read More

The formula of the resistance of a conductor is-$\boxed{R=\rho(\frac{L}{A})}$Here, $R\rightarrow$ resistance of the conductor$\rho\rightarrow$ resistivity of the conductor$L\rightarrow$length of the conductor$A\rightarrow$area of the cross-section of the conductorWe know the formula of the area of a circle, $A=\pi r^2$Or $A=\pi (\frac{d}{2})^2$            [$d$ is the diameter of the circle and $d=2r$]Therefore, resistance of a conductor $R=\rho(\frac{L}{\pi(\frac{d}{2})^2})$Or $R=\rho(\frac{4L}{\pi d^2})$    ........ $(i)$If both the length and diameter are halved, then $R'=\rho(\frac{4(\frac{L}{2})}{\pi (\frac{d}{2})^2})$Or $R'=\rho(\frac{8L}{\pi d^2})$      .......$(ii)$On comparing $(i)$ and $(ii)$, we get to know that $R'=2R$Therefore, the resistance of a conductor gets doubled if the length and diameter ... Read More

Here given,Current $i=5\ A$Voltage $V=220\ Volt$Time $t=2\ hour=2\times60\times60=7200\ s$So, the power of the motor $P=Vi$$=220\ Volt\times 5\ A$$=1100\ Watt$$=1.1\ kW$                 [we know that $1\ kW=1000\ Watt$]So, energy consumed by the motor $E=Pt$$=1.1\ kW\times 2\ hour$ $=2.2\ kWh$Therefore, the electric motor consumed $2\ kWh$ energy in $2\ hours$.

Given:An observer \( 1.5 \) metres tall is \( 20.5 \) metres away from a tower 22 metres high. To do:We have to determine the angle of elevation of the top of the tower from his eye.Solution:Let $AB$ be the tower and $CD$ be the observer who is $20.5\ m$ away from the tower.From the figure, $AB = 22\ m, CD=1.5\ m, AC=20.5\ m$This implies, $AE=CD=1.5\ m, DE=AC=20.5\ m$ and $BE=22-1.5=20.5\ m$Let \( \theta \) be the angle of elevation of the top of the tower from the eye of the observer.In $\Delta \mathrm{BDE}$, $\tan \theta=\frac{\text { Perpendicular }}{\text { Base ... Read More