MCQ Questions for Class 11 Engineering Physics Units And Measurement Quiz 13

The dimension of $\cfrac {1}{2}\epsilon_o E^2$ permittivity of free space and E:
Intensity of electric field) is

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

Explanation

Energy density of an electric field

$E$ is ,

$\mu_E=\dfrac 12 \epsilon_0 E^2$

where

$\epsilon_0$ is permitivity of free space

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

Hence, the dimension of

$\dfrac 12 \epsilon_0 E^2$ is

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

The radius of a sphere is measured as

$(10 \pm 0.02\ \% )$ cm. The error in the measurement of its volume is

0%
$25.1$ cc
0%
$25.12$ cc
0%
$2.51$ cc
0%
$251.2$ cc

Explanation

Given,

$r=10\pm 0.02$ %

Volume,

$V=\dfrac{4}{3}\pi r^3$ . . . . . . . . .(1)

$\dfrac{\Delta V}{V}=3\dfrac{\Delta r}{r}$

$\Delta V=3V\dfrac{\Delta r}{r}$

$\Delta V=\dfrac{4}{3}\pi r^3\times 3\times \dfrac{\Delta r}{r}$

$\Delta V=\dfrac{4}{3}\times 3.14\times 10\times 10\times 10\times 3\times \dfrac{0.02}{10}=25.12cc$

The correct option is B.

A particle is moving with a velocity

$35\ m/s$ along positive x-axis.It's acceleration is towards negative X-axis with the magnitude

$4m/s^ {2}$ then the distance covered by the particle in the

$9th$ second:-

0%
$1m$
0%
$\dfrac {9}{8}m$
0%
$\dfrac {5}{4}m$
0%
$153m$

Explanation

$\begin{array}{l}\vec{u}=35 \mathrm{~m} /\mathrm{s}\hat{\imath} \\\vec{a}=4 \mathrm{~m} /\mathrm{s}^{2}\hat{\imath} \\S_{a}=u+\frac{1}{2} a(2 n-1)\end{array}$

$\begin{aligned}S_{a} &=u+\frac{1}{2} a(2 \times 9-1) \\&\left.=35+\frac{1}{2} \times(4\right)\times(18-1) \\&=35+\frac{1}{2} \times(-68) \\&=35-34\\&=1\mathrm{~m}\mathrm{~}\mathrm{~}\end{aligned}$

The vector sum of three forces having magnitudes

$| \overrightarrow F_1 | = 100 N$ ,

$| \overrightarrow F_2 | = 80 N$ &

$| \overrightarrow F_3 | = 60 N$ acting on a particle is zero. the angle between

$\overrightarrow F_1$ &

$\overrightarrow F_2$ is nearly:-

0%
$53^0$
0%
$143^0$
0%
$37^0$
0%
$127^0$

Explanation

$\text { For equilibrium:} \\\begin{array}{c}\left|\vec{F}_{1}+\vec{F}_{2}\right|=\left|\vec{F}_{3}\right| \\\end{array}$

$\begin{array}{c}\text { Squaring both side }\\\Rightarrow\left|\vec{F}_{1}+\vec{F}_{2}\right|^{2}=\left|\vec{F}_{3}\right|^{2}\\\Rightarrow\left|\vec{F}_{1}\right|^{2}+\left|\vec{F}_{2}\right|^{2}+2\left|F_{1}\right|\left|F_{2}\right|\cos\theta\\=\left|\vec{F}_{3}\right|^{2} \\\Rightarrow 100^{2}+80^{2}+2\times 100 \times 80\times\cos\theta\\=60^{2}\end{array}$

$\begin{aligned}\Rightarrow \cos\theta&=\frac{12800}{16000} \\\Rightarrow \cos \theta&=\frac{4}{5}\end{aligned}$

$\theta=37^{\circ}$

1992324_1383157_ans_c5ddb8aeb83a4552ae5133b877eca014.png

Two constant force

$\overrightarrow F_1 and \overrightarrow F_2$ acts on a body.these forces displaces the body from point P(1, -2, 3) to Q (2, 3, 7 ) in 2s starting from rest.force

$\overrightarrow F_1$ is of magnitude 9 N and acting along vector

$( 2 \hat i - 2 \hat j + \hat k )$ . the positions are in meter. find work done by

$\overrightarrow F_1$ .

