MCQ Questions for Class 11 Engineering Physics Units And Measurement Quiz 5

The extent to which a given measurement agrees with the standard value for that measurement is known as :

0%
Accuracy
0%
Precision
0%
Accuracy and Precision
0%
None of these

Explanation

Accuracy is defined as the extent upto which a given measurement agrees with the standard value for that measurement.
For example : Actual weight of the given substance is

$10.0 \ kg$ and the measured value of the mass is

$9.9 \ kg$ . This measurement is said to be accurate.

The ratio of dimensions electric field and magnetic field is,

0%
$MT^{1}L^{-1}$
0%
$MT^{-1}L^{-2}$
0%
$MT^{2}L^{-1}$
0%
$LT^{-1}$

Explanation

Electric field

$=$ force/charge or

$E=F/q$

So,

$[E]=\dfrac{[MLT^{-2}]}{[IT]}=[MLT^{-3}I^{-1}]$ where

$[I]=$ dimension of current.

We know that magnetic force

$F=qvB$ or

$B=\dfrac{F}{qv}$

So,

$[B]=\dfrac{[MLT^{-2}]}{[IT][LT^{-1}]}=[MT^{-2}I^{-1}]$

Thus,

$\dfrac{[E]}{[B]}=\dfrac{[MLT^{-3}I^{-1}]}{MT^{-2}I^{-1}}=[LT^{-1}]$

On dividing dimension of gravitational constant by the dimension of surface tension we get

0%
$M$
0%
$ML^{2}$
0%
$M^{-2}L^{3}$
0%
None of he above

Explanation

By Newton's law of gravitation,

$F=\dfrac{Gm_1m_2}{r^2}$ or

$G=\dfrac{Fr^2}{m_1m_2}$

So,

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

Surface tension

$=$ force/length or

$S=F/l$

So,

$[S]=\dfrac{[MLT^{-2}]}{[L]}=[MT^{-2}]$

Thus,

$\dfrac{[G]}{[S]}=\dfrac{[M^{-1}L^3T^{-2}]}{[MT^{-2}]}=[M^{-2}L^3]$

The mass of an object is measured by two students (true weight is 12 kg)
Student 1: He measures 12 kg
Student 2: She measures 12.5 kg
Measurement of which student is more accurate.

0%
Student 1
0%
Student 2
0%
Both are equally accurate
0%
Cannot be compared.

Explanation

Accuracy defines nearness of a measured value to the standard or true value. Since value measured by student 1 is nearer to the actual value

$12 \ kg$ compared to that measured by student 2. So measurement of student 1 is more accurate.

In lab one student obtained a weight measurement of

$3.2 kg$ for a given substance, but the actual or known weight is

$10 kg$ , then his measurement is :

0%
accurate
0%
not accurate
0%
precised
0%
not presiced

The dimensions

$[ML^{-1}T^{-2}]$ may be of

0%
pressure
0%
energy per unit volume
0%
force
0%
linear momentum

Explanation

Pressure

$P=$ force/ area

$=F/A$

$[P]=[MLT^{-2}]/[L^2]=[ML^{-1}T^{-2}]$

Energy per unit volume ,

$E=$ work done/volume

$=$ (force

$\times$ displement)/volume

$[E]=\dfrac{[MLT^{-2}][L]}{[L^3]}=[ML^{-1}T^{-2}]$

The dimension of force,

$[F]=[MLT^{-2}]$ and linear momentum,

$[p]=[mv]=[MLT^{-1}]$

The dimensions of

$\dfrac{QV}{C}$ is,(

$Q$ =charge,

$V$ =voltage,

$C$ =capacitance,take dimension of current as

$k$ )

0%
$M^{2}L^{-4}T^{-6}k^{-1}$
0%
$M^{2}L^{4}T^{-6}k^{-2}$
0%
$M^{2}L^{-4}T^{6}k^{-2}$
0%
$M^{2}L^{4}T^{-6}k^{1}$

