MCQ Questions for Class 11 Engineering Physics Units And Measurement Quiz 12

The dimensional formula

$[ML^1T^{-3}]$ is more closely associated with

0%
power
0%
energy
0%
intensity
0%
velocity gradient

A balloon of mass

$1$ g has

$100$ g of water in it. If it is completely immersed in water, its mass will be

0%
$0.5$ g
0%
$1.2$ g
0%
$1$ g
0%
$1.5$ g

Dimensional formula of self inductance is:

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

Explanation

The magnetic flux is given by

$\phi=LI$

The dimensional formula of self inductance,

$[L]=\dfrac{[BA]}{[I]}$ . . . . . .(1)

Force,

$f=qvB$

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

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

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

From equation (1), we get

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

The correct option is C.

While measuring

$g$ using simple pendulum method

$\%$ error in measurement of

$\ell$ is

$\pm\ 2\%$ and in measurement time is

$\pm\ 1\%$ then relative error in

$g$ is:

0%
$\pm 0.04$
0%
$0.00$
0%
$\pm 0.03$
0%
$\pm 0.01$

Explanation

$\begin{aligned} \text { Given }-& * \% \text { error } \operatorname{ in} \ell=2 \% \\ & * \% \text { errole in time } T=1 \% \end{aligned}$

$\begin{array}{l} \text { we Know that - } \\ \qquad T=2 \pi \cdot \sqrt{\frac{\ell}{g}} \\ \Rightarrow \text { on rearanging the formula } \\ \qquad g=4 \pi^{2} \cdot \frac{\ell}{T^{2}} \end{array}$

$\begin{array}{l} \text { Taking logarithm in above eq- } \\ \log (g)=\log (l)-2 \cdot \log (T)+\log \left(4 \pi^{2}\right) \end{array}$

$\begin{array}{l} \text { Ditterentiating both Sides - } \\ \qquad \frac{d g}{g}=\frac{d l}{l}-2 \cdot \frac{d T}{T} \end{array}$

$\frac{d g}{g}=200+2 \cdot(1 \%)=4 \%=\frac{4}{100}=0.04$

$\begin{array} \Rightarrow \text { Relative error in } g=\pm 0.04 \end{array}$

Suppose an attractive nuclear force acts between two protons, which may be written as

$F=\dfrac { { Ae }^{ -kr } }{ { r }^{ 2 } }$ . The dimensional formula for

$\dfrac {A}{K}$ is

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

Explanation

Given,

$F=\dfrac{Ae^{-kr}}{r^2}$

$[F]=\dfrac{[A]}{[r^2]}$

$\implies {[A]}=[F][r^2]$

$\implies [A]=[MLT^{-2}]{[L^2]}$

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

We know that

$Kr$ is unitless.

Therefore, S.I unit of

$K=\dfrac 1m$

Then,

$[K]=\dfrac{1}{[L]}=[L^{-1]}$

So,

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

The dimensions of magnetic flux are?

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

Explanation

The force experienced by a moving charge in a magnetic field is given bu

$F=q(\vec{v}\times \vec{B})$

Here,

$\vec{B}$ is the magnetic field or magnetic flux density.

Thus, dimensions of magnetic flux,

$\phi=[B]\times [Area]=\dfrac{[F][A]}{[q][v]}$

We know the dimension of the following as,

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

$[A]=[L^2]$

$[q]=[AT]$

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

Thus,

$\phi=\dfrac{[MLT^{-2}][L^2]}{[AT][LT^{-1}]}=[ML^2T^{-2}A^{-1}]$

Option D

Find the dimension of

$\dfrac {G}{4\pi \epsilon_{0}}$ , where

$G$ is universal gravitation constant and

$\epsilon_{0}$ is permittivity

of free space.

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

Explanation

We know that ,

Gravitational force,

$F_g= \dfrac{Gm_1m_2}{r^2}$

Dimension of

$G= \dfrac{F_g r^2}{m_1 m_2} = \dfrac{MLT^{-2}\times L^2}{M^2}=M^{-1}L^3T^{-2}$

Also

Elecrostatic force ,

$F_e= \dfrac{q_1 q_2}{4\pi \epsilon_0 r^2}$

Dimension of

$\dfrac1{4\pi \epsilon_0 }= \dfrac{F_er^2}{q_1 q_2}=\dfrac{MLT^{-2}\times L^2}{AT\times AT}= ML^3T^{-4}A^2$

So, Dimension of

$\dfrac G{4\pi \epsilon_0 }= ML^3T^{-4}A^2 \times M^{-1}L^3T^{-2}= M^0 L^6T^{-6}A^{-2}$

A man of mass

$m$ is suspended in air by holding the rope of a balloon of mass

$M$ . As the man climbs up the rope, the balloon.

