Ex 12.2, 5 - Chapter 12 Class 10 Surface Areas and Volumes

Last updated at April 16, 2024 by Teachoo

Transcript

Ex 12.2, 5
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Number of lead shots = (𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒘𝒂𝒕𝒆𝒓 𝒇𝒍𝒐𝒘𝒏 𝒐𝒖𝒕)/(𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝟏 𝒍𝒆𝒂𝒅 𝒔𝒉𝒐𝒕)
Volume of water flown out
Volume of water flown out = 𝟏/𝟒 × Volume of cone
Given
Radius of cone = r = 5 cm
Height of cone = h = 8 cm
Volume of cone = 1/3 𝜋𝑟^2 ℎ
= 1/3×22/7×5×5×8
= 𝟒𝟒𝟎𝟎/𝟐𝟏 cm3
Thus,
Volume of water flown out = 1/4 × Volume of cone
= 𝟏/𝟒 × 𝟒𝟒𝟎𝟎/𝟐𝟏
= 1100/21 cm3
Volume of 1 lead shot
Lead shot is in from of sphere with
Radius = r = 0.5 cm
Volume of 1 lead shot = Volume of sphere
= 𝟒/𝟑 𝛑𝐫𝟑
= 4/3×22/7×(0.5)3
= 𝟏𝟏/𝟐𝟏 cm3
Now,
Number of lead shots = (𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑓𝑙𝑜𝑤𝑛 𝑜𝑢𝑡)/(𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 1 𝑙𝑒𝑎𝑑 𝑠ℎ𝑜𝑡)
= (𝟏𝟏𝟎𝟎/𝟐𝟏)/(𝟏𝟏/𝟐𝟏)
= 1100/21 × 21/11
= 100
Hence there are 100 such lead shots.

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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