MCQ Questions for Class 11 Engineering Physics Units And Measurement Quiz 2

A pair of physical qunatities having the same dimensional formula are:

0%
momentum and impulse
0%
momentum and energy
0%
energy and pressure
0%
force and power

Explanation

Momentum

$\rightarrow mv=M^{1}L^{1}T^{-1}$
Impulse

$\rightarrow f\times t=M^{1}L^{1}T^{-1}$
Energy

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

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

$=ML^{2}T^{-3}$
Hence, option A is correct.

A pair of physical quantities having the same dimensional formula are:

0%
Force and Work
0%
Work and Energy
0%
Force and Torque
0%
Work and Power

Explanation

Force

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

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

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

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

$=ML^{3}T^{-3}$
Hence, option B is correct.

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

Explanation

$\dfrac{1}{\sqrt{\mu _{0}\epsilon _{0}}}=\sqrt{L^{2}T^{-2}}=LT^{-1}$
Speed

$=LT^{-1}$
Stress, Young's modulus

$=\dfrac{MLT^{-2}}{L^{2}}=ML^{-1}T^{-2}$
Momentum

$=MLT^{-1}$
Torque, Work

$=ML^{2}T^{-2}$
Planck's constant

$=\dfrac{ML^{2}T^{-2}}{\left ( LT^{-1} \right )}\times L=ML^{2}T^{-1}$
Hence, option B is correct.

The dimensional formula for universal gravitational constant is:

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

Explanation

The correct option is $D.$

Formula for Gravitational force is:

$F= \dfrac{G M m}{r^{2}}$

So, Formula for G will be,

$G = \dfrac{F r^{2}}{M m}$

Thus, the dimensional formula for G is,

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

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

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

Explanation

For calculating coefficient of viscosity we can use the formula,

$F=\dfrac{\eta vA}{S}$
Dimension of coefficient of viscosity will be

$=\dfrac{Dimension \ of \ force \times Distance}{Velocity \times Area}$

$=\dfrac{MLT^{-2}\times L}{LT^{-1}L^2}$

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

Dimensions of

$\dfrac { 1 }{ { \mu }_{ 0 }{ \epsilon }_{ 0 } }$ , where symbols have their usual meaning are :

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

The pair of physical quantities having the same dimensional formula is:

0%
angular momentum and torque
0%
torque and strain energy
0%
entropy and power
0%
power and angular momentum

Explanation

Angular momentum

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

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

$=\frac{1}{2} \times ML^{2}T^{-2}$
Entropy

$=ML^{2}T^{-2}K^{-1}$
Power

$=ML^{2}T^{-3}$
Hence torque and strain energy has same dimensions,

Dimensional formula of Torque is:

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

Explanation

$\textbf {Hint :}$ Use Torque

$T= \vec{r} \times \vec{F}$

${\textbf{Explanation:}}$

Torque

$T= \vec{r} \times \vec{F}$

F is force and r is perpendicular distance

Dimensional formula of

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

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

Planck's constant has the dimensions as that of:

0%
Energy
0%
Power
0%
Linear momentum
0%
Angular momentum

Explanation

Plank's constant

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

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

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

$=MLT^{-1}$
Angular momentum

$=ML^{2}T^{-1}$
Hence, option D is correct.

The dimensional formula

${ M }^{ -1 }{ L }^{ 3 }{ T }^{ -2 }$ refers to :

0%
Force
0%
Power
0%
Gravitational constant
0%
Energy

Explanation

Force

$=MLT^{-2}$
Power

$=ML^{2}T^{-3}$
Energy

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

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

The dimensional formula for angular velocity is:

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

Explanation

$Angular \ velocity, W=\dfrac{1}{Time}=T^{-1}M^{0}L^{0}$

The SI unit of a physical quantity having the dimensional formula of

${ ML }^{ 0 }{ T }^{ -2 }{ A }^{ -1 }$ is:

0%
$tesla$
0%
$weber$
0%
$amp- meter$
0%
$amp$ - ${ m }^{ 2 }$

Explanation

$amp$

$m^{2}=A^{1}L^{2}$

$amp$

$meter$

$=A^{1}L^{1}$

$weber$

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

$A\rightarrow$ ampere

$Tesla$

$=M^{1}L^{0}T^{-2}A^{-1}$
Hence, option A is correct answer.

