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MAT Exam 2021 Paper

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Question : 83 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 open cistern is lined only on the bottom and its four side walls, not on the top.

Explanation:

Let the height of the cistern be

h

.

Volume

= 4.5 \times 3 \times h = 50

So,

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

m (approx).

Area to be lined

=

bottom area

+

area of four walls

= 4.5 \times 3 + 2(4.5 + 3) \times 3.704

= 13.5 + 2 \times 7.5 \times 3.704

= 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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