0%
$-12J$
0%
$+12J$
0%
$36J$
0%
$-36J$

Explanation

$\begin{aligned}\vec{F}_{1}&=k(2\hat{i}-2\hat{\jmath}+\hat{k}) \\\vec{d}=&(2-1) \hat{1}+(3-(-2))\hat{j}\\&+(7-3) \hat{k} \\=&(1) \hat{1}+(5)\hat{j}+(4)\hat{k}\\\vec{W}_{1}=&\vec{F}_{1}\cdot\vec{d}\\&=k(2\hat{i}-2\hat{j}+\hat{k})\cdot(i+5\hat{j}+4\hat{k})......(i)\end{aligned}$

$|\vec{F}_1|=9$

$g=\sqrt{4 k^{2}+4 k^{2}+k^{2}}$

$9=3 k$

$k=3...(ii)$

$\text{putting (ii) in (i) we get}$

$\overrightarrow{W_{1}}=-12 \mathrm{~J}$

A force is given by

$F = at + b{t^2}$ , where

$t$ is time, the dimensions of

$a$ and

$b$ are respectively :

0%
$\left[ {ML{T^{ - 4}}} \right]$ and $\left[ {ML{T^{ - 1}}} \right]$
0%
$\left[ {ML{T^{ - 1}}} \right]$ and $\left[ {ML{T^0}} \right]$
0%
$\left[ {ML{T^{ - 3}}} \right]$ and $\left[ {ML{T^{ - 4}}} \right]$
0%
$\left[ {ML{T^{ - 3}}} \right]$ and $\left[ {ML{T^0}} \right]$

Explanation

$\Rightarrow \$ Since

$F=at+bt^2---(1)$

$\therefore$ Dimension of

$at$ and

$bt^2$ must be equal to force only.

Hence

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

$F$ from

$(1)$ and

$(2)$

$\Rightarrow \ [at]=a[T] =[F]$

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

$\therefore \ [a]=[M^{1} L^{1} T^{-3}]$

and per

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

$\therefore \ \boxed {[b]= [M^{1} L^{1} T^{-4}]}$

The radius of a sphere is measured to be

$(4.0 +2.0)$ cm the maximum percentage error in the measurement of the volume of the sphere is

0%
$10$ %
0%
$15$ %
0%
$20$ %
0%
$25$ %

Explanation

Given,

$r=(4\pm0.2)cm$

$\dfrac{\Delta r}{r}=\dfrac{0.20}{4}=0.05$

Volume,

$V=\dfrac{4}{3}\pi r^3$

$\dfrac{\Delta V}{V}\times 100=3\dfrac{\Delta r}{r}\times 100=3\times 0.05\times 100=15$ %

The correct option is B.

If the constant of gravitation

$(G)$ , Planck's constant

$(h)$ and the velocity of light

$(c)$ be chosen as fundamental units. The dimension of the radius of gyration is

0%
$h^{-1/2} c^{1/2} G^{1/2}$
0%
$h^{1/2} c^{-3/2} G^{1/2}$
0%
$h^{1/2} c^{-3/2} G^{-1/2}$
0%
$h^{-1/2} c^{-3/2} G^{1/2}$

In the equation

${ S }_{ nth }=u+\dfrac { a }{ 2 } \left( 2n-1 \right)$ , the letters have their usual meanings. The dimensional formula of

$S_{nth}$ is

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

Explanation

The Dimensional Formula of

${S}_{nth}$ is same as that of the Dimensional formula of

$u$ (2 variables with same units can be added)

So Dimensional Formula of

${S}_{nth}$ = DImensional Formula of

${u}$ =

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

A physical quantity x depends on quantities y and Z as follows : x= Ay+B tan (Cz), where A, B and C are constants . Which of the following do not have same dimensions ?

