Match the following Column-I Column-II A. Pressure E. $$ML^{-1} T^{-2}$$ B. Stress F. $$Nm^{-2}$$ C. Energy per unit volume G. $$M^{0} L^{0} T^{0}$$ D. Strain H. $$Jm^{-3}$$

A - E, F, H : B - E, F, H : C - E, F, H : D - G

A - E, F : B - E,H : C - E, F, H : D - G

A - E, H : B - E, F : C - E, F, H : D - G

A - E, F, H : B - E, F : C - E, F, H : D - G

MCQ Questions for Class 11 Engineering Physics Units And Measurement Quiz 6

The correct dimensional formula for impulse is given by

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

Explanation

$[Impulse]=[MLT^{-1}]$

Mass has both magnitude and direction.

0%
True
0%
False

Select the correct alternative
A metre ruler is graduated in:

0%
$m$
0%
$cm$
0%
$mm$
0%
$km$

The expression

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

0%
Pressure
0%
Kinetic energy
0%
Momentum
0%
Power

Explanation

The dimension

$[ML^2T^{-2}]$ is the dimension of the energy.

So, option

$B$ is correct.

The dimensions of pressure are

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

Explanation

We know that the pressure is defined as :

$Pressure=\dfrac{Force}{Area}$

So, the dimension of pressure is:

[Pressure] =

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

$MLT^{-1}$ represents the dimensional formula of

0%
Power
0%
Momentum
0%
Force
0%
Couple

Explanation

We know that momentum is defined as:

$p=mv$

Momentum =

$[MLT^{-1} ]$

The dimensional formula of wave number is

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

Explanation

Wave number is given as =

$\dfrac{1}{\lambda}$

$\therefore$ dimension is

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

The foundations of dimensional analysis were laid down by

0%
Gallileo
0%
Newton
0%
Fourier
0%
Joule

The dimensions of stress are equal to

0%
Force
0%
Pressure
0%
Work
0%
$\dfrac{1}{Pressure}$

Explanation

We know that the stress is defined as :

$Stress=\dfrac{Force}{Area}$

It is same as that of pressure.

So, [Pressure] =[stress] =

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

Dimensional formula of heat energy is

0%
$ML^2 T^{-2}$
0%
$ML T^{-1}$
0%
$M^0L^0 T^{-2}$
0%
None of these

Explanation

Heat is a form of energy and all form of energy have the same dimension as that of work done.

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

Dimensions of frequency are

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

Explanation

Frequency is defined as the reciprocal of the time period.

Frequency=

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

Dimensions of kinetic energy are

0%
$ML^2T^{-2}$
0%
$M^2LT^{-1}$
0%
$ML^2T^{-1}$
0%
$ML^3T^{-1}$

Explanation

The formula for the kinetic energy is written as:

Kinetic energy=

$\dfrac{1}{2}mv^2$

$=M[LT^{-1}]^2$

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

Dimensions of time in power are

0%
$T^{-1}$
0%
$T^{-2}$
0%
$T^{-3}$
0%
$T^{-0}$

Explanation

The power has the dimension of

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

So, the dimension of time in power is

$T^{-3}$ .

Match the physical quantities given in Column I with suitable dimensions expressed in Column II.

Column I	Column II
a) Angular momentum	1) ${ M }^{ -1 }{ L }^{ 3 }{ T }^{ -2 }$
b) Torque	2) ${ MLT }^{ -2 }$
c) Gravitational constant	3) ${ ML }^{ 2 }{ T }^{ -2 }$
d) Tension	4) ${ ML }^{ 2 }{ T }^{ -1 }$

0%
a-1, b-3, c-4, d-2
0%
a-1, b-4, c-3 d-2
0%
a-4, b-3, c-1, d-2
0%
a-1, b-2, c-4, d-3

Explanation

Angular momentum is given by

$\vec L =\vec r \times \vec p$ .

Thus, it's dimensions are

$[L]=L.MLT^{-1}=ML^{2}T^{-1}$
Torque is given as

$\vec r \times \vec F$ .

Thus, dimensions are

$L.MLT^{-2}=ML^2T^{-2}$ .
Tension is force and hence its dimensions are

$MLT^{-2}$ .
Gravitational constant relationship is

$F=G\dfrac{Mm}{r^2}$
Thus,

$[G]=\dfrac{[F][r^2]}{[Mm]}=\dfrac{MLT^{-2}.L^2}{M^2}=M^{-1}L^3T^{-2}$
Thus, option C is correct.

