12. A right circular cylindrical flask has outer diameter 6 cm and inner diameter as 5 cm. and height without a lid is 28cm. A lid of the flask is of height 5cm with the same radius so as to give the flask a shape of right circular cylinder. a. Find the capacity of the flask (the lid does not hold water.) b. Find amount of paper needed to wrap it for gifting.
Pls answer
Answers
Given,
Radius of the spherical ball = 3 cm
We know that,
The volume of the sphere = 4/3 πr3
So, it’s volume (V) = 4/3 πr3
That the ball is melted and recast into 3 spherical balls.
Volume (V1) of first ball = 4/3 π 1.53
Volume (V2) of second ball = 4/3 π23
Let the radius of the third ball = r cm
Volume of third ball (V3) = 4/3 πr3
Volume of the spherical ball is equal to the volume of the 3 small spherical balls.
R D Sharma Solutions For Class 10 Maths Chapter 16 Surface Areas And Volumes ex 16.1 - 1
Now,
Cancelling out the common part from both sides of the equation we get,
(3)3 = (2)3 + (1.5)3 + r3
r3 = 33– 23– 1.53 cm3
r3 = 15.6 cm3
r = (15.6)1/3 cm
r = 2.5 cm
As diameter = 2 x radius = 2 x 2.5 cm
= 5.0 cm.
Thus, the diameter of the third ball is 5 cm
Hope it helps you ☺️❤️
The diameter of the third ball is 5 cm
Step-by-step explanation:
Given,
Radius of the spherical ball = 3 cm
We know that,
The volume of the sphere = 4/3 πr³
So, it’s volume (V) = 4/3 πr³
That the ball is melted and recast into 3 spherical balls.
➳Volume (V1) of first ball = 4/3 π 1.5³
➳Volume (V2) of second ball = 4/3 π2³
= Let the radius of the third ball = r cm
Volume of third ball (V3) = 4/3 πr³
Volume of the spherical ball is equal to the volume of the 3 small spherical balls.
Now
Cancelling out the common part from both sides of the equation we get,
➳(3)³ = (2)³ + (1.5)³ + r³
➳r3 = 3³– 2³– 1.5³ cm
➳r3 = 15.6 cm
➳r = (15.6)1/3 cm
➳r = 2.5 cm
As diameter = 2 x radius = 2 x 2.5 cm
= 5.0 cm.
Thus, the diameter of the third ball is 5 cm