MCQ Questions for Class 11 Engineering Physics Units And Measurement Quiz 3

$C$ denotes the capacity and

$L$ denotes the inductance, the dimensions of '

$LC$ ' are :

0%
${ M }^{ 0 }{ L }^{ 0 }{ T }^{ 2 }$
0%
${ M }^{ 1 }{ L }^{ 0 }{ T }^{ 2 }$
0%
${ M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 }$
0%
${ M }^{ 0 }{ L }^{ 1 }{ T }^{ 2 }$

Explanation

$L=ML^{2}I^{-2}T^{-2}$

$C=M^{-1}L^{-2}I^{2}T^{4}$

$LC=M^{0}L^{0}I^{0}T^{2}$

The dimensional formula for strain is same as that of

0%
thrust
0%
angle
0%
modulus of elasticity
0%
stress

The dimensions of the coefficient of viscosity are :

0%
$ML^{-1}T^{-1}$
0%
$MLT$
0%
$M^{-1}L^{-1}T^{-1}$
0%
$M^0L^0T^0$

The division of energy by time is

$X$ . The dimensional formula of

$X$ is same as that of

0%
momentum
0%
power
0%
torque
0%
electric field

An athlete's coach told him that muscle times speed equals power.
Then according to him the dimensions of muscle are

0%
$[MLT^{-2}]$
0%
$[ML^2T^{-2}]$
0%
$[MLT^{+2}]$
0%
$[MLT^{-3}]$

Explanation

Let us represent the dimension of muscle as X and we are given

$X \times speed=power$ .

$X \times speed = power = \dfrac{Work}{Time}$

$=\dfrac{\left[ML^{2}T^{-2}\right]}{\left[T\right]}$

Now,

$X=\dfrac{\left[ML^{2}T^{-3}\right]}{\left[LT^{-1}\right]}$ .

The dimension of muscle is

$\left[MLT^{-2}\right]$ .

A quantity

$X$ is given by

$\epsilon_0 L\dfrac {\Delta V}{\Delta t}$ , where

$\epsilon_0$ is the permittivity of fee space,

$L$ is length,

$\Delta V$ is a potential difference and

$\Delta t$ is a time interval. The dimensional formula for

$X$ is the same as that of

0%
resistance
0%
charge
0%
voltage
0%
current

Explanation

Dimensional formula for

$\epsilon_0 = [M^{-1} L^{-3} T^{4} I^{2}]$
Dimensional formula for

$L = [L]$
Dimensional formula for

$\Delta V = [M L^2 T^{-3} I^{-1}]$
Dimensional formula for

$\Delta t = [T^{1}]$
Therefore, dimensional formula for

$X$ is:

$\dfrac{ [M^{-1} L^{-3} T^{4} I^{2}] [L] [M L^2 T^{-3} I^{-1}] }{ [T^{1}] }=[I]$

Therefore, its the same as that of current.

The correct statement about Poisson's ratio is:

0%
Its unit is $N/m^2$
0%
It is dimensionless
0%
Its unit is Newton
0%
Its dimensions are $MLT^{-2}$

The dimensions of quantity

$\dfrac{1}{\mu_0} (\vec{E} \times \vec{B})$ are:

$[\mu_0=$ permeability of free space,

$\displaystyle E^1=$ electric field strength,

$\displaystyle B^1=$ magnetic field induction

$]$

0%
$[MT^{-3}]$
0%
$[ML^2T^{-3}]$
0%
$[ML^{-1}T^{-1}]$
0%
$[MT^{-3}A^2]$

Explanation

$\dfrac { \left[ E \right] \left[ B \right] }{ { \left[ { \mu }_{ 0 } \right] } } =\dfrac { \left[ ML{ T }^{ -3 }{ I }^{ -1 } \right] \left[ M{ T }^{ -2 }{ I }^{ -1 } \right] }{ \left[ ML{ T }^{ -2 }{ I }^{ -2 } \right] }$

$={ M }^{ 1+1-1 }{ L }^{ 1-1 }{ T }^{ -3-2+2 }{ I }^{ -1-1+2 }$

$=M{ T }^{ -3 }$

The dimensional formula of latent heat is identical to that of

0%
internal energy
0%
angular momentum
0%
gravitational potential
0%
electric potential

Explanation

Latent heat is given by:

$Q=mL$

$\Rightarrow L=\displaystyle \frac {Q}{m}$ i.e Energy per unit mass.
a) internal energy has the units of energy
b) angular momentum has the units Joule-sec, i.e Energy.second
c) Gravitational potential is the potential energy per unit mass, i.e Energy per mass
d) Electric potential is the potential energy per unit charge.