Which is not a dimensionless quantity is :

0%
Moment of Momentum
0%
Moment of force
0%
Moment of inertia
0%
All of the above

Explanation

Hint: a dimensionless quantity is a quantity to which no physical dimension is assigned.

${\textbf{Explanation:}}$

A. Moment of Momentum

$= m v r$

Dimension:

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

B. Moment of force

$= r \times F$

Dimension:

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

C. Moment of Inertia

$= Mr^{2}$

Dimension:

$= M^{1} L^{2}$

Hence, none of the quantity is dimensionless.

${\textbf{Correct option: D}}$

A physicist performs an experiment and takes

$200$ readings. He repeats the same experiment and now takes

$800$ readings. By doing so :

0%
the probable error remains same.
0%
the probable error is four times.
0%
the probable error is halved.
0%
the probable error is reduced by a factor $\dfrac{1}{4}$ .

Explanation

More the number of experiments we do, improvement in our output will be achieved. And on an average our expected value of output will become more accurate. Therefore, probable error which is determined on average basis will get reduced by

$\dfrac{1}{4}$ in this case.

Consider the following two statements A and B. Identify the correct answer.
A) The quantity

$\dfrac { { e }^{ 2 } }{ { \epsilon }_{ 0 }ch }$ is dimensionless.
B)

$\dfrac { 1 }{ \sqrt { { \mu }_{ 0 }{ \epsilon }_{ 0 } } }$ has the dimensions of velocity and is numerically equal of velocity of light.

0%
A is true but B is false
0%
B is true but A is false
0%
Both A and B are true
0%
Both A and B are false

Explanation

Dimensions of permittivity

${ \epsilon }_{ 0 }$ :

${ A }^{ 2 }{ T }^{ 4 }{ M }^{ -1 }{ L }^{ 3 }$

Dimensions of

${ e }^{ 2 }$ :

${ A }^{ 2 }{ T }^{ 2 }$

Dimensions of

$c$ :

$L{ T }^{ -1 }$

Dimensions of

$h$ :

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

Therefore, dimensions of the given expression:

$\dfrac { { A }^{ 2 }{ T }^{ 2 } }{ ({ A }^{ 2 }{ T }^{ 4 }{ M }^{ -1 }{ L }^{ 3 })\times (L{ T }^{ -1 })\times (M{ L }^{ 2 }{ T }^{ -1 }) } =1$ which is dimensionless.

Hence, Statement A is true.

In statement B, the expression given is for the speed of light.Hence, it has dimensions of velocity.

Choose the false statement from given statements.
I. Relative permittivity is dimensionless variable.
II. Angular displacement has neither units nor dimensions.
III. Refractive index is dimensionless variable.
IV. Permeability of vacuum is dimensional constant.

0%
Only I and II
0%
Only II
0%
Only III
0%
Only IV

Explanation

Statements 1 and 3 are true because both these are ratios of two quantities with same dimensions, thus they are dimensionless.

Statement 4 is also correct because permeability of vacuum is a dimensionless constant with value equal to

$8.85\times 10^{-12}$ .
But angular displacements has unit (although no dimensions). It has the unit 'radian'.

The dimensional formula for strain energy density is:

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

Explanation

Strain energy density

$= \dfrac{Energy}{Volume} \\= \dfrac{M^{1}L^{2}T^{-2}}{L^{3}} \\= M^{1}L^{-1}T^{-2}$

Some physical constants are given in List - I and their dimensional formulae are given in List - II. Match the following :

List - I	List - II
A) Planck's constant	e) $\left[ { ML }^{ -1 }{ T }^{ -2 } \right]$
B) Gravitational constant	f) $\left[ { ML }^{ 1 }{ T }^{ -1 } \right]$
C) Bulk modulus	g) $\left[ { ML }^{ 2 }{ T }^{ -1 } \right]$
D) Linear momentum	h) $\left[ { M}^{ -1 }{ L }^{ 3 }{ T }^{ -2 } \right]$

0%
A-h, B-g, C-e ,D-f
0%
A-f, B-e, C-g, D-h
0%
A-g, B-f, C-e, D-h
0%
A-g, B-h, C-e, D-f

Explanation

Planck's constant

$\displaystyle =\frac{\left ( energy \right )}{\left ( c/\lambda \right )}=\frac{ML^{2}T^{-2}}{\dfrac{LT^{-1}}{L}}=ML^{2}T^{-1}$
Gravitational constant

$\displaystyle =\frac{force}{\left ( M_{1}M_{2} \right )}\times r^{2}=\frac{ML^{1}T^{-2}\times L^{2}}{M^{2}}$

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

Bulk modulus

$\displaystyle =\frac{dp}{dv/v}=\frac{M^{1}L^{-1}T^{-2}}{L^{3}/L^{3}}=M^{1}L^{-1}T^{-2}$

Linear momentum

$={ M }{ L }{ T }^{ -1 }$

Hence, option D is correct.

