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Question: In a new system, the unit of mass is \(10\) times, the unit of length is \(10\) times and the unit o...

In a new system, the unit of mass is 1010 times, the unit of length is 1010 times and the unit of time is 100100 times the SI unit. Then the value of 1J1 J in the new system if unit is 10a{10^a}, then aa is
A. Zero
B. 11
C. 22
D. −1-1

Explanation

Solution

The dimensional formula for 1 Joule is [ML2T−2][M{L^2}{T^{ - 2}}]. The dimensional formula for the new system with the given conditions will 10 times the unit of mass and length and 100 times the unit of time, that is[(10M)(10L)2(100T)−2][(10M){(10L)^2}{(100T)^{ - 2}}]. Dividing the dimensional formula of the new system with the old system will give the value of 1joule.

Formula Used:
The formula for work done is W=∫F.dsW = \int {F.ds} , that is force times distance.

Complete step by step answer:
Work done can be given by the integral of force times the distance. Joule is a unit of work or energy. Joule can also be written as Newton meter, that isN.mN.m. 1 Newton is also written as kg.ms2\dfrac{{kg.m}}{{{s^2}}}. Therefore the unit of 1 Joule is kgm2s−2kg{m^2}{s^{ - 2}}. Then the dimensional formula will be [ML2T−2][M{L^2}{T^{ - 2}}].

Consider two systems: the old system and the new system. The dimensional formula in the old system will be [ML2T−2][M{L^2}{T^{ - 2}}]. It is given in the problem that the unit of mass and length is changed to 10 times and the unit of time is changed to 100 times the SI unit. Therefore, the new system of units will be [(10M)(10L)2(100T)−2][(10M){(10L)^2}{(100T)^{ - 2}}]

Now, consider the division of both the cases:

\Rightarrow\dfrac{{1{J_{{\text{new system}}}}}}{{1{J_{{\text{old system}}}}}}= [(10){(10)^2}{(100)^{ - 2}}] \\\ \therefore\dfrac{{1{J_{{\text{new system}}}}}}{{1{J_{{\text{old system}}}}}}= 10 $$ Therefore, The SI unit of the value of 1 J in the new system is 10. Then, the value of $$a$$ in$${10^a}$$will be $$a = 1$$. **Hence, option B is the correct answer.** **Note:** The formula for work done is $$W = \int {F.ds} $$. The integral shows the force $$F$$ applied on an object or the distance covered by an object to do the work. This integral has no effect on the dimensional formulas of force and displacement. Work done per second is also equal to power. The force acting on an object times the velocity of the object is the power being delivered to the object by the force. The unit of power is given as Joule per second.