Tag: standardized measurement

Questions Related to standardized measurement

State whether given statement is True or False.
The standard quantity used for comparison is called fundamental quantity.

physics measurement of physical quantities kinds of units fundamental quantities standardized measurement

  1. True

  2. False

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Correct Option: B
Explanation:
The standard quantity, used for comparison, is called unit with which unknown quantities are compared.
While the fundamental quantity is a quantity which is independent of any other physical quantity. For example, length, time, mass and temperature are fundamental physical quantities. The units used to measure the fundamental quantities is called fundamental units.
Hence, the given statement is false.

One second is defined to be equal to:

physics measurement of physical quantities kinds of units fundamental quantities standardized measurement

  1. 1650763.73 periods of krypton clock

  2. 652189.63 periods of krypton clock

  3. 1650755.73 periods of cesium clock

  4. 9, 19, 26, 31, 770 periods of cesium clock

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Correct Option: D
Explanation:

One second is defined to be equal to $9,19,26,31,770$ periods of cesium clock.

Which of the following are dimensionless quantities,(symbols have their usual meaning)
[$\eta$=viscocity,$\rho$=density,$r$=radius,$k$=thermal conductivity,$c$=heat capacity.]

physics measurement of physical quantities kinds of units fundamental quantities standardized measurement

  1. $\dfrac{k}{\rho\times c}$

  2. $\dfrac{\rho \times v \times r}{\eta}$

  3. Specific gravity

  4. Rate of change of angle (in radians) of rotation.

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Correct Option: A,B,C
Explanation:

The dimesnion of $k $ is $M^1T^1L^{-3}\theta^{-1}$

$\eta =  ML^{-1}T^{-1}$
$c= L^2MT^{-2}\theta^{-1}$
$\rho = ML^{-3}$
$r = L$
Specific gravity is the ratio of the density of a substance to the density of a reference substance, so the specific gravity is dimensionless.
$\dfrac{k}{\rho\times c} = M^0L^0T^0$
$\dfrac{\rho \times v \times r}{\eta}=M^0L^0T^0$

Which of the following is dimensionally  correct? ($\rho$=density,$\eta$=coefficient of viscosity,$P$=pressure,$S$=surface tension,$r$=radius,$g$=gravitational constant)

physics measurement of physical quantities kinds of units fundamental quantities standardized measurement

  1. $h=\dfrac{2Scos\theta}{\rho \times rg}$

  2. $v=\dfrac{P}{\rho}$

  3. $V=\dfrac{Pr^{4}t}{\eta}$

  4. none of the above

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Correct Option: A
Explanation:

Dimensional formula of the following are given as-

$S =MT^{-2}$,  $g = LT^{-2}$,  $\rho = ML^{-3}$,  $r = L$,  $h = L$,  $v = LT^{-1}$,  $P = ML^{-1}T^{-2}$,  $t = T$,  $\eta = ML^{-1}T^{-1}$
Equating LHS and RHS of the following options :
(A) :  LHS = $  L$
RHS $=\dfrac{[MT^{-2}]}{[ML^{-3}][L][LT^{-2}]} = [L]$ 
$\implies$  LHS  = RHS

(B) :  LHS = $  LT^{-1}$
RHS $=\dfrac{[ML^{-1}T^{-2}]}{ML^{-3}} = [L^2 T^{-2}]$ 
$\implies$  LHS  $\neq$ RHS

(C) :   LHS = $  LT^{-1}$
RHS $=\dfrac{[ML^{-1}T^{-2}] [L^4] [T]}{[ML^{-1}T^{-1}]} = [L^4]$ 
$\implies$  LHS  $\neq$ RHS
Thus option A is correct.

Consider the following equation which gives a hypothetical physical quantity mutual dynamic constant $\psi$ as,
$\dfrac{2Scos\theta}{\rho \times rg}$+$\dfrac{1}{2\pi}\dfrac{mgl}{I}$
($I$=moment of inertia ,$S$=surface tension,others symbol have usual meanings)

physics measurement of physical quantities kinds of units fundamental quantities standardized measurement

  1. $\psi$ may exist

  2. $\psi$ will never exist

  3. Such physical quantity is a standard result of electromagnetism ,hence it exists.

  4. none of the above

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Correct Option: B
Explanation:

Dimensional formula of the following are given as-

$S =MT^{-2}$,  $g = LT^{-2}$,  $\rho = ML^{-3}$,  $r = L$,  $I = ML^2$,  $m = M$,  $l = L$
Dimensions of  $\dfrac{2S\cos\theta}{\rho g r} = \dfrac{[MT^{-2}]}{[ML^{-3}][L][LT^{-2}]} =[L]$
Dimensions of  $\dfrac{mgl}{2\pi I} = \dfrac{[M][LT^{-2}][L]}{ML^2} =[T^{-2}]$
Since dimensions of the two additive terms are not same, thus $\psi$ can never exist.

Which of the following is not the fundamental quantity,

physics measurement of physical quantities kinds of units fundamental quantities standardized measurement

  1. mass

  2. length


  3. velocity

  4. time

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Correct Option: C
Explanation:

We know that the fundamental quantities are mass, length and time. 

velocity = $\dfrac{displacement}{time}$
hence velocity is derived quantity

The most basic rule of dimensional analysis is that of dimensional homogeneity. Only commensurable quantities  may be

physics measurement of physical quantities kinds of units fundamental quantities standardized measurement

  1. compared

  2. equated

  3. added or subtracted 

  4. all of the above 

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Correct Option: D
Explanation:

In dimensional analysis, only commensurable quantities may be compared, equated and added or subtracted.

Light year is a unit of:

physics measurement of physical quantities kinds of units fundamental quantities standardized measurement

  1. time

  2. distance

  3. light

  4. intensity of light

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Correct Option: B
Explanation:

Light year is a unit of $distance$.

Which one can be represented as the symbol of time?

physics measurement of physical quantities kinds of units fundamental quantities standardized measurement

  1. $m$

  2. $F$

  3. $t$

  4. None

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Correct Option: C
Explanation:

Time is generally represented by the symbol $t$.
Mass is represented by symbol $m$ and force by the symbol $F$.

The unit of electric field intensity is:

physics measurements and units kinds of units fundamental quantities standardized measurement

  1. Newton/ metre

  2. Coulomb/ newton

  3. Newton/ coulomb

  4. Joule/ newton

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Correct Option: C
Explanation:

Electric field due to a charge is given by  $E = \dfrac{F}{q}$
Unit of force is Newton $(N)$ and that of charge is Coulomb $(C)$.
So, unit of electric field is  Newton/coulomb  i.e  $N/C$