The pair of physical quantities that have same dimensions are:
a) Reynold's number and coefficient of friction
b) Latent heat and gravitational potential
c) Curie and frequency of light wave
d) Planck constant and torque

0%
b and c are correct
0%
a and b are correct
0%
a, b and c are correct
0%
all are correct

Explanation

Coefficient of friction and Reynolds no. have no dimensions.
Planck's constant

$=ML^{2}T^{-1}$
Torque

$=ML^{2}T^{-2}$
Latent heat

$=\dfrac{energy}{mass}=\dfrac{M^{1}L^{2}T^{-2}}{M}=L^{2}T^{-2}$
Gravitational potential

$=\dfrac{GM}{r}=\dfrac{force}{mass}\times r$

$=\dfrac{M^{1}L^{1}T^{-2}}{M^{1}}\times L^{1}$

$=L^{2}T^{-2}$
Curie

$=$ freq

$=M^{0}L^{0}T^{-1}$

Which of the following pairs have same dimensions.
a) Torque and work
b) Angular momentum and work
c) Energy and Youngs modulus
d) Light year and wavelength

0%
a and b are correct
0%
b and c are correct
0%
c and d are correct
0%
d and a are correct

Explanation

Torque $=$ Work $=$ Energy $=[ML^{2}T^{-2}]$
Angular momentum $=mvr=[ML^{2}T^{-1}]$
Young's modulus $=[ML^{-1}T^{-2}]$
Light year and wavelength $=[M^{0}L^{1}T^{0}]$
Hence option D is correct.

Names of units of some physical quantities are given in List - I and their dimensional formulae are given in List - II. Match the correct pair of the lists.

List - I	List - II
a) Pa-s	e) $\left[ { L }^{ 2 }{ T }^{ -2 }{ K }^{ -1 } \right]$
b) ${ NmK }^{ -1 }$	f) ${ MLT }^{ -3 }{ K }^{ -1 }$
c) ${ J kg }^{ -1 }{ k }^{ -1 }$	g) ${ ML }^{ -1 }{ T }^{ -1 }$
d) ${ Wm }^{ -1 }{ k }^{ -1 }$	h) $\left[ { ML }^{ 2 }{ T }^{ -2 }{ K }^{ -1 } \right]$

0%
a-h, b-g, c-e, d-f
0%
a-g, b-f, c-e, d-h
0%
a-g, b-e, c-h, d-f
0%
a-g, b-h, c-e, d-f

Explanation

$Pa-s=M^{1}L^{-1}T^{-2}\times T^{1}=ML^{-1}T^{-1}$

$Nmk=M^{1}L^{1}T^{-2}\times L^{1}\times K^{1}=M^{1}L^{2}T^{-2}K^{-1}$

$Jkg^{-1}K^{-1}=ML^{2}T^{-2}M^{-1}K^{-1}=M^{0}L^{2}T^{-2}K^{-1}$

$Wm^{-1}K^{-1}=ML^{2}T^{-3}L^{-1}K^{-1}=MLT^{-3}K^{-1}$
Hence option D is correct.

Watt per meter kelvin is the unit of

0%
Stress
0%
Period of revolution of satellite
0%
Angular displacement
0%
Coefficient of thermal conductivity

Explanation

$\dfrac { Watt }{ \left( meter \right) \times \left( kelvin \right) } =\dfrac { M{ L }^{ 2 }{ T }^{ -3 } }{ LK } =M{ L }{ T }^{ -3 }{ K }^{ -1 }$
Coefficient of thermal conductivity

$\left( k \right) =\dfrac { \left( \dfrac{q}{A} \right) \times \ t }{ dT }$

$\left( q/A \right) =$ heat transfer per unit area

$\left( W/{ m }^{ 2 } \right)$

$dT=$ temperature difference

$t=$ thickness

$\left[ k \right] =\dfrac { \left[ M{ L }^{ 2 }{ T }^{ -3 } \right] \left[ { L }^{ -2 } \right] \left[ L \right] }{ \left[ K \right] } =M{ L }{ T }^{ -3 }{ K }^{ -1 }$

Match List I with List II and select the correct answer.