0%
Moves upward
0%
Moves downward
0%
Remain stationary
0%
Cannot say

If mass (M), velocity (V) and time (T) are taken as fundamental units, then the dimensions of force (F) are

0%
$[MVT]$
0%
$[MVT^{-1}]$
0%
$[M^2VT]$
0%
$[M^{-1}V^{-1}T]$

Order of

$\frac { 1 } { 8 \times 10 ^ { 9 } }$ is:

0%
$1.77\times {{10}^{-10}}$
0%
$1.25\times {{10}^{-10}}$
0%
$1.55\times {{10}^{-10}}$
0%
$1.95\times {{10}^{-10}}$

Explanation

$\dfrac{1}{8\times 10^{9}}=\left(\dfrac{1}{8}\right)\times 10^{-9}=(0.125)\times 10^{-9}$

$=1.25\times 10^{-10}$

Order of

$\dfrac{1}{8\times 10^{9}}$ is

$10^{-10}$

Option

$B$ is correct

The time of 25 oscillations of a simple pendulum is measured to be

$50.0 s$ by a watch of least count

$0.1 s$ . The percentage error in time is

0%
$0.2 \%$
0%
$0.02 \%$
0%
$0.002 \%$
0%
$2 \%$

Explanation

$\begin{array}{l} \text { Given- } \text { Time of } 25 \text { oscillations }=50.0 \mathrm{~s} \\ \text { Least count of the watch }=0.1 \mathrm{~s} \\ \text { Percentage error in Time } \\ \text {%err }=\frac{\Delta T}{T}=\frac{0.1}{50}=\frac{0.1}{50} \times 100 \%=0.2 \% \\ \text { % error in Time is } 0.2 \% \end{array}$

'The parallax angle in radians is:

$\theta = \left( 1 + \frac { 54 } { 60 } \right) \times \frac { \pi } { 180 } = 0.03316 \mathrm { rad }$
Hence, the distance between moon and earth:

0%
$\dfrac { \text { Diameter of Earth } } { \theta }$
0%
$\dfrac { 1.276 \times 10 ^ { 7 } m } { 0.03316 }$
0%
$3.84 \times 10 ^ { 8 } m$
0%
nnone of these

Explanation

Given,

$\theta =0.03316\,\,rad$

Also $b = \mathrm { AB } =$ diameter of earth $= 1.276 \times 10 ^ { 7 } \mathrm { m }$

Now d $=\dfrac{b}{\theta }=\dfrac{1.276\times {{10}^{7}}}{0.03316\,}=3.84\times {{10}^{8}}\text{m}$

Hence, distance between earth and moon is $3.84\times {{10}^{8}}\text{m}$

If C is the capacity and V is the potential across the condenser, what will be the dimensional representation of

$\dfrac1 2 C V ^ { 2 }$ ?

0%
$M L ^ { 2 } T ^ { -2 } A ^ { 2 }$
0%
$\mathrm { ML } ^ { - 2 } \mathrm { T } ^ { - 2 }$
0%
$M L ^ { 2 } T ^ { - 3 } A ^ { - 2 }$
0%
$\mathrm { ML } ^ { 2 } \mathrm { T } ^ { - 2 }$

Explanation

Let $X$ be the quantity represented.

$X = \dfrac 12 CV^2$

$X = \dfrac 12 \dfrac QV \times V^2$

$X = \dfrac12 Q.V$

$[X] = [Q][V]=[IT][ML^2T^{-3}I^{-1}]=[ML^2T^{-2}]$

The length of a rod is (11.05

$\pm$ 0.2) cm. What is the length of the two rods?

0%
(22.1 $\pm$ 0.05) cm
0%
(22.1 $\pm$ 0.1) cm
0%
(22.10 $\pm$ 0.05) cm
0%
(22.10 $\pm$ 0.4) cm

Explanation

Given,

$l=(11.05\pm 0.2)cm$

The length of two rod,

$l'=l+l$

$l'=(11.05\pm0.2)+(11.05\pm0.2)$

$l'=(22.10\pm 0.4)cm$ (error must be added.)

