UGC NET CS 2010 June-Paper-2
October 16, 2023Digital-Logic-Design
October 16, 2023Digital-Logic-Design
| Question 414 |
The decimal number has 64 digits.The number of bits needed for its equivalent binary representation is?
| 200 | |
| 213 | |
| 246 | |
| 277 |
Question 414 Explanation:
Consider three digits(1,2,3) of decimal numbers.Maximum number, we can generate by that three digits are 103-1 which is 999.
Then, Decimal number has 64 digits, so maximum number is 1064-1
Similarly, in the binary representation with “n” bits the maximum number is 2n-1
So we can write 1064 –1 = 2n – 1 —>1064 = 2n
After taking log2 on both sides
log22n=log21064
n log22=64 log 210
n=64*(3.322) [ log22=1 & log210 =3.322]
n=212.608
n=213
Then, Decimal number has 64 digits, so maximum number is 1064-1
Similarly, in the binary representation with “n” bits the maximum number is 2n-1
So we can write 1064 –1 = 2n – 1 —>1064 = 2n
After taking log2 on both sides
log22n=log21064
n log22=64 log 210
n=64*(3.322) [ log22=1 & log210 =3.322]
n=212.608
n=213
Correct Answer: B
Question 414 Explanation:
Consider three digits(1,2,3) of decimal numbers.Maximum number, we can generate by that three digits are 103-1 which is 999.
Then, Decimal number has 64 digits, so maximum number is 1064-1
Similarly, in the binary representation with “n” bits the maximum number is 2n-1
So we can write 1064 –1 = 2n – 1 —>1064 = 2n
After taking log2 on both sides
log22n=log21064
n log22=64 log 210
n=64*(3.322) [ log22=1 & log210 =3.322]
n=212.608
n=213
Then, Decimal number has 64 digits, so maximum number is 1064-1
Similarly, in the binary representation with “n” bits the maximum number is 2n-1
So we can write 1064 –1 = 2n – 1 —>1064 = 2n
After taking log2 on both sides
log22n=log21064
n log22=64 log 210
n=64*(3.322) [ log22=1 & log210 =3.322]
n=212.608
n=213
Subscribe
Login
0 Comments