MCQ Questions for Class 11 Engineering Physics Units And Measurement Quiz 1

Dimensions of

$\displaystyle \frac{1}{\mu_{0}{\varepsilon_{0}}}$ , where symbols have their usual meaning, are :

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

Explanation

Vaccum permittivity:

${ F }_{ c }=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \times \dfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } }$

Where

$\varepsilon_0$ is permttivity of vaccum.

Dimension will be $[M^{-1}L^{-3}T^{4}I^{2}]$

vaccum permeability:

${ \mu }_{ 0 }=4\pi \times { 10 }^{ -7 }\quad N/{ A }^{ 2 }$

Dimensions will be $[MLT^{-2}I^{-2}]$

Dimensions for

$\dfrac { 1 }{ { \mu }_{ 0 }{ \varepsilon }_{ 0 } }$ will be

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

In the following

$I$ refers to current and other symbols have their usual meaning. Choose the option that corresponds to the dimensions of electrical conductivity:

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

Explanation

Conductivity has relation,

$\sigma =\dfrac { l }{ AR } =\dfrac { lI }{ AV }$ .

V has dimension

$[M{ L }^{ 2 }{ T }^{ -3 }{ I }^{ -1 }]$ . Hence dimension of

$\sigma$ is

$[{ M }^{ -1 }{ L }^{ -3 }{ T }^{ 3 }{ I }^{ 2 }]$ .

The physical quantities not having same dimensions are:

0%
torque and work
0%
momentum and Plancks constant
0%
stress and Youngs modulus
0%
speed and $(\mu_{0}\varepsilon _{0})^{-1/2}$

Explanation

The and the work have the same dimensions because they both are defined as the product of the force and the distance.

The stress and the young's modulus have the same dimensions because the young's modulus is defined as the ratio of stress and strain and strain is a dimensionless quantity.

The speed of an electromagnetic wave is given by $\dfrac{1}{\mu_0\epsilon_0}$ . so, both have the same dimension.

Dimensions of momentum $=$ kg $\mathrm{m}/\sec=[\mathrm{M}\mathrm{L}\mathrm{T}^{-2}]$

Dimensions of PIanck's constant $=$ joule $\sec=[\mathrm{M}\mathrm{L}^{2}\mathrm{T}^{-1}]$

$\therefore$ Dimensions of momentum $\neq$ dimensions of PIanck's constant.

The dimension of magnetic field in

$\mathrm{M},\ \mathrm{L},\ \mathrm{T}$ and

$\mathrm{C}$ (Coulomb) is given as :

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

Explanation

Magnetic force $\mathrm{F}=\mathrm{i}\mathrm{B}l$

$\implies$ Magnetic field $B = \dfrac{F}{il}$

Dimensions of magentic field $[B] = \dfrac{MLT^{-2}}{[CT^{-1}][L]}$

$\Rightarrow [\mathrm{B}]=\lfloor \mathrm{M}\mathrm{T}^{-1}\mathrm{C}^{-1}\rfloor$

Which of the following does not have the same dimension?

0%
Electric flux, Electric field, Electric dipole moment
0%
Pressure, stress, Youngs modulus
0%
Electromotive force, Potential difference, Electric voltage
0%
Heat, Potential energy, Work done

Explanation

B) Pressure:

$\dfrac { force }{ Area }$

Stress :

$\dfrac { force }{ Area }$

Youngs modulus :

$\dfrac { stress }{ strain }$ , here strain is dimension less So, all three has same dimension.

C) Electromotive force:

$\dfrac { Energy }{ charge }$

Potential difference:

$\dfrac { W}{ Q }$ , with is same as Electromotive

Electric voltage: If you see in a capacitor, then Voltage is given by

$\dfrac { Energy }{ charge }$

D) Heat:

$ML^2T^{-2}$

Potential energy:

$mgh \Rightarrow ML^2T^{-2}$

Work done

$=f.d\Rightarrow ML^2T^{-2}$

A) Electric flux:

$M^{-2}SA$

Electric field:

$M^2S^{-3}A^{-1}$

Hence, it has quantities with different units.

