A)
B)
C)
D)
Correct Answer: B
Solution :
[b]| We know that \[\left| {\vec{\tau }} \right|=\left| \frac{d\vec{L}}{dt} \right|\] | ||
| Where \[L=I\omega \] | ||
| \[\therefore \tau =\frac{d}{dt}\left( I\omega \right)=\omega \frac{dI}{dt}\] | ?(i) | |
| From the situation it is clear that the moment of inertia for (road +insect) system is increasing. | ||
| Let at any instant of time \['t',\]the insect is at a distance \[x\]from \[O.\]at this instant, the moment of inertia of the system is | ||
| \[I=\frac{1}{3}M{{L}^{2}}+m{{x}^{2}}\] | ?(ii) | |
| Where \[M=\]mass of the rod | ||
| \[L=\]Length of the rod | ||
| \[m=\]Mass of the insect | ||
| From (i) & (ii) | ||
| \[\tau =\omega \frac{d}{dt}\left[ \frac{1}{3}M{{L}^{2}}+m{{x}^{2}} \right]=\omega \,m\frac{d}{dt}({{x}^{2}})\] | ||
| \[=2\omega mx\frac{dx}{dt}=2\omega mxv=2\omega m{{v}^{2}}t[\therefore x=vt]\] | ||
| \[\therefore \tau \propto t\] (till t=T) | ||
| When the insect stops moving, \[\overrightarrow{L}\] does not change and therefore \[\tau \]becomes constant. | ||
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