The internal surface area (in m2) of an open cuboidal tank whose internal length, internal breadth and internal height are given as 3 m, 11 m, and 9 m, respectively, is:

The correct answer is 285.

Step 1: Understand the dimensions of the tank
The internal dimensions of the open cuboidal tank are given as follows:
Length ($l$) = 3 m
Breadth ($b$) = 11 m
Height ($h$) = 9 m

Step 2: Formula for the internal surface area of an open cuboidal tank
Since the tank is open at the top, its total internal surface area consists of the area of the base plus the areas of the 4 vertical walls (lateral surface area).
Area of base = $l\×b$
Area of 4 walls = $2\×h\×(l+b)$

Therefore, the total internal surface area ($A$) is given by:

$A=(l\×b)+2\×h\×(l+b)$

Step 3: Substitute the given values into the formula

$A=(3\×11)+2\×9\×(3+11)$

Step 4: Perform the step-by-step calculations

Calculate the area of the base:

$3\×11=33$

Calculate the sum of length and breadth:

$3+11=14$

Calculate the area of the 4 walls:

$2\×9\×14=18\×14=252$

Add the base area and the wall area together:

$A=33+252=285$

Thus, the internal surface area of the open cuboidal tank is 285 m².