MCQ Questions for Class 11 Engineering Physics Units And Measurement Quiz 7

$M^{-1}L^{-2}T^3\theta^1$ are the dimensions of :

0%
coefficient of thermal conductivity
0%
coefficient of viscosity
0%
modulus of rigidity
0%
thermal resistance

Explanation

Thermal resistance is a heat property and a measurement of a temperature difference by which an object or material resists a heat flow (heat per unit time or thermal resistance).
Hence, the thermal resistance is ratio of temperature difference by power. Dimension of power is

$ML^{2}T^{-3}$ .
Therefore, unit of thermal resistance is

$kg^{-1} m^{-2}s^{3}K$
Hence, the dimensional formula is

$[M^{-1} L^{-2} T^{3} \theta^{1}]$

5.74 gm of a substance occupies a volume of

$1.2\;cm^3$ . Calculate its density with due regard for significant figures.

0%
$4.8\; g/cm^3$
0%
$2.3\; g/cm^3$
0%
$2.4\; g/cm^3$
0%
$3.8\; g/cm^3$

Explanation

Density

$\rho=M/V=5.74/1.2=4.78333 g/cm^3$

As voulme has least two significant figures so the answer must be expressed in two singnificant figures.

Thus density will be

$4.8 g/cm^3$

The number of significant figures in 3400 is :

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

Explanation

Trailing zeros in a whole number are not significant.Hence

$3400$ has 2 significant digits only.

Therefore, option D is correct.

In an experiment to determine acceleration due to gravity by simple pendulum, a student commits 1 % positive error in the measurement of length and 3% negative error in the measurement of time period. The percentage error in the value of g will be :

0%
7%
0%
10%
0%
4%
0%
3%

Explanation

We know that the acceleration due to gravity

$g= 4 \pi^{2}\left(\dfrac{l}{t^{2}}\right)$ .
Here, this student commits 1 % +ve error in the measurement of length and 3% +ve in the measurement of time period.

$\therefore \dfrac{\Delta g}{g} \times 100 =\left(\dfrac{\Delta l}{l}\times 100 +2 \dfrac{\Delta t}{t} \times 100 \right)$
Thus, the percentage error in the value of

$g=\pm \left ( 1+2 \times 3 \right )100$ =7%.

The dimensions of intensity of energy are :

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

Explanation

The energy flowing per unit area per unit time is called the intensity of energy.
The intensity of energy is (by formula):

$I = \dfrac{E}{At}$

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

$I = [M^1L^0T^{-3}]$
Hence, the dimensional formula of intensity of energy is

$[MT^{-3}]$ .

The dimensions of

$\dfrac {L}{RCV}$ (

$L=$ inductance,

$R=$ resistance,

$C=$ capacitance,

$V=$ potential difference) are

0%
[ $A$ ]
0%
$[A^{-1}]$
0%
[ $AT$ ]
0%
dimensionless

Explanation

We know that:

$[V]=[iR]=[L\dfrac{di}{dt}]$

$[i][R]=\dfrac{[L][i]}{[t]}$

Thus,

$[\dfrac{L}{R}]=[t]=T$

$[L]=\dfrac{[V][t]}{[i]}$

Also,

$i=C\dfrac{dV}{dt}$ , or

$[CV]={[i][t]}={A}{T}$

Thus,

$[\dfrac{L}{RCV}]=\dfrac{T}{AT}={A}^{-1}$

The dimensions of

$(h/e) [h=$ Planck's constant,

$e= electronic \ charge]$ are same as that of

0%
Magnetic field
0%
Magnetic flux
0%
Electric field strength
0%
Electric flux

Explanation

Plancks constant (h) has joule-seconds. dimensions are =

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

electron charge (e) will have

$[TI]$

Dimension for h/e will be

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

From the given options,

Dimension for magnetic moment will be $[L^2A]$

Dimension for magnetic flux will be $[ML^2T^{-2}I^{-1}]$

Dimension for electric flux will be $[ML^3T^{-3}I^{-1}]$

Dimension for electric field strength will be $[MLT^{-3}I^{-1}]$

Both magnetic flux and h/e will have same dimensions.