The dimensional formula of

$\dfrac {L}{R}$ is same as that of (where

$L$ is inductance and

$R$ is resistance)

0%
(frequency) $^2$
0%
time period
0%
frequency
0%
(time period) $^2$

Explanation

Self inductance is given by:

$\displaystyle v(t)=L\dfrac {di}{dt}$
From this we have:

$\displaystyle L=v\dfrac {dt}{di}$

$\displaystyle \dfrac {v}{di}$ has the dimensions of

$R$

So,

$L=Rdt$ hence

$\displaystyle \dfrac {L}{R}=dt$ , same units as that of time.

Which of the following combinations has dimensions?

0%
$\dfrac{rpv}{\eta}$
0%
$\mu_0\in_0c^2$
0%
$\dfrac{t}{RC}$
0%
$\dfrac{\sigma}{\lambda_mT}$

Explanation

All the first three options are dimensionless quantity except the last one.
Since

$\left[ \sigma \right] =\left[ { M }{ T }^{ -3 }{ K }^{ -4 } \right] ,\quad \left[ { \lambda }_{ m } \right] =\left[ L \right] \quad \& \quad \left[ T \right] =\left[ K \right]$

How many significant figures are there in

$0.030100\times 10^6$ ?

0%
1
0%
3
0%
5
0%
7

Explanation

As per the following rules of significant numbers:

1) ALL non-zero numbers (1,2,3,4,5,6,7,8,9) are ALWAYS significant.

2) ALL zeroes between non-zero numbers are ALWAYS significant.

3) Zeroes placed after other digits but behind a decimal point are significant 4)Zeroes placed before other digits are not significant.

Hence in given number i.e.

$0.030100 \times 10^6$ can be further simplified as

$30100.00$

According to the rule 1, 3 and rule 4, the total significant numbers are 5.

The dimensional formula for relative density is

0%
$[ML^{-3}]$
0%
$[M^0L^{-3}]$
0%
$[M^0L^0T^{-1}]$
0%
$[M^0L^0T^0]$

Explanation

Relative density is the ratio of the density of a substance to the density of a given reference material.
Dimension of Relative density

$=\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } \right]$

The ratio of the dimensions of Planck's constant and that of the moment of intertia is the dimensions of

0%
time
0%
frequency
0%
angular momentum
0%
velocity

Explanation

Planck constant,

$h=[ML^2T^{-1}]$
Moment of inertia,

$I=[ML^2T^{0}]$
So,

$\dfrac { \left[ h \right] }{ \left[ I \right] } =\dfrac { \left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -1 } \right] }{ \left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ 0 } \right] } =\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ -1 } \right] \Rightarrow frequency$

Which of the following is a dimensional constant?

0%
Refractive index
0%
Dielectric constant
0%
Relative density
0%
Gravitational constant

Explanation

Gravitational constant also known as universal gravitational constant has a symbol

$G$ and has a dimension

$\left[ { M }^{ -1 }{ L }^{ 3 }{ T }^{ -2 } \right]$ while others are dimensionless constant.

The dimension of length for coefficient of viscosity is

0%
-1
0%
+1
0%
-2
0%
+2

Explanation

Units of coefficient of viscosity is

$kgm^{-1}s^{-1}$ i.e.

$ML^{-1}T^{-1}$
Therefore the dimension of length(L) is

$-1$ .

Place the odd unit out.

0%
kilowatt hour
0%
joule second
0%
watt second
0%
electron volt

Explanation

$\left[ kilowatt \ hour \right] =\left[ watt \ second \right] =\left[ electron \ volt \right] =\left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -2 } \right]$

And they have the unit of energy while joule second is the odd one.

$\sqrt {\mu_0\epsilon_0}$ has the dimensions

0%
$[M^0LT^{-1}]$
0%
$[M^0L^{-1}T]$
0%
$[M^2L^3T^{-1}]$
0%
$[M^{-1}L^3T^{-2}]$

Explanation

Both

$\mu _o$ and

$\epsilon _o$ are related by

$\displaystyle c=\frac {1}{\sqrt {\mu _o\epsilon _o}}$ ; c-speed of light.

So, the units of

$\sqrt {\mu _o\epsilon _o}$ is inverse of speed, i.e.

$M^oL^{-1}T^1$

Which is most accurate?

0%
3 m
0%
3.0 m
0%
3.00 m
0%
3.000 m

How many significant figures are there in 30100?

0%
1
0%
3
0%
5
0%
none of these

Which of the following have the same dimensions?

