Question Detail

A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weights 8g/cm cube, then find the weight of the pipe.

3.696 kg
3.686 kg
2.696 kg
2.686 kg

Answer: Option A

Explanation:

In this type of question, we need to subtract external radius and internal radius to get the answer using the volume formula as the pipe is hollow. Oh! line become a bit complicated, sorry for that, lets solve it.

External radius = 4 cm
Internal radius = 3 cm [because thickness of pipe is 1 cm]

\begin{aligned}
\text{Volume of iron =}\pi r^2h\\
= \frac{22}{7}*[4^2 - 3^2]*21 cm^3\\
= \frac{22}{7}*1*21 cm^3\\
= 462 cm^3 \\
\end{aligned}
Weight of iron = 462*8 = 3696 gm
= 3.696 kg

1. The surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each. The radius of the sphere is:

4 cm
6 cm
8 cm
10 cm

Answer: Option B

Explanation:

\begin{aligned}
\text{Curved surface area of sphere =}\\
\frac{4}\pi r^2 \\
\text{Surface area of cylinder =} \\
2\pi rh \\
=> \frac{4}\pi r^2 = 2\pi rh \\
=> r^2 = \frac{6*12}{2} \\
=> r^2 = 36 \\
=> r = 6
\end{aligned}

Note: Diameter of cylinder is 12 so radius is taken as 6.

2. There are bricks with 24 cm x 12 cm x 8 cm dimensions. Find the total number of bricks required to construct a wall 24 m long, 8 m high and 60 m thick with 10% of wall filled with mortar.

35000
40000
45000
50000

Answer: Option C

Explanation:

So as per question,
\begin{aligned}
\text{Volume of wall} = (2400 * 800 * 60 ) cm^3 \\
\text{Volume of bricks} = \text { 90% of volume of wall } \\
= [ \frac{90}{100} * 2400 * 800 * 60 ] cm^3 \\
\text{ Volume of 1 brick = } (24 * 12 * 8) cm^3 \\
\text{Number of Bricks required} \\
= \frac{ (\frac{90}{100}) * (2400 * 800 * 60)}{24 * 12 * 8} \\
= 45000
\end{aligned}

3. A cistern 6 m long and 4 m wide contains water up to a breadth of 1 m 25 cm. Find the total area of the wet surface.

42 m sqaure
49 m sqaure
52 m sqaure
64 m sqaure

Answer: Option B

Explanation:

Area of the wet surface =
2[lb+bh+hl] - lb = 2 [bh+hl] + lb
= 2[(4*1.25+6*1.25)]+6*4 = 49 m square

4. How many cubes of 10 cm edge can be put in a cubical box of 1 m edge.

10000 cubes
1000 cubes
100 cubes
50 cubes

Answer: Option B

Explanation:

\begin{aligned}
\text{Number of cubes =}\frac{100*100*100}{10*10*10} \\
= 1000
\end{aligned}

Note: 1 m = 100 cm

5. The maximum length of a pencil that can he kept is a rectangular box of dimensions 8 cm x 6 cm x 2 cm, is

\begin{aligned} 2\sqrt{17} \end{aligned}
\begin{aligned} 2\sqrt{16} \end{aligned}
\begin{aligned} 2\sqrt{26} \end{aligned}
\begin{aligned} 2\sqrt{24} \end{aligned}

Answer: Option C

Explanation:

In this question we need to calculate the diagonal of cuboid,
which is =
\begin{aligned}
\sqrt{l^2+b^2+h^2} \\
= \sqrt{8^2+6^2+2^2} \\
= \sqrt{104} \\
= 2\sqrt{26}
\end{aligned}