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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= $kg ms^{-2}$ 
                 =$10^3 g*10^2 cms  s^{-2}$
                 =$10^5 g cm s^{-2}$
                 =$10^5 dyne$
Multiple choice electrostatic and magnetic analogy magnetism and matter magnetic effects of current and magnetism physics

Two equal poles repel each other with a force of 10$^{-3}$ N. When placed 2cm apart in air, the pole strength of each is (in amp-m).

  1. 4$\pi $

  2. 2

  3. 4
  4. 2$\pi $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$F=10^{-3}N$
$d=2cm$
$m _{1}=m _{2}=m$
$F=\dfrac{\mu _{0}m _{1}m _{2}}{4\pi d^{2}}$

$10^{-3} = \dfrac{10^{-7}\times m^{2}}{(2\times 10^{-2})^2}$
$m=2Am$

Multiple choice electrostatic and magnetic analogy magnetism and matter magnetic effects of current and magnetism physics

Force between two identical bar magnets whose centres are  $r meters $ apart is $4.8 N,$ when their axis are in the same line. If the separation is increased to $2r$ meters, the force between them is reduced to:

  1. 2.4 N

  2. 1.2 N

  3. 0.6 N

  4. 0.3 N
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$F=\dfrac{\mu _{0}m _{1}m _{2}}{4\pi d^{2}}$
$F\propto \dfrac{1}{d^{2}}$
$Fd^{2}=Constant$
$F _{1}d _{1}^{2}=F _{2}d _{2}^{2}$
$4.8\times r^2=F _{2}(2r)^2$
$F _{2}=1.2  N$

Multiple choice electrostatic and magnetic analogy magnetism and matter magnetic effects of current and magnetism physics

The magnetic induction at a distance d from the magnetic pole of the unknown strength m is B. If an identical pole is now placed at a distance of 2d from the first pole, the force between the two poles is          

  1. mB
  2. $\frac{mB}{2}$
  3. $\frac{mB}{4}$

  4. 2mB
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$B=\dfrac{\mu _{0}m}{4\pi d^{2}}$
$r=2d$
$F=\dfrac{\mu _{0}m _{1}m _{2}}{4\pi r^{2}}$
$F=\dfrac{\mu _{0}\times m\times m}{4\pi (d)^{2}}$
$F=\dfrac{\mu _{0}m^{2}}{4\pi \times 4d^{2}}$
$F=\dfrac{mB}{4}N$