The dimensional formula for impulse is:

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

Explanation

$Impulse=force\times time=ML^{1}T^{-2}\times T^{1}$ $=MLT^{-1}$

The number of significant figures in

$6.023\times { 10 }^{ 23 }$ is

0%
$4$
0%
$3$
0%
$2$
0%
$23$

Explanation

The correct option is A.

Every experimental observation value is having some amount of uncertainty associated with it. Significant figures are the number of meaningful digits having certainty. All non – zero digits are significant. Thus in $6.023 \times 10^{23}$ , we can see $4$ significant figures.

The dimensions of calorie are :

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

Explanation

Its a unit of energy whose dimensions can be calculated as

$F.dr=MLT^{-2}L=ML^2T^{-2}$

The correct order in which the dimensions of "length" decreases in the following physical quantities is
a) Coefficient of viscocity
b) Thermal capacity
c) Escape velocity
d) Density

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

Explanation

a) Coefficient of viscosity

$=\dfrac{MLT^{-2}\times L}{\left ( L^{2} \right )\left ( LT^{-2}\right )}=ML^{-1}T^{-3}$
b) Thermal capacity

$=ML^{2}T^{-2}T^{-1}\rightarrow$ (temperature)
c) Escape velocity

$=LT^{-2}$
d) Density

$=ML^{-3}$
So, dimension of length in

$b>c>a>d$ .

Match List I with II and select the correct answer:

List-I	List-II
A.Spring constant	${ M }^{ 1 }{ L }^{ 2 }{ T }^{ -2 }$
B.pascal	2. ${ M }^{ 0 }{ L }^{ 0 }{ T }^{ -1 }$
C.hertz	3. ${ M }^{ 1 }{ L }^{ 0 }{ T }^{ -2 }$
D.joule	4. ${ M }^{ 1 }{ L }^{ -1 }{ T }^{ -2 }$

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

Explanation

Spring constant dimensions will be

$K=\dfrac{F}{X}=\dfrac{MLT^{-2}}{L}=ML^0T^{-2}$

Pascal is unit of pressure

$\dfrac{F}{A}=\dfrac{MLT^{-2}}{L^2}=ML^{-1}T^{-2}$

Hertz is frequency

$=\dfrac{1}{t}=T^{-1}$

Joule is unit of energy so dimension is

$ML^2T^{-2}$

When the number 0.046508 is reduced to 4 significant figures, then it becomes :

0%
$0.0465$
0%
$4650.8 \ \times$ ${ 10 }^{ -5 }$
0%
$4.651 \times { 10 }^{ -2 }$
0%
$4.650 \times { 10 }^{ -2 }$

Explanation

$0.046508\approx 4.6508\times 10^{-2}$

$\downarrow$
5 significant digits
To have 4 significant digits

$\rightarrow 4.6508\approx 4.651$
(If the digit to be dropped is more than 5, then the preceding digit is raised by 1)
So,

$0.046508\approx 4.651\times 10^{-2}$

Which of the following pair does not have the same dimensions?

0%
Moment of inertia and Torque
0%
Linear Momentum and impulse
0%
Angular Momentum and Plank's constant
0%
Work and internal energy

Explanation

Moment of inertia

$={mr}^{2}={ML}^{2}$
Torque

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

$=mv=MLT^{-1}$
Impulse

$=MLT^{-1}$
Angular momentum

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

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

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

$=ML^{2}T^{-2}$
Hence, option A is correct.

Let

${ \epsilon}_{0 }$ denote the permittivity of the vacuum and

$\mu _o$ is permeability of vacuum.
If M

$=$ mass, L

$=$ length, T

$=$ time and I

$=$ electric current, then
a)

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

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

${ \mu}_{0 }={ MLT }^{ -2 }{ I }^{ -2 }$
d)

${ \epsilon}_{0 }={ ML }^{ 2 }{ T }^{ -1 }I$

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

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

B and C are correct.