The dimension of the expression

$\dfrac { 1 }{ MB }$ is

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

Explanation

$\left[ \ { 1 } /MB \right] =\dfrac { 1 }{ \left[ M \right] \left[ B \right] } =\dfrac { 1 }{ \left[ M \right] \left[ M{ I }^{ -1 }{ T }^{ -2 } \right] }$

$=\left[ { M }^{ -2 }{ I }^{ 1 }{ T }^{ 2 } \right]$

The dimensions of

$\dfrac{{v}^{2}}{\mu g}$ where

$\mu=$ coefficient of friction,

$v=$ the velocity,

$g=$ acceleration due to gravity is same as that of

0%
$\dfrac {F}{Bq}$
0%
$\mu F$
0%
$\dfrac {F}{m}$
0%
None

Explanation

Units of

$\dfrac {v^2}{\mu g}$ are

$m$
A) Units of

$\dfrac {F}{Bq}=\dfrac {qvB}{Bq}=v$ are

$ms^{-1}$
B) Units of

$\mu F$ are

$kgms^{-2}$
C) Units of

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

$ms^{-2}$
D) Units of

$qvB=F$ are

$kgms^{-2}$

The dimensional formula for magnetic permeability

$\mu$ is

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

Explanation

Magnetic permeability is a constant of proportionality that exists between magnetic induction and magnetic field intensity.

In SI units, permeability is measured in henries per meter

$H/m$ or

$H m^{-1}$ .

Henry has the dimensions of

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

Dimensions for magnetic permeability will be

$[ML^2T^{-2}A^{-2}]/[L]=$

$[MLT^{-2}A^{-2}]$

The dimensional formula of modulus of rigidity is

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

Explanation

Modulus of rigidity is the ratio of shear stress to the shear strain.

$\eta =\displaystyle \frac {stress}{strain}$
Strain is dimensionless, where as Stress has the dimensions of Pressure.

$Stress=\displaystyle \frac {F}{a}=\frac {MLT^{-2}}{L^2}=ML^{-1}T^{-2}$

What are the number of significant figures in the measurement

$0.0040\times10^{-15}m$ ?

0%
1
0%
2
0%
4
0%
5

$L$ and

$R$ denote inductance and resistance respectively, then the dimensions of

$L/R$ are

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

Explanation

The ratio

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

Area of a square is

$(100\pm 2)m^2$ . Its side is:

0%
$(10\pm 1)m$
0%
$(10\pm 0.1)m$
0%
$(10\pm \sqrt 2)m$
0%
$10\pm \sqrt 2$ %

Explanation

$Area=(Length)^2$

$Length =(Area)^{1/2}$

$=(100\pm 2)^{1/2}$

$=(100)^{1/2}\pm \dfrac {1}{2}\times 2$

$=(10\pm 1)m$

The deBroglie wavelength associated with a particle of mass

$m$ and energy

$E$ is

$\dfrac{h}{\sqrt {2mE}}$ . The dimensional formula of Planck's constant h is

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

Explanation

Planck constant

$=h=\lambda \sqrt {2mE}=L\sqrt {M(ML^2T^{-2)}}=[ML^2T^{-1}]$

Match List I with List II and select the correct answer using the codes given below the list

List I	List II
P. Boltzmann constant	$[ML^2T^{-1}]$
Q. Coefficient of viscosity	2. $[ML^{-1}T^{-1}]$
R. Planck constant	3. $[MLT^{-3}K^{-1}]$
S. Thermal conductivity	4. $[ML^2T^{-2}K^{-1}]$

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

Explanation

Boltzmann constant

$=\displaystyle \frac {R}{N}=\frac {PV}{nTN}=\frac {ML^{-1}T^{-2}L^3}{K}=ML^2T^{-2}K^{-1}$

Coefficient of viscosity

$=\displaystyle \frac {F}{6\pi rv}=\frac {MLT^{-2}}{L.LT^{-1}}=ML^{-1}T^{-1}$

Planck constant

$=\displaystyle \frac {E}{v}=\frac {ML^2T^{-2}}{T^{-1}}=ML^{2}T^{-1}$

Thermal conductivity

$=\displaystyle \frac {Hl}{tA\Delta T}=\frac {ML^2T^{-2}L}{TL^2K}=MLT^{-3}K^{-1}$

Let

$[\epsilon_0]$ denote the dimensional formula of the permittivity of vacuum. If

$M=mass, L=length, T=time$ and

$A=electric$ current, then

0%
$\epsilon_0 = [M^{-1}L^{-3}T^2A^2]$
0%
$\epsilon_0 = [M^{-2}L^{-3}T^4A^2]$
0%
$\epsilon_0 = [M^{1}L^{2}T^1A^2]$
0%
$\epsilon_0 = [M^{1}L^{2}T^1A]$