Explanation

We know that,

$Q=CV$ so

$\dfrac{QV}{C}=\dfrac{(CV)V}{C}=V^2$

So we have to find out only dimension of

$V^2$

Volatge

$V=Ed=\dfrac{F}{Q}d$

We know that dimension of force,

$[F]=[MLT^{-2}]$ , distance

$[d]=[L]$ and charge

$[Q]=[kT]$ (as charge

$=$ current

$\times$ time)

Now,

$[V]=\dfrac{[MLT^{-2}]}{[kT]}[L]=[ML^2T^{-3}k^{-1}]$

Thus,

$[V^2]=[M^2L^4T^{-6}k^{-2}$ ]

Dimension of energy in terms of force (

$F$ ), length(

$L$ ) and time (

$T$ ) is

0%
$F$ $L$
0%
$L$ $T$
0%
$LF\times\dfrac{1}{T}$
0%
none of the above

Explanation

We know that, energy

$E=$ work done

$(W)=$ force

$\times$ displacement

$=F\times L$

Thus,

$[E]=[FL]$

________ refers to the degree of conformity and correctness of something when compared to a true or absolute value. Fill the appropriate answer.

0%
Accuracy
0%
Precision
0%
Accuracy and Precision
0%
None of these

________ reflects quality of measuring instrument.

0%
Precision
0%
Accuracy
0%
Accuracy and Precision
0%
None of these

The degree to which several measurement provide answer very close to each other is known as:

0%
Accuracy
0%
Precision
0%
Accuracy and Precision
0%
None of these

Which of the following pair of quantities do not have the same dimensions?

0%
Light year,time period
0%
Light year,distance
0%
Impedance,resistance
0%
None of these

Explanation

Light year is the unit of astronomical distance. It is equal to the distance travels by light in a year. So its dimension is

$[L]$ .

The dimension of time period is

$[T]$

As impedance and resistance both have same unit, ohm so they will have same dimension.

Match the following.

	List I		List II
A.	Angular momentum	1.	$[M^{-1}L^2T^{-2}]$
B.	Torque	2.	$[M^1T^{-2}]$
C.	Gravitational constant	3.	$[M^1L^2T^{-2}]$
D.	Tension	4.	$M^1L^2T^{-1}]$

0%
C-2, D-1
0%
A-4, B-3
0%
A-3, C-2
0%
B-2, A-1

Explanation

$\left[ L \right] ={ M }^{ 1 }{ L }^{ 2 }{ T }^{ -1 }\\ \left[ \tau \right] ={ M }^{ 1 }{ L }^{ 2 }{ T }^{ -2 }\\ \left[ G \right] ={ M }^{ -1 }{ L }^{ 3 }{ T }^{ -2 }\\ \left[ T \right] ={ M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 }$

Which of the following is not a unit of mass?

0%
Slug
0%
Kilogram
0%
Ton
0%
Newton

Explanation

The slug is the unit of mass in the US common system of units.

It is a mass that accelerates by

$1\ ft/s^2$ when

$1$ pound force is applied.

The kilogram (kg) is the SI unit of mass.

Ton is also represent the unit of mass and

$1$ ton is equal to

$1000$ kg.

Newton is the SI unit of force.

The dimensions of universal gravitational constant are ___________.

0%
$\left[ { L }^{ 1 }{ M }^{ 0 }{ T }^{ 0 } \right]$
0%
$\left[ { L }^{ 2 }{ M }^{ 1 }{ T }^{ 0 } \right]$
0%
$\left[ { L }^{ -1 }{ M }^{ 1 }{ T }^{ -2 } \right]$
0%
$\left[ { L }^{ 3 }{ M }^{ -1 }{ T }^{ -2 } \right]$

Explanation

$\textbf{Step 1: Dimensions of standard quantities}$

For mass

$\rightarrow M$

For length

$(r) \rightarrow L$

Force

$(F) \rightarrow MLT^{-2}$

$\textbf{Step 2: Dimensions of universal gravitational constant}$

From

$F=\dfrac{GMm}{r^2}$

$G=\dfrac{Fr^2}{Mm}$

$=\dfrac{MLT^{-2}[L^2]}{[M][M]}$

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

Hence Option

$D$ is correct.