List - I	List - II
A) Spring constant	I) [ ${ M }^{ 1 }{ L }^{ 2 }{ T }^{ -2 }$ ]
B) Pascal	II) [ ${ M }^{ 0 }{ L }^{ 0 }{ T }^{ -1 }$ ]
C) Hertz	III)[ ${ M }^{ 1 }{ L }^{ 0 }{ T }^{ -2 }$ ]
D) Joule	IV) [ ${ M }^{ 1 }{ L }^{ -1 }{ T }^{ -2 }$ ]

0%
A-III, B-IV, C-II, D-I
0%
A-IV, B-III, C-I, D-II
0%
A-IV, B-III, C-II, D-I
0%
A-III, B-IV, C-I, D-II

Explanation

$Spring \ constant, \ k=\dfrac{F}{x}=\dfrac{MLT^{-2}}{L}=[M^1L^0T^{-2}$ ]

$Hertz=\dfrac{1}{t}=[M^0L^0T^{-1}$ ]

$Pascal=\dfrac{F}{A}=\dfrac{MLT^{-2}}{L^2}=[M^1L^{-1}T^{-2}$ ]

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

Study the following.

List - I	List - II
a) Same negative dimensions of mass	I) Pressure, Rydberg constant
b) Same negative dimensions of length	II) Mangnetic induction field, Potential
c) Same dimensions	III) Capacity, Universal gravitational constant
d) Same dimension of current	IV) Energy density, Surface tension

0%
a-III, b-I, c-IV, d-II
0%
a-III, b-IV, c-I, d-II
0%
a-I, b-II, c-III, d-IV
0%
a-II, b-I, c-IV, d-III

Explanation

Pressure=

$\dfrac{force}{area}=\dfrac{mass\times acc^{n}}{area}=\dfrac{MLT^{-2}}{L^{2}}=ML^{-1}T^{-2}$

Unit of Rydberg constant

$=L^{-1}$ ( These 2 quantities have same -ve dimension of length)

Surface tension

$=M^{1}L^{0}T^{-2}$
Energy density

$=$ Energy per unit area

$=\dfrac{ML^{2}T^{-2}}{L^{2}}=ML^{0}T^{-2}$

$\rightarrow$ (same dimensional formula)

Potential

$=ML^{2}T^{-3}A^{-1}$
Magnetic induction field

$=ML^{2}A^{-1}T^{-4}$ (same dimension for current)

Match the physical quantities given in Column I with suitable dimensions expressed in Column II.

Column I	Column II
a) Angular momentum	g) ${ ML }^{ 2 }{ T }^{ -2 }$
b) Latent heat	h) ${ ML }^{ 2 }{ Q }^{ -2 }$
c) Torque	i) ${ ML }^{ 2 }{ T }^{ -1 }$
d) Capacitane	j) ${ ML }^{ 3 }{ T }^{ -1 }{ Q }^{ -2 }$
e) Inductance	k) ${ M }^{ -1 }{ L }^{ -2 }{ T }^{ 2 }{ Q }^{ 2 }$
f) Resistivity	l) ${ L }^{ 2 }{ T }^{ -2 }$

0%
a-i, b-l, c-g, d-k, e-h, f-j
0%
a-l, b-i, c-k, d-g, e-j, f-h
0%
a-i, b-l, c-h, d-j, e-g, f-k
0%
a-h, b-j, c-g, d-k, e-i, f-l

Explanation

Angular momentum =

$mvr=ML^2T^{-1}$

Latent Heat,

$Q=ML, \dfrac{Q}{M}=L, [L]=M^0L^2T^{-2}$

Torque

$=F\times perpendicular \ distance= MLT^{-2}L=ML^2T^{-2}$

Inductance

$=\dfrac{2E}{I^2}=L=\dfrac{ML^2T^{-2}}{A^2}=ML^2T^{-2}A^{-2}$

Capacitance

$C=\dfrac{Q^2}{2E}=\dfrac{A^2T^2}{ML^2T^{-2}}$

Resistivity

$=ML^{3}T^{-1}Q^{-2}$

Option A is the appropriate choice.