0%
impulse and momentum
0%
specific heat and latent heat
0%
moment of inertia and force
0%
thrust and surface tension

Explanation

Impulse is defined as the change in momentum.

Impulse

$I = \Delta p = mv - mu$
Hence, both Impulse and momentum have the same units/dimensions.

Dimensions of

$\dfrac {1}{\mu_o\epsilon_o}$ , where symbols have their usual meaning, are

0%
$[L^{-1}T]$
0%
$[L^{-2}T^{2}]$
0%
$[L^{2}T^{-2}]$
0%
$[LT^{-1}]$

Explanation

We have:

$\displaystyle c=\frac {1}{\sqrt {\mu _o\epsilon _o}}$

$\displaystyle c^2=\frac {1}{\mu _o\epsilon _o}$

$c=m/s=LT^{-1}$

$\Rightarrow c^2=L^2T^{-2}$

The dimensions of electromotive force in terms of current A are

0%
$[MT^{-2}A^{-2}]$
0%
$[ML^{2}T^{-2}A^2]$
0%
$[ML^{2}T^{-2}A^{-2}]$
0%
$[ML^{2}T^{-3}A^{-1}]$

Explanation

Electromotive force, also called emf (measured in volts), is the voltage developed by any source of electrical energy such as a battery or dynamo.
Electromotive force

$=$ potential difference

$=V=\dfrac {W}{q}=\dfrac {ML^2T^{-2}}{AT}=[ML^2T^{-3}A^{-1}]$

The dimensions of pressure gradient are

0%
$[ML^{-2}T^{-2}]$
0%
$[ML^{-2}T^{-1}]$
0%
$[ML^{-1}T^{-1}]$
0%
$[ML^{-1}T^{-2}]$

Explanation

Pressure gradient is given as,

$P=\dfrac {\text {Change in Pressure}}{changedistance}$

$\left[ P \right] =\dfrac {ML^{-1}T^{-2}}{L}$

$=[ML^{-2}T^{-2}]$

The dimensions of universal gas constant are

0%
$[L^2M^1T^{-2}K^{-1}]$
0%
$[L^1M^2T^{-2}K^{-1}]$
0%
$[L^1M^1T^{-2}K^{-1}]$
0%
$[L^2M^2T^{-2}K^{-1}]$

Dimensions of impulse are

0%
$[MLT^{-1}]$
0%
$[MLT^{2}]$
0%
$[MT^{-2}]$
0%
$[ML^{-1}T^{-3}]$

Explanation

Impulse

$=$ Force

$\times$ time

$=MLT^{-2}T=MLT^{-1}$

Which of the following set have different dimensions?

0%
Pressure, Young's modulus, Stress
0%
Emf, Potential difference, Electric potential
0%
Heat, Work done, Energy
0%
Electric dipole moment, Electric flux, Electric field

Explanation

Electric dipole moment=charge

$\times$ distance between charges

$=[AT]\times [L]=[ALT]$
Electric flux

$=$ Electric field strength x area

$=[MLT^{-3}A^{-1}][L^2]=[ML^3T^{-3}A^{-1}]$
Electric field

$=$ Force per unit charge

$=[MLT^{-3}A^{-1}]$

Which one of the following represents the correct dimensions of the coefficient of viscosity?

0%
$[ML^{-1}T^{-1}]$
0%
$[MLT^{-1}]$
0%
$[ML^{-1}T^{-2}]$
0%
$[ML^{-2}T^{-2}]$

Explanation

From Stokes' law:

$F_d=6\pi \eta Rv$

$\displaystyle \eta =\frac {F_d}{6\pi Rv}=\frac {MLT^{-2}}{(L)(LT^{-1})}=ML^{-1}T^{-1}$

Which of these is dimensionless?

0%
Force $/$ acceleration
0%
Velocity $/$ acceleration
0%
Volume $/$ area
0%
Energy $/$ work

Explanation

$\displaystyle \frac {Force}{acceleration}=\frac {kg.m.s^{-2}}{m.s^{-2}}=Kg$

ii)

$\displaystyle \frac {Velocity}{acceleration}=\frac {m.s^{-1}}{m.s^{-2}}=s$

iii)

$\displaystyle \frac {Volume}{Area}=\frac {m^3}{m^2}=m$

iv)

$\displaystyle \frac {Energy}{Work}=\frac {J}{J}=1$

The dimensions of Rydberg's constant are

0%
$[M^0L^{-1}T]$
0%
$[MLT^{-1}]$
0%
$[M^0L^{-1}T^0]$
0%
$[ML^0T^2]$

Explanation

Using the relation,

$\dfrac {1}{\lambda}=R\left (\dfrac {1}{n_1^2}-\dfrac {1}{n_2^2}\right )$ where

${ n }_{ 1 }\& { n }_{ 2 }$ denote orbit number and so are dimensionless
Therefore dimensions of

$R=\frac {1}{L}=L^{-1}=[M^0L^{-1}T^0]$

Identify the pair whose dimensions are equal.