The relation between

$[\in_0]$ and

$[\mu_0]$ is

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$[\mu_0] = [\in_0] [L]^2[T]^{-2}$
0%
$[\mu_0] = [\in_0] [L]^{-2}[T]^{2}$
0%
$[\mu_0] = [\in_0]^{-1} [L]^2[T]^{-2}$
0%
$[\mu_0] = [\in_0]^{-1} [L]^{-2}[T]^{2}$

Explanation

We have,

$C = \dfrac{1}{\sqrt{\mu_0 \in_0}}$

$\therefore [C^2] = \left[\dfrac{1}{\mu_0 \in_0}\right]$

$\Rightarrow L^2T^{-2} = \dfrac{1}{[\mu_0][\in_0]}$

$[\mu_0] = [\in_0]^{-1} [L]^{-2} [T]^2$

The dimensions of

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

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

Explanation

Since expression for speed of light

$c=\dfrac { 1 }{ \sqrt { { \mu }_{ 0 }{ \varepsilon }_{ 0 } } }$ , the given expression has the dimensions of velocity i.e.

$L{ T }^{ -1 }$ .

Hence, option B is correct.

The work done to raise a mass m from the surface of the earth to a height h, which is equal to the radius of the earth, is :

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mgR
0%
2 mgR
0%
$\dfrac{1}{2} mgR$
0%
$\dfrac{3}{2} mg R$

Explanation

Initial potential energy at earths surface is

$U_i = \dfrac{-GMm}{R}$
Final potential energy at height h = R

$U_f = \dfrac{-GMm}{2R}$
As work done = change in PE

$\therefore W = U_f - U_i$

$= \dfrac{GMm}{2R}$

$(\because GM = gR^2)$

$= \dfrac{gR^2m}{2R} = \dfrac{mgR}{2}$

1259381_1618833_ans_ff63c06b4e034645bba0381c10201b32.png

If force

$(F)$ , velocity

$(V)$ and time

$(T)$ are taken as fundamental units, the dimensions of mass are

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

Explanation

Hint: Find the formula of the mass in terms of force, velocity, and time.

Step 1: Relate the formula of mass to the given quantities.

We know,

$F=ma$

Acceleration is velocity per unit time.

$a=\dfrac{V}{T}$

Hence,

$F=\dfrac{mV}{T}$

Step 2: Find out the dimensional formula of mass.

Now, from the above equation, we can write,

$m=\dfrac{FT}{V}$

$m=[FV^{-1}T]$

Hence, the dimension of mass is $[FV^{-1}T]$

The correct answer is option (D).

Planck's constant (

$h$ ), speed of light in vacuum (

$c$ ) and Newton's gravitational constant (

$G$ ) are three fundamental constants. Which of the following combinations of these has the dimension of length?

0%
$\sqrt{\dfrac{Gc}{h^{3/2}}}$
0%
$\dfrac{\sqrt{hG}}{c^{3/2}}$
0%
$\dfrac{\sqrt{hG}}{c^{5/2}}$
0%
$\sqrt{\dfrac{hc}{G}}$

Explanation

Hint : Write the Units of all given physical quantities put these values in all option to verify which expression have same unit or dimension of length

Step 1: Write the SI units of given physical quantities

The SI unit of h =J.s or N.m.s

SI unit of c =m/s

S.I unit of

$G=\dfrac{N.m^{2}}{Kg^{2}}$

Step 2: Finding the units of all options

For A:

$\sqrt{\dfrac{(N.m^2/kg^2)(m/s)}{(N.m.s)^{3/2}}}=\dfrac{kg.m^3}{N.s^5}=\dfrac{kg.m^3}{(kg.m/s^2)(s^5)}=m^2/s^3$

For B:

$\dfrac{\sqrt{(N.m.s)(N.m^2/kg^2)}}{(m/s)^{3/2}}=\dfrac{Ns^2}{kg}=\dfrac{(kg.m/s^2)(s^2)}{kg}=m$ , it is the unit length.

For C:

$\dfrac{\sqrt{(N.m.s)(N.m^2/kg^2)}}{(m/s)^{5/2}}=\dfrac{Ns^3}{m.kg}=\dfrac{(kg.m/s^2)(s^3)}{m.kg}=s$

For D:

$\sqrt{\dfrac{(N.m.s)(m/s)}{N.m^2/kg^2}}=kg$

Hence option B is correct

The dimensions of specific resistance are:

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

Explanation

Resistance of a substance=

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

$\implies \rho=\dfrac{RA}{l}$

$=\dfrac{V}{I}\dfrac{A}{l}$

Dimension of

$V$ is

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

Dimension of

$I$ is

$[A]$

Dimension of

$A$ is

$[L^2]$

Dimenstion of

$l$ is

$[L]$
Thus dimension of

$\rho$ is

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

The dimensional formula for electric flux is?