What are the dimensions of

$K=\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } ?$ (Given, N is the unit of force, with dimensions

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

0%
${ C }^{ 2 }{ N }^{ -1 }{ M }^{ -2 }$
0%
${ NM }^{ 2 }{ C }^{ -2 }$
0%
${ NM }^{ 2 }{ C }^{ 2 }$
0%
Unitless

Explanation

$F=\dfrac{KQ^2}{R^2}$

$K=\dfrac{FR^2}{Q^2}=\dfrac{NM^2}{C^2}$
Hence, option B is correct.

According to theory of significant figures the value of

$\sqrt { 3.0 }$ is:

0%
$1.732$
0%
$1.7$
0%
$1.73$
0%
$1.8$

Explanation

$\sqrt{3.0}=1.732$
Since 3.0 has 2 significant figure,
So, our answer should have 2 significant figure
i.e.

$1.732\approx 1.7$

The dimensional formula for Areal velocity is :

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

Explanation

It is the area covered per unit time

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

The number of significant figures in the numbers

$672.9$ and

$2.520\times { 10 }^{ 7 }$ are :

0%
$4, 4$
0%
$3, 4$
0%
$4, 3$
0%
$3, 3$

Explanation

$672.9 \rightarrow 6, 7, 2, 9 \rightarrow$ 4 significant digits since all non zero digits are significant.
$2.520\times 10^{7} \rightarrow 2, 5, 2, 0 \rightarrow 4$ significant digits, since all the zero on the right side of last non zero digit in the decimal part are significant.

When

$24.25\times { 10 }^{ 3 }$ is rounded off to three significant figures, it becomes :

0%
242
0%
243
0%
$244\times { 10 }^{ 2 }$
0%
$24.2\times { 10 }^{ 3 }$

Explanation

$24.25\times 10^{3} \rightarrow 2,4,2,5\rightarrow$ present by it has 4 significant digits.
If the digit to be dropped is 5 or 5 followed by zeros, then preceding digit is left unchanged, if it is even.
So, up to three significant figures, the number

$=24.2\times 10^{3}$

Universal gravitational constant is given by

$6.67\times { 10 }^{ -11 }N{ m }^{ 2 }{ kg }^{ -2 }$ . Then the no. of significant figures in it is:

0%
$14$
0%
$2$
0%
$3$
0%
$11$

Explanation

$6.67\times 10^{-11}$ has

$3$ significant digits

$\rightarrow6,6,7$

When we express the velocity of light,

$30,00,00,000/ ms^{-1}$ , in standard form up to three significant figures, it's value is:

0%
$3\times { 10 }^{ 8 }\ { ms }^{ -1 }$
0%
$3.00\times { 10 }^{ 8 }\ { ms }^{ -1 }$
0%
$3\times { 10 }^{ 10}\ { cms }^{ -1 }$
0%
$3\times { 10 }^{ 6}\ { ms }^{ -1 }$

Explanation

Since all the zeros on the right of last non zero digit in the decimal part are significant.

$\Rightarrow 300000000=3.00\times 10^{8}\ m/s$

$'m$ ' is the mass of a body,

$'a'$ is amplitude of vibration, and

$'w '$ is the angular frequency then

$\frac { 1 }{ 2 } { ma }^{ 2 }{ w }^{ 2 }$ has same dimensional formula as :

0%
work
0%
moment of force
0%
energy
0%
all the above

Explanation

Dimension of

$\dfrac{1}{2}ma^{2}w^{2}=ML^{2}\left ( \dfrac{1}{T^{2}} \right )=ML^{2}T^{-2}$
Work

$=F.d\left ( force\times distance \right )=\left ( \dfrac{ML}{T^{2}} \right )\times L=ML^{2}T^{-2}$
Moment of force