Explanation

Using Coulomb law,

$F=\dfrac {1}{4\pi \epsilon_0}\dfrac {q_1q_2}{R^2}$

$\Rightarrow \epsilon_0=\dfrac {q_1q_2}{4\pi FR^2}$

Substituting the units in the equation:

So,

$\epsilon_0=\dfrac {C^2}{N.m^2}$

$=\dfrac {[AT]^2}{MLT^{-2}.L^2}=[M^{-1}L^{-3}T^4A^2]$

$L, C, R$ represent physical quantities inductance, capacitance and resistance respectively. The combinations which have the dimensions of frequency are

0%
$1/RC$
0%
$R/L$
0%
$1/\sqrt {LC}$
0%
$C/L$

Explanation

The formula of R can be written as,

$R=\frac{V}{I}$

Dimensional formula of the potential is,

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

Dimensional formula of the current is,

$[I]= [ M^{0}L^{0}T^{0}A^{1}]$

Therefore the dimensional formula of R will be,

$[R]= [ M^{1}L^{2}T^{-3}A^{-2}]$ .......(1)

Now the dimension of R/L is,

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

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

Hence, option (B) is also correct.

If g is the acceleration due to gravity and

$\lambda$ is wavelength then which physical quantity does

$\sqrt {\lambda g}$ represent?

0%
Velocity
0%
Length
0%
Time
0%
Charge

Explanation

Since

$\lambda$ is wavelength, hence its dimension is that of length =

$L$ .
Dimension of g is

$LT^{-2}$
Thus, dimension of

$\sqrt{\lambda g}=\sqrt{L^2T^{-2}}=LT^{-1}$ , which represents velocity.

$e$ is the charge,

$V$ the potential difference,

$T$ the temperature, then the units of

$\dfrac {eV}{T}$ are the same as that of

0%
planck's constant
0%
stefan's constant
0%
boltzmann constant
0%
gravitational constant

Explanation

$\displaystyle \frac {eV}{T}=\frac {Workdone}{T}=\frac {PV}{T}=\frac { R }{ N } =k=Boltzmann \ constant$

(By using ideal gas equation

$PV=\dfrac { RT }{ N }$ )

The dimensions of

$\dfrac {1}{\epsilon_0}\dfrac {e^2}{hc}$ are

0%
$M^{-1}L^{-3}T^4A^2$
0%
$ML^{3}T^{-4}A^{-2}$
0%
$M^0L^0T^0A^0$
0%
$M^{-1}L^{-3}T^2A$

Explanation

Here,

$h$ =planck's constant,

$c$ = light velocity
Using Coulomb's law we have

$F=\dfrac {1}{4\pi \epsilon_0}\dfrac {e^2}{r^2}$

$\therefore \dfrac {e^2}{\epsilon_0}=4\pi Fr^2$ (dimensionally)

$[\dfrac {e^2}{\epsilon_0hc}]=[\dfrac {4\pi Fr^2}{hc}]=\dfrac {(MLT^{-2})L^2}{ML^2T^{-1}[LT^{-1}]}=[M^0L^0T^0A^0]$
The term

$\dfrac {e^2}{\epsilon_ohc}$ is called fine structure constant & has the value

$=\dfrac {1}{137}$ .

The length, breadth and thickness of a block are given by

$l = 12\ cm$ ,

$b = 6\ cm$ and

$t = 2.45 cm$ . The volume of the block according to the idea of significant figures should be :

0%
$\displaystyle 1\times 10^{2}\:cm^{3}$
0%
$\displaystyle 2\times 10^{2}\:cm^{3}$
0%
$\displaystyle 1.763\times 10^{2}\:cm^{3}$
0%
None of these

Explanation

Since question contains atleast 1 significant number in one of the value, therefore, we should write the answer in such a form that it contains only one significant figure.

$V=lbt=12\times 6\times 2.45=2\times 10^2\ cm^3$

Using mass

$(M)$ , length

$(L)$ , time

$(T)$ and electric current

$(A)$ as fundamental quantities the dimensions of permittivity will be

0%
$[MLT^{-1}A^{-1}]$
0%
$[MLT^{-2}A^{-2}]$
0%
$[M^{-1}L^{-3}T^{+4}A^{2}]$
0%
$[M^{-2}L^{-2}T^{-2}A^{2}]$

Explanation

Using Coulomb's law, Force

$F=\dfrac {1}{4\pi \epsilon_0}\dfrac {q_1q_2}{r^2}\Rightarrow \epsilon_2=\dfrac {q_1q_2}{4\pi Fr^2}$
So dimension of

