An iron pillar consists of a cylindrical portion 2.8 m high and 20 cm in diameter and a cone 42 cm high is surmounting it. Find the weight of the pillar, given that 1 cubic cm of iron weighs 7.5 gm.
Answers
Answer:
The weight of the pillar is 693 kg.
Step-by-step explanation:
SOLUTION :
Given :
Diameter of a Cylinder = 20 cm.
Radius of a cylinder = 20/2 = 10 cm.
Height of cylinder ,h = 2.8 m = 2.8 × 100 = 280 cm.
[1 m = 100 cm]
Height of cone , H = 42 cm.
Volume of pillar,V = Volume of cylindrical portion + Volume of conical portion
V = (πr²h + 1/3πr²H)
V = πr²(h + 1/3H)
V = (22/7) × (10)² × (280 + 1/3 × 42)
V = (22/7 × 100) × ( 280 + 14)
V = (2200/7) × (280 + 14)
V = (2200/7) × 294
V = 2200 × 42
Volume of pillar ,V = 92400 cm³
Weight of 1 cm³ of iron = 7.5 g (Given)
Weight of 92400 cm³ iron pillar = (92400 × 7.5)/1000 kg = 693000/1000 = 693 kg
Hence, the weight of the pillar is 693 kg.
HOPE THIS ANSWER WILL HELP YOU….
ANSWER:----------------
H() = 2.8m = 280cm
R () = diameter/2 = 20/2cm = 10cm
see as follows:--
[note/ ·…{volume of cylinder }[=]=>{ πr²h}
={ 22/7 <> (10)² <>. 280}
={ 22 <> 100 40}
={ 88000cm³}
radius and night ={the follows}
(R) = 10cm
(H) = 42cm
volume of cone = 1/3{=(πR²H)
= {1/3 [ (22/7 ] (10)² { 42)
= {1/3 {22 .. 100 .. 6)
= 1/3 6 * 2200
={ 2 ={2200}
= {4400cm³}
,{weight of iron }{=}{ volume of cylinder + volume of cone}
= {88000cm³[] +[] 4400cm³}
={ 92400cm³}
{ 1cm³ = 7.5g]
={ 92400 = 7.5g}
={ 693000g}
if the weigh for [1kg = 1000g]
{weight of iron = 693kg}
hope it helps:--
T!—!ANKS!!!