Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

Multiple choice physics measurements and units kinds of units fundamental quantities standardized measurement

$\left( Coulomb \right) ^{ 2 }{ J }^{ -1 }$ can be the unit of :

  1. electric resistence

  2. electric energy

  3. electric capacity

  4. electric power
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Electric energy $=ML^{2}T^{-2}$
Electric power $=ML^{2}T^{-3}$
Electric resistance $= ML^{2}I^{-2}T^{-3}$
Electric capacity $=M^{-1}L^{-2}I^{2}T^{4}$$=\left ( \dfrac{I^{2}T^{2}}{(M^{1}L^{2}T^{-2})} \right )=\left ( \dfrac{Q^{2}}{J} \right )$
Hence, option C is correct.

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

The units which can neither derived from one another nor resolved into any thing more basic are called 

  1. Fundamental unit

  2. Scale

  3. Derived unit

  4. Standard unit
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The units which can neither derived from one another nor resolved into any thing more basic are called fundamental unit. It is independent of any other unit.

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

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
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.
Multiple choice physics measurement of physical quantities kinds of units fundamental quantities standardized measurement

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

  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.
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
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.
$\dfrac{k}{\rho\times c} = M^0L^0T^0$
$\dfrac{\rho \times v \times r}{\eta}=M^0L^0T^0$

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

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

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

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

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)

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.