$[\epsilon_0]$

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

The dimensional formula of farad is

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

Explanation

One farad is defined as the capacitance

$C$ of a capacitor across which, when charged with one coulomb

$Q$ of electricity, there is a potential difference

$V$ of one volt.
So,

$\displaystyle [C]=\left [\frac {Q}{V}\right ]=\left [\frac {Q^2}{W}\right ]=\left [\frac {Q^2}{M^{1}L^{2}T^{-2}}\right ]=[M^{-1}L^{-2}T^2Q^2]$

If time

$T$ , acceleration

$A$ and force

$F$ are regarded as base units, then the dimensional formula of work is

0%
$[FA]$
0%
$[FAT]$
0%
$[FAT^2]$
0%
$[FA^2T]$

Explanation

$Work=Force \times distance=F\times L$
We need distance in terms of acceleration and time.

$A=LT^{-2}$

$L=AT^2$

$W=FAT^2$

The dimensional formula for magnetic flux is

0%
$\displaystyle \left [ ML^2T^{-2}A^{-1} \right ]$
0%
$\displaystyle \left [ ML^{3}T^{-4}A^{-2} \right ]$
0%
$\displaystyle \left [ M^{0}L^{-2}T^{-2}A^{-2} \right ]$
0%
$\displaystyle \left [ ML^{2}T^{-1}A^{2} \right ]$

Explanation

Formula for magnetic flux will be

$B \times A$

$B$ =magnetic field. Dimensions will be

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

$A$ =Area. Dimensions will be

$[L^2]$

Dimension for magnetic flux will be $[ML^2T^{-2}I^{-1}]$

Which one of the following has the dimensions of pressure?

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

Explanation

Dimension of pressure is given as

$\displaystyle \frac{|F|}{|A|}=\frac{|ma|}{|A|}=\frac{M.L{T}^{-2}}{{L}^{2}}=M{L}^{-1}{T}^{-2}$

The dimensions of voltage in terms of mass

$(M)$ , length

$(L)$ and time

$(T)$ and ampere

$(A)$ are

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

Explanation

Voltage(V) is a representation of the electric potential energy(W) per unit charge(Q).
So,

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

Consider the following pairs of quantities:
Young's modulus; pressure
Torque; energy
Linear momentum; work
Solar day; light year
In which cases are the dimensions, within a pair, same?

0%
1 and 3
0%
1 and 4
0%
1 and 2
0%
2 and 4

Explanation

Young's Modulus

$=$ Pressure

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

[Torque]

$=$ [Energy]

$=\left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -2 } \right]$
Linear momentum

$=\left[ { M }^{ 1 }{ L }^{ 1 }{ T }^{ -1 } \right]$ , work

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

Solar day

$=[T]$ and Light year

$=[L]$

The physical quantity having the dimensions

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

0%
resistance
0%
resistivity
0%
electrical conductivity
0%
electromotive force

Explanation

Electrical conductivity is given as,

$\sigma = \dfrac{Gl}{a}$

where

$G$ is the conductance i.e.

$G = \dfrac{I}{V}$

So,

$[\sigma]=[\dfrac{Gl}{a}]=[\dfrac{Il}{Va}]=[\dfrac{AL}{ML^2T^{-2}L^2/AT}]$

$\implies [\sigma]=[M^{-1}T^3L^{-3}A^2]$

metres

0%
$10^{-1}$
0%
$10^{-2}$
0%
$10^1$
0%
$10^2$

Explanation

1 mm is

$1\times { 10 }^{ -3 }m$ .
So,

$326mm = 3.26\times1\times { 10 }^{ 2 }\times { 10 }^{ -3 }m$ .
Hence its order of magnitude in metres is

${ 10 }^{ -1 }m$

The dimensional formula for Planck's constant and angular momentum are

0%
$\displaystyle \left [ ML^{2}T^{-2} \right ]and\left [ MLT^{-1} \right ]$
0%
$\displaystyle \left [ ML^{2}T^{-1} \right ]and\left [ ML^{2}T^{-1} \right ]$
0%
$\displaystyle \left [ ML^{3}T^{1} \right ]and\left [ ML^{2}T^{-2} \right ]$
0%
$\displaystyle \left [ MLT^{-1} \right ]and\left [ MLT^{-2} \right ]$

Explanation

We know that

$E=h\nu$
where dimensions of Energy, E are

$M{L}^{2}{T}^{-2}$ and dimensions of frequency are inverse of Time Period, i.e.

