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A cistern 6┬аm long and 4┬аm wide contains water up to a depth of 1┬аm┬а25┬аcm. The total area of the wet surface is:
50┬аm2
53.5┬аm2
55┬аm2
49┬аm2
49┬аm2
Calculate the area of the base (6├Ч4=24) and add the perimeter of the base (2├Ч(6+4)=20) multiplied by the depth (1.25); result is 24+20├Ч1.25=24+25=49.
Length of cistern l = 6 m, width b = 4 m, depth of water h = 1 m 25 cm = 1.25 m.
AreawetтАЛ=(l├Чb)+2(l├Чh)+2(b├Чh)
Calculate the area of the base (6├Ч4=24) and add the perimeter of the base (2├Ч(6+4)=20) multiplied by the depth (1.25); result is 24+20├Ч1.25=24+25=49.
Students often mistakenly calculate the surface area of a closed box (using 2(lb+lh+bh)) instead of the wet surface area (which excludes the top surface).
Convert units
Convert the depth from meters and centimeters to meters: 1┬аm┬а25┬аcm=1+10025тАЛ=1.25┬аm.
h=1.25┬аm
Identify surface components
The wet surface consists of the base area plus the area of the four side walls up to the water level.
AreawetтАЛ=(l├Чb)+2(lh)+2(bh)
Calculate total area
Substitute the values l=6, b=4, and h=1.25 into the formula: 6├Ч4+2(6├Ч1.25)+2(4├Ч1.25).
AreawetтАЛ=24+15+10=49┬аm2
D is correct because the total wet surface area is 6├Ч4+2(6├Ч1.25)+2(4├Ч1.25)=49┬аm2.
This logic is identical to finding the inner paintable surface area of an open tank or the lateral surface area of a prism, frequently asked in RRB and SSC exams.
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