71Infact, not a rotational problem, But simple maths you can use.
it was found the the winding radius of a tape on a cassette was reduced by half in time t1 = 20. in what time will the winding radius will be reduced by half again?
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1in 20 more seconds
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4)Hi Anirudh:
The volume rate of removal of the tape is constant (assume this). Now, loss in volume in the 2nd case is 1/4 of the volume loss in the 2nd case. This yields the answer 1/4*t.
341The reasonable assumption is that the length of tape is reduced at a constant rate.
Since the thickness of the spool is also constant, this means the area is reduced at a constant rate.
In 20 s the area is reduced from A to A/4 , a reduction of 3A/4
Next, the reduction in area is 3A/16 which will take 1/4 X 20 = 5s