${T}^{-1}$
Thus,

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

$[h]=M{L}^{2}{T}^{-1}$
Now, angular momentum,

$L=I\omega$
where units of omega are inverse of time (rad/sec, where radian is dimensionless).
Units of I are

$M{L}^{2}$ (because I is of the form

$kM{x}^{2}$ )
Thus

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

Which of the following is the dimension of the coefficient of friction?

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

Explanation

Coefficient of friction is unitless and dimensionless.

Thus,

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

option B is correct.

Dimensions of linear impulse are

0%
$\displaystyle \left [ ML^{-2}T^{-3} \right ]$
0%
$\displaystyle \left [ ML^{-2} \right ]$
0%
$\displaystyle \left [ MLT^{-1} \right ]$
0%
$\displaystyle \left [ MLT^{-2} \right ]$

Explanation

Linear impulse is given as the change in linear momentum,

$I=m\Delta v$
Thus,

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

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

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

Dimension of velocity gradient is

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

Explanation

Velocity gradient =

$\dfrac{Velocity}{Distance}$
Thus

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

$C$ and

$R$ denote capacitance and resistance, then dimensions of

$CR$ will be

0%
$\displaystyle \left [ M^{0}L^{0}TA^{0} \right ]$
0%
$\displaystyle \left [ ML^{0}TA^{-2} \right ]$
0%
$\displaystyle \left [ ML^{0}TA^{2} \right ]$
0%
$\displaystyle \left [ ML^{0}T^{2}A^{-2} \right ]$

Explanation

We know that value

$RC$ gives time constant in an

$R-C$ circuit and that

$V(t)={V}_{0}{e}^{\frac{-t}{RC}}$
Exponents are always dimensionless, thus

$[\dfrac{-t}{RC}]={M}^{0}{L}^{0}{T}^{0}{A}^{0}$ or

$[RC]=[t]=[{M}^{0}{L}^{0}{T}^{1}{A}^{0}]$

It takes time

$8$ min for light to reach from sun to the earth surface. If speed of light is taken to be

$3 \times 10^8$ m

$s^{-1}$ , find the order of magnitude of distance from the sun to the earth in km.

0%
$10^7 km$
0%
$15\times10^8 km$
0%
$10^8 km$
0%
$10^9 km$

Explanation

Time

$t=8\ min=480\ sec$

Speed of light

$c=3\times10^8\ ms^{-1}$

$Distance=t\times c=1.44\times10^{11}m=1.44\times10^8km$

So order of magnitude os distance is

$10^8km$

Write the dimensions of

$\cfrac{d{\phi}_{B}}{dt}$ where

$\phi_B$ is magnetic flux through a coil.

0%
$[ M{ L }^{ 2 }{ A }^{ -1 }{ T }^{ -3 }]$
0%
$[ M{ L }^{ 3 }{ A }^{ -1 }{ T }^{ -2 }]$
0%
$[ M{ L }^{ 2 }{ A }^{ -2 }{ T }^{ -3 }]$
0%
None of these

Explanation

$\cfrac{d{\phi}_{B}}{dt}=$ [potential or EMF]

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

The length, breadth and height of a glass slab are respectively

$50 cm$ ,

$20 cm$ and

$25 cm$ . Find the order of magnitude of volume of slab in Sl unit.

0%
$10^{-2} m^3$
0%
$10^{-4} m^3$
0%
$10^{-3} m^3$
0%
$10^{-3} cm^3$

Explanation

$L=0.5 m , b=0.2 m ,h=0.25 m$

Volume

$=$ length

$\times$ breadth

$\times$ height

Volume

$=l\times b\times h=0.025 m^{3 }\ = \ 2.5\times 10^{-2}\ m^3$

So, the order of magnitude is

$10^{-2} m^3$ .

The radius of a hydrogen atom is

$0.5\mathring{A}$ . Find the order of magnitude of volume of

$1$ mole hydrogen in

$m^3$ . Given that 1 mole of hydrogen has

$6.02 \times 10^{23}$ hydrogen atoms.

0%
$2\times10^{-4} m^3$
0%
$10^{-3} m^3$
0%
$10^{-7} m^3$
0%
$10^{-2} m^3$

Explanation

Volume of 1 hydrogen atom,

$V=4\pi r^3/3; r=0.5\times10^{-10}m$

So,

$V=0.5245\times10^{-30}m^3$

So, volume of 1 mole (

$6.02\times10^{23}$ ) is

$V\times6.02\times10^{23}=3.157\times10^{-7}m^3$

So, order of magnitude is

$10^{-7}m^3$ .