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

Explanation

Electric flux ,

$\phi= BA= \dfrac F q A= \dfrac{MLT^{-2}\times L^2}{AT}= [ML^3A^{-1}T^{-3}]$

Which of the following pairs does not have same dimensions?

0%
impulse and momentum
0%
moment of inertia and moment of force
0%
angular momentum and Planck's constatn
0%
work and torque

Explanation

Moment of force

$(\tau)$ is equal to the product of moment of inertia

$(I)$ and the angular acceleration

$(\alpha)$ of the rotating body.

$\therefore$

$\tau = I \alpha$

Hence the dimensions of moment of force and moment of inertia must be different.

What is the dimensions of magnetic field B in terms of C (=coulomb), M, L, T?

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

Explanation

We know that force on a particle moving perpendicular to magnetic field is

$F=qvB$

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

Dimension of

$F$ is

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

Dimension of

$q$ is

$[C^1]$

Dimension of

$v$ is

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

Hence, the dimensions of

$B$ is

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

Dimensional formula of angular momentum is

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

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

Explanation

Both quantities are dimensionless.

Resistance

$\times$ Conductance=

$R\times\sigma=R\times \dfrac{1}{R}=1\approx [M^0L^0T^0]$

Dielectric constant

$K$ is dimensionless.

Dimensional formula of

$\Delta Q$ heat supplied to the system is given by

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

Explanation

Heat has the same dimesnsions as energy which has the same dimensions as force times distance. Hence the answer is

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

The dimensional formula of Planck's constant is:

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

Explanation

$\textbf{Hint:}$ Use the dimensional formula of energy and frequency.

$\textbf{Step 1:}$ Energy of photon is given as

$E=h\nu$

$h=\dfrac{E}{\nu}$

$\textbf{Step 2:}$ Calculating the dimension of the Planck's constant,

Dimension of

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

Dimension of

$\nu$

$=$

$[T^{-1}]$

So, the dimension of plank's constant

$h$ is

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

$\textbf{Option A is correct.}$

A cube has a side of length

$\displaystyle 1.2\times { 10 }^{ -2 }m$ . Calculate its volume.

0%
$\displaystyle 1.7\times { 10 }^{ -6 }{ m }^{ 3 }$
0%
$\displaystyle 1.73\times { 10 }^{ -6 }{ m }^{ 3 }$
0%
$\displaystyle 1.70\times { 10 }^{ -6 }{ m }^{ 3 }$
0%
$\displaystyle 1.732\times { 10 }^{ -6 }{ m }^{ 3 }$

Explanation

Volume,

$\displaystyle V={ l }^{ 3 }={ \left( 1.2\times { 10 }^{ -2 }m \right) }^{ 3 }$

$\displaystyle =1.728\times { 10 }^{ -6 }{ m }^{ 3 }$
Since length (l) has two significant figure, the volume (V) will also have two significant figure.
Therefore, the correct answer is

$\displaystyle V=1.7\times { 10 }^{ -6 }{ m }^{ 3 }$

If the force is given by

$F$ =

$at+bt^2$ with

$t$ as time. The dimensions of a and b are:

0%
$[MLT^{-4}], [MLT^{-2}]$
0%
$[MLT^{-3}], [MLT^{-4}]$
0%
$[ML^2T^{-3}], [ML^2T^{-2}]$
0%
$[ML^2T^{-3}], [ML^3T^{-4}]$

Explanation

Dimension of force F :

$[MLT^{-2}]$

$F=at+bt^2$

$\implies$ Dimension of

$a:[MLT^{-2}]\times [T^{-1}]$

$=[MLT^{-3}]$

Dimensions of

$b:[MLT^{-2}]\times [T^{-2}]$

$=[MLT^{-4}]$

The dimensional formula for inductance is:

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

Linear momentum and angular momentum have the same dimensions in:

0%
mass and length
0%
length and time
0%
mass and time
0%
mass, length and time

Explanation

Linear momentum

$= mv = [M^{1}L^{1}T^{-1}]$
Angular momentum

$= mvr = [M^{1}L^{2}T^{-1}]$
Hence, linear and angular momenta have same dimensions in mass and time.

Which of the following does not give the unit of energy ?

