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 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 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 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 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 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}