2 Marks Questions

Question 12 Marks

A spherical glass vessel has a cylindrical neck $8\ cm$ long, $2\ cm$ in diameter the diameter of the spherical part is $8.5\ cm$. By measuring the amount of water it holds, a child finds its volume to be $345\ cm^3$. Check whether she is correct, taking the above as the inside measurements and $\pi $ $= 3.14$.

Answer

Amount of water it holds $ = \frac{4}{3}\pi {r^3} + \pi {r^2}h$
$ = \frac{4}{3}\pi {\left( {\frac{{8.5}}{2}} \right)^3} + \pi {\left( {\frac{2}{2}} \right)^2} \times 8$
$ = \frac{4}{3} \times 3.14 \times 4.25 \times 4.25 \times 4.25 + 8 \times 3.14$
$= 321.39 + 25.12$
$= 346.51\ cm^3$
Hence, she is correct. The correct volume is $346.51\ cm^3$.

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Question 22 Marks

A solid iron pole consists of a cylinder of height $220$ cm and base diameter $24$ cm is surmounted by another cylinder of height $60\ cm$ and radius $8\ cm$. Find the mass of the pole, given that $1\ cm^3$ of iron has approximately $8$ g mass. $($Use $\pi = 3.14)$

Answer

Radius of lower cylinder $= R = 12\ cm$
Radius of upper cylinder $= r = 8\ cm$
Height of upper cylinder $= h = 60\ cm$
Height of lower cylinder $= H = 220\ cm$
Volume of solid iron pole $= \pi R ^ { 2 } H + \pi r ^ { 2 } h$
$= 3.14 \times ( 12 ) ^ { 2 } \times 220$$+ 3.14 \times ( 8 ) ^ { 2 } \times 60$
$= 111532.8 \mathrm { cm } ^ { 3 }$
Mass of the pole $= 111532.8 \times 8 g$
$= 892.2624 \mathrm { kg }$

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Question 32 Marks

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are $15\ cm$ by $10\ cm$ by $3.5\ cm$. The radius of each of the depressions is $0.5\ cm$ and the depth is $1.4\ cm$. Find the volume of wood in the entire stand.

Answer

Depth $(h)$ of each conical depression $= 1.4\ cm$
Radius $(r)$ of each conical depression $= 0.5\ cm$
Volume of wood = Volume of cuboid $- 4 × $ Volume of cones
$= lbh – 4 \times$ $\frac{1}{3} \pi r^{2} h$
$= 15\times$ $10$ $\times$ $3.5 – 4$ $\times$ $\frac{1}{3}$ $\times$ $\frac{22}{7}$ $\times$ $\left(\frac{1}{2}\right)^{2}$ $\times$ 1.4
$= 525 – 1.47$
$= 523.53\ cm^3$
Hence, the volume of wood in the entire stand $= 523.53\ cm^3$

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Question 42 Marks

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder as shown in figure. If the height of the cylinder is $10\ cm$ and its base is of radius $3.5\ cm$, Find the total surface area of the article.

Answer

$TSA$ of the article $=2 \pi \mathrm{rh}+2\left(2 \pi \mathrm{r}^2\right) $
$ =2 \pi(3.5)(10)+2\left[2 \pi(3.5)^2\right] $
$ =70 \pi+49 \pi $
$ =119 \pi $
$ =119 \times \frac{22}{7} $
$ =374 \mathrm{~cm}^2$

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Question 52 Marks

Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end (see Fig.). The height of the cylinder is $1.45\ m$ and its radius is $30\ cm$. Find the total surface area of the bird-bath.
(Take $\pi = \frac{22}{7}$ )

Answer

Let $h$ be height of the cylinder, and $r$ the common radius of the cylinder and hemisphere.
Then, the total surface area of the bird-bath $= CSA$ of cylinder + $CSA$ of hemisphere
$=2 \pi r h+2 \pi r^{2}=2 \pi r(h+r)$
$=2 \times \frac{22}{7} \times 30(145+30) \mathrm{cm}^{2}$
$= 33000\ cm^2= 3.3\ m^2$

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Question 62 Marks

2 cubes each of volume 64 cm³are joined end to end. Find the total surface are the resulting cuboid.

Answer

The volume of a cube is $V=a^3$.
Given $V=64 cm^3$ the side length (a) is:
$a=\sqrt[3]{64}=4 cm$
When two cubes of side 4 cm are joined end-to-end, the new dimensions are:
Length (L) = 4 cm+4 cm=8 cm
Width (W) = 4 cm
Height (H) = 4 cm
Calculate the total surface area:
The total surface area (TSA) of a cuboid is 2(LW+LH+WH).
$\begin{array}{l}T S A=2(8 \times 4+8 \times 4+4 \times 4) \\ T S A=2(32+32+16) \\ T S A=2(80)=160 cm^2\end{array}$

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