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Question 56
A cistern, open at the top, is to be lined with lead sheet which weighs 27 kg/m$$^2$$. The cistern is 4.5 m long and 3 m wide and holds 50 m$$^3$$. The weight of lead required is ............
Solution
Given that the volume of the container is 50, length = 4.5, and breadth = 3.
=> height of the container = 50 / 4.5 * 3 = 3.7 [nearly]
We need to find the surface area of the container [open top]
=> lb + 2(lb + bh) = 69.
Now the weight of lead required = 69 *Â 27 = 1863 [Option-B].
