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06 May 2019, 21:29
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
64% (01:38) correct
36%
(01:59)
wrong
based on 25
sessions
History
Date
Time
Result
Not Attempted Yet
A cylindrical rod of iron, whose height is equal to its radius, is melted and cast intospherical balls whose radius is half the radius of the rod. Find the number of balls?
a. 2
b. 3
c. 4
d. 5
e. 6
Posted from my mobile device
a. 2
b. 3
c. 4
d. 5
e. 6
Posted from my mobile device
08 May 2019, 08:30
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Hello,
Greetings for the day!
This is a fairly simple problem on Mensuration. The concept being tested in this question is that of calculation of Volume of 3-D objects. You need to be familiar with the formulas related to the Volume and Surface Area of 3-D objects, if you want to be able to solve this question.
The volume of a cylinder is given by π*r^2*h, where ‘r’ is the radius of the base of the cylinder and ‘h’ is the height of the cylinder.
The volume of a sphere is given by (4/3) * π * r^3 where r is the radius of the sphere.
Let us have a look at the question now. We have been told that the height of the cylinder is equal to its radius. Therefore, h = r. Hence,
Volume of the cylindrical rod = π* r^2*r = π * r^3.
Let us assume the radius of each spherical ball as R. Then, the volume of one spherical ball will be (4/3) π *R^3. We also know that R = (r/2).
On melting the cylindrical rod and casting them into spherical balls, the one thing that does not change is the volume of the material. Therefore, we can say,
Volume of cylindrical rod = Volume of ONE spherical ball * Number of spherical balls.
So, π * r^3 = (4/3) π * R^3 * N (let’s say).
Substituting the value of R and simplifying, we get N = 6.
The correct answer option, therefore, is E.
In mensuration questions like these, it is important to figure out the one thing that DOES NOT change and then use this idea to frame an equation. Once this happens, everything else falls into place. As said earlier, it’s also a good idea to have the basic formulas memorized so that you don’t struggle in framing the equations.
Hope this helps!
Cheers!
Arvind,
CrackVerbal Prep Team
Greetings for the day!
This is a fairly simple problem on Mensuration. The concept being tested in this question is that of calculation of Volume of 3-D objects. You need to be familiar with the formulas related to the Volume and Surface Area of 3-D objects, if you want to be able to solve this question.
The volume of a cylinder is given by π*r^2*h, where ‘r’ is the radius of the base of the cylinder and ‘h’ is the height of the cylinder.
The volume of a sphere is given by (4/3) * π * r^3 where r is the radius of the sphere.
Let us have a look at the question now. We have been told that the height of the cylinder is equal to its radius. Therefore, h = r. Hence,
Volume of the cylindrical rod = π* r^2*r = π * r^3.
Let us assume the radius of each spherical ball as R. Then, the volume of one spherical ball will be (4/3) π *R^3. We also know that R = (r/2).
On melting the cylindrical rod and casting them into spherical balls, the one thing that does not change is the volume of the material. Therefore, we can say,
Volume of cylindrical rod = Volume of ONE spherical ball * Number of spherical balls.
So, π * r^3 = (4/3) π * R^3 * N (let’s say).
Substituting the value of R and simplifying, we get N = 6.
The correct answer option, therefore, is E.
In mensuration questions like these, it is important to figure out the one thing that DOES NOT change and then use this idea to frame an equation. Once this happens, everything else falls into place. As said earlier, it’s also a good idea to have the basic formulas memorized so that you don’t struggle in framing the equations.
Hope this helps!
Cheers!
Arvind,
CrackVerbal Prep Team
