The data shows volume and height of 4 right circular tanks P, Q, R and S. The given data also depicts time taken to fill or empty the tank by inlet pipe A, B and outlet pipe C.

1. Ratio between time taken by pipe B and that by pipe C is different because pipes were used by the operator in different ways for different tanks.
2. Radius of tank S is 10.5 m.

Tank	Volume	Height	Time taken by pipe A to fill tank (in hours)	Ratio of time taken by pipe B to fill and pipe C to empty the tank
P	38808	28	10	2: 3
Q	--------	10	12	3: 2
R	5390	35	18	1: 3
S	19250	--------	---------	2: 5

Find respective ratio of curved surface area of tank P and total surface area of tank R.

The correct option is 2: 1.

Step 1: Find the radius and Curved Surface Area (CSA) of Tank P
We are given the volume of Tank P as 38808 m³ and its height ($h_P$) as 28 m.
The formula for the volume of a right circular cylinder is:
$\mathrm{Volume}=\pi r_P^2{h}_P$
Substituting the given values:
$38808=\frac{22}{7}$
$38808=22\times 4\times r_P^2$
$38808=88\times r_P^2$
$r_P^2=\frac{38808}{88}=441$
$r_P=\sqrt{441}=21\text{m}$

Now, calculate the Curved Surface Area (CSA) of Tank P using the formula:
${\mathrm{CSA}}_P=2\pi r_P{h}_P$
${\mathrm{CSA}}_P=2\times \frac{22}{7}$
${\mathrm{CSA}}_P=2\times 22\times 3\times 28=3696{\text{m}}^2$

Step 2: Find the radius and Total Surface Area (TSA) of Tank R
We are given the volume of Tank R as 5390 m³ and its height ($h_R$) as 35 m.
Using the volume formula:
$5390=\frac{22}{7}$
$5390=22\times 5\times r_R^2$
$5390=110\times r_R^2$
$r_R^2=\frac{5390}{110}=49$
$r_R=\sqrt{49}=7\text{m}$

Now, calculate the Total Surface Area (TSA) of Tank R using the formula:
${\mathrm{TSA}}_R=2\pi r_R(h_R+r_R)$
${\mathrm{TSA}}_R=2\times \frac{22}{7}$
${\mathrm{TSA}}_R=44\times 42=1848{\text{m}}^2$

Step 3: Calculate the required ratio
We need to find the ratio of ${\mathrm{CSA}}_P$ to ${\mathrm{TSA}}_R$:
$\mathrm{Ratio}=\frac{{\mathrm{CSA}}_P}{{\mathrm{TSA}}_R}=\frac{3696}{1848}$
Simplifying the fraction:
$\mathrm{Ratio}=\frac{2}{1}=2:1$