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

Answer: Option C

Explanation:

Volume will be length * breadth * height, but in this case two heights are given so we will take average,

\begin{aligned}
Volume = \left(12*9*\left(\frac{1+4}{2}\right)\right)m^3 \\
12*9*2.5 m^3 = 270 m^3
\end{aligned}

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

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.

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}

Answer: Option A

Explanation:

Let the edges be a and b of two cubes, then

\begin{aligned}
\frac{a^3}{b^3} = \frac{27}{1} \\
=> \left( \frac{a}{b} \right)^3 = \left( \frac{3}{1} \right)^3 \\
\frac{a}{b}=\frac{3}{1} \\
=> a:b = 3:1
\end{aligned}