Tag: need of unit for measurement

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Multiple choice standardized measurement measurement of physical quantities need of unit for measurement measurements and experimentation physics

Units of Planck's constant in CGS system are:

  1. Erg per second

  2. Second per erg

  3. Erg second

  4. Erg per second per second
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Planck's constant, symbolized h, relates the energy in one quantum (photon) of electromagnetic radiation to the frequency of that radiation.  In the centimeter-gram-second (CGS) or small-unit metric system, it is equal to approximately $6.626176\times 10^{-27}\,$Erg Second.

Multiple choice standardized measurement measurement of physical quantities need of unit for measurement measurements and experimentation physics

If force (F), work (W) and velocity (V) are taken as fundamental quantities then the dimensional formula of time (T) is

  1. $\left[ { W }^{ 1 }{ F }^{ 1 }{ V }^{ 1 } \right] $
  2. $\left[ { W }^{ 1 }{ F }^{ 1 }{ V }^{ -1 } \right] $
  3. $\left[ { W }^{ -1 }{ F }^{ -1 }{ V }^{ -1 } \right] $
  4. $\left[ { W }^{ 1 }{ F }^{ -1 }{ V }^{ -1 } \right] $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
We know,

$[W]=ML^2T^{-2}$

$[F]=MLT^{-2}$

$[V]=LT^{-1}$

Let,
$W^aF^bV^c=M^0L^0T$

$a+b=0$

$2a+b+c=0$, $a+c=0$

$-2a-2b-c=1$

$c=-1,a=1,b=-1$

Hence , $[T]=[WF^{-1}V^{-1}]$

Option $\textbf D$ is the correct answer
Multiple choice standardized measurement measurement of physical quantities need of unit for measurement measurements and experimentation physics

The ratio of SI unit to CGS unit of G is

  1. $10^{3}$
  2. $10^{2}$
  3. $10^{-2}$
  4. $10^{-3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
SI unit of G is  $\dfrac{N m^2}{kg^2}$.
CGS unit of G is  $\dfrac{dyne \ cm^2}{gm^2}$
We know that  $1 \ N = 10^5 \ dyne$ and $1 \ m = 10^2 \ cm$ and $1 \ kg = 10^3 \ gm$
So ratio of SI unit to CGS unit   $ = \dfrac{\dfrac{N  \ m^2}{kg^2}}{\dfrac{dyne \ cm^2}{gm^2}} = \dfrac{\dfrac{10^5 \ dyne \ (10^2 \ cm)^2}{(10^3 \ gm)^2}}{\dfrac{dyne \ cm^2}{gm^2}} = 10^3$
Correct answer is option A.
Multiple choice standardized measurement measurement of physical quantities need of unit for measurement measurements and experimentation physics

1 Newton $=$

  1. $10^4 dyne$
  2. $10^5 dyne$
  3. $10^6dyne$
  4. $10^7 dyne$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
S.I. unit of force is Newton and CGS unit of force is done.

We know $F=ma$
so, force can be expresses in S.I. Units as $Kg m s^{-2}$
and dyne can be expressed as $gcms^{-2}$
1 Newton= $kg ms^{-2}$ 
                 =$10^3 g*10^2 cms  s^{-2}$
                 =$10^5 g cm s^{-2}$
                 =$10^5 dyne$