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

Answer: Option D

Explanation:

To solve this type of question, simply divide the volume of wall with the volume of brick to get the numbers of required bricks
So lets solve this

Number of bricks =
\begin{aligned}
\frac{\text{Volume of wall}}{\text{Volume of 1 brick}} \\
= \frac{800*600*22.5}{25*11.25*6} \\
= 6400
\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}

Answer: Option B

Explanation:

Let the radius of the base be r km. Then,
\begin{aligned}
\pi r^2 = 1.54 \\
r^2 = \frac{1.54*7}{22} = 0.49\\
= 0.7 km \\
\text{Now l=2.5 km, r = 0.7 km} \\
h = \sqrt{2.5^2 - 0.7^2} km \\
=\sqrt{6.25 - 0.49}\\
=\sqrt{5.76} km \\
= 2.4 km
\end{aligned}

Answer: Option A

Explanation:

In this type of question, just equate the two volumes to get the answer as,
\begin{aligned}
\text{Volume of cylinder =}\pi r^2h\\
\text{Volume of sphere =} \frac{4}{3}\pi r^3\\
=> 12*\frac{4}{3}\pi r^3 = \pi r^2h \\
=> 12*\frac{4}{3}\pi r^3 = \pi *8*8*2 \\
=> r^3 = \frac{8*8*2*3}{12*4} \\
=> r^3 = 8 \\
=> r = 2 cm \\
=> \text{Diameter =}2*2 = 4 cm
\end{aligned}

Answer: Option A

Explanation:

Let the drop in the water level be h cm, then,

\begin{aligned}
\text{Volume of cylinder= }\pi r^2h \\
=> \frac{22}{7}*\frac{35}{2}*\frac{35}{2}*h = 11000 \\
=> h = \frac{11000*7*4}{22*35*35}cm\\
= \frac{80}{7}cm\\
= 11\frac{3}{7} cm
\end{aligned}