Match the physical quantities given in Column I with dimensions expressed in terms of mass (M), length (L) time (T), and charge (Q) given in Column II. Column I Column II i. Angular momentum a. $$ML^2T^{-2}$$ ii. Torque b. $$ML^2T^{-1}$$ iii. Inductance c. $$M^{-1}L^{-2}T^2Q^2$$ iv. Latent heat d. $$ML^2Q^{-2}$$ v. Capacitance e. $$ML^3T^{-1}Q^{-2}$$ vi. Resistivity f. $$L^2T^{-2}$$

$$i \rightarrow b., ii \rightarrow a., iii \rightarrow d., iv \rightarrow f., v \rightarrow c., vi \rightarrow e$$

$$i \rightarrow c., ii \rightarrow a., iii \rightarrow d., iv \rightarrow f., v \rightarrow b., vi \rightarrow e$$

$$i \rightarrow b., ii \rightarrow e., iii \rightarrow d., iv \rightarrow f., v \rightarrow c., vi \rightarrow a$$

$$i \rightarrow b., ii \rightarrow a., iii \rightarrow f., iv \rightarrow d., v \rightarrow c., vi \rightarrow e$$

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

Multiple Correct Answers Type
Which of the following pairs has the same dimensions?

0%
Torque and work
0%
Angular momentum and Planck's constant
0%
Energy and Young's modulus
0%
Light year and wave length

Explanation

$\ast \quad$ Torgue

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

$\left[ i \right] =\left[ { ML }^{ 2 }{ T }^{ -2 } \right]$

Work

$=$ Force

$\times$ displacement

$\left[ W \right] =\left[ { ML }^{ 2 }{ T }^{ -2 } \right]$

$\ast \quad$ Planks constent

$-$ Joule second

$\left[ h \right] =\left[ { ML }^{ 2 }{ T }^{ -1 } \right]$

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

$\ast \quad$ Light year is a distance

$\ast \quad$

$Y=\dfrac { Fl }{ \Delta DI }$ While Energy

$=$

$F$

$\times$ displacement.

A vernier callipers has its main scale of

$10 cm$ equally divided into

$200$ equal parts. Its vernier scale of

$25$ divisions coincides with

$12 mm$ on the main scale. The least count of the instrument is -

0%
$0.020 cm$
0%
$0.002 cm$
0%
$0.010 cm$
0%
$0.001 cm$

Explanation

$10\ cm$ of Main Scale

$=200$ div of Main scale

$\Rightarrow IMSD=0.05\ cm$

given

$25$ vernier scale division coincide with

$12\ mm$ on main scale

$\Rightarrow 25VSD=12\ mm$ on main scale

$=\dfrac{12}{0.5}=24\ MSD$

$\Rightarrow 25VSD=24MSD$

$\Rightarrow 1VSD=\dfrac{24}{25}MSD=0.048\ cm$

Least Count

$\Rightarrow 1MSD-1VSD$

$\Rightarrow (0.050-0.048)cm$

$\Rightarrow 0.002\ cm$

The potential energy of a particle varies with distance

$x$ as

$U = \dfrac{Ax^{1/2}}{x^2 + B}$ , where A and B are constants. The dimensional formula for A x B is:

0%
$M^1L^{7/2}T^{-2}$
0%
$M^1L^{11/4}T^{-2}$
0%
$M^1L^{5/2}T^{-2}$
0%
$M^1L^{9/2}T^{-2}$

Explanation

Given,

$U=\dfrac { { Ax }^{ 1/2 } }{ x^{ 2 }+B }$

${ x }^{ 2 }+B$ should have some dimension

$\Rightarrow$

${ L }^{ 2 }=dim\left( B \right)$

$dim\left( A \right) =dim\left( \dfrac { U\times \left( { x }^{ 2 }+B \right) }{ { x }^{ 1/2 } } \right) =\dfrac { \left[ { ML }^{ 2 }{ T }^{ -2 } \right] \left[ { L }^{ 2 } \right] }{ { L }^{ 1/2 } }$

$={ ML }^{ 7/2 }{ T }^{ -2 }$

$dim\left( A\times B \right) =\left[ { ML }^{ 7/2 }{ T }^{ -2 } \right] \left[ { L }^{ 2 } \right] =\left[ { ML }^{ 11/2 }{ T }^{ -2 } \right]$

In terms of resistance

$R$ and time

$T$ , the dimensions of ratio

$\dfrac {\mu}{\epsilon}$ of the permeability

$\mu$ and permittivity

$\epsilon$ is:

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

Explanation

The electrostatic force can be given by:

$F_e=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{r^2}$

So, permittivity is:

