UGC NET CS 2018DEC Paper2
October 5, 2023UGC NET CS 2018DEC Paper2
October 5, 2023UGC NET CS 2018DEC Paper2
Question 12

If the frame buffer has 10bits per pixel and 8bits are allocated for each of the R,G and B components then what would be the size of the color lookup table(LUT)
(2 ^{10} +2^{ 11} ) bytes


(2 ^{10} +2^{ 8} ) bytes


(2 ^{10} + 2^{ 24} ) bytes


(2 ^{8} + 2^{ 9} ) bytes

Question 12 Explanation:
10bits per pixel it means we will have 2^{ 10} entries in color lookup table.
8bits are allocated for each of the R,G and B components, means each entry in color lookup table is of 24bits( 8bits for each of the R, G and B component).
24bits = 3 Bytes
So the size of lookup table is = Number of entries * size of each entry
So the size of lookup table is = (2^{ 10} * 3) Bytes
the size of lookup table is = 3072 Bytes = (2 ^{10} + 2^{ 11} ) Bytes
8bits are allocated for each of the R,G and B components, means each entry in color lookup table is of 24bits( 8bits for each of the R, G and B component).
24bits = 3 Bytes
So the size of lookup table is = Number of entries * size of each entry
So the size of lookup table is = (2^{ 10} * 3) Bytes
the size of lookup table is = 3072 Bytes = (2 ^{10} + 2^{ 11} ) Bytes
Correct Answer: A
Question 12 Explanation:
10bits per pixel it means we will have 2^{ 10} entries in color lookup table.
8bits are allocated for each of the R,G and B components, means each entry in color lookup table is of 24bits( 8bits for each of the R, G and B component).
24bits = 3 Bytes
So the size of lookup table is = Number of entries * size of each entry
So the size of lookup table is = (2^{ 10} * 3) Bytes
the size of lookup table is = 3072 Bytes = (2 ^{10} + 2^{ 11} ) Bytes
8bits are allocated for each of the R,G and B components, means each entry in color lookup table is of 24bits( 8bits for each of the R, G and B component).
24bits = 3 Bytes
So the size of lookup table is = Number of entries * size of each entry
So the size of lookup table is = (2^{ 10} * 3) Bytes
the size of lookup table is = 3072 Bytes = (2 ^{10} + 2^{ 11} ) Bytes
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