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MAT Exam 2020 Paper

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Question : 113 of 200

Marks: +1, -0

A cistern, open at the top, is to be lined with sheet lead which weights

27 \mathrm{~kg} / \mathrm{m}^{3}

. The cistern is 4.5 m long and 3 m wide, and holds

50 \mathrm{~m}^{3}

. The weight of lead required is

1764.60 kg
1864.62 kg
1660.62 kg
1860.62 kg

Solution:

Concept:

The cistern is open at the top, so the lead lining covers the bottom and the four side walls only.

Explanation:

Given: length

= 4.5

m, breadth

= 3

m, and capacity

= 50

m

^3

.

Height of cistern:

h = \dfrac{50}{4.5 \times 3} = \dfrac{100}{27} \approx 3.704

m

Area to be lined:

Bottom area

= 4.5 \times 3 = 13.5

m

^2

Side walls area

= 2h(l+b) = 2 \times 3.704 \times (4.5+3) = 55.56

m

^2

Total lining area

= 13.5 + 55.56 = 69.06

m

^2

Weight of lead required:

= 69.06 \times 27 = 1864.62

kg

Answer:

Option B:

1864.62

kg

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