The correct order in which the dimensions of length increases in the following physical quantities is:
a) permittivity
b) resistance
c) magnetic permeability
d) stress

0%
a, b, c, d
0%
d, c, b, a
0%
a, d, c, b
0%
c, b, d, a

Explanation

a) Permittivity

$=M^{-1}L^{-3}T^{4}I^{2}={\epsilon}_{0}$
b) Resistance

$=ML^{2}T^{-1}Q^{-2}$
c) Magnetic permeability

$=\mu_{0}=MLT^{-2}I^{-2}$
d) Stress

$=F/A=ML^{-1}T^{-2}$
Hence, dimensions of length in a<d<c<b.

The dimensional formula for pressure gradient is :

0%
${ ML }^{ -1 }{ T }^{ -2 }$
0%
${ M }^{ 1 }{ L }^{ -2 }{ T }^{ -2 }$
0%
${ M }^{ 1 }{ L }^{ 2 }{ T }^{ -2 }$
0%
${ M }^{ 1 }{ L }^{ -1 }{ T }^{ -3 }$

Explanation

Pressure gradient

$=\dfrac{dp}{dx}$

$\displaystyle p=\frac{force}{Area}=\left ( kg\frac{m}{s^{2}} \right )\left ( \frac{1}{m^{2}} \right )=ML^{-1}T^{-2}$

Where,

$x=L\left ( M^{0} L^{1}T^{0}\right )$

So,

$\displaystyle \frac{dp}{dx}=\frac{ML^{-1}T^{-2}}{M^{0}L^{1}T^{0}}=ML^{-2}T^{-2}$

L, C and R represents inductance, capacitance and resistance respectively. The combinations that have the dimensions of frequency are :

0%
$\dfrac { 1 }{ CR }$
0%
$\dfrac { R }{ L }$
0%
$\dfrac { 1 }{ \sqrt { LC } }$
0%
All of the above

Explanation

We know that unit for the frequency is per second i.e.

$[T]^{ -1 }$

Now let us consider the dimensions of the given physical quantities.

We know that Capacitance i.e.

$C=Q/V$

where Q is electric charge and V is electric potential.

Now we know that Q= It where I is current and t is time.

Also that V=IR where R is resistance.

So,

$C= It/IR$

Thus,

$C= t/R$

also

$1/t=1/RC$

Thus dimension of

$1/RC$ will be

$[T^{-1}]$ that is same as that of frequency.

Now Inductance(L) can be given as

$L= Vt/I$

And also

$V=IR$

Therefore

$L=IRt/I$

Thus,

$R/L = 1/t$

It means both A and B are correct now checking for the third option

$LC= Rt * t/R = t^2$

That gives

$1/\sqrt { LC } = 1/t$

Thus dimension of

$1/\sqrt{LC}$ will be

$T^{-1}$ that is same as that of frequency.

The dimensions of resistivity in terms of M, L, T and Q, where Q stands for the dimensions of charge is :

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

Explanation

$R=\dfrac{\rho l}{A}$
Power

$=I^2R$
Dimension of R is

$\dfrac{ML^2{T}^{-1}}{Q^2T^{-2}}=ML^2TQ^{-2}$
Now by equation,

$\rho=\dfrac{RA}{L}=ML^2TQ^{-2}L=ML^3TQ^{-2}$
Hence, option A is correct.

Dimension of specific heat is:

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

Explanation

$Q=ms\delta\theta$
Dimension for specific heat will be

$=\dfrac{ML^2T^{-2}}{MK}$

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

The dimensional formula for latent heat is :

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

Explanation

$\textbf {Hint :}$ Use latent heat $L$ = $\dfrac{Q}{m}$

${\textbf{Explanation:}}$

Latent heat, $L$ = $\dfrac{Q}{m}=\dfrac{Energy\ released\ or\ absorbed}{Mass}$

Dimensional formula of $L$ $=\dfrac{M^{1}L^{2}T^{-2}}{M^{1}}$

$=M^{0}L^{2}T^{-2}$

${\textbf{Correct option: C}}$

Match the following

Column-I	Column-II
A. Pressure	E. $ML^{-1} T^{-2}$
B. Stress	F. $Nm^{-2}$
C. Energy per unit volume	G. $M^{0} L^{0} T^{0}$
D. Strain	H. $Jm^{-3}$

0%
A - E, F, H : B - E, F, H : C - E, F, H : D - G
0%
A - E, F : B - E,H : C - E, F, H : D - G
0%
A - E, H : B - E, F : C - E, F, H : D - G
0%
A - E, F, H : B - E, F : C - E, F, H : D - G

