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Momet of Inertia (M.I) of four bodies having same mass $\mathrm{'M'}$ and radius $\mathrm{'2R'}$ are os follows :

$\mathrm{I_{1}=M.I}$ of solid spehere about it's diameter

$\mathrm{I_{2}=M.I}$ of solid cylinder about it's axis

$\mathrm{I_{3}=M.I}$ of solid circular disc about it's daimeter

$\mathrm{I_{4}=M.I}$ of thin circular ring about it's diameter

If $\mathrm{2\left ( I_{2}+I_{3} \right )+I_{4}=x\cdot I_{1}}$ then the value of $\mathrm{x}$ will be____________.
Option: 1
5

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{I_{1}=\frac{2}{5} M(2 R)^{2}=\frac{8 M R^{2}}{5}} \\$

$\mathrm{I_{2}=\frac{M(2 R)^{2}}{2}=2 M R^{2}} \\$

$\mathrm{I_{3}=\frac{M(2 R)^{2}}{4}= MR^{2}}$

$\mathrm{I_{4}=\frac{M(2 R)^{2}}{2}=\frac{4 M R^{2}}{2}=2 M R^{2}}$

$\mathrm{2\left(I_{2}+I_{3}\right)+I_{4}=x I_{1}} \\$

$\mathrm{6 M R^{2}+2 M R^{2}=x \frac{\left(8 M R^{2}\right)}{5}} \\$

$\mathrm{8 M R^{2}=x\frac{\left(8 M R^{2}\right)}{5}} \\$

$\mathrm{x=5}$

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Riya

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