
A right circular cone with a base diameter of 3 cm and height of 6 cm is cut from a solid cylinder of diameter 5 cm and height of 12cm. Find the position of the center of gravity of the rest of the body.
Answer
163.5k+ views
Hint: In order to solve this problem we need to know about the center of gravity of the cylinder. For a cylinder center of gravity lies at the midpoint of the axis of the cylinder. That is nothing but the total surface area of the remaining solid.
Formula used: To find the slant height of the conical part the formula is,
\[l = \sqrt {{r^2} + {h^2}} cm\]
Where,
r is radius
h is height
Complete step by step solution: Height of the cylindrical part = height of the conical part \[ = 12cm\]
The diameter of the cylindrical part \[ = 5cm\]
Therefore, the radius of the cylindrical part \[ = \dfrac{5}{2}\]\[ = 2.5cm\]
The slant height of the conical part \[ = \sqrt {{r^2} + {h^2}} cm\]
\[ = \sqrt {{{\left( {2.5} \right)}^2} + {{\left( {12} \right)}^2}} \]
\[ = \sqrt {({0.25}) + ({144})} \]
\[ = \sqrt {150.25} \]
\[ = 12.257cm\]
We need to find the position of center of gravity for the rest of the body. For a cylinder center of gravity lies at the midpoint of the axis of the cylinder. That is nothing but the total surface area of the remaining solid.
Then the total surface area of the remaining solid = CSA of cylindrical part + CSA of conical part + base area of the circular part.
\[ = 2\pi rh + \pi rl + \pi {r^2}\]
Here, \[r = 2.5cm\], \[h = 12cm\], \[l = 12.25cm\]
Substitute the value in above equation we get,
\[ = \left( {2\pi \times 2.5 \times 12} \right) + \left( {\pi \times 2.5 \times 12.25} \right) + \pi {\left( {2.5} \right)^2}\]
\[ = \left( {2 \times 3.142 \times 2.5 \times 12} \right) + \left( {3.142 \times 2.5 \times 12.25} \right) + 3.142 \times {\left( {2.5} \right)^2}\]
\[ = 188.52 + 96.223 + 19.637\]
\[ = 304.308c{m^2}\]
Therefore, the total surface area of the remaining solid is \[304.308c{m^2}\]
Note: Here in the given problem it is important to remember that the equation for the center of gravity of a cylinder. That is nothing but the total surface area of the remaining solid. So, we need to find the total surface area of remaining solid.
Formula used: To find the slant height of the conical part the formula is,
\[l = \sqrt {{r^2} + {h^2}} cm\]
Where,
r is radius
h is height
Complete step by step solution: Height of the cylindrical part = height of the conical part \[ = 12cm\]
The diameter of the cylindrical part \[ = 5cm\]
Therefore, the radius of the cylindrical part \[ = \dfrac{5}{2}\]\[ = 2.5cm\]
The slant height of the conical part \[ = \sqrt {{r^2} + {h^2}} cm\]
\[ = \sqrt {{{\left( {2.5} \right)}^2} + {{\left( {12} \right)}^2}} \]
\[ = \sqrt {({0.25}) + ({144})} \]
\[ = \sqrt {150.25} \]
\[ = 12.257cm\]
We need to find the position of center of gravity for the rest of the body. For a cylinder center of gravity lies at the midpoint of the axis of the cylinder. That is nothing but the total surface area of the remaining solid.
Then the total surface area of the remaining solid = CSA of cylindrical part + CSA of conical part + base area of the circular part.
\[ = 2\pi rh + \pi rl + \pi {r^2}\]
Here, \[r = 2.5cm\], \[h = 12cm\], \[l = 12.25cm\]
Substitute the value in above equation we get,
\[ = \left( {2\pi \times 2.5 \times 12} \right) + \left( {\pi \times 2.5 \times 12.25} \right) + \pi {\left( {2.5} \right)^2}\]
\[ = \left( {2 \times 3.142 \times 2.5 \times 12} \right) + \left( {3.142 \times 2.5 \times 12.25} \right) + 3.142 \times {\left( {2.5} \right)^2}\]
\[ = 188.52 + 96.223 + 19.637\]
\[ = 304.308c{m^2}\]
Therefore, the total surface area of the remaining solid is \[304.308c{m^2}\]
Note: Here in the given problem it is important to remember that the equation for the center of gravity of a cylinder. That is nothing but the total surface area of the remaining solid. So, we need to find the total surface area of remaining solid.
Recently Updated Pages
Trigonometry Formulas: Complete List, Table, and Quick Revision

Difference Between Distance and Displacement: JEE Main 2024

IIT Full Form

Uniform Acceleration - Definition, Equation, Examples, and FAQs

Difference Between Metals and Non-Metals: JEE Main 2024

Newton’s Laws of Motion – Definition, Principles, and Examples

Trending doubts
JEE Main Marks Vs Percentile Vs Rank 2025: Calculate Percentile Using Marks

JEE Mains 2025 Cutoff: Expected and Category-Wise Qualifying Marks for NITs, IIITs, and GFTIs

NIT Cutoff Percentile for 2025

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025 CutOff for NIT - Predicted Ranks and Scores

Other Pages
NCERT Solutions for Class 10 Maths Chapter 13 Statistics

NCERT Solutions for Class 10 Maths Chapter 11 Areas Related To Circles

NCERT Solutions for Class 10 Maths Chapter 12 Surface Area and Volume

NCERT Solutions for Class 10 Maths Chapter 14 Probability

NCERT Solutions for Class 10 Maths In Hindi Chapter 15 Probability

Total MBBS Seats in India 2025: Government and Private Medical Colleges