$=\underset{f}{\rightarrow}\times \underset{r}{\rightarrow}=\left [ \dfrac{ML}{T^{2}} \right ]\times L=ML^{2}T^{-2}$
Energy = work =

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

The dimension of time in Electric field intensity is :

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

Explanation

$F=QE$

$E=F/Q$

$\Rightarrow \dfrac{MLT^{-2}}{AT}=MLT^{-3}A^{-1}$

The radius of a sphere is

$5$ cm. Its volume will be given by (according to the theory of significant figures) :

0%
$523.33\ { cm }^{ 3 }$
0%
$5.23\ { \times\ { 10 }^{ 2 }cm }^{ 3 }$
0%
$5.0\ { \times\ { 10 }^{ 2 }cm }^{ 3 }$
0%
$5\ { \times\ { 10 }^{ 2 }cm }^{ 3 }$

Explanation

Volume

$=\dfrac{4}{3}{\pi r}^{3}=\dfrac{4}{3}\times \pi \left ( 5 \right )^{3}=523.33\ {cm}^{3}$

$=5.2333\times 10^{2}\ cm^{3}$

$\downarrow$
5 significant figures
Since radius has single significant figure, so, volume should also have single significant figure.

For single significant figure, we have to drop all after decimal.

$\Rightarrow$ Volume

$= 5\times 10^{2}\ cm^{2}$ (if the digit to be dropped is less than 5, preceding digit is left unchanged)

Two numbers are given as i) 20.96 and ii) 0.0003125. Rounding off above numbers to 3 significant figures will lead to :

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

Explanation

a) 20.96
We have to round off for 3 significant digits. Last digit has to be rounded off.
If the digit to be dropped (6) is more than 5, then preceding digit is raised by 1.
Hence

$20.96\approx 21.0$

$0.0003125=3.125\times 10^{-4}$
For 3 significant digits

$\Rightarrow 3.12\times 10^{-4}$

The dimension of magnetic induction in

$M, L, T$ and

$C$ (Coloumb) is given as:

0%
${ MT }^{ -1 }{ C }^{ -1 }$
0%
${ MT }^{ -2 }{ C }^{ -1 }$
0%
${ MLT }^{ -1 }{ C }^{ -1 }$
0%
${ MT }^{ 2 }{ C }^{ -2 }$

Explanation

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

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

$C = Charge = I^{1}T^{1}$

$So, B = M^{1}C^{-1}T^{-1}$

$\left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -3 }{ A }^{ -2 } \right]$ is the dimensional formula of :

0%
Electric resistance
0%
Capacity
0%
Electric potential
0%
Specific resistance

Explanation

$Power=I^2R$

$[R]=\dfrac{ML^2T^{-3}}{A^2}$
Option A is correct.

The ratio

$\dfrac { L }{ R }$ has the dimensions of:
[ L:Inductance R:Resistance]

0%
Velocity
0%
Acceleration
0%
Time
0%
Force

Explanation

$L$ has the diemnsions of

$ML^2T^{-2}I^{-2}$ .
Resistance has the dimensions

$ML^2T^{-3}I^{-2}$ .

$\dfrac{L}{R}=\dfrac{ML^2T^{-2}I^{-2}}{ML^2T^{-3}I^{-2}}=T$

The physical quantity having dimensions

$2$ in length is:

0%
Power
0%
Acceleration
0%
Force constant
0%
Stress

Explanation

$Power = ML^{2} T^{-3}$

$Acceleration = LT^{-2}$

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

$Stress = F/A = ML^{-1}T^{-2}$
Hence, option A is correct.

The fundamental physical quantities that have the same dimension in the dimensional formula of Torque and Angular Momentum are:

0%
mass, time
0%
time, length
0%
mass, length
0%
time, mole

Explanation

Torque

$\rightarrow M^{1}L^{2}T^{-2}$
Angular momentum

$\rightarrow M^{1}L^{2}T^{-1}$
Mass and length have same dimensions.