_______ refers to a state of strict exactness.Fill the appropriate answer.

0%
Accuracy
0%
Precision
0%
Accuracy and Precision
0%
None of these.

If we weigh a given substance five times, and get 5 kg each time, then your measurement is very __________.

0%
Accurate
0%
Precised
0%
Accurate and Precised
0%
Cannot be judged

Accuracy of potentiometer can be easily increased by :

0%
increasing resistance of wire
0%
decreasing resistance of wire
0%
increasing the length of wire
0%
decreasing the length of wire

The relation between force '

$F$ ' and density '

$d$ ' is

$F=\displaystyle\frac{x}{\sqrt{d}}$ . The dimensions of

$x$ are:

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

Explanation

The quantity is

$x = F \sqrt{d}$

Dimensions of force

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

Dimensions of density

$d = [M^1 L^{-3}]$

Dimensions of

$x$ are

$[x] = [M^1 L^1 T^{-2}] [M^1 L^{-3}]^{1/2}$

$\implies$

$[x] = [M^{3/2} L^{-1/2} T^{-2}]$

Dimensions

$[ML^{-1}T^{-1}]$ are related to

0%
Torque
0%
Work
0%
Energy
0%
Coefficient of viscosity

Explanation

We know that

Coefficient of viscosity=

$\displaystyle \frac{force/area}{velocity\,\,gradient}$

$\displaystyle [\eta] =\frac{[MLT^{-2}]/[L^2]}{[LT^{-1}]/[L]}$

$\displaystyle =[ML^{-1}T^{-1}]$

If acceleration (

$A$ ), mass (

$M$ ), velocity (

$V$ ), time (

$T$ ) are the fundamental units, then the dimensions of height will be

0%
$V{ T }^{ -1 }{ M }^{ 2 }\quad$
0%
$AM{ T }^{ -1 }$
0%
${ A }^{ -1 }{ M }^{ 3 }{ V }^{ -1 }$
0%
$A{ T }^{ 2 }$

Explanation

Given, acceleration

$(A)$ , mass

$(m)$ , velocity

$(v)$ , time

$(T)$
Hence,

$\quad A=\left[ L{ T }^{ -2 } \right] \quad \Rightarrow \left[ L \right] =\left[ A{ T }^{ 2 } \right]$

The dimension of universal gravitational constant is :

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

Explanation

$F=\displaystyle\frac{Gm_1m_2}{r^2}$

$\Rightarrow \displaystyle G=\frac{Fr^2}{m_1m_2}=\frac{[MLT^{-2}][L^2]}{[M^2]}$

$=\displaystyle [M^{-1}L^3T^{-2}]$ .

The pressure on a square plate is measured by measuring the force on the plate. If the maximum error in the measurement of force and length are respectively

$4\%$ and

$2\%$ , then the maximum error in the measurement of pressure is :

0%
$1\%$
0%
$2\%$
0%
$4\%$
0%
$8\%$

Explanation

As per question,

Error in the measurement of the force

$=4\%$

Error in the measurement of the length

$=2\%$

Pressure

$=\dfrac{force}{area}=\dfrac{F}{L^2}$

$\therefore \dfrac{\triangle p}{p}=\dfrac{\triangle F}{F}+\dfrac{2\triangle L}{L}=4\% +2(2\%)$

$=8\%$

So the maximum error in the measurement of the pressure is

$=8\%$

Are senses reliable for accurate measurement?

