Tag: measurements and experimentation

Questions Related to measurements and experimentation

Which system of units has been taken as standard?

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

  1. M.K.S

  2. C.G.S

  3. Both

  4. None

Show answer
Correct Option: A
Explanation:

In S.I system (standard international system), the units of mass,length and time are same,as that of M.K.S system. However , it is an enlarged system encompassing all fundamental units.

MKS system means

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

  1. Millimeter, kilometre, seconds

  2. Metre, kilogram, seconds

  3. Millisecond, kilolitre, seconds

  4. Milligram, kilogram, seconds

Show answer
Correct Option: B
Explanation:

In MKS system,M stands for metre (length),K stands for kilogram(mass) and S stands for seconds(time).

Units of Planck's constant in CGS system are:

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

  1. Erg per second

  2. Second per erg

  3. Erg second

  4. Erg per second per second

Show answer
Correct Option: C
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.

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

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

  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] $

Show answer
Correct Option: D
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

The ratio of SI unit to CGS unit of G is

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

  1. $10^{3}$

  2. $10^{2}$

  3. $10^{-2}$

  4. $10^{-3}$

Show answer
Correct Option: A
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.

Which of the following represents the magnitude of a temperature correctly?

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

  1. 10 k

  2. 10 Kelvins

  3. 10 Ks

  4. 10 K

Show answer
Correct Option: D
Explanation:
Temperature is measured in Kelvin which is represented by $K$
Thus, $10 \ K$ is the correct representation of temperature.

1 Newton $=$

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

  1. $10^4 dyne$

  2. $10^5 dyne$

  3. $10^6dyne$

  4. $10^7 dyne$

Show answer
Correct Option: B
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= $kg ms^{-2}$ 
                 =$10^3 g*10^2 cms  s^{-2}$
                 =$10^5 g cm s^{-2}$
                 =$10^5 dyne$

The frequency of a seconds pendulum is equal to :

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

  1. 0.5 Hz

  2. 1 Hz

  3. 2 Hz

  4. 0.1 Hz

Show answer
Correct Option: A
Explanation:

The frequency of seconds pendulum is $= \dfrac{1}{2} = 0.5$ HZ

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

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

  1. $20s$

  2. $50s$

  3. $40s$

  4. $5s$

Show answer
Correct Option: C
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.

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

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

  1. True

  2. False

Show answer
Correct Option: A
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,

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