The physical quantity which has the dimensional formula

${ M }^{ 1 }{ T }^{ -3 }$ is:

0%
surface tension
0%
solar constant
0%
density
0%
compressibility

Explanation

Surface tension

$\displaystyle \rightarrow \frac{F}{l}=\frac{MLT^{-2}}{L}=MT^{-2}$

Solar constant

$=MT^{-3}$

Density

$=\dfrac{Mass}{Vol}=ML^{-3}$
Compressibillity

$=\dfrac{1}{Bulk \ Modulus}=\dfrac{\dfrac{dv}{v}}{dp}=\dfrac{L^{3}/L^{3}}{ML^{-1}T^{-2}}=M^{-1}L^{1}T^{2}$

Hence, option B is correct.

The unit of impulse per unit area is same as that of:

0%
coefficient of viscosity
0%
surface tension
0%
bulk modulus
0%
none of the above

Explanation

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

$Viscocity = \dfrac{stress}{\dfrac{dv}{dy}} = [M{ L }^{ -1 }{ T }^{ -1 }]\quad$

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

$Bulk \ Modulus = \dfrac { dP }{ dV/v } =[ { M }^{ 1 }{ L }^{ -1 }{ T }^{ -2 }]\quad$

Hence, option A is correct.

Given

$M$ is the mass suspended from a spring of force constant

$k$ . The dimensional formula for

$\left[ M/k \right] ^{ 1/2 }$ is same as that for:

0%
frequency
0%
time period
0%
velocity
0%
wavelength

Explanation

$K = \dfrac{f}{x} = \dfrac { { MLT }^{ -2 } }{ L } ={ MT }^{ -2 }$

$\displaystyle \Rightarrow \frac { M }{ K } = \frac { { M }^{ 1 } }{ { M }^{ 1 }{ T }^{ -2 } } = { T }^{ 2 }$

$\Rightarrow \sqrt { M/K } \quad =\quad { T }^{ 1 }$

Hence, option B is correct.

$\dfrac { 1 }{ \sqrt { Capacitance\times Inductance } }$ has the same unit as:

0%
time
0%
velocity
0%
velocity gradient
0%
none of the above

Explanation

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

$I=Inductance =M{ L }^{ 2 }{ I }^{ -2 }{ T }^{ -2 }$

$\dfrac { 1 }{ \sqrt { CI } } = \dfrac { I }{ \sqrt { \left( { M }^{ -1 }{ L }^{ -2 }{ I }^{ 2 }{ T }^{ 4 } \right) \times \left( M{ L }^{ 2 }{ I }^{ -2 }{ T }^{ -2 } \right) } }$

$\displaystyle \frac { 1 }{ \sqrt { CI } } =\frac { 1 }{ T } = { T }^{ -1 }$

$Time = { T }^{ 1 }, Velocity = { LT }^{ -1 }$

$Velocity _{grad} = \dfrac{dv}{dx} = \dfrac { { LT }^{ -1 } }{ L } = { T }^{ -1 }$

Hence, option C is correct.

Dimensions of solar constant are:

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

Explanation

Solar constant(S) is defined as energy due to sun falling on an unit area per unit time .

$\Rightarrow \displaystyle S=\left ( \frac{E}{A} \right )\times \frac{1}{T}=\left ( \frac{ML^{2}}{T^{2}} \right )\times \frac{1}{L^{2}}\times \frac{1}{T}=MT^{-3}$

If the unit of work is

$100J$ , the unit of power is

$1KW$ , the unit of time in second is:

0%
$10^{ -1 }$
0%
$10$
0%
$10^{ -2 }$
0%
$10^{ -3 }$

Explanation

Power

$=1000=M^{1}L^{2}T^{-3}$

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

Work $=power \times time$

$\displaystyle time=\frac{work}{power}=T^{1}=\frac{100}{1000}=\frac{1}{10}$

Hence, Option A is correct.