0%
x and B
0%
C and $z^{ -1 }$
0%
y and B/A
0%
x and A

The term

$(1/2)pv^{ 2 }$ occurs in Bernoulli's equation , with

$\beta$ being the density of fluid and V its speed . The dimensions of this term are

0%
$\left[ { M }^{ -1 }{ L }^{ 5 }{ T }^{ 2 } \right]$
0%
$\left[ { M }{ L }{ T }^{ 2 } \right]$
0%
$\left[ { M }{ L }^{ -1 }{ T }^{ -2 } \right]$
0%
$\left[ { M }^{ -1 }{ L }^{ 9 }{ T }^{ -2 } \right]$

Two resistance are measured in ohm and is given as:-

${ R }_{ 1 }=3\Omega \neq 1%$

${ R }_{ 2 }=6\Omega \neq 2%$
When they are connectrd in parallel, the percentage error in equivalent resistance is

0%
$3.33\%$
0%
$4.5\%$
0%
$0.67\%$
0%
$1.33\%$

Explanation

$\text { Given: } R_{1}=3 \Omega \pm 0 \cdot 1 \quad R_{2}=6 \Omega \pm 0.2$

$\begin{array}{l}\text { For parallel } \\\frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}=\frac{1}{3}+\frac{1}{6}=\frac{2+1}{6}=\frac{3}{6}=\frac{1}{2}\end{array}$

$\begin{array}{l}R_{e q}=2 \Omega \\\text { Differentiating eq }^{n} \text { (i) both side } \\-\frac{d R_{e q}}{R_{e q}^{2}}=-\frac{d R_{1}}{R_{1}^{2}}-\frac{d R_{2}}{R_{2}^{2}}\end{array}$

$\begin{array}{l}\text { Error is always added }\\\qquad \begin{aligned}\frac{d R_{e q}}{R_{e q}^{2}} &=\frac{d R_{1}}{R_{1}^{2}}+\frac{d R_{2}}{R_{2}^{2}} \\\frac{d R_{e q}}{R_{e q}}&=\left(\frac{d R_{1}}{R_{1}^{2}}+\frac{d R_{2}}{R_{2}^{2}}\right)R_{0}\mathrm{~V}\\&=\left(\frac{0.1}{3^{2}}+\frac{0.2}{6^{2}}\right){2}\\&=\left(\frac{0.1}{9}+\frac{0.2}{36}\right){2}\end{aligned}\end{array}$

$\begin{aligned}\frac{d \operatorname{Req} x}{\operatorname{Reg}} \times100\% &=\left(\frac{4+0.2}{36}\right)^{2} \times 100 \% \\&=\frac{0.6}{36}\times 2 \times 100 \\&=\frac{100}{3}\\&=3.33\%\end{aligned}$

The pair of physical quantities, that have difference dimensions is

0%
Planck's constant and angular momentum
0%
Energy density and pressure
0%
Relative density and plane angle
0%
Specific heat capacity and latent heat

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 dimension of the ratio of magnetic flux and the resistance is equal to that of:

0%
induced emf
0%
charge
0%
inductance
0%
current

Explanation

Dimension of magnetic flux

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

Dimension of resistance

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

Therefore,

$\dfrac{[Magnetic\,\,flux]}{[Resistance]}=\dfrac{[ML^2T^{-2}A^{-1}]}{[ML^2T^{-3}A^{-2}]}=[AT]=Charge$

If pressure is (P), velocity (V), and time (T) are taken as the fundamental quantities, then the dimensional formula of force is