Explanation

Stress

$= Pressure = \dfrac{force}{area}=\dfrac{N}{M^{2}}=\dfrac{M^{1}L^{1}T^{-2}}{L^{2}}=M^{1}L^{-1}T^{-2}$

Energy per unit volume

$= \dfrac{M^{1}L^{2}T^{-2}}{L^{3}}=M^{1}L^{-1}T^{-2}$

Strain

$= \dfrac{\Delta l}{l}=M^{0}L^{0}T^{0}$
Hence, A, B, & C each has E, F, & H options dimensionally
while D has only G.
Hence, option A is correct.

$L$ is the inductance,

$'i'$ is current in the circuit,

$\dfrac { 1 }{ 2 } { Li }^{ 2 }$ has the dimensions of:

0%
Work
0%
Power
0%
Pressure
0%
Force

Explanation

$L \rightarrow M{ L }^{ 2 }{ I }^{ -2 }{ T }^{ -2 }$

$I\rightarrow I$

${ LI }^{ 2 } = (M{ L }^{ 2 }{ I }^{ -2 }{ T }^{ -2 }){ I }^{ 2 }= M{ L }^{ 2 }{ T }^{ -2 } \rightarrow (unit \ of \ work)$

$Power = f\times v = ML{ T }^{ -2 } \times L{ T }^{ -1 }\quad =\quad M{ L }^{ 2 }{ T }^{ -3 }$

$Pressure = f/A = \dfrac { ML{ T }^{ -2 } }{ { L }^{ 2 } } = { ML }^{ -1 }{ T }^{ -2 }$

$force = { M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 }$

Dimensional formula of magnetic field of induction (B) is:

Given Force on a charged particle is given by: $F = qvBsin \theta$ where $\theta$ is angle between velocity and magnetic field.

0%
$ML^{2}T^{2}I^{-1}$
0%
$MLT^{2}I^{-1}$
0%
$MT^{-2}I^{-1}$
0%
$MT^{2}I^{1}$

Explanation

It is given:

$F=qvBsin \theta$

Hence,

$B= \dfrac{F}{qvsin\theta}$

Dimensional formula of magnetic field is:

$[B] = [F/qv]$ as

$sin \theta$ is dimensionless

$[B] = [MLT^{-2}][I^{-1}T^{-1}][L^{-1}T]$

$[B] = [MT^{-2}I^{-1}]$

The units of force, velocity and energy are 100 dyne, 10

${cm s }^{ -1 }$ and 500 erg respectively. The units of mass, length and time are:

0%
5 g, 5 cm, 5 s
0%
5 g, 5 cm, 0.5 s
0%
0.5 g, 5 cm, 5 s
0%
5 g, 0.5cm, 5 s

Explanation

Force

$= 100 dyne= M^{1}L^{1}T^{-2}$

$velocity = 10 \dfrac{cm}{s}=LT^{-1}$

$Energy = 500 erg =M^{1}L^{2}T^{-2}\Rightarrow \dfrac{energy}{force}=L^{1}=\dfrac{500}{100}=5cm$

$\Rightarrow \dfrac{energy}{(velocity)^{2}}=M^{1}=\dfrac{500}{(10)^{2}} =5gm \Rightarrow T=0.5\ sec$

Resistance of an ideal Ammeter is zero and that of an ideal Voltmeter is infinite. A Voltmeter and an Ammeter are connected in the circuit as shown. Resistance of ammeter is say

$R/10$ and that of Voltmeter is

$10 R$ . Then:

0%
Percentage error in the reading of Ammeter (compared to that measured, if both Ammeter and voltmeter were ideal) is $1.0$ %.
0%
Percentage error in the reading of voltmeter is $10.0$ %
0%
Both (a) and (b) are correct
0%
Both (a) and (b) are wrong

Explanation

As per the given question, actual current flowing through the circuit is,

$I=\dfrac {E}{R}$
Now, the voltmeter is connected in parallel to resistance R, hence

$\dfrac{1}{R_v} = \dfrac{1}{R} + \dfrac{1}{10R} = \dfrac{11}{10R}$

$R_v = \dfrac{10R}{11}$
The voltmeter and ammeter both are ideal hence, the measured current is,

$I = \dfrac{E}{R_a+R_v}$
where,

$R_a$ and

$R_v$ are resistances of ammeter and voltmeter respectively.