According to theory of significant figures

$\left( 2.0 \right) ^{ 10 }$ is :

0%
1024
0%
$1.024\times { 10 }^{ 10 }$
0%
$1.0\times { 10 }^{ 3 }$
0%
1000

Explanation

$\left ( 2.0 \right )^{10}=1024$
But since 2.0 contains 2 significant figure.
So,

$1024=1.0\times 10^{3}$

Dimensions of '

$ohm$ ' are same as that of:

(Given :

$h$ is Planck's constant and

$e$ is charge)

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

Explanation

$a = ohm = ML^{2}I^{-2}T^{-3}$

$b = h = ML^{2}T^{-1}$

$c = e = I^{1}T^{1}$

$a = b^{\alpha } c^{\beta }$

$ML^{2}I^{-2}T^{-3} = (ML^{2}T^{-1})^{\alpha } (I^{1}T^{1})^{\beta}$

$\Rightarrow \beta = -2, \alpha = 1$

$\Rightarrow ohm = \dfrac{h}{e^{2}}$

$m$ is the mass,

$Q$ is the charge and

$B$ is the magnetic induction,

$\dfrac{m}{BQ}$ has the same dimensions as :

0%
Frequency
0%
Time
0%
Velocity
0%
Acceleration

Explanation

$F=BQV$

$F=ma$

$ma=BQV$

$\dfrac{m}{BQ}=\dfrac{V}{a}=\dfrac{LT^{-1}}{LT^{-2}}=T$

The length, breadth and thickness of a rectangular lamina are

$1.024\ m$ ,

$0.56\ m$ , and

$0.0031\ m$ . The volume, in

${ m }^{ 3 }$ , is:

0%
${ 1.8\times 10 }^{ -3 }$
0%
${ 1.80\times 10 }^{ -3 }$
0%
${ 0.180\times 10 }^{ -4 }$
0%
$0.00177$

Explanation

Volume

$= l\times b\times h=1.777664\times 10^{-3}\ m^3$
Since, the minimum number of significant figures is

$2$ (i.e

$0.56$ ) out of all quantities, the answer has to be reported in

$2$ significant digits.

$\Rightarrow V = 1.8\times 10^{-3} m^{3}$

Dimensions of

$\dfrac { L }{ RCV }$ are:

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

Explanation

Here

$\dfrac{L}{R} = time$

and

$CV = Q$

$\dfrac{t}{Q} = current^{-1}$

$A^{-1}$ is Correct answer

The edge of a cube is a

$=1.2\times 10^{-2}m$ . Then its volume will be recorded as :

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$1.7\times 10^{-6} m^3$
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$1.70\times 10^{-6} m^3$
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$1.70\times 10^{-7} m^3$
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$1.78\times 10^{-6} m^3$

Explanation

Given that the edge of a cube is

$a=1.2 \times 10^{-2}m$ .
The volume of a cube

$= a^{3}$ .
Volume

$=[1.2 \times 10^{-2} ]^{3}$
The given volume of the cube is

$1.728 \times 10^{-6}m^{3}$

$\approx 1.7 \times 10^{-6}m^{3} \ \ \ (upto \ two \ significant \ digits)$

The physical quantity which has the dimensional formula as that of

$\dfrac { energy }{ mass \times length }$ is:

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Force
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Power
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Pressure
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Acceleration

Explanation

$\dfrac{energy}{mass\times length}=\dfrac{ML^{2}T^{-2}}{M\times L}=LT^{-2}$

$\downarrow$
(acceleration)
Force

$=MLT^{-2}$
Power

$=ML^{2}T^{-3}$
Pressure

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

If the unit of mass is

$\dfrac { 1 }{ 2 } kg$ and that of length is

$2$ m and the unit of time is one second, the unit of pressure is:

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2 pascal
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0.5 pascal
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0.25 pascal
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1.0 pascal

Explanation

$\displaystyle Pressure=\frac{force}{Area}$

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

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

$\displaystyle =\frac{1}{2}\times\frac{1}{2}\times \frac{1}{(1)^{2}}\frac{kg}{ms^{2}}$

$=\dfrac{1}{4} =0.25 \ pascal$

CBSE Questions for Class 11 Engineering Physics Units And Measurement Quiz 2 - 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 2 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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