0%
$\left[ P ^ { 1 } V ^ { 1 } T ^ { 1 } \right]$
0%
$\left[ P ^ { 1 } V ^ { 2 } T ^ { 1 } \right]$
0%
$\left\lceil P ^ { 1 } V ^ { 1 } T ^ { 2 } \right\rceil$
0%
$\left[ P ^ { 1 } V ^ { 2 } T ^ { 2 } \right]$

Explanation

$Force=P^xV^yT^z$

$[MLT^{-2}]=[MLT^{-2}L^{-2}]^x[LT^{-1}]^y[T]^z$

Comparing the dimensions on both side, we get,

$x=1$ , $y=2$ , $z=2$

So, $Force=[P^1V^2T^2]$

In a system of units if force (F), acceleration (A) and time, (T) are taken as fundamental units then the dimensional formula of energy is -

0%
$FA^{2}T$
0%
$FAT^{2}$
0%
$F^{2}AT$
0%
FAT

Explanation

Consider that the energy is expressed as:

$E=KF^aA^bT^c$

Substituting the dimensional formulas:

$[ML^2T^{−2}]=[MLT^{−2}]^a[LT^{−2}]^b[T]^c$

$[ML^2T^{−2}]=[M^aL^{a+b}T^{−2a−2b+c}]$

$\Rightarrow a=1,a+b=2\Rightarrow b=1$

and

$−2a−2b+c=−2\Rightarrow c=2$

$\therefore E=KFAT^2$

If the radius of the earth were to shrink by 1% its mass remaining the same, the acceleration due to gravity on the earth's surface would

0%
Decreased by 2%
0%
Increased by 2%
0%
Remain unchanged
0%
Increased by 1%

Dimensional formula

$[{ M }^{ 1 }{ L }^{ 0 }{ T }^{ -2 }]$ is equal to dimensional formula of

0%
Surface tension
0%
surface energy
0%
Both A and B
0%
None of these

Dimensional formula of momentum is :

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

Dimensions of rate of change of flux are equivalent to those of

0%
voltage
0%
current
0%
1/charge
0%
charge

$S = \dfrac { d } { t } + \dfrac { e } { ( f - t ) ^ { 2 } } ,$ where

$S$ is the displacement of the body in time

$t .$ Find the dimensions of

$e$ and

$f$

0%
$\mathrm { LT } , \mathrm { T }$
0%
$\mathrm { LT } ^ { 2 } , \mathrm { T }$
0%
$\mathrm { L } ^ { 2 } \mathrm { T } , \mathrm { T } ^ { - 3 }$
0%
$LT, T ^ { 2 }$

The dimensional formula for the acceleration, velocity and length are

$\alpha { \beta }^{ -2 }, \alpha { \beta }^{ -1 }$ and

$\alpha y$ . What is the dimensional formula for the coefficient of friction ?

0%
$\alpha \beta y$
0%
${ \alpha }^{ -1 }{ \beta }^{ 10 }{ y }^{ 0 }$
0%
${ \alpha }^{ 0 }{ \beta }^{ -1 }{ y }^{ 0 }$
0%
${ \alpha }^{ 0 }{ \beta }^{ 0 }{ y }^{ 0 }$

The dimensions of the ratio of angular momentum to linear momentum:

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

The dimensions of universal gas constant is

0%
$\left[ { M }^{ }{ L }^{ 2 }{ T }^{ -2 }{ \theta }^{ -1 } \right]$
0%
$\left[ { M }^{ 2 }{ L }^{ }{ T }^{ -2 }{ \theta }^{ } \right]$
0%
$\left[ { M }^{ }{ L }^{ 3}{ T }^{ -1 }{ \theta }^{ -1 } \right]$
0%
None of these

Explanation

The dimension of universal gas constant is

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

where,

$\theta$ is the temperature.

Velocity gradient has same dimensional formula as

0%
Frequency
0%
angular momentum
0%
velocity
0%
none of these

The dimension of electric permittivity is

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

Find dimension formula of polarisation?

