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Question: If the average life of a person is taken as \(100s\), the age of universe on this scale is of order ...

If the average life of a person is taken as 100s100s, the age of universe on this scale is of order
(A) 1010s{10^{10}}s
(B) 108s{10^8}s
(C) 1017s{10^{17}}s
(D) 109s{10^9}s

Explanation

Solution

To solve this question, we need to use the average values of the universe and humans. We have to take the ratio of these two to get the scaling factor of the given scale. Multiplying the given age of humans with the scaling factor will give us the final answer.

Complete step-by-step solution:
We know that the age of the universe is approximately equal to 2020 billion years. So the age of the universe can be written as
AU=20×109y{A_U} = 20 \times {10^9}y
Now, we know that there are 365365 days in a year. So the age of the universe becomes
AU=20×109×365 days{A_U} = 20 \times {10^9} \times 365{\text{ days}}
Now, there are 2424 hours in a day. Therefore the age of the universe now becomes
AU=20×109×365×24h{A_U} = 20 \times {10^9} \times 365 \times 24h
Finally, there are a total of 36003600 seconds in an hour. Thus, the age of the universe in seconds is given by
AU=20×109×365×24×3600s{A_U} = 20 \times {10^9} \times 365 \times 24 \times 3600s
On solving we get
AU=6.31×1017s{A_U} = 6.31 \times {10^{17}}s........................(1)
Therefore, the age of the universe is of the order of 1017s{10^{17}}s.
Now, we know that the average life of a person is approximately 7979 years. So the average human life can be written as
AH=79y{A_H} = 79y
By carrying out the same conversion as above, we get the average human life in seconds as
AH=79×365×24×3600s{A_H} = 79 \times 365 \times 24 \times 3600s
On solving we get
AH=2.49×109s{A_H} = 2.49 \times {10^9}s..........................(2)
Therefore, the average life of a person is of the order of 109s{10^9}s.
Now, we calculate the scaling factor of the given scale.
Dividing (1) by (2) we get
AUAH=2.53×108\dfrac{{{A_U}}}{{{A_H}}} = 2.53 \times {10^8}
As we are only considering the order of the age, so we consider the order of this ratio. Therefore we get the scaling factor as
AUAH=108\dfrac{{{A_U}}}{{{A_H}}} = {10^8}
Now, according to the question, AH=100s{A_H} = 100s. Substituting it above, we get
AU100s=108\dfrac{{{A_U}}}{{100s}} = {10^8}
AU=1010s\Rightarrow {A_U} = {10^{10}}s
Therefore, the age of the universe on this scale is of the order 1010s{10^{10}}s.

Hence, the correct answer is option A.

Note: Even if we do not remember the exact value of the average ages of humans and the universe, we can still attempt this question correctly. We just need to be familiar with the order of the ages, that’s it. From the order only we can get the correct scaling factor, from which the required order of age can be calculated.