$I =\dfrac {E}{\dfrac {R}{10}+ \dfrac {10R}{11}}$

$I =\dfrac {E}{\dfrac {111R}{110}}$

$I =\dfrac {110E}{111R} = 0.99 \dfrac{E}{R}$
Therefore, percentage error in the reading of ammeter is:

$(\frac{\dfrac{E}{R}-0.99 \dfrac{E}{R}}{\dfrac{E}{R}}) \times 100 = \dfrac{1-0.99}{1} \times 100 = 0.01 \times 100 = 1$ %
Also, Actual voltage through R is E = IR
And measured voltage is

$V = I(\dfrac {10R}{11})$

$V = 0.9 IR = 0.9E$
Therefore percentage error in the reading of voltmeter is:

$(\dfrac{E-0.9E}{E}) \times 100 = \dfrac{1-0.9}{1} \times 100 = 0.1 \times 100 =10$ %

The unit of an electrical parameter whose formula is

$[M^{1}L^{2}T^{-3}A^{-2}]$ is:

0%
ohm
0%
ampere
0%
volt
0%
newton

Explanation

$V = IR$

So,

$[R] = [\dfrac{V}{I}]$

$= [\dfrac{W}{QI} ]= [\dfrac{W}{I^{2} T}] = [\dfrac{F \times S}{I^{2} \times T}] = [\dfrac{m \times a \times S}{I^{2} \times T}]$

$= [\dfrac{M(\dfrac{L}{T^{2}}) \times L}{A^{2} \times T}]$

$\Rightarrow [\Omega]= [ML^{2}T^{-3}A^{-2}]$

$e, { \epsilon}_{0}, h$ and

$c$ respectively represents electric charge, permittivity of free space, Planck's constant and speed of light. The dimension of

$\dfrac{e^2 }{ 4\pi \epsilon_0 hc}$ is :

0%
$[M^0 L^0 T^0]$
0%
$[M^1 L^0 T^0]$
0%
$[M^0 L^1 T^0]$
0%
$[M^0 L^0 T^1]$

Explanation

Planck's constant (h) has joule-seconds unit, dimensions are

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

Electron charge (e) will have

$[TI]$

Dimension for

${ \varepsilon }_{ 0 }$ is $[M^{-1}L^{-3}T^{4}I^{2}]$

Dimensions for

$\dfrac { e^{ 2 } }{ 4\pi { \varepsilon }_{ 0 }hc }$ will be

$[M^0L^0T^0]$ .

SI unit of a physical quantity whose dimensional formula is

${ M }^{ -1 }{ L }^{ -2 }{ T }^{ 4 }{ A }^{ 2 }$ is :

0%
$ohm$
0%
$volt$
0%
$siemen$
0%
$farad$

Explanation

Capacitance

$C=\dfrac{Q^2}{2E}=\dfrac{A^2T^2}{ML^2T^{-2}}=M^{-1}L^{-2}T^4A^2$

Option D: farad is unit for capacitance.

The product of energy and time is called action. The dimensional formula for action is same as that for

0%
$force \times velocity$
0%
$impulse \times distance$
0%
$power$
0%
$angular \times energy$

Explanation

$Energy \times Time = \left ( M^{1}L^{2}T^{-2} \right )\times \left ( T^{1} \right )= M^{1}L^{2}T^{-1}$

$Force \times Velocity = \left ( M^{1}L^{1}T^{-2} \right )\times \left ( L^{1}T^{-1} \right ) = M^{1}L^{2}T^{-3}$

$Impulse \times distance = \left ( M^{1}L^{1}T^{-1} \right )\times \left ( L^{1} \right ) = M^{1}L^{2}T^{-1}$

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

$Angular \ Energy = M^{1}L^{1}T^{-2}$

A highly rigid cubical block

$A$ of small mass

$M$ and side

$L$ is fixed rigidly on to another cubical block of same dimensions and of low modulus of rigidity

$\eta$ such that the lower face of

$A$ completely covers the upper face

$B$ . The lower face of

$B$ is rigidly held on a horizontal surface. A small force

$F$ is applied perpendicular to one of the side face of

$A$ . After the force is withdrawn, block

$A$ executes small oscillations, the time period of which is given by :

0%
$2\pi \sqrt{M\eta L}$
0%
$2\pi \sqrt{\dfrac{M\eta}{ L}}$
0%
$2\pi \sqrt{\dfrac{ML}{\eta}}$
0%
$2\pi \sqrt{\dfrac{M}{L\eta}}$

