12 spheres of the same size are made from melting a solid cylinder of 16 cm diameter and 2 cm height. Find the diameter of each sphere.

Answer: Option B

Explanation:

Area of the wet surface =
2[lb+bh+hl] - lb = 2 [bh+hl] + lb
= 2[(4*1.25+6*1.25)]+6*4 = 49 m square

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}

Answer: Option B

Explanation:

Please note we are talking about "Hollow" ball. Do not ignore this word in this type of question in a hurry to solve this question.
If we are given with external radius and thickness, we can get the internal radius by subtracting them. Then the volume of metal can be obtained by its formula as,

External radius = 3 cm,
Internal radius = (3-0.5) cm = 2.5 cm

\begin{aligned}
\text{Volume of sphere =}\frac{4}{3}\pi r^3 \\
= \frac{4}{3}*\frac{22}{7}*[3^2 - 2.5^2]cm^3 \\
= \frac{4}{3}*\frac{22}{7}*\frac{91}{8}cm^3 \\
= \frac{143}{3} cm^3 \\
= 47\frac{2}{3}cm^3
\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:

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}