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Question: In a system called star system, \(1\) star Kg \( = {10^{23}}Kg\), \(1\) star meter \( = {10^8}m,\) \...

In a system called star system, 11 star Kg =1023Kg = {10^{23}}Kg, 11 star meter =108m, = {10^8}m, 11 star second =1023seconds = {10^{23}} seconds, then calculate the value of 11 Joule in this system.

Explanation

Solution

Here a new system is introduced and its relation with the existing system is given in the question. Find the equivalent Kg, meter, and second from the given values using the given equation. Use the dimension of Joule to find the value of joule. Use the dimensional formula of Joule for the calculation.

Formula Used:
The dimensional formula of Joule =ML2T−2 = M{L^2}{T^{ - 2}}

Complete step by step solution:
11 Star Kg=1023Kg = {10^{23}}Kg
11Star meter =108 = {10^8}meter
11 Star second = 10−23{10^{ - 23}}star second
These are unit values of mass, length, and time in the star system. The equivalent mass, length, and time in our existing system can be obtained from these values.
1Kg=110231Kg = \dfrac{1}{{{{10}^{23}}}}Star Kg (∵  \because \; 11 star Kg=1023Kg = {10^{23}}Kg)
1Kg=10−231Kg = {10^{ - 23}}Star Kg
11 Meter =1108 = \dfrac{1}{{{{10}^8}}}star meter (∵  \because \; 11 star meter =108 = {10^8}meter)
11 Meter =10−8 = {10^{ - 8}} star meter
11 Second = 11023\dfrac{1}{{{{10}^{23}}}}star second (∵  \because \; 11 star second = 10−23{10^{ - 23}}star second)
11 Second = 10−23{10^{ - 23}}star second
We need to calculate 11 Joule in the star system.
To find the star Joules corresponding to 11 Joule, we need to use the dimensional formula of Joule
It is given by ML2T−2M{L^2}{T^{ - 2}}
We need to put the values of M, L and T in this formula
M= 10−23{10^{ - 23}}star Kg
L = 10−8{10^{ - 8}} star meter
T = 10−23{10^{ - 23}} star second
ML2T−2M{L^2}{T^{ - 2}}= (10−23)(10−8)2(10−23)−2\left( {{{10}^{ - 23}}} \right){\left( {{{10}^{ - 8}}} \right)^2}{\left( {{{10}^{ - 23}}} \right)^{ - 2}}

11 Joule = 107{10^7} star Joules.

Note: Similar kinds of questions can be asked with different values of mass, length, and time. Same method can be used to solve such kinds of questions. If you are not familiar with the dimensions of some physical quantity then the units of that physical quantity can be taken to solve the problem. For example in this case, the SI base unit, Joule= Kgm2s−2Kg{m^2}{s^{ - 2}}. We can solve the problem by substituting in these units also.