4 Marks Questions

Question 14 Marks

Two solid spheres made of the same metal have weights 5920g and 740g, respectively. Determine the radius of the larger sphere, if the diameter of the smaller one is 5cm.

Answer

Two side spheres made of the same matal have weights 5920g and 740g, respectively.Mass per unit volume is called the density.
Density (D) $=\frac{\text{Mass}}{\text{Volume}}\ \ \Rightarrow\ \ \text{Volume}=\frac{\text{Mass}}{\text{Density}}$
Here density is same because the spheres are of same matal.
$\therefore\ \ \frac{\text{V}_1}{\text{V}_1}=\frac{\frac{5920}{\text{D}}}{\frac{740}{\text{D}}}\Rightarrow\frac{\frac{4}{3}\pi\text{r}^3_1}{\frac{4}{3}\pi\text{r}^3_2}=\frac{5920}{740}$
$\Rightarrow\frac{\text{r}^3_1}{\text{r}^3_1}=\frac{5920}{740}\Rightarrow\frac{\text{r}^3_1}{\Big(\frac{5}{2}\Big)^3}=\frac{5920}{740}\ \ \ [\because\text{r}_2=\frac{1}{2}\times5]$
$\Rightarrow\text{r}^3_1=\frac{5920}{740}\times\frac{125}{8}=125$
$\Rightarrow\text{r}_1=(125)^{\frac{1}{3}}=5\text{cm}$
Hence, the radius of the larger sphere = 5cm.

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Question 24 Marks

The water for a factory is stored in a hemispherical tank whose internal diameter is $14\ m$. The tank contains $50$ kilolitres of water. Water is pumped into the tank to fill to its capacity. Calculate the volume of water pumped into the tank.

Answer

Internal diameter of hemispherical tank $= 14\ m$
$\therefore$ Internal radius of hemispherical tank $= 14\ m \div 2 $
$= 7\ m$
Volume of hemispherical tank $=\frac{2}{3}\pi\text{r}^3$
$=\frac{2}{7}\times\frac{22}{7}\times(7)^3$
$=\frac{44\times49}{3}$
$=718.66\text{ m}^3.$
The tank contains $50$ kilolitres of water $= 50,000$ litres
$=\frac{50,000}{1,000}\text{ m}^3$
$=50\text{ m}^3$
Volume of water pumped into the tank $= 718.66\ m^3 - 50\ m^3$
$ = 668.66\ m^3.$

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Question 34 Marks

A cylindrical tube opened at both the ends is made of iron sheet which is 2cm thick. If the outer diameter is 16cm and its length is 100cm, find how many cubic centimeters of iron has been used in making the tube?

Answer

A cylinder tube is made iron sheet. Which is 2cm thick.Its outer radius (R) = 16 ÷ 2 = 8cm
Thickness of iron sheet = 2cm
And its inner radius (r) = 8cm - 2cm = 6cm.
Height of cylinder = 100cm
Quantity of iron used in making the = Volume of hollow cylinder $=\pi(\text{R}^2-\text{r}^2)\text{h}$
$=\frac{22}{7}\times(8^2-6^2)\times100=\frac{22}{7}(64-36)\times100$
$=\frac{22}{7}\times28\times100=8800\text{cm}^2$

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Question 44 Marks

30 circular plates, each of radius 14cm and thickness 3cm are placed one above the another to form a cylindrical solid. Find:

The total surface area.
Volume of the cylinder so formed.

Answer

Radius of one circular plate = 14cm.Thickness of one circular plate = 3cm.
As the plates are placed one above the other, so the height of the cylinder formed by placing 30 plates = 30 × 3 = 90cm

Total surface area of circular $=2\pi\text{rh}+2\pi\text{r}^2$

$=2\pi\text{r}(\text{r+h})=2\times\frac{22}{7}\times14(14+90)=44\times208=9152\text{cm}^2$

Volume of the cylinder $=\pi\text{r}^2\text{h}=\frac{22}{7}\times14\times14\times90=55440\text{cm}^3$

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Question 54 Marks

A cube of side $4\ cm$ contains a sphere touching its sides. Find the volume of the gap in between.

Answer

Side of cube $= 4\ cm$
Volume of cube $= (4)^3$
$ = 4 \times 4 \times 4 $
$= 64\ cm^3$
As cube contains a sphere touching its sides, so the diameter of the sphere $= 4\ cm.$
Radius of sphere $=\frac{4}{3}\pi\text{r}^3$
$=\frac{4}{3}\times\frac{22}{7}\times(2)^3$
$=\frac{88\times8}{21}$
$=\frac{704}{21}$
$=33.52\text{ cm}^3$
Volume of gap in between them $= 64\ cm^3 - 33.52\ cm^3$
$ = 30.48\ cm^3.$

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Question 64 Marks

A semi-circular sheet of metal of diameter $28\ cm$ is bent to form an open conical cup. Find the capacity of the cup.

Answer

Given, diameter of a semi$-$circular sheet $= 28\ cm$
$\therefore$ Radius of a semi-circular sheet, $\text{r}=\frac{28}{2}=14\text{ cm}$
Since, a semi$-$cicular sheet of metal is bent to form an open cup.
Let the radius of a conical be $R$.

$\therefore$ Circumference of base of cone $=$ Circumference of semi$-$circle $2\pi\text{R}=\pi\text{r}$
$\Rightarrow\ \ 2\pi\text{R}=\pi\times14$
$\Rightarrow\text{R}=7\text{ cm}$
Now, $\text{h}=\sqrt{\text{l}^2-\text{R}^2}$
$=\sqrt{14^2-7^2}\ \ \ \ [\therefore\text{l}^2=\text{h}^2+\text{R}^2]$
$=\sqrt{196-49}$
$=\sqrt{147}$
$=12.1243\text{ cm}$
Volume $($capacity$)$ of conical cup $=\frac{1}{3}\pi\text{R}^2\text{h}$
$=\frac{1}{3}\times\frac{22}{7}\times7\times7\times12.1243$
$=622.38\text{ cm}^3$
Hence, the capacity of an open conical cup is $622.38\ cm^3.$

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