A) \[{{l}_{Y}}=32\,{{l}_{X}}\]
B) \[{{l}_{Y}}=16\,{{l}_{X}}\]
C) \[{{l}_{Y}}=\,{{l}_{X}}\]
D) \[{{l}_{Y}}=\,64{{l}_{X}}\]
Correct Answer: D
Solution :
| [d] Mass of disc (X), \[{{m}_{x}}=\pi {{R}^{2}}t\rho \] |
| \[(m=v\rho =At\rho =\pi {{R}^{2}}-\rho )\] |
| where, \[\rho =\] density of material of disc |
| \[\therefore \] \[{{l}_{x}}=\frac{1}{2}{{m}_{x}}\,{{R}^{2}}=\frac{1}{2}\pi {{R}^{2}}t\rho {{R}^{2}}\] |
| \[{{l}_{x}}=\frac{1}{2}\pi \rho t{{R}^{4}}\] ... (i) |
| Mass of disc (Y) |
| \[{{m}_{Y}}=\pi {{(4R)}^{2}}\frac{t}{4}\rho =4\pi {{R}^{2}}t\rho \] |
| and \[{{l}_{Y}}=\frac{1}{2}{{m}_{Y}}{{(4R)}^{2}}=\frac{1}{2}4{{R}^{2}}t\rho .\,16{{R}^{2}}\] |
| \[\Rightarrow \] \[{{l}_{Y}}\,=32\,\pi t\rho {{R}^{4}}\] ... (ii) |
| \[\therefore \] \[\frac{{{l}_{Y}}}{{{l}_{X}}}=\frac{32\,\pi t\rho {{R}^{4}}}{\frac{1}{2}\pi \rho t{{R}^{4}}}=64\] |
| \[\therefore \] \[{{l}_{Y}}\,=64\,{{l}_{X}}\] |
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