UGC NET CS 2010 JunePaper2
October 16, 2023DigitalLogicDesign
October 16, 2023DigitalLogicDesign
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 10^{3}1 which is 999.
Then, Decimal number has 64 digits, so maximum number is 10^{64}1
Similarly, in the binary representation with “n” bits the maximum number is 2^{n}1
So we can write 10^{64} –1 = 2^{n} – 1 —>10^{64} = 2^{n}
After taking log_{2} on both sides
log_{2}2^{n}=log_{2}10^{64}
n log_{2}2=64 log_{ 2}10
n=64*(3.322) [ log_{2}2=1 & log_{2}10 =3.322]
n=212.608
n=213
Then, Decimal number has 64 digits, so maximum number is 10^{64}1
Similarly, in the binary representation with “n” bits the maximum number is 2^{n}1
So we can write 10^{64} –1 = 2^{n} – 1 —>10^{64} = 2^{n}
After taking log_{2} on both sides
log_{2}2^{n}=log_{2}10^{64}
n log_{2}2=64 log_{ 2}10
n=64*(3.322) [ log_{2}2=1 & log_{2}10 =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 10^{3}1 which is 999.
Then, Decimal number has 64 digits, so maximum number is 10^{64}1
Similarly, in the binary representation with “n” bits the maximum number is 2^{n}1
So we can write 10^{64} –1 = 2^{n} – 1 —>10^{64} = 2^{n}
After taking log_{2} on both sides
log_{2}2^{n}=log_{2}10^{64}
n log_{2}2=64 log_{ 2}10
n=64*(3.322) [ log_{2}2=1 & log_{2}10 =3.322]
n=212.608
n=213
Then, Decimal number has 64 digits, so maximum number is 10^{64}1
Similarly, in the binary representation with “n” bits the maximum number is 2^{n}1
So we can write 10^{64} –1 = 2^{n} – 1 —>10^{64} = 2^{n}
After taking log_{2} on both sides
log_{2}2^{n}=log_{2}10^{64}
n log_{2}2=64 log_{ 2}10
n=64*(3.322) [ log_{2}2=1 & log_{2}10 =3.322]
n=212.608
n=213
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