0%
True
0%
False

The dimension of the universal constant of gravitation

$G$ is

0%
$\left[ M{ L }^{ 2 }{ T }^{ -1 } \right]$
0%
$\left[ { M }^{ -1 }{ L }^{ 3 }{ T }^{ -2 } \right]$
0%
$\left[ { M }^{ -1 }{ L }^{ 2 }{ T }^{ -2 } \right] \quad$
0%
$\left[ M{ L }^{ 3 }{ T }^{ -2 } \right] \quad$

Explanation

$\left[ G \right] =\left[ \cfrac { F{ r }^{ 2 } }{ { m }_{ 1 }{ m }_{ 2 } } \right] =\left[ { M }^{ -1 }{ L }^{ 3 }{ T }^{ -2 } \right]$

Which one of the following pair of quantities has same dimension?

0%
Force and work done
0%
Momentum and impulse
0%
Pressure and force
0%
Surface tension and stress

Which of the following physical quantities has a unit but no dimensions?

0%
Relative velocity
0%
Relative density
0%
Angle
0%
Energy

Explanation

Among the given physical quantities angle has a unit but no dimensions. Angie =

$[M^0L^0T^0]$
The SI unit of angle is radian.

Possible errors with screw gauge are

0%
Systematic error
0%
Backlash error
0%
Personal error
0%
All these

The dimension of permittivity (

${ \varepsilon }_{ 0 }$ ) are ______. Take

$Q$ as the dimension of charge.

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

Explanation

We know that electrostatic force is given as:

$F=\cfrac { K{ Q }^{ 2 } }{ { r }^{ 2 } }$

We know that the dimension of force is

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

$\therefore ML{ T }^{ -2 }=\cfrac { K{ Q }^{ 2 } }{ { r }^{ 2 } } \\ [K]=M{ L }^{ 3 }{ T }^{ -2 }{ Q }^{ -2 }\\ K=\cfrac { 1 }{ 4\pi { \varepsilon }_{ o } } \\ [{ \varepsilon }_{ o }]={ M }^{ -1 }{ L }^{ -3 }{ T }^{ 2 }{ Q }^{ 2 }$

The error which follows a particular rule is

0%
Personal error
0%
Systematic error
0%
Random error
0%
Gross error

The order of magnitude of

$379$ is

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

Order of magnitude of

$379$ is

$\dfrac { 379 }{ { 10 }^{ 2 } } =3.79<{ 5 }^{ -1 }$ i.e,

$379\approx { 10 }^{ 2 }$

order of magnitude

$=2$ .

Which of the following is dimensionless?

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

Explanation

$\textbf{Correct option: Option (D)}$
$\textbf{Hint:}$ If two quantity have same dimension then their ratio will be dimensionless.

$\textbf{Explanation:}$ A dimensionless quantity means the quantity which has no physical dimension or has a dimension one, they are also referred to as scalar quantities. They are often obtained as the ratio of two or more quantities.

Since the dimension of work and energy is the same, i.e., $M{{L}^{2}}{{T}^{-2}}$ , when we take their ratio both the dimensions will cancel out and the result will be a dimensionless quantity.

$\textbf{Thus, option (D) is correct.}$

If the speed of light (c), acceleration due to gravity (g) and pressure (p) are taken as fundamental units, the dimensions of gravitational constant (G) are:

0%
$c^0gp^{-3}$
0%
$c^2g^3p^{-2}$
0%
$c^0g^2p^{-1}$
0%
$c^2g^2p^{-2}$

Explanation

Dimension of

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

$[c] = [LT^{-1}]$

$[g] = [LT^{-2}]$

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

Equate dimensions

$[M^{-1}L^3T^{-2}] = [LT^{-1}]^\alpha [LT^{-2}]^\beta [ML^{-1}T^{-2}]^\gamma$

$-1 = \gamma$

$3 = \alpha + \beta - \gamma$

$\alpha + \beta =2$ (1)

And,

$-2 = -\alpha -2\beta -2\gamma$

$\alpha + 2\beta =4$ (2)

Solving (1) and (2) we get,

$\alpha =0, \beta =2$

$\therefore Dimension = c^0g^2p^{-1}$

The number of significant figures in

$0.00210$ is

0%
five
0%
six
0%
four
0%
three

Explanation

In the given number

$0.00210$ , only last digits

$210$ are considered as significant figures. Thus the given number has three significant figures.