The period of osciflation of a simple pendulum in the recorded as

$2.63 \,sec, 2.42 \,sec, 2.42 \,sec, 2.71 \,sec$ and

$2.80 \,sec$ respectively. The average absolute error is

0%
$0.1 \,sec$
0%
$0.14 \,sec$
0%
$0.01 \,sec$
0%
$1.0 \,sec$

Explanation

Average value

$=\dfrac{2.63+2.42+2.42+2.71+2.80}{5}$

$=2.6$ sec

Now for

Average absolute error

$|\Delta T_1|=|2.63-2.42|=0.21$

$|\Delta T_2|=2.42-2.42|=0.00$

$|\Delta T_3|=|2.42-2.71|=0.29$

$|\Delta T_4|=|2.71-2.80|=0.09$

Mean absolute error

$\Delta T=\dfrac{|\Delta T_1|+|\Delta T_2|+|\Delta T_3|+|\Delta T_4|}{4}$

$=\dfrac{0.59}{4}$

$=0.1475$ sec.

$X = 3YZ^{2}$ find dimension of

$Y$ in

$(MKSA)$ system, if

$X$ and

$Z$ are the dimension of capacity and magnetic field respectively.

0%
$M^{-3}L^{-2}T^{-4}A^{-1}$
0%
$ML^{-2}$
0%
$M^{-3}L^{-2}T^{4}A^{4}$
0%
$M^{-3}L^{-2}T^{8}A^{4}$

The quantity

$X$ is given by

${\varepsilon _o}L\frac{{\Delta V}}{{\Delta T}}$ , where

${\varepsilon _o}$ is the permittivity of free space,

$L$ is a length,

${\Delta V}$ is a potential difference and

${\Delta T}$ is a time interval. The dimensional formula for

$X$ is the same as that of

0%
resistance
0%
charge
0%
voltage
0%
current

Explanation

We know than, from coulomb's law

$E_0 =\dfrac{1}{4\pi}\,\,\,\,\,\, \dfrac {Q_1 Q_2}{F R^2}$

$\Rightarrow [E_0]= \dfrac {[Q_1][Q_2]}{[F][R^2]}$

$\Rightarrow [E_0]=\dfrac {A^{2}T^{2}}{mLT^{-2}L^2}=\left [ \dfrac {A^2T^2}{mLT^{3}L^{-2}} \right ]$

$V = - \int \vec {E}. \vec{d\pi}$

$[V]= [E][L]$

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

$[T]= T$

$[L]= L$

$X = \epsilon _0L\dfrac {\Delta V}{\Delta T}$

$[X]= \left [ \dfrac {A^2T^2}{mLT^{3}L^{-2}} \right ] [L]\left [ \dfrac {mL^2A^{-1}T^{-3}}{T} \right ]$

$[X]= [A]$

$[A]$ is the dimension of current.

The accuracy in the measurement of the diametre of a hydrogen atom as ${ 1.06\times 10 }^{ -10 }$ m is

0%
$0.01$
0%
${ 106\times 10 }^{ -10 }$
0%
$\dfrac { 1 }{ 106 }$
0%
${ 0.01\times 10 }^{ -10 }$

The physical quantity having the dimensions

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

0%
resistance
0%
resistivity
0%
electric cunductivity
0%
electromotive force

Explanation

$\textbf{Explanation:-}$

We know that,

Resistivity,

$\rho = \dfrac{m}{ne^2t}$

$\Rightarrow \rho = \dfrac{[M]}{[L^{-3}][AT]^{2}[T]}$

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

Now,

Electrical conductivity,

$\sigma = \dfrac{1}{\rho }$

$\Rightarrow \sigma = \dfrac{1}{[ML^{3}A^{-2}T^{-3}]}$

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

$\textbf{Hence the correct option is C}$

The diameter of wire measured with a varnier calliper with keast count 0.1 mm is Scm, then the percentage error in measurement is

0%
$0.2\%$
0%
$2\%$
0%
$5\%$
0%
$0.5\%$

A boy wants to buy new shoes. To find out his correct shoe size the ________ of his feet should be measured.

0%
Length
0%
Thickness
0%
Area
0%
Volume

If error in measurement of radius of a sphere is

$1\%$ , what will be the error in measurement volume?

0%
$1\%$
0%
$\dfrac{1}{3}\%$
0%
$3\%$
0%
None of these

Explanation

$Volume=\dfrac { 4πR^3 }{ 3 }$

$\dfrac { ΔV }{ V } \times 100=3\dfrac { ∆R }{ R } \times 100$

$=3\times \dfrac { 1 }{ 100 } \times 100$

$=3$

∴Percentage error in volume is

$3$ %

Which of the following measurements is most accurate?

