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2¹²*20*/10³ micro seconds

Here we don't need to calculate the no of chips used as all the chips will be refreshed in parallel. We will calculate only for 1 chip i.e no of rows * 1 refresh operation time

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chirudeepnamini asked Jan 28, 2019
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Consider a disk containing 10 equidistance tracks, the inner track capacity is 20 MB and diameter is 1 cm;