Tag: measurements and experimentation

Questions Related to measurements and experimentation

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 physics measurements and experimentation a few applications of linear shm simple pendulum example of simple harmonic motion

The time taken to complete $20$ oscillations by a seconds pendulum is: 

  1. $20s$
  2. $50s$
  3. $40s$
  4. $5s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that the time period of a seconds pendulum is $T=2$ sec. One second for a swing in one direction and one second for the return swing. 

Thus, time taken to complete one oscillation is $2$ sec.
Hence, time taken to complete 20 oscillations is $2\times 20=40$ sec.

Multiple choice physics measurements and experimentation a few applications of linear shm simple pendulum example of simple harmonic motion

The length of a second's pendulum on the surface of the earth is equal to 99.49 cm. True or false.

  1. True

  2. False

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The time period of seconds pendulum T = 2 seconds, acceleration due to gravity at earth g= 980 $\dfrac { cm }{ { s }^{ 2 } } $,It '$l$' is the length of pendulum,

$l=\dfrac { { T }^{ 2 }g }{ 4{ \pi  }^{ 2 } } \ \Rightarrow l=\dfrac { 4\times 980 }{ 4\times \left( \dfrac { 22 }{ 7 }  \right) ^{ 2 } } =\dfrac { 4\times 980\times 49 }{ 4\times 489 } =99.49$