0%
$0.005 mm$
0%
$5.00 mm$
0%
$50.00 mm$
0%
$5.0 mm$

The dimensions of latent heat are -

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

Explanation

$Latent\, heat (L)=\dfrac{Heat}{Mass}$ . . . . . . (1)

The dimensional formula of mass $=[M^1L^0T^0]$ . . . . (2)

Also, the dimensions of heat $=$ dimensions of energy $=$ dimensions of work

Since, $work = force\times displacement$ . . . . (3)

And, the dimensional formula of,

Displacement $= [M^0L^1T^0]$ . . . (4)

Force $= m \times a = [M^1L^1T^{-2}]$ . . . (5)

On substituting equation (4) and (5) in equation (3) we get,

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

Therefore, the dimensions of work or heat $=[M^1L^2T^{-2}]$ . . . . (6)

On substituting equation (2) and (6) in equation (1) we get,

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

Therefore, latent heat is dimensionally represented as $[M^0L^2T^{-2}]$

The dimensions of electric potential are:

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

Explanation

The electric potential is expressed as:

$V=\dfrac {Work}{charge}$

So, the dimensional formula will be

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

The dimensions of elelctrical permittivity are

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

If pressure

$P$ , area

$A$ and length

$L$ are taken to be fundamental quantities in a system of units, then energy has dimensional formula.

0%
$P^{1}A^{1}L^{1}$
0%
$P^{1}A^{3/2}L^{0}$
0%
$P^{1}A^{0}L^{3}$
0%
Cannot take $P, A$ and $L$ as fundamental quantities

Explanation

We know that

$P = \dfrac {P}{A} = \dfrac {MLT^{-2}}{L^{2}}$

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

$A = L^{2}$

$L = L$

Let

$E = P^{x} A^{y} L^{z}$

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

$= (ML^{-1}T^{-2})^{x} (L^{2})^{y} (L)^{2}$

$= M^{x} L^{-x + 2y + z} T^{-2x}$

$x = 1, -2x = -2$

$x = 1$

$-x + 2y + z = 2$

$z = 0, 2y = 2 + 1$

$y = \dfrac {3}{2}$

$\therefore E = P^{1} A^{3/2} L^{0}$ .

The error in the measurement of length of a simple pendulum is

$0.1\%$ and error in the time period is

$2 \%$ . The possible maximum error in the quantity having dimensional formula

$LT^{-2}$ is

0%
$1.1 \%$
0%
$2.2\%$
0%
$4.1\%$
0%
$6.1\%$

Explanation

Given,

Error

$\%$ in time

$=2\%=\dfrac{2}{100}=\dfrac{1}{50}=2\%$

Error

$\%$ in length

$=0.1\%$

Find

$\%$ error in

$\dfrac{1}{12}$

$\therefore$ as

$\dfrac{\triangle A}{A}=\left( \dfrac{\triangle 2}{2}\right) +\left( \dfrac{\triangle Q}{Q}\right)$ if

$A=\dfrac{t}{Q}$

Thus,

$\Rightarrow \%$ error

$=\left( \dfrac{\triangle L}{L}\right) +2\left( \dfrac{\triangle T}{T}\right)$

$=\left( 2\%\right) +2\left( .1\%\right)$

$=2\%+.2\%$

$=2.2\%$

Hence, the answer is

$2.2\%.$

Of the following quantities , which one has dimensions different from the remaining three?

0%
Energy per unit volume
0%
Force per unit area
0%
Product of voltage and charge per unit volume
0%
Angular momentum

Explanation

Energy density

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

Force per unit area

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

Product of the charge per unit volume and voltage

$=\dfrac{Work}{Charge}\times \dfrac{Charge}{Volume}=\dfrac{Work}{Volume}$

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

Angular momentum per unit mass

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

So correct answer is option (D).

The dimensional formula of

$\frac{GM_{e}}{gRe^{2}}$ is ---------

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

Explanation

Dimensional formula

$\dfrac{Gm_2}{gR^2_e}$

for

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

$M_e=M$

$g=LT^{-2}$

$R_e=L$

$\therefore \dfrac{m^{-1}L^3T^{-2}}{LT^{-2}\cdot L}M$

$\Rightarrow M^0LT^0$

$\Rightarrow L$

$\Rightarrow M^0L^1T^0$ .

