Solveeit Logo
Login

Question

Question: In a new system, the unit of mass is \(10kg\), the unit of length is \(5m\) and the unit of time is ...

In a new system, the unit of mass is 10kg10kg, the unit of length is 5m5m and the unit of time is 10s10s. Then 10Pa=10Pa =
a. 500500 new units
b. 10001000 new units
c. 15001500 new units
d. 20002000 new units

Explanation

Solution

Dimensional analysis is the very easy method to convert one system of units to another system of units. So, when the charge unit is varying, then the numerical values also may change. We can write the conversion of units like length, mass and time.

Complete step by step answer:
This can be solved by dimensional analysis method:
Given, SI unit of pressure is PascalPascal (Pa)(Pa) or Nm−2N{m^{ - 2}}
Let dimensional formula for pressure is [ML−1T−2]\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]
In SI system of units M1=1kg,L1=1m,T1=1s{M_1} = 1kg,{L_1} = 1m,{T_1} = 1s
In SI system of units M2=10kg,L2=5m,T2=10s{M_2} = 10kg,{L_2} = 5m,{T_2} = 10s
To find 10Pa10Pa
We have, n1u1=n2u2{n_1}{u_1} = {n_2}{u_2}
⇒10Pa=n2\Rightarrow 10Pa = {n_2}…………………….(1)
⇒n2=u1u2n1\Rightarrow {n_2} = \dfrac{{{u_1}}}{{{u_2}}}{n_1}
⇒n2=SInew system×10\Rightarrow {n_2} = \dfrac{{SI}}{{{\text{new system}}}} \times 10
Now apply dimensional formula we get,
⇒n2=[M1L1−1T1−2][M2L2−1T2−2]×10\Rightarrow {n_2} = \dfrac{{\left[ {{M_1}{L_1}^{ - 1}{T_1}^{ - 2}} \right]}}{{\left[ {{M_2}{L_2}^{ - 1}{T_2}^{ - 2}} \right]}} \times 10
M1,L1,T1{M_1},{L_1},{T_1} in terms of SI system
M2,L2,T2{M_2},{L_2},{T_2} in New system of units.
Then, n2=10×[1kg10kg][1m5m]−1[1s−210s−2]{n_2} = 10 \times \left[ {\dfrac{{1kg}}{{10kg}}} \right]{\left[ {\dfrac{{1m}}{{5m}}} \right]^{ - 1}}\left[ {\dfrac{{1{s^{ - 2}}}}{{10{s^{ - 2}}}}} \right]
Seconds’ term, a kg and m term cancels out. Then we have,
⇒n2=10×[110][15]−1[110]−2\Rightarrow {n_2} = 10 \times \left[ {\dfrac{1}{{10}}} \right]{\left[ {\dfrac{1}{5}} \right]^{ - 1}}{\left[ {\dfrac{1}{{10}}} \right]^{^{ - 2}}}
⇒n2=[0.2]−1[0.1]−2\Rightarrow {n_2} = {\left[ {0.2} \right]^{ - 1}}{\left[ {0.1} \right]^{^{ - 2}}}
⇒n2=5×100=500\Rightarrow {n_2} = 5 \times 100 = 500
Now equation (1) becomes,
10Pa=500 new units10Pa = 500{\text{ }}n{\text{ew units}}

Hence, the correct answer is option (A).

Additional information:
Dimensional formula is the expression of showing the powers to which the physical quantity must be raised to represent fundamental quantities.
Dimensions for the fundamental physical quantities

  1. Mass| [ M ]
    ---|---
    Length| [ L ]
    Time| [ T ]
    Temperature| [ K ]
    Electric current| [ A ]
    Luminous intensity| [ cd ]
    Amount of substance| [ mol ]

Constants having dimensional formulas are called Dimensional constants.
Eg : Planck’s constant, speed of light, Universal gravitational constant..
Physical quantities having no dimensional formula are called Dimensionless quantities.
Eg : Angle, Strain, Relative Density.
Limitations of Dimensional analysis
Proportionality constants cannot be determined by dimensional analysis.
Formulae containing non- algebraic functions like sin, cos, log, exponential etc., cannot be derived.

Note: Square bracket [ ] is used to represent dimension of the physical quantity Dimensional formulae can be used to convert one system of unit to another system of unit. And also it can be used to derive the relationship among different physical quantities.