A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. From each of its corners, a square is cut off so as to make an open box. If the length of the square is 8 m, the volume of the box (in m cube) is:

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 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:

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 C

Explanation:

Volume of the largest cone = Volume of the cone with diameter of base 7 and height 7 cm

\begin{aligned}
\text{Volume of cone =}\frac{1}{3}\pi r^2h \\
= \frac{1}{3}*\frac{22}{7}*3.5*3.5*7 \\
= \frac{269.5}{3}cm^3 \\
= 89.8 cm^3
\end{aligned}

Note: radius is taken as 3.5, as diameter is 7 cm

Answer: Option D

Explanation:

We will first subtract the cone volume from wood volume to get the wood wasted.
Then we can calculate its percentage.

\begin{aligned}
\text{Sphere Volume =}\frac{4}{3}\pi r^3 \\
\text{Cone Volume =}\frac{1}{3}\pi r^2h\\

\text{Volume of wood wasted =}\\
\left(\frac{4}{3}\pi *9*9*9\right)-\left(\frac{1}{3}\pi *9*9*9\right) \\
= \pi *9*9*9 cm^3 \\
\text{Required Percentage =} \\
\frac{\pi *9*9*9}{\frac{4}{3}\pi *9*9*9}*100 \% \\
= \frac{3}{4}*100 \% \\
= 75\%
\end{aligned}