0%
Watt second
0%
Kilowatt hour
0%
Newton meter
0%
Pascal metre

Explanation

Energy

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

$= M^{1} L^{2} T^{-3}\times T = M^{1} L^{2} T^{-2}$
Kilowatt hr

$= M^{1} L^{2} T^{-3}\times T^{1} = M^{1} L^{2} T^{-2}$
Newton mt

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

$= M^{1} L^{-1} T^{-2}\times L^{1} = M^{1} L^{0} T^{-2}$
Hence, option D is correct.

Magnetic flux and magnetic induction differ in the dimensions of:

0%
mass
0%
length
0%
time
0%
all of these

Explanation

Magnetic flux = Magnetic Induction

$\times$ Area

$=[ ML^{2}I^{-1}T^{-2}]$
Magnetic Induction field strength

$= [MI^{-1}T^{-2}]$
Hence, dimensions differ in length.

The dimensional formula for potential energy is:

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

Explanation

Potential energy

$= [ mgh] \\= [(M^{1}) (LT^{-2}) (L)]$

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

The mathematical operation in which the accuracy is limited to least accurate term is :

0%
addition
0%
subtraction
0%
multiplication and division
0%
both A and B

The SI unit of a physical quantity is [

$Jm^{-2}$ ]. The dimensional formula for that quantity is:

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

Explanation

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

Impulse and angular velocity have the same dimensions in

0%
mass
0%
length
0%
time
0%
mass, length and time

Explanation

Impulse

$= F\times t = [M^{1}L^{1}T^{-2} \times T^{1}] = [M^{1}L^{1}T^{-1}]$
Angular velocity

$= \omega = [M^{0}L^{0}T^{-1}]$
Hence, time has the same dimension in both the quantities.

The dimensional formula for magnetic moment of a magnet is :

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

Explanation

We know that magnetic moment for a solenoid is NIS, where N is number of loops, I is current and S is the vector area.
Thus, dimensions of

$\mu$ are

$[AL^2]=[M^0L^2T^0A]$

The dimensional formula for moment of couple is :

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

Explanation

Moment of a couple is calculated by multiplying the size of one of the force (F) by the perpendicular distance between the two forces.
Thus the units are that of

$Nm$ and dimensions are

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

The measure of accuracy is :

0%
absolute error
0%
relative error
0%
percentage error
0%
Both B and C

Explanation

${\textbf{Explanation:}}$

$\bullet$ Accuracy is measured as the deviation from true reading compared with the observed reading. Accuracy of a measured value tells how close the measured value is to the actual value.

$\bullet$ Absolute error is the deviation of a reading from the actual or true reading.

$\bullet$ Relative error is defined on true error, as the difference of a value of the true or actual value divided by the actual value.

$\bullet$ Whereas percentage error is expressed in terms of percentage, i.e. by multiplying the relative error by 100, it is called percentage error. Both relative and percentage error give a quantitative analysis of the deviation from the true value.

$\bullet$ For example in 2 separate experiments with true value 1000m and 1m have absolute error as 0.1m. Now the relative error in these cases will be 0.999 and 0.9. So absolute error fails to tell us the correct measure of accuracy and relative or percentage error takes the win.

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

If the unit of tension is divided by the unit of surface tension the derived unit will be same as that of

0%
mass
0%
length
0%
area
0%
work

Explanation

Tension

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

$\displaystyle = \frac{f}{l} = \frac{M^{1} L^{1} T^{-2}}{L^{1}} = M T^{-2}$

$\displaystyle \frac{Tension}{Surface Tenstion} = \frac{M^{1} L^{1} T^{-2}}{M T^{-2}} = L^{1}$

$\left[ \dfrac { Permeability }{ Permittivity } \right]$ will have the dimensions of :

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

Explanation

Permeability

$= \mu _{0} = M L A^{-2} T^{-2}$ .........(1)
Permittivity

$= M^{-1} L^{-3} A^{2} T^{4}$ .................(2)
So,

$\dfrac{eq^{n}(1)}{eq^{n}(2)} = \dfrac{M L A^{-2} T^{-2}}{M^{-1} L^{-3} A^{2} T^{4}} = M^{2} L^{+4} A^{-4} T^{-6}$
Hence, option C is correct.

Which of the following is/are dimensionless ?
a) Boltzman’s constant b) Planck’s constant c) Poisson’s ratio d) Relative constant

0%
a and b
0%
c and b
0%
c and d
0%
d and a

Explanation

Poisson's ratio is ratio of young modulus, hence dimensionless.

Relative constant is also a constant.

Boltzman's constant

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

Plancks constant $=ML^{2}T^{-1}$

Hence, option C is correct.