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

$\eta$ represents the coefficient of viscosity and

$T$ the surface tension, then the dimension of

$\cfrac{T}{\eta}$ is same as that of:

0%
length
0%
mass
0%
time
0%
speed

Dimensional formula of modulus of elasticity is

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

Given

$x = a + b t + c t ^ { 2 }$ where

$x$ in metre and

$t$ in second. Find the dimensional formula of

$b$

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

The radius of nucleus is

$r = r_0 A^{{1}/{3}}$ , where

$A$ is mass number. The dimensions of

$r_0$ is

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

If L be the inductance, R resistance and C be the capacitance of a capacitor, then dimensional formula of

$\cfrac {L}{R}$ and RC are

0%
$M^oLT^{-1}ML ^oT^{-1}$
0%
$M^oL^oT,MLT^o$
0%
$M^oL^oT, 1$
0%
$M^oL^oT, M^oL^oT$

The dimensions of

$(\mu_{0}\epsilon_{0})^{-\dfrac {1}{2}}$ are

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

If C be the capacitance and V be the electric potential,then the dimensional formula of

$CV^2$ is

0%
$[ML ^{-3} TA]$
0%
$[M^oLT^{-2} A^o]$
0%
$[ML ^1 A^{-1 }]$
0%
$[ML^2T^{-2} A^o]$

Given

$x = a + b t + c t ^ { 2 }$ where

$x$ in metre and

$t$ in second. Find the dimensional formula of

$C$

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

Given that m is the mass suspended from a spring force constant k. The dimension of the formula for

$\sqrt{m/k}$ is same as that for

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

Velocity gradient has dimension of

0%
Force
0%
Frequency
0%
Surface tension
0%
momentum

The Earth's radius is

$6371\ km$ . The order of magnitude of the Earth's radius is

0%
$10^3$
0%
$10^2$
0%
$10^7$
0%
$10^5$

Explanation

Order of magnitude is usually written as

$10$ to the

$n$ th power. The represents the order of magnitude.

e.g., if we write a number

$X$ in such a way that

$X = m × 10^n$ then n is order of magnitude.

Where

$\dfrac{\sqrt{10}}{10} \leq m \leq \sqrt{10}$

here,

$6371 km = 6371000 m$

$6371000 = 6.371 × 10^6$

we see,

$6.371 ≥ \sqrt{10}$

so,

$0.6371 × 10^7$

now,

$\dfrac{\sqrt{10}}{10} \leq 0.6371 \leq \sqrt{10}$

$\therefore 7$ is our order of magnitude.

The diagram shows a cuboid block made from a metal of density

$2.5g/{cm}^{3}$
What is the mass of the block?

0%
$8.0kg$
0%
$16kg$
0%
$50g$
0%
$100g$

Explanation

Volume of cuboid =

$l \times b \times h$

$10 \times 2 \times 2$

$cm^{3}$

$40 cm^3$

Now , Mass =

$V \times d$

$40 \times 2.5$

$g$

$100g$

The dimensional formula of pressure is

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

Explanation

$Pressure (P) = \dfrac{Force }{ Area}$ . . . . . (1)

Since, $Force = Mass \times Acceleration$

And, $Acceleration = \dfrac{Velocity}{Time} =\dfrac{[LT^{-1}]}{[T]}=[LT^{-2}]$

$\therefore$ The dimensional formula of force $=[M][LT^{-2}]=[MLT^{-2}]$ . . . . (2)

The dimensional formula of area is $[M^0L^2T^0]$ . . . . (3)

On substituting equation (2) and (3) in equation (1) we get,

$Pressure (P) = \dfrac{Force }{ Area}$

Or,

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

Therefore, the pressure is dimensionally represented as $[ML^{-1}T^{-2}]$

The dimension of

$\frac {1}{2} \varepsilon _ o E^2 ( \varepsilon _ o$ is the permitivity of free space and E is electric field) is

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

In the density measurement of a cube, the mass an edge length are measured as

$\left( 10.00\pm 0.10 \right) kg$ and

$\left( 0.10\pm 0.01 \right) m$ , respectively. The error in the measurement of density:

