Tag: kinds of units

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

The SI unit of specific latent heat is 

  1. $Jkg^{-1}K^{-1}$
  2. $J^{0}K^{-1}$
  3. $J kg^{-1}$
  4. $J kgK^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The specific latent heat of a substance is the quantity of heat energy required to change the state of a unit mass of a substance.

${E} _{L}= ml$

here ${E} _{L}$ is the heat transferred, in joules,m is the mass, in kilograms, and l is the latent heat in joules per kilogram.

The SI unit for specific latent heat is $J{kg}^{-1}$.

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

The specific heat capacity of water in SI unit is :

  1. $4.2 Jg^{-1}K^{-1}$
  2. $42 Jg^{-1}K^{-1}$
  3. $420 J kg^{-1}K^{-1}$
  4. $4200 J kg^{-1}K^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The specific heat capacity  of a substance is the heat needed to raise

the temperature of 1 kg of the substance by 1K (or by ${1}^{0}C$).
The specific heat capacity of water is $4200 J{kg}^{-1}{K}^{-1}$.

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

The SI unit of thermal capacity is ;

  1. $J\ kg^{-1}\ K^{-1}$
  2. $J^{-1}\ kg\ K^{-1}$
  3. $JK^{-1}$
  4. $J\ kg\ K^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation


The heat or thermal capacity of a defined system is the amount of heat required to raise the system's temperature by one degree. It is expressed in units of thermal energy per degree temperature.

$q= C\times t$

$C=\dfrac{q}{\triangle t}$

here,$C=$heat or thermal capacity
        $q=$heat required (J)
        $\triangle t={t} _{1}-{t} _{2}$ (K)

so,The S.I. unit of heat or thermal capacity is $J{K}^{-1}$

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

Column I gives three physical quantities. Select the appropriate units for these from the choices given in column II. Some of the physical quantities may have more than one choice.

Column-I Column - II
a) Capacitance d) Ohm second
b) Inductance e) Coulomb$^{2}$ joule$^{ -1 }$
c) Magnetic induction f) Coulomb volt$^{ -1 }$
g) Newton (ampere/metre)$^{ -1 }$
h) Volt second (ampere)$^{ -1 }$

  1. a-e,f; b-d,h; c-g

  2. a-h; b-d; c-e

  3. a-e; b-g,h; c-g

  4. a-e,f; b-d,h; c-d,e
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Capacitance is charge per unit volt. 

Volt has unit of work per unit chage. 
So capacitance has units of charge square per unit energy. 
Inductance is defined as volt per unit rate of change of current and voltage per unit current is resistance, so inductance is ohm second. 

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

The SI unit of magnetic permeability is

  1. ${ Am }^{ -1 }$
  2. ${ Am }^{ -2 }$
  3. ${ Hm }^{ -2 }$
  4. ${ Hm }^{ -1 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The correct option is D.

The magnetic permeability of any material is its property for allowing the magnetic line of force to pass through it. Therefore, the magnetic material can support the development of the magnetic field. This capability is measured by the magnetic permeability. Also, the magnetic line of force is directly proportional to the conductivity of the material. Its SI unit is Henry per meter i.e. $Hm^{-1}$ or newton per ampere square.

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.