$m,e,{\varepsilon _0},$ and

$c$ denote mass electron, charge of electron Plank's constant and speed of light, respectively. The dimensions of

$\frac{{m{e^4}}}{{\varepsilon _0^2{h^3}c}}$ are

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

Which of the following pair does not have identical dimension ?

0%
Moment of inertia and moment of force
0%
Angular momentum and plank's constant
0%
Impulse and momentum
0%
Work and torque.

Explanation

Planck's constant, symbolized h, relates the energy in one quantum (photon) of electromagnetic radiation to the frequency of that radiation.

$h=\dfrac Ev=\dfrac{[ML^2T^{-2}]}{[T^{-1}]}=[ML^2T^{-1}]$

Angular momentum

$=$ Moment of inertia

$\times$ Angular velocity (Angular momentum)

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

So, here we can see that both the Planck's constant and Angular momentum have the same dimensions.

The dimensions of

$\cfrac{1}{2}{\in }_{0}{E}^{2}$ , where

${\in }_{0}$ is permittivity of free space and

$E$ is electric field, is:

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

Out of the following the more precise value is .....

0%
$1.200 \times 10 ^ { 3 } k m$
0%
$1.20 \times 10 ^ { 3 } k m$
0%
$1.2 \times 10 ^ { 3 } k m$
0%
$1200 k m$

A uniform rod undergoes a longitudinal strain of

$2$ . find percentage change in its volume if poisson's ratio of the material is

$0.50$

0%
0.4%
0%
0.7%
0%
0%
0%
0.3%

Explanation

We know that realtion between moduler of elasticity

$\left( \varepsilon \right),$ possion's ratio

$\left( u\right)$ and balls modules elasticity

$\left(h\right).$

Given by,

$\Rightarrow \varepsilon =3h\left( 1-2u\right)$

here given

$u=0.3$

$h\rightarrow \alpha$

and we also know

$\rightarrow h=\dfrac{\dfrac{-\triangle p}{\triangle v}}{v}$

$h$ tends to infinite then

$\triangle v$ tends to zero.

$\%$ change in volume will be zero.

Hence, solve.

Which of the following pairs of physical quantities have same dimension

$?$

0%
force and work
0%
torque and energy
0%
torque and power
0%
force and torque

Explanation

we know

$,$ force

$=$ mass

$×$ acceleration

energy

$=$ Workdone

$=$ force

$×$ displacement

power

$=$ energy/time

torque

$=$ force

$×$ perpendicular distance between force and axis of rotation

$.$

now

$,$ by these formula

$,$ find dimension of all terms

$.$

e.g., dimension of force

$= [MLT⁻²]$

dimension of energy

$= [ML²T⁻²]$

dimension of power

$= [ML²T⁻³]$

dimension of torque

$= [ML²T⁻²]$

Here it is clear that

$,$ dimension of energy and dimension of torque are equal

$.$

Hence,

option

$(b)$ is correct

$.$

An equation is given by

$\dfrac{d}{dt}[\int\vec{v}.\vec{dS}]$ -

$k[\vec{v}.\vec{\theta}]$ . If

$\vec{v}$ represents velocity,

$\vec{ds}$ is small dosplacement,

$t$ is time and a is acceleratation. The dimension of

$k$ is

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

The dimensions of

$\frac{1}{{{\mu _0}{ \in _0}}}$ is (where symbols have their usual menaing)

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

Explanation

Velocity of light in vacuum is

${\frac{1}{{\sqrt {\left( {{\mu _0}{ \in _0}} \right)} }}}$

$\left[ {L{T^{ - 1}}} \right] = \left[ {\frac{1}{{\sqrt {\left( {{\mu _0}{ \in _0}} \right)} }}} \right]$

$\left[ {{L^2}{T^{ - 2}}} \right] = \left[ {\frac{1}{{\left( {{\mu _0}{ \in _0}} \right)}}} \right]$

Hence,

option

$(A)$ is correct answer.

$\left[M^{1}L^{2}T^{-3}\right]$ are the dimensions of ....

0%
pressure
0%
power
0%
force
0%
work

Explanation

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

If % error in length, diameter, current and voltage are same than which of the following affects % error in measurement of resistivity, the most:

0%
length measurement
0%
voltage measurement
0%
current measurement
0%
diameter measurement

Explanation

$R = \frac{{\rho l}}{\Delta } \Rightarrow \rho = \frac{{R\Delta }}{l}$

$\rho = \frac{{R\left( {M{r^2}} \right)}}{l}$

$\therefore$ Diameter measurement is most affect the error of

$\rho$

$($ Resistivity

$)$ because it has power of two

$.$

Hence,

option

$(D)$ is correct answer.