So,

$\varepsilon_0=\dfrac{q^2}{4\pi r^2F_e}$

So, dimensions :

$\varepsilon_0=\dfrac{[Q^2]}{[L^2][MLT^{-2}]}$

The dimensions of the permittivity

$\varepsilon$ is

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

The magnetic force can be given by:

$F_m=\dfrac{\mu_0}{4\pi}\dfrac{q_{m_1}q_{m_2}}{r^2}$

So, permeability is:

So,

$\mu_0=\dfrac{q_{m_1}q_{m_2}}{4\pi r^2F_m}$

The dimensions of the permeability

$\mu$ is

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

So, the ratio of the two dimensions is:

$\dfrac{\mu}{\varepsilon}=\dfrac{[M^{1}L^{1}Q^{-2}]}{[M^{-1}L^{-3}T^2Q^2]}$

$\Rightarrow M^2L^4T^{-2}Q^{-4}$

The dimension of the resistance is

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

comparing the dimension with the dimesion of resistance:

$\therefore \dfrac{\mu}{\varepsilon}=[R^2]$

Match the physical quantities given in Column I with dimensions expressed in terms of mass (M), length (L) time (T), and charge (Q) given in Column II.

Column I	Column II
i. Angular momentum	a. $ML^2T^{-2}$
ii. Torque	b. $ML^2T^{-1}$
iii. Inductance	c. $M^{-1}L^{-2}T^2Q^2$
iv. Latent heat	d. $ML^2Q^{-2}$
v. Capacitance	e. $ML^3T^{-1}Q^{-2}$
vi. Resistivity	f. $L^2T^{-2}$

0%
$i \rightarrow c., ii \rightarrow a., iii \rightarrow d., iv \rightarrow f., v \rightarrow b., vi \rightarrow e$
0%
$i \rightarrow b., ii \rightarrow a., iii \rightarrow d., iv \rightarrow f., v \rightarrow c., vi \rightarrow e$
0%
$i \rightarrow b., ii \rightarrow e., iii \rightarrow d., iv \rightarrow f., v \rightarrow c., vi \rightarrow a$
0%
$i \rightarrow b., ii \rightarrow a., iii \rightarrow f., iv \rightarrow d., v \rightarrow c., vi \rightarrow e$

Multiple Correct Answers Type
Which of the following pairs have different dimensions?

0%
Frequency and angular velocity
0%
Tension and surface tension
0%
Density and energy density.
0%
Linear momentum and angular momentum.

Explanation

$\rightarrow$

$W=2\pi f$

$\rightarrow$ Tension is a force but

[surface tension]

$=$ [Force

$\times$ length]

$\rightarrow$ Density

$=\dfrac { mass }{ volume }$

Energy density

$=\dfrac { Energy }{ Volume }$

$\rightarrow$ Linear momentum

$=$ mass

$\times$ velocity

Angular momentum

$=$ mass

$\times$ velocity

$\times$ distance

The dimension of electrical conductivity is _________.

0%
$M^{-2}L^{-4}T^3A^2$ .
0%
$M^{-1}L^{-2}T^3A^2$ .
0%
$M^{-1}L^{-3}T^4A^2$ .
0%
$M^{-2}L^{-3}T^6A^2$ .

The dimension of permittivity of vacuum

0%
$[M^{ -2} L^ {-3} T^4 A^2]$
0%
$[M^{ -1} L^ {-3} T^4 A^3]$
0%
$[M^{ -1} L^ {-3} T^4 A^2]$
0%
$[M^{ -1} L^ {-5} T^4 A^2]$

Explanation

The permittivity of the vacuum can be given by:

$F_e=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{r^2}$

$permitivity\ (\varepsilon_0)= \dfrac{(q)^2}{(F_e)\times(r)^2}$

$=\dfrac{[TA]^2}{[MLT^ {-2}] [L]^2}$

$=\dfrac{[TA]^2}{[ML^3T^ {-2}]}$

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

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

This is the required solution.

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%
$[LT^{-1}]$

Explanation

We know that the speed of light is represented by:

$c=\dfrac{1}{\sqrt{\mu_0\epsilon_0}}$

So, we can write:

$\dfrac {1}{\mu_{0}\epsilon_{0}} = c^{2}$

The dimension of speed of light is represented as

$[LT^{-1}]$

So, for

$\dfrac{1}{\mu_0\epsilon_0}= [L^{2}T^{-2}]$ .

The dimension of magnetic field in

$M, L, T$ and

$C$ (Coulomb) is given as:

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

Explanation

$[B] = \dfrac {F}{il} = \dfrac {Ft}{ql} = \dfrac {MLT^