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.

Answer: Option C

Explanation:

\begin{aligned}
Volume = \pi r^2h \\
Volume = \left(\frac{22}{7}*1*1*14\right)m^3 \\
= 44 m^3
\end{aligned}

Answer: Option A

Explanation:

Let their radii be 2x, x and their heights be h and H resp.
Then,
\begin{aligned}
\text{Volume of cone =}\frac{1}{3}\pi r^2h \\
\frac{\frac{1}{3}*\pi *{(2x)}^2*h}{\frac{1}{3}*\pi *{x}^2*H} \\
=> \frac{h}{H} = \frac{1}{4} \\
=> h:H = 1:4
\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}

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}

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}