Explanation
Two vectors \overline{A} and \overline{B} lie in a plane. Another vector \overline{C} lies outside this plane, then the resultant of the three vectors (\overline{A}+\overline{B}+\overline{C}) :
The observer will see only four spokes if the difference of angular
displacements of wheel 1 \& 2 is
n\dfrac { 2\pi }{ 4 } =n\dfrac { \pi }{ 2 } \\ Difference in angular displacement = (8rev/min - 6rev/min)\times t = 2\times 2\pi \times t=4\pi \times t (t in minutes)\\ The least time after which observer will see only four spokes is given by
4\pi \times t = (1)\frac { \pi }{ 2 } \\ \therefore t= 1/8 min Therefore no of times observer will see only four spokes in a min= 8We can also think of this in following way: No of times observer will see only four spokes in a time period will be the value of n as it will tell us the number of (\dfrac { \pi }{ 2 } ) degrees rotations that the wheels have
taken relative to each other. And everytime they takes a \frac { \pi }{ 2 } degree rotation relative to each other
observer will see only four spokes once.
In 1 min 2\times 2\pi =4\pi =8\frac { \pi }{ 2 } . Therefore 8 times a min.
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