Find the order of magnitude of the mass of the star, whose radius is

$384 \times 10^6$ m and average density is

$4 \times 10^3$ kg

$m^{-3}$ :

0%
30
0%
29
0%
24
0%
21

Explanation

Radius of star

$r=384\times10^6m$

So, Volume of star is

$V=4\pi r^3/3=2.371\times10^{26}m^3$

Density of star is

$\rho=4\times10^3kgm^{-3}$

Mass of star is

$M=V\times\rho\approx9.5\times10^{29}kg=0.95\times10^{30}kg$

So the mass is of order of

$10^{30}kg$

The electric potential existing in space is

$V(x, y, z) = A (xy+ yz + zx)$ . Write the dimensional formula of

$A$ :

0%
$MT^{-2} I^{-1}$
0%
$MT^{-3} I^{-1}$
0%
$MT^{-2} I^{-2}$
0%
$MT^{-3} I^{-2}$

Explanation

Since

$x, y$ and

$z$ have units of length i.e.

$m$ .
Thus Unit of A =unit of potential

$\times m^{-2}$ = joules per coulomb

$\times m^{-2}$

Dimension of joule

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

Dimension of coulomb

$= I T$ where

$I$ is the dimension of current.
Dimension of A = dimension of joule per coulomb

$\times L^{-2}$
=

$M L^2 T^{-2} (I T)^{-1} \times L^{-2} = M T^{-3} I^{-1}$

The value due regard to significant figure

$4.0\times 10^{-4}-2.5\times 10^{-6}$

0%
$\displaystyle 4.0\times 10^{-4}$
0%
$\displaystyle 3.975\times 10^{-4}$
0%
$\displaystyle 3.9\times 10^{-4}$
0%
none of the above

Explanation

The value of this case is,

$4.0\times 10^{-4}-2.5\times 10^{-6}$

$=3.975\times 10^{-4}$ .

Dimension of relative density is -

0%
$kg m^{-3}$
0%
$ML^{-3}$
0%
Dimensionless
0%
$M^2L^{-6}$

Round off 2.0082 to four significant figures.

0%
2.008
0%
2.0082
0%
2.009
0%
2

Explanation

$2.0082=2.008$ ,

Here the digit

$2$ is less than

$5$ so

$2.008$ is the closest value in four significant figures.

The dimensions of universal gravitational constant

$G$ are :

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

Explanation

$\textbf{Hint:}$ Dimensional formula of force is $ML{{T}^{-2}}$ .

$\textbf{Solution:}$

$\textbf{Step 1: Calculating the gravitational constant.}$

We know that Gravitational force exerted on a body of mass m is given by,

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

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

$\textbf{Step 2: Calculating dimensions of universal gravitational constant.}$

$G=\dfrac{[ML{{T}^{-2}}][{{L}^{2}}]}{[{{M}^{2}}]}$

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

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

Write the dimensional formula of charge.

0%
$\left [ AT \right ]$
0%
$\left [ AT^{-1} \right ]$
0%
$\left [ AT^{-2} \right ]$
0%
$\left [ AT^{2} \right ]$

Explanation

Since current is rate of flow of charge, i.e.

$I=\dfrac { q }{ t }$ . Therefore

$q=It$ . Hence,

$[q]=[AT]$ .

Which of the following groups have the same dimensions ?

0%
velocity, speed
0%
pressure, stress
0%
force, impulse
0%
work, energy

Explanation

Velocity and speed have same units i.e

$m/s$ . So, both have same dimensions.

Pressure and stress have same units i.e.

$N/{m}^{2}$ . So, both of them have the same dimensions.

Work and energy have same units i.e.

$N-m$ . So, both have same dimensions.

The units of Force is

$N$ and units of impulse is

$N-s$ . So, they will have different dimensions.

Hence, options A, B & D are correct.

Round off following number upto four significant figures.

$45.689$

0%
$45.689$
0%
$45.69$
0%
$45.68$
0%
$45.7$

Explanation

The given number has

$5$ significant figures.

$9>5$ , thus the digit

$8$ gets changed to

$9$ after rounding off upto four significant figures.

$\therefore$ After rounding off,

$45.689$ becomes

$45.69$

CBSE Questions for Class 11 Engineering Physics Units And Measurement Quiz 7 - 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 7 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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