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UGC NET CSE | December 2018 | Part 2 | Question: 41

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If the frame buffer has $10$-bits per pixel and $8$-bits are allocated for each of the $R, G$, and $B$ components, then what would be the size of the color lookup table (LUT)?

  1. $(2^8+2^9)$ bytes
  2. $(2^{10}+2^8)$ bytes
  3. $(2^{10}+2^{24})$ bytes
  4. $(2^{10}+2^{11})$ bytes
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There are 2^10=1024 entries in the lookup table.

Each entry size(R,G,B each 8 bit)=3X8=24 bits

Hence total size of lookup table =(1024 X 24)/8 =3072 bytes

= 2^10 + 2^11
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