Explanation

Modulus of rigidity

$\eta =\dfrac { E }{ A\theta }$

We know,

$A={ L }^{ 2 }$

For small

$\theta$ ,

$\theta =\dfrac { x }{ L }$

Restoring force

$F=-\eta AB$

$F=-\eta Lx$

Acceleration,

$a=-\dfrac { \eta L }{ M } x\qquad \rightarrow (1)\\$

$a=\dfrac { F }{ m }$

Here,

$\alpha \propto -x\\$

$\therefore$ here we have SHM,

Hence the time period is given by

$T=2x\sqrt { \left| \dfrac { x }{ a } \right| } \\ T=2\pi \sqrt { \dfrac { M }{ \eta L } }$

where,

$a$ =acceleration and

$x$ =small deformation.

The dimensions of

$\left ( \mu _{0}\varepsilon _{0} \right )^{-1/2}$ are:

0%
$\left [ L^{-1}T \right ]$
0%
$\left [ LT ^{-1}\right ]$
0%
$\left [ L^{1/2} T^{1/2}\right ]$
0%
$\left [ L^{1/2} T^{-1/2}\right ]$

Explanation

We know that the speed of an electromagnetic wave is given by :

$c=\dfrac{1}{\sqrt{\mu _{0}\varepsilon _{0}}}$

Therefore,

$(\mu_0\varepsilon_0)^{-1/2}$ has the dimension the same as that of the speed.

$\therefore dimension \,\,LT^{-1}$

A is a tank filled to its 75% with water, B is a weighing balance and C is a stone hung from a stand. If fig. 1 is correct, what do you expect to be the position of needle in fig. 2?

A person was weighing

$102.1\ kg$ last week and gained

$0.28 \ kg$ this week. His weight as of now is correctly expressed as :

0%
$102.38\ kg$
0%
$102.3\ kg$
0%
$102.4\ kg$
0%
$102\ kg$

Explanation

We know that number of significant figures after addition will be equal to least number of significant figures present in any of the operand.
Therefore total weight after weight gain is

$102.1+0.28=102.38 kg$
Now least number of significant figure present after the decimal in the operand is 1, present in

$102.1\ kg$ .
Therefore, the number of significant figures present after the calculation should also be

$1$ after the decimal. Hence, when we round the answer

$102.38$ becomes

$102.4\ kg$
Therefore correct answer is option (C)

The string of a kite is 100 m long and it makes

an angle of $60^{0}$ with the horizontal. If there is no slack in

the string, find the height of the kite from the ground.

0%
$50 \sqrt{3} m$
0%
$100 \sqrt{3} m$
0%
$50 \sqrt{2} m$
0%
100 m

The velocity

$v$ of a particle at time

$t$ is given by

$v = at + \frac{b} {t + c}$ , where

$a, b\ and\ c$ are constants.The dimensions of

$a, b\ and\ c$ are respectively:

0%
LT $^{-2}$ , L and T
0%
L $^{2}$ , T and LT $^{2}$
0%
LT $^{2}$ , LT and L
0%
L, LT and T $^{2}$

Explanation

Two terms can be added or subtracted if their dimensions are same.

Thus

$c$ has the dimension of time i.e.

$[T]$

Dimension of

$v$ and

$at$ must be same.

Thus dimensions of

$[a] = \dfrac{[v]}{[t]} = \dfrac{[LT^{-1}]}{[T]} = [LT^{-2}]$

Dimensions

$[b] = [v][t] = [LT^{-1}][T] = L$

Of the four figures given below,

$x_0$ is known value of the outcome in an experiment. Which of the following data plot a precise but not an accurate measurement?

The dimension of the ratio of angular momentum to linear momentum is

0%
$L^0$
0%
$L^1$
0%
$L^2$
0%
$MLT$

Explanation

Angular momentum is given by:

$L = r \times mv$
Linear momentum is given by:

$p = mv$
Ratio of angular momentum to the linear momentum is:

$\dfrac{L}{p} = \dfrac{r\times mv}{mv} = r$
Unit of

$r$ is meter. Hence, dimension of

$r$ is

$[L^1]$
Therefore, the dimension of the ratio of angular momentum and linear momentum is:

$\dfrac{L}{p} = [L^1]$

CBSE Questions for Class 11 Engineering Physics Units And Measurement Quiz 6 - 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 6 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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