0%
Torque and work
0%
Stress and energy
0%
Force and stress
0%
Force and work

$L$ denotes the inductance of an inductor through which a current

$i$ is flowing, the dimensions of

$Li^2$ are

0%
$[ML^2T^{-2}]$
0%
$[MLT^{-2}]$
0%
$[M^2L^2T^{-2}]$
0%
not expressible in $M, L, T$

Explanation

Energy stored in an inductor

$=\frac {1}{2}Li^2=[Energy]=[ML^2T^{-2}]$

The dimensions of Hubble's constant are

0%
$[T^{-1}]$
0%
$[M^0L^0T^{-2}]$
0%
$[MLT^4]$
0%
$[MT^{-1}]$

Explanation

Hubble's constant,

$H=\dfrac {velocity}{distance}=\dfrac {[LT^{-1}]}{[L]}$
The Hubble Constant is the unit of measurement used to describe the expansion of the universe.

$[ML^2T^{-2}]$ are dimensions of

0%
force
0%
moment of force
0%
momentum
0%
power

Explanation

Force

$=[ML^1T^{-2}]$
Moment of force

$=r\times F=[L][MLT^{-2}]=[ML^2T^{-2}]$ .....correct
Momentum

$=[ML^1T^{-1}]$
Power

$=[ML^2T^{-3}]$

The dimensions of magnetic moment are

0%
$[L^2A^1]$
0%
$[L^2A^{-1}]$
0%
$[L^2/A^3]$
0%
$[LA^2]$

Explanation

Magnetic moment is given as

$M=current\times area-AL^2=[L^2A^1]$

The unit and dimensions of impedance in terms of charge Q are

0%
$mho, [ML^2T^{-2}Q^{-2}]$
0%
$ohm, [ML^2T^{-1}Q^{-2}]$
0%
$ohm, [ML^2T^{-2}Q^{-1}]$
0%
$ohm, [MLT^{-1}Q^{-1}]$

Explanation

Unit of impedance is

$ohm$ and
Dimension of Impedance

$=\dfrac {V}{I}=\dfrac {W}{QI}=\dfrac {ML^2T^{-2}}{QQT^{-1}}$

$=[ML^2T^{-1}Q^{-2}]$

$L$ and

$R$ denote inductance and resistance then dimension of

$L/R$ is

0%
$[M^0L^0T^0]$
0%
$[M^0L^0T]$
0%
$[M^2L^0T^2]$
0%
$[MLT^2]$

Explanation

The ratio

$\dfrac {L}{R}=\dfrac {[ML^2T^{-2}A^{-2}]}{[ML^2T^{-3}A^{-2}]}=T=[M^0L^0T^1]$

The dimensional formula of current density is

0%
$[M^0L^{-2}T^{-1}Q]$
0%
$[M^0L^{2}T^{1}Q^{-1}]$
0%
$[MLT^{-1}Q]$
0%
$[ML^{-2}T^{-1}Q^2]$

Explanation

Current density

$=\dfrac {Current}{area}=\dfrac {Q}{area\times t}=[M^0L^{-2}T^{-1}Q]$

The dimensional formula for relative refractive index is

0%
$[M^0L^1T^{-1}]$
0%
$[M^0L^0T^0]$
0%
$[M^0L^1T^{1}]$
0%
$[MLT^{-1}]$

Explanation

Relative refractive index

${ \mu }_{ r }$ is a ratio of refractive index of medium

$\mu$ to refractive index of vacuum

${ \mu }_{ o }$ .

Hence dimensionless

$\mu_r=\dfrac {\mu}{\mu_0}=[M^0L^0T^0]$

The dimensions of Wien's constant are

0%
$[ML^0TK]$
0%
$[M^0LT^0K]$
0%
$[M^0L^0TK]$
0%
$[MLTK]$ 4

Explanation

Here

${ \lambda }_{ m }$ is maximum wavelength at temperature

$T$
Wien's constant

$b$ is given as

$b=\lambda_mT=LK=[M^0L^1T^0K^1]$

Which physical quantities have same dimensions?

0%
Moment of couple and work
0%
Force and power
0%
Latent heat and specific heat
0%
Work and power

Explanation

Moment of couple

$=$ force

$\times$ perpendicular distance

$=[M^1L^2T^{-2}]$

Work

$=$ force

$\times$ distance

$=[M^1L^2T^{-2}]$

They have the same dimension.

