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 $5\ cm.$

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 $= 14m ÷ 2 = 7m$
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
$= 718.66m^3 - 50m^3 = 668.66m^3.$

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Question 34 Marks

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

Answer

A cylinder tube is made iron sheet. Which is $2\ cm$ thick.Its outer radius $(R) = 16 ÷ 2 = 8\ cm$
Thickness of iron sheet $= 2\ cm$
And its inner radius $(r) = 8\ cm - 2\ cm = 6\ cm.$
Height of cylinder $= 100\ cm$
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 $14\ cm$ and thickness $3\ cm$ are placed one above the another to form a cylindrical solid. Find:
$i.$ The total surface area.
$ii.$ Volume of the cylinder so formed.

Answer

Radius of one circular plate $= 14\ cm$.Thickness of one circular plate $= 3\ cm.$
As the plates are placed one above the other, so the height of the cylinder formed by placing $30$ plates $= 30 \times 3 = 90\ cm$
$i.$ Total surface area of circular $=2\pi\text{rh}+2\pi\text{r}^2$
$=2\pi\text{r}(\text{r+h})=2\times\frac{22}{7}\times14(14+90)=44\times208=9152\text{ cm}^2$
$ii.$ 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 semi-circular sheet of metal of diameter $28cm$ 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 $622.38cm^3$.

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Question 64 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 $= (4)^3 = 4 \times 4 \times 4 = 64cm^3$
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 $= 64cm^3 - 33.52cm^3 = 30.48cm^3.$

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