A) 9.34cm
B) 2.5cm
C) 11.0cm
D) 8.5cm
Correct Answer: A
Solution :
Consider all the force belonging to the system and work done by all the forces with proper sign. Then, apply work-energy theorem. Given,\[J=\frac{i}{\pi {{a}^{2}}}\] Force of kinetic friction,\[\oint{B.dl}={{\mu }_{0}}.{{i}_{enclosed}}\] Mass of block\[x=2\times {{10}^{-2}}\] Initial velocity of the block\[cos\text{ }\pi t\] Suppose block gets the state-.of rest after compressing the spring. Applying work-energy theorem, i.e., work done by all the forces = change in kinetic energy \[E={{E}_{0}}\] \[sin\text{ }\omega t\]\[I={{I}_{0}}\sin \left( \omega t-\frac{\pi }{2} \right)\] \[p=\frac{{{E}_{0}}{{I}_{0}}}{\sqrt{2}}\]\[p=\frac{{{E}_{0}}{{I}_{0}}}{2}\] \[P=\sqrt{2}{{E}_{0}}{{I}_{0}}\]\[{{10}^{-3}}\mu C\] Thus, is quadratic in terms of\[(\sqrt{2},\sqrt{2})\]which on solving gives the value of\[\frac{1}{4}\] As \[x=-15\pm \,\sqrt{\frac{225-4\times 5000\times -16}{2\times 5000}}\] \[5\,\Omega \] \[(1-e)\] \[e-1\] \[(1-{{e}^{-1}})\]
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