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The main memory unit with a capacity of $4$ $\text{megabytes}$ is built using $1\text{M} \times \text{1-bit}$ DRAM chips. Each DRAM chip has $1\text{K}$ rows of cells with $1\text{K}$ cells in each row. The time taken for a single refresh operation is $100\; \text{nanoseconds}$. The time required to perform one refresh operation on all the cells in the memory unit is

  1. $100$ nanoseconds
  2. $100\times 2^{10}$ nanoseconds
  3. $100\times 2^{20}$ nanoseconds
  4. $3200\times 2^{20}$ nanoseconds
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There are $4\ast 8 = 32$ DRAM chips to get $4\;\text{MB}$ from $1\text{M} \times \text{1-bit}$ chips. Now, all chips can be refreshed in parallel so do all cells in a row. So, the total time for refresh will be number of rows times the refresh time

$= 1\text{K} \times 100$

$= 100 \times 2^{10}$ nanoseconds

Reference: http://www.downloads.reactivemicro.com/Public/Electronics/DRAM/DRAM%20Refresh.pdf

Correct Answer: $B$

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