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 B

Explanation:

In this type of question, first we will calculate the volume of water displaces then will multiply with the density of water.
Volume of water displaced = 3*2*0.01 = 0.06 m cube

Mass of Man = Volume of water displaced * Density of water

= 0.06 * 1000 = 60 kg

Answer: Option C

Explanation:

Let the radius of hemisphere and cone be R,
Height of hemisphere H = R.
So the height of the cone = height of the hemisphere = R
Slant height of the cone
\begin{aligned}
= \sqrt{R^2+R^2} \\
= \sqrt{2}R \\
\frac{\text{Hemisphere Curved surface area}}{\text{Cone Curved surface area}} = \\
\frac{2\pi R^2}{\pi *R*\sqrt{2}R} \\
= \sqrt{2}:1
\end{aligned}

Answer: Option C

Explanation:

Let the length of the wire be h
\begin{aligned}
Radius = \frac{1}{2}mm = \frac{1}{20}cm\\
\pi r^2h = 66 \\
\frac{22}{7}*\frac{1}{20}*\frac{1}{20}*h = 66 \\
=> h = \frac{66*20*20*7}{22} \\
= 8400 cm \\
= 84 m
\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}

Answer: Option C

Explanation:

We will first calculate the Surface area of cube, then we will calculate the quantity of paint required to get answer.
Here we go,

\begin{aligned}
\text{Surface area =}6a^2 \\
= 6 * 8^2 = 384 \text{sq feet} \\
\text{Quantity required =}\frac{384}{16} \\
= 24 kg\\
\text{Cost of painting =} 36.50*24 \\
= Rs. 876
\end{aligned}