The fundamental unit which has the same power in the dimensional formula of surface tension and coefficient of viscosity is:

0%
mass
0%
length
0%
time
0%
none

Explanation

Surface tension

$= \dfrac{F}{l} = \dfrac{M^{1}L^{1}T^{-2}}{L^{1}} = M^{1}T^{-2}$
Coeff of viscosity

$= \dfrac{k}{qA\dfrac{dv}{dz}} = \dfrac{M^{1}L^{1}T^{-2}}{\dfrac{L^{2}LT^{-1}}{L}} = M^{1}L^{-1}T^{-1}$
Hence, mass has the same power.

$\mu$ is the permeability and

$\epsilon$ is the permittivity then

$\dfrac { 1 }{ \sqrt { \mu \epsilon } }$ is equal to

0%
speed of sound
0%
speed of light in vacuum
0%
speed of sound in medium
0%
speed of light in medium

Explanation

We know that speed of light in vacuum

$c=\dfrac{1}{\sqrt{\mu_0\epsilon_0}}=\dfrac{1}{\sqrt{(4\pi\times 10^{-7})(8.314\times 10^{-12})}}\sim 2.98\times 10^8 m/s$

Thus,

$\dfrac{1}{\sqrt{\mu\epsilon}}$ will also represent the speed of light in a medium whose permittivity is

$\epsilon$ and permeability,

$\mu$ .

The physical quantity having the same dimensional formula as that of entropy is:

0%
Latent heat
0%
Thermal capacity
0%
Heat
0%
Specific heat

Explanation

Entropy

$= \dfrac{Energy}{temp} = \dfrac{M^{1} L^{2} T^{-2}}{K} = M^{1} L^{2} T^{-2} K^{-1}$

Latent heat

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

Heat

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

Specific heat

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

Thermal capacity

$= \dfrac{Energy}{temp} = \dfrac{M^{1} L^{2} T^{-2}}{K^{1}} = M^{1} L^{2} T^{-2} K^{-1}$

The dimensional equation for magnetic flux is:

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

Explanation

Magnetic flux

$=$ Magnetic Field (B) x Area (A)
B's dimensions

$=$ Tesla

$= MI^{-1}T^{-2}$
A's dimensions

$= L^2$
Therefore,

$BA= ML^2T^{-2}I^{-1}$

"Impulse per unit area" has same dimensions as that of

0%
coefficient of viscosity
0%
surface tension
0%
bulk modulus
0%
gravitational potential

Explanation

Impulse per unit area

$\displaystyle = \frac{f\times t}{L^{2}} = \frac{M^{1} L^{1} T^{-2}\times T^{1}}{L^{2}} =M^{1} L^{-1} T^{-1}$

Coeff. of viscosity

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

Surface Tension

$= \dfrac{f}{l} = \dfrac{M^{1} L^{1} T^{-2}}{L} = MT^{-2}$

Bulk Modulus

$= \dfrac{dp}{\dfrac{dv}{v}} = M^{1} L^{-1} T^{-2}$

Gravitational poential

$\displaystyle = \frac{GM}{r} = \frac{foru\times r}{Mass} = \frac{M^{1} L^{1} T^{-2}\times L^{1}}{M^{1}} = L^{2} T^{-2}$

Hence, option A is correct.

Which of the following pair does not have same dimensions ?

0%
Pressure, Modulus of Elasticity
0%
Angular velocity, Velocity gradient
0%
Surface tension, Force constant
0%
Impulse, Torque

Explanation

${\textbf{Explanation of options A, B and C:}}$

A. Pressure

$=$ Modulus of Elasticity

$= \dfrac{F}{A} = M^{1} L^{-1} T^{-2}$
B. Angular velocity

$= \omega = T^{-1}$

Velocity gradient

$= \dfrac{dv}{dz} = \dfrac{LT^{-1}}{L} = T^{-1}$
C. Surface Tenstion

$= \dfrac{F}{l} =$ Force constant

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

${\textbf{Explanation of option D:}}$
Impulse

$= F\times t = M^{1} L^{1} T^{-2} \times T^{1} = M^{1} L^{1} T^{-1}$
Torque

$= \vec{r} \times \vec{F} = L^{1} \times M^{1} L^{1} T^{-2} = M^{1} L^{2} T^{-2}$

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

Arrange the following physical quantities in the decreasing order of dimension of length.
I. Density

II. Pressure

III. Power

IV. Impulse

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I, II, III, IV
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III, II, I, IV
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IV, I, II, III
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III, IV, II, I

Explanation

$Density = \dfrac{Mass}{length} = M^{1} L^{-3} T^{0}$

$Pressure = \dfrac{force}{Area} = \dfrac{M^{1}L^{1}t^{-2}}{L^{2}} = M^{1}L^{-1}T^{-2}$

$Power = F\times v = M^{1}L^{1}T^{-2} \times LT^{-1} = M^{1}L^{2}T^{-3}$

$Impulse = f\times t = M^{1}L^{1}T^{-2}\times T^{1} = M^{1}L^{1}T^{-1}$

Hence, option D is correct.

