###### UGC NET CS 2010 June-Paper-2

October 16, 2023###### Digital-Logic-Design

October 16, 2023# Digital-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 10**

Then, Decimal number has 64 digits, so maximum number is 10

Similarly, in the binary representation with “n” bits the maximum number is 2

So we can write 10

After taking log

log

n log

n=64*(3.322) [ log

n=212.608

n=213

^{3}-1 which is 999.Then, Decimal number has 64 digits, so maximum number is 10

^{64}-1Similarly, in the binary representation with “n” bits the maximum number is 2

^{n}-1So we can write 10

^{64}–1 = 2^{n}– 1 —>10^{64}= 2^{n}After taking log

_{2}on both sideslog

_{2}2^{n}=log_{2}10^{64}n log

_{2}2=64 log_{ 2}10n=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**

Then, Decimal number has 64 digits, so maximum number is 10

Similarly, in the binary representation with “n” bits the maximum number is 2

So we can write 10

After taking log

log

n log

n=64*(3.322) [ log

n=212.608

n=213

^{3}-1 which is 999.Then, Decimal number has 64 digits, so maximum number is 10

^{64}-1Similarly, in the binary representation with “n” bits the maximum number is 2

^{n}-1So we can write 10

^{64}–1 = 2^{n}– 1 —>10^{64}= 2^{n}After taking log

_{2}on both sideslog

_{2}2^{n}=log_{2}10^{64}n log

_{2}2=64 log_{ 2}10n=64*(3.322) [ log

_{2}2=1 & log_{2}10 =3.322]n=212.608

n=213

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