Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The solid cylinder of length 80 cm and mass M has a radius of 20 cm. Calculate the density of the material used if the moment of inertia of the cylinder about an axis CD parallel to AB as shown in figure is 2.7 kg m2 . <img alt="" src="h

The solid cylinder of length 80 cm and mass M has a radius of 20 cm. Calculate the density of the material used if the moment of inertia of the cylinder about an axis CD parallel to AB as shown in figure is 2.7 kg m2 .

Option: 1 1\cdot 49\times 10^{2} \frac{kg}{m^{3}}
Option: 2 7\cdot 5\times 10^{1} \frac{kg}{m^{3}}
Option: 3 14\cdot 9 \frac{kg}{m^{3}}
Option: 4 7\cdot 5\times 10^{2} \frac{kg}{m^{3}}

Answers (1)

best_answer

I_{CD}= I_{AB}+Md^{2}
         = \left ( \frac{MR^{2}}{2} \right )+M\left ( \frac{L}{2} \right )^{2}
2\cdot 7=\left (\frac{M\times0\cdot 04}{2} \right )+\left (\frac{M\times0\cdot 64}{4} \right )
10\cdot 8=0\cdot 08 \, M+0\cdot 64 \, M
10\cdot 8=0\cdot 72 \, M
M= \frac{10\cdot 8}{0\cdot 72}= 15\, kg
M= \rho \times\left ( \pi r^{2} L\right )
s= \frac{15}{\frac{22}{7}\times0\cdot 04\times0\cdot 8}
s= \frac{105}{22\times0\cdot 04\times0\cdot 8}= \frac{105\times 10^{+3}}{22\times 32}
= 0\cdot 149\times 10^{3}
s= 1\cdot 49\times 10^{2}\frac{kg}{m^{3}}  
The correct option is (1)

Posted by

vishal kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

73 or 99 percentile? JEE Main result questioned, J-K candidate says, ‘Don’t create hype’
anam-khan

JEE Main 2025: 11 from Kota coaching centres among 24 students with 100 percentile
anam-khan

Low JEE Main rank? Explore alternative paths for engineering admission

Latest Question