0%
$0.10kg/{m}^{3}$
0%
$0.31kg/{m}^{3}$
0%
$0.07kg/{m}^{3}$
0%
$0.01kg/{m}^{3}$
0%
No answer

Explanation

$\rho =\cfrac { m }{ v }$

$\cfrac { \Delta \rho }{ \rho } \times 100percent=\cfrac { \Delta m }{ m } \times 100percent+3\left( \cfrac { \Delta L }{ L } \right) \times 100percent....(i)$
which is only possible when error is small which is not the case in this question
Yet if we apply equation (i) we get

$\Delta \rho =3100kg/{ m }^{ 3 }\quad$
Now we will calculate error, without using approximation

${ \rho }_{ min }=\cfrac { { m }_{ min } }{ { v }_{ max } } =\cfrac { 9.9 }{ { \left( 0.11 \right) }^{ 3 } } =7438kg/{ m }^{ 3 }$
and

$\quad { \rho }_{ max }=\cfrac { { m }_{ max } }{ { v }_{ min } } =\cfrac { 10.1 }{ { \left( 0.09 \right) }^{ 3 } } =13854kg/{ m }^{ 3 }$

$\Delta \rho =6416.6kg/{ m }^{ 3 }$
No option is matching

The dimensional formula for the magnetic fields is

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

Explanation

We have the equation,

$F=qvB\,sin\theta$

Dimension of Force,

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

Dimension of Velocity,

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

Dimension of Charge,

$q=[AT]$

$Sin\theta$ is dimensionless

Then,

Magnetic Field,

$B=\dfrac{Force}{Velocity \times Charge}=\dfrac{[MLT^{-2}]}{[LT^{-1}][AT]}=[MT^{-2}A^{-1}]$

The dimension of Magnetic flux.

0%
$ML^{2}I^{-1}T^{-3}$
0%
$ML^{2}I^{-4}T^{-2}$
0%
$ML^{2}I^{-1}T^{-2}$
0%
None of these

Which of the following does not have the dimension of force?

0%
Potential gradient
0%
Energy gradient
0%
Weight
0%
Rate of change of momentum

$P$ and

$Q$ have different non-zero dimensions, which of the following operations is possible?

0%
$P+Q$
0%
$PQ$
0%
$P-Q$
0%
$1-\dfrac{P}{Q}$

The pairs of physical quantities that have the same dimensions in (are):

0%
Reynolds number and coefficient of friction
0%
Curie and frequency of a light wave
0%
Latent heat and gravitational potential
0%
Planck's constant and torque

Taking frequency

$f$ , velocity

$v$ and density

$\rho$ to be the fundamental quantities then the dimensional formula for momentum will be?

0%
$\rho v^4f^{-3}$
0%
$\rho v^3f^{-1}$
0%
$\rho v f^2$
0%
$\rho^2v^2f^2$

If x,v and a denote the displacement, the velocity and the acceleration of a particle
executing simple harmonic motion of time period T, then which of the following do not
change with time?

0%
$\dfrac{aT}{v}$
0%
$aT+2\pi v$
0%
$a^2T^2+4\pi^2 v^2$
0%
$aT$
0%
$vT$

Explanation

$\because$ In option

$(a),\dfrac{aT}{v}=\dfrac{\dfrac{m}{s^2}\times s}{\left(\dfrac{m}{s}\right)}=\dfrac{m\times s^2}{m\times s^2}=[M^0L^0T^0]$

So, according to the result above relation does not depend on time

The dimensions of time constant are:

0%
$[M^0L^0T^0]$
0%
$[M^0L^0T]$
0%
$[MLT]$
0%
None of these

CBSE Questions for Class 11 Engineering Physics Units And Measurement Quiz 13 - 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 13 - MCQExams.com

0:0:2

Practice Class 11 Engineering Physics Quiz Questions and Answers

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