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

Share

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 * 27 = 1863 [Option-B].

Was this solution helpful?

Get AI Help

View More Questions From Geometry

SRCC Quant Questions | SRCC Quantitative Ability

SRCC DILR Questions | LRDI Questions For SRCC

SRCC Verbal Ability Questions | VARC Questions For SRCC

Free SRCC Quant Questions

Book Free CAT Mentorship

Get personalized CAT strategy from a 99%iler

500+ students mentored
CAT mentor

Join CAT 2026 course by 5-Time CAT 100%iler

Start your IIM journey with the right preparation and crack CAT 2026.

Enroll Now
Ask AI