le 14: An open metal bucket is in the
shape of a frustum of a cone, mounted on a
hollow cylindrical base made of the same me-
tallic sheet (see Fig. 13.23). The diameters of
the two circular ends of the bucket are 45 cm
and 25 cm, the total vertical height of the bucket
is 40 cm and that of the cylindrical base is
6 cm. Find the area of the metallic sheet used
to make the bucket, where we do not take into
account the handle of the bucket. Also, find
the volume of water the bucket can hold.
(Take n = 22)
Answers
Solution:
Given: height of bucket=40 cm.
height of frustum of cone=40-6=34 cm
slant height(l)= √h^2+(R-r)^2
Here R=22.5 cm
and, r=12.5 cm
h=34 cm
l= √34^2+(22.5-12.5)^2
l=√34^2+10^2
l= 35.44 cm
Area of metallic sheet used
= CSA of frustum+area of circular base
={π×35.44(22.5+12.5)+π×(12.5)^2+2π×12.5×6}
=22/7(1240.4+156.25+150)cm^2
=4860.9 cm^2
Now, volume of water that the bucket can hold= π×h/3×(R^2+r^2+R×r)
= 22/7×34/3×{(22.5)^2 + (12.5)^2 +22.5×12.5}cm^3
=22/7 × 34/3 × 943.75
=33615.48 cm^3 i.e 33.62litres(approx.)
Hence,
Area of metallic sheet used =4860.9cm^2
Volume of water that bucket holds
=33.62l
Hope it helps you❤❤❤
Answer:
Step-by-step explanation:
Solution:
Given: height of bucket=40 cm.
height of frustum of cone=40-6=34 cm
slant height(l)= √h^2+(R-r)^2
Here R=22.5 cm
and, r=12.5 cm
h=34 cm
l= √34^2+(22.5-12.5)^2
l=√34^2+10^2
l= 35.44 cm
Area of metallic sheet used
= CSA of frustum+area of circular base
={π×35.44(22.5+12.5)+π×(12.5)^2+2π×12.5×6}
=22/7(1240.4+156.25+150)cm^2
=4860.9 cm^2
Now, volume of water that the bucket can hold= π×h/3×(R^2+r^2+R×r)
= 22/7×34/3×{(22.5)^2 + (12.5)^2 +22.5×12.5}cm^3
=22/7 × 34/3 × 943.75
=33615.48 cm^3 i.e 33.62litres(approx.)
Hence,
Area of metallic sheet used =4860.9cm^2
Volume of water that bucket holds
=33.62l
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