The dimensional formula of couple is

0%
$[ML^{2}T^{-2}]$
0%
$[MLT^{-2}]$
0%
$[ML^{-1}T^{-3}]$
0%
$[ML^{-2}T^{-2}]$

Explanation

Dimensionally couple

$=Torque = Force\times distance=[MLT^{-2}][L]=[ML^{2}T^{-2}]$

The dimensional formula of arial velocity is

0%
$[M^0L^{-2}T^{1}]$
0%
$[M^0L^{-2}T^{-1}]$
0%
$[M^0L^{2}T^{-1}]$
0%
$[M^0L^{2}T^{1}]$

Explanation

In classical mechanics, areal velocity is the rate at which area is swept out by a particle as it moves along a curve.
Areal velocity

$=\frac {area}{time}$

$=[M^0L^{2}T^{-1}]$

$I$ is the moment of inertia and

$\omega$ the angular velocity, what is the dimensional formula of rotational kinetic energy,

$\dfrac {1}{2}I\omega^2$ ?

0%
$[ML^{2}T^{-1}]$
0%
$[M^2L^{-1}T^{-2}]$
0%
$[ML^{2}T^{-2}]$
0%
$[M^2L^{-1}T^{-1}]$

Explanation

Dimensionally rotational kinetic energy

$=Work$

$done=Energy$

$=[ML^{2}T^{-2}]$

Which of the following is the unit of molar gas constant?

0%
$JK^{-1} mol^{-1}$
0%
$J$
0%
$JK^{-1}$
0%
$Jmol^{-1}$

Explanation

Molar gas constant is given as

$R=\dfrac {PV}{nT}$
So, unit of

$R=\dfrac {Joule}{mol K}=JK^{-1} mol^{-1}$

The dimensional formula of velocity gradient is

0%
$[M^0L^0T^{-1}]$
0%
$[MLT^{-1}]$
0%
$[ML^0T^{-1}]$
0%
$[M^0LT^{-2}]$

Explanation

Velocity Gradient

$V$ is defined as rate of change in velocity per unit of distance. Mathematically, Velocity Gradient= velocity/distance.
Dimensionally

$V=\dfrac { \left[ { M }^{ 0 }{ L }^{ 1 }{ T }^{ -1 } \right] }{ \left[ L \right] } =\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ -1 } \right]$

Dimensions of '

$ohm$ ' are same as (where

$h$ is Planck's constant and

$e$ is charge)

0%
$\dfrac {h}{e}$
0%
$\dfrac {h^2}{e}$
0%
$\dfrac {h}{e^2}$
0%
$\dfrac {h^2}{e^2}$

Explanation

'ohm' is the unit of resistance and having dimension

$=[ML^2T^{-3}A^{-2}]$
From option C

$\Rightarrow \dfrac {h}{e^2}=\dfrac {ML^2T{-1}}{(AT)^2}=[ML^2T^{-3}A^{-2}]=Resistance (ohm)$

$C$ and

$L$ denote the capacitance and inductance, the dimensions of

$LC$ are

0%
$[M^0L^0T^{-1}]$
0%
$[M^0L^{-1}T^0]$
0%
$[M^{-1}L^{-1}T^0]$
0%
$[M^0L^0T^{2}]$

Explanation

The frequency

$v$ of a circuit consisting of capacitance and inductance is given as

$v=\dfrac {1}{2\pi \sqrt {LC}}$
So,

$[LC]=\dfrac {1}{(2\pi v)^2}=\dfrac {1}{(T^{-1})^2}=[T^2]=[M^0L^0T^2]$

$I$ is regarded as the fourth dimension, then the dimensional formula of charge in terms of current

$I$ is

0%
$[IT^2]$
0%
$[I^0T^0]$
0%
$[I^{-1}T^0]$
0%
$[IT]$

Explanation

An electric current is defined as a rate of flow of electric charge q.
So,

$I=\dfrac {q}{t}$ or

$[q]=[IT]$

Specific gravity has ___ dimensions in mass, ___ dimensions in length and ___ dimensions in time.

0%
0, 0, 0
0%
0, 1, 0
0%
1, 0, 0
0%
1, 1, 3

Explanation

Specific gravity is the ratio of density of substance and density of water at

${ 4 }^{ o }C$ . Hence it is the ratio of like quantities, therefore dimensionless.

CBSE Questions for Class 11 Engineering Physics Units And Measurement Quiz 3 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Units And Measurement Quiz 3 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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