Dimensional formula for capacitance is:

0%
${ M }^{ -1 }{ L }^{ -2 }{ T }^{ 4 }{ I }^{ 2 }$
0%
${ M }^{ 1 }{ L }^{ 2 }{ T }^{ 4 }{ I }^{ -2 }$
0%
${ M }^{ 1 }{ L }^{ 2 }{ T }^{ 2 }$
0%
${ MLT }^{ -1 }$

Explanation

$E=\dfrac{Q^2}{2C}$
Dimensions of C will be dimension of

$\dfrac{Q^2}{E}$

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

$=M^{-1}L^{-2}T^4I^2$
Hence, option A is correct.

The dimensional formula for coefficient of kinematic viscosity is :

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

Explanation

Kinematic viscocity $=\dfrac{coeff \ of \ viscocity}{ density}$

$=\dfrac{ML^{-1}T^{-1}}{ML^{-3}}$

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

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

0%
Stress
0%
Young's Modulus
0%
Pressure
0%
All the above

Explanation

Stress and pressure both are defined as

$\dfrac{F}{A}=\dfrac{MLT^{-2}}{L^2}=ML^{-1}T^{-2}$
Young's modulus is calculated by formula

$\dfrac{stress}{strain}$ . As strain is dimensionless the Young's modulus will also have dimensions of stress only.
Hence, option D is correct.

Dimensions of impulse are :

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

Explanation

$Impulse=F\times t=M^{1}L^{1}T^{-2}\times T^{1}=M^{1}L^{1}T^{-1}$

The dimensional formula for Magnetic induction is :

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${ MT }^{ -1 }{ A }^{ -1 }$
0%
${ MT }^{ -2 }{ A }^{ -1 }$
0%
${ MLA }^{ -1 }$
0%
${ MT }^{ -2 }A$

Explanation

$F=BQV$
Dimension of B will be

$\dfrac{MLT^{-2}}{ATLT^{-1}}=ML^0T^{-2}A^{-1}$

Dimensions of

$C\times R$ (Capacity x Resistance) is:

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frequency
0%
energy
0%
time period
0%
current

Explanation

C = Capacity

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

$=ML^{2}I^{-2}T^{-3}$
So,

$C\times R=T^{1}M^{0}L^{0}I^{0}$

Dimensional formula for Angular momentum is:

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

Explanation

$\textbf {Hint :}$ Use angular momentum

$L=mvr$

${\textbf{Explanation:}}$

Angular Momentum

$L=mvr$

m is mass, v is velocity and r is radius
We know that dimension formula of velocity is

$LT^{-1}$

Dimensional formula of

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

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

The pair of physical quantities not having the same dimensional formula are:

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Acceleration, Gravitational field strength
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Torque, Angular momentum
0%
Pressure, Modulus of Elasticity
0%
All the above

Explanation

Acceleration

$=LT^{-2}=$ Gravitational field strength
Torque

$=ML^{2}T^{-2}$
Angular momentum

$=ML^{2}T^{-1}$
Modulus of elasticity

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

Hence, option B is correct.

Modulus of Elasticity is dimensionally equivalent to:

0%
Stress
0%
Surface tension
0%
Strain
0%
Coefficient of viscosity

Explanation

Modulus of elasticity

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

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

$=\dfrac{f}{l}=M^{1}L^{0}T^{-2}$
Strain

$=\dfrac{\Delta L}{L}=M^{0}L^{0}T^{0}$
Coeff of viscosity

$=F/A\dfrac{dv_{x}}{dx}=\dfrac{MLT^{-2}}{L^{2}\dfrac{LT^{-1}}{L}}=ML^{-1}T^{-1}$
Hence, option A is correct.

CBSE Questions for Class 11 Engineering Physics Units And Measurement Quiz 1 - 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 1 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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