Question

Match Event (given in the first column) with Order of the time interval for happening of the event (given in the second column) and select the correct option

(a)	Rotation period of the earth	(i)	\[{10^5}\]
(b)	Revolution period of earth	(ii)	${10^7}$
(c)	Period of a light wave	(iii)	\[{10^{ - 15}}\]
(d)	Period of a sound wave	(iv)	\[{10^{ - 5}}\]

(A). (a)-(i),(b)-(ii),c-(iii),d-(iv)
(B). (a)-(ii),(b)-(i),c-(iv),(d)-(iii)
(C). (a)-(ii),b-(ii),c-(iv),d-(iii)
(D). (a)-(ii),b-(ii),c-(iii),d-(iv)

Answer 1

Hint : The time taken for one rotation of earth is approximately 1 day. The time taken for a complete revolution around its orbit is 1 year \[(365.25\] days). The frequency of sound varies from 20Hz to 20KHz and the frequency of visible light varies from $4.3 \times {10^{14}}to7.3 \times {10^{14}}Hz$.

Complete step by step answer:
Starting with rotation of earth, we know it is the time taken to complete a single rotation in its own axis. The time taken for this is 23 hours 56 minutes and 4 seconds which is approximately considered a day .We will convert the time required completely into seconds.
\[{T_{rot}} = (23 \times 60 \times 60) + (56 \times 60) + 4\]
\[ = 82800 + 3360 + 4\]
\[ = 86164s\]
We have been asked to choose the order of magnitude which for this case is ${10^5}$($0.86164 \times {10^5}$)
The revolution of earth is defined as the time taken by the earth to revolve around the sun and coming back to its initial position. This is equal to $365.256$ days. Since the options are in seconds, we will find the time taken in seconds.
\[365.256\] days \[ = 365.256 \times 24 \times 60 \times 60s\]
\[ = 31558118.4s\]
The order of magnitude of the revolution of earth is ${10^7}$ $3.1558118 \times {10^7}$
The frequency range of sound is 20 Hz to 20KHz. We know
\[T = \dfrac{1}{f}\]
where, $T$ is the time period if wave and $f$ is the frequency.
Using this equation when we substitute the value of frequency of sound wave we get that time period of sound ranges from $\dfrac{1}{{20Hz}}$ to $\dfrac{1}{{20KHz}}$ i.e. $5 \times {10^{ - 2}}s$ to $5 \times {10^{ - 5}}s$.
Now,this has a order of magnitude of ${10^{ - 5}}$ ($5 \times {10^{ - 5}}$)
Using the same equation we can find the frequency range of light by substituting its frequency $4.3 \times {10^{14}}to7.3 \times {10^{14}}Hz$in the equation \[T = \dfrac{1}{f}\] we get that its time period ranges from $\dfrac{1}{{4.3 \times {{10}^{14}}}}s$ to $\dfrac{1}{{7.3 \times {{10}^{14}}}}s$.
Here the order of magnitude is ${10^{ - 15}}$.($1.369 \times {10^{ - 15}}$ to $2.32 \times {10^{ - 15}}$)
Therefore we can see that the order of magnitude of rotation period of earth ,revolution period of earth ,period of light wave and period of sound wave are \[{10^5}\],\[{10^7}\], \[{10^{ - 15}}\],\[{10^{ - 5}}\]respectively.

We can conclude that the correct option is A

Note: We can simply arrange the time period of the following cases in ascending or descending order . By general knowledge we know the least time period is required for light and the highest time period among the given is for revolution. By arranging them in order gives the range of answers without calculation and saves a lot of time.