Sign in
Please select an account to continue using cracku.in
↓ →
A cistern, open at the top. is to be lined with sheet lead which weights $$27 kg/m^3$$". The cistern is 4.5 m long and 3 m wide and holds 50 $$m^3$$'. The weight of lead required is
Let the depth of the cistern be h metersΒ
ThenΒ 4.5Γ3ΓhΒ = 50Β
So, h =Β 50/13.5
h = 100/27
Area of sheet required = lb + 2 (bh + lh)Β
= lb + 2h(l + b)
Β =Β [4.5Γ3+2Γ100/27(4.5+3)]Β sq mΒ
Β =Β (13.5+200/27Γ7.5)
Β =Β (27/2+500/9)
Β =Β 1243/18
Therefore weight of lead =Β (27Γ1243/18)
=Β Β 3729/2Β kgΒ
= 1864.5
Create a FREE account and get:
Educational materials for CAT preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.