The resultant of two equal forces acting at right angles to each other is

$1414$ dyne. Find the magnitude of either force.

0%
$960$
0%
$1000$
0%
$1200$
0%
$1414$

Explanation

Given,

$F_1=F_2=F(say)$

$\theta=90^0$

$R=1414dyne$

Resultant force,

$R=\sqrt{F_1^2+F_2^2+2F_1F_2cos\theta}$

$1414dyne=\sqrt{F^2+F^2+2F^2cos90^0}=\sqrt{2F^2}$

$1414dyne=\sqrt{2}F$

$F=\dfrac{1414}{\sqrt{2}}=999.84 \approx 1000dyne$ (by rounding off)

The correct option is B.

The sum of the numbers

$436.32$ ,

$227.244$ and

$0.301$ in appropriate significant figures is:

0%
$663.821$
0%
$664$
0%
$663.8$
0%
$663.82$

Explanation

$\begin{array}{l}\text{ According to the rule of addition more decimal places }\\\text { will be preferred in case of addition. }\end{array}$

$\begin{array}{l}=436.32+227.244+0.301\\=663.865\mathrm{}\end{array}$

$\therefore \text{the most appropriate number out of all the options is option C i.e. 663.8}$

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

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

Explanation

Planck's constant, symbolized h, relates the energy in one quantum (photon) of electromagnetic radiation to the frequency of that radiation.

$h=\dfrac Ev=\dfrac{[ML^2T^{-2}]}{[T^{-1}]}=[ML^2T^{-1}]$

Angular momentum

$=$ Moment of inertia

$\times$ Angular velocity (Angular momentum)

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

So, correct answer is option (B).

The velocity of a particle varies with distance x from a fixed origin as

$v=Ax+\dfrac { { Bx }^{ 2 } }{ C+x }$ , where A,B and C are dimensional constant then the dimensional formula of

$\dfrac { AB }{ C }$ is

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

Given

$m$ is the mass per unit length,

$T$ is the tension

$\& L$ is the length of the wire the dimensional formula for

$L \left( \cfrac { m } { T } \right) ^ { 1 / 2 }$ is same as that for:-

0%
Wavelength
0%
velocity
0%
Time period
0%
Frequency

The physical quantity having the dimensions

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

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

Dimensions of electrical resistance are :

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

Explanation

Since according to Ohm's Law :

$V=IR$

Hence,

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

Where

$V=\dfrac {W}{q}=\dfrac {work\ done}{charge}$

Hence; Potential

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

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

From above equations,

$\therefore \$ Resistance

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

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

The dimension of

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

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

The electric field in a certain region is given by

$\overrightarrow E=( \frac {K}{x^3}) \hat i$ . the dimension of K are:-

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

Explanation

The electric field in the given region is

$\vec E=\dfrac{K}{x^3}\hat{i}$

The dimension for the electric field is

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

In order to obtain the dimension of

$K$ , simplify the above expression:

$K=E\cdot x^3$

Here

$X$ denotes the position. So, dimensions of

$x$ are

$[L]$ .

Simplify the above expression.

$K=[M^1\ L^1\ T^{-3}\ A^{-1}][L]^3$

$\implies[M^1\ L^4\ T^{-3}\ A^{-1}]$

Option

$C$ is correct.

A thermometer reads 0

$^{o}$ C as 10

$^{o}$ C and 100

$^{o}$ C as 90

$^{o}$ C then
Find the correct temperature if the thermometer reads 20

$^{o}$ C

0%
$20^{o}$ C
0%
$12.5^{o}$ C
0%
$25^{o}$ C
0%
$30^{o}$ C

Explanation

Temperature on any scale can be easily converted into different scale by the following formula

$\dfrac{T - T_f}{T_b - T_f} = \text constant$

where

$T$ = Reading of thermometer

$T_f$ = Freezing point of water

$T_b$ = Boiling Point of water

So here we will equate the reading of two scale namely, thermometer and actual scale.

$\dfrac{20 - 10}{90 - 10} = \dfrac{x - 0}{100 - 0}$

$\dfrac{10}{80} = \dfrac{x}{100}$

$x = 12.5 ^oC$

CBSE Questions for Class 11 Engineering Physics Units And Measurement Quiz 12 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Units And Measurement Quiz 12 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.