Calculate area enclosed by a circle of diameter

$1.06\ m$ to correct number of significant figures.

0%
$0.883 m^2$
0%
$0.0883 m^2$
0%
$0.88333 m^2$
0%
$0.8830 m^2$

Explanation

Area =

$\pi (\dfrac{d}{2})^2$

$\Rightarrow Area = \dfrac{22}{7}\times 0.53^2 = 0.882828$

But, no of significant in diameter should be equal to the no. of significant in Area, so Ans is

$0.883 m^2$

Subtract

$2.5 \times {10}^{4}$ from

$3.9 \times {10}^{5}$ and give the answer to correct number of significant figures.

0%
$3.6\times 10^6$
0%
$3.6\times 10^5$
0%
$36\times 10^5$
0%
$36\times 10^6$

Explanation

We have

$3.9 \times 10^5- 2.5 \times 10^4$

$= 3.9 \times 10^5- 0.25 \times 10^5$

$=3.65 \times 10^5$

But our answer should be rounded off upto two significant digits.

So the ans is

$3.6\times 10^5$

The order of magnitude of 0.00701 is:

0%
-3
0%
-1
0%
2
0%
1

Explanation

the order of magnitude of

$0.00701$ is

$-3$

we know that,

$N=a\times {{10}^{b}}$

$0.00701=7.01\times {{10}^{-3}}$

Where b is the order of magnitude

So, the order of magnitude is -3

Note: For taking the order of magnitude, the number should always be written in the scientific notation first.

The amplitude of a damped oscillator of mass

$'m'$ varies with the time 't' as

$A=A_0e^{-at/m}$ . The dimension of

$'a'$ are:

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

Explanation

Given,

$Mass=m,time=t.A=A_0e^{\dfrac{-at}{m}}$

Since the term

$\dfrac{-at}{m}$ is the exponential term so it is a dimension less

We know that,

The dimension of

$T=[M^0L^0T^1]$ The dimension of

$m=[M^1L^0T^0]$

So,

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

$y$ represents the pressure and

$x$ represents the position then dimensional expression of

$\int { \dfrac { dx }{ \sqrt { p^{ 2 }- y^2} } }$

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

Explanation

Dimension of pressure =

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

Dimensions of p and y has to be equal

Dimensions of

$dx = L^{1}$

$\therefore$ Dimensions of

$\int { \dfrac { dx }{ \sqrt { p^2 - y^2 } } }$

Dimension of

$\cfrac{dx}{y} = \cfrac{L^{1}}{ML^{-1}T^{-2}}$

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

If 'muscle times speed equals power', then what is the ratio of the

$SI$ unit and the

$CGS$ unit of muscle?

0%
$10^{5}$
0%
$10^{3}$
0%
$10^{7}$
0%
$10^{-5}$

Explanation

Given :

Muscles

$\times$ Speed

$=$ Power

and we know that

Force

$\times$ Speed

$=$ Power

comparing both the equations;

Muscles

$=$ Force

S.I. unit of force

$=$ N

$=$ Newton

CGS unit of force

$=$ dyne

and

$1N = 10^5 dynes$

So the SI unit of muscles will be

$N/ dyne=10^5$

Dimensional formula of

$\dfrac{megnetic\, flux}{electric\, flux}$ is

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

Explanation

Dimensional formula of magnetic flux

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

Dimensional formula for electric flux

$=ML^3T^{-3}A^{-1}$

Dimensional formula of

$\dfrac{Magnetic\, flux}{Electric flux}=\dfrac{[ML^2T^{-2}A^{-1}]}{[ML^3T^{-3}A^{-1}]}=[TL^{-1]}$

The dimensions of a modulus of elasticity are:

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

Explanation

Young modulus, $\gamma =\dfrac{stress}{strain}=\dfrac{Force\,/\,Area}{\Delta L/L}=\dfrac{\left[ ML{{T}^{-2}} \right]/\,\left[ {{M}^{o}}{{L}^{2}}{{T}^{o}} \right]}{\left[ {{M}^{o}}L{{T}^{o}} \right]/\left[ {{M}^{o}}L{{T}^{o}} \right]}=\left[ M{{L}^{-1}}{{T}^{-2}} \right]$

Hence, Dimension is $\left[ M{{L}^{-1}}{{T}^{-2}} \right]$ .

Dimensional formula for pressure head is

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

Explanation

$[Presure\,head]=\dfrac{[Pressure]}{[Density]\times [g]}$

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

But,

$Force=Mass \times Accelaration=[M][LT^{-2}]=[MLT^{-2}]$

Then,

$[Pressure]=\dfrac{[MLT^{-2}]}{[L^2]}=[ML^{-1}T^{-2}]$

$[Density]=\dfrac{[Mass]}{[Volume]}=\dfrac{[M]}{[L^{3}]}=ML^{-3}$

Now,

$[Presure\,head]=\dfrac{[Pressure]}{[Density]\times [g]}=\dfrac{[ML^{-1}T^{-2}]}{[ML^{-3}][LT^{-2}]}=[M^0L^1T^0]$

Round off to

$3$ significant figures,

$20.96$
b)

$0.0003125$

0%
$21.0, 312\times 10^{-4}$
0%
$21.0, 3.12\times 10^{-4}$
0%
$2.10, 3.12\times 10^{-2}$
0%
$210, 3.12\times 10^{-4}$

Explanation

$\begin{array}{l}\text { According to the rule }\\\text { of rounding off:}\end{array}$

$\text { (a) } 20.96^{}=21.0$

$\text{ (b) }0.0003125=3.125 \times 10^{-4} = 3.12 \times 10^{-4}$

Which one of the following does not have the same dimensions

0%
Work and energy
0%
Angle and strain
0%
Relative density and refractive index
0%
Plank constant and energy

Explanation

The work and energy have the same dimensions whereas the option

$B$ and

$C$ have the quantities which are dimensionless.

For option

$D$ :-

[Planck constant]=

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

and [Energy]=

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

So, option

$D$ is correct.

Dimension of work done is

0%
$\left[ L ^ { 2 } M ^ { 1 } T ^ { -2} \right]$
0%
$\left[ L ^ { 2 } M ^ { 2 } T ^ { 2 } \right]$
0%
$\left[ L ^ { 2 } M ^ { 0 } T ^ { 0 } \right]$
0%
$\left[ L ^ { 2 } M ^ { 2 } T ^ { -3 } \right]$

Explanation

We know that

Work done = Force

$\times$ Displacement

$W= MLT^{-2} \times L= [ML^2T^{-2}]$

$L^{-1}M^{1}T^{-2}$ are the dimension for________.

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

Explanation

$Pressure=\dfrac{Force}{Area}=\dfrac{Mass\times Acceleration}{Area}$

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

So, correct answer is option (A).

Of the following pairs of physical quantities, which one is dimensionally odd?

0%
Stress and pressure
0%
Energy and work
0%
Young's modulus and stress
0%
Frequency and angle

Explanation

All other pair have same dimensions while frequency and angle have different dimensions. The dimension of frequency is

$[T^{-1}]$ while angle is a dimensionless quantity.

Dimensional formula of pressure is :

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

Velocity gradient has same dimensional formula as:

0%
Angular frequency
0%
Angular momentum
0%
Velocity
0%
None of these

CBSE Questions for Class 11 Engineering Physics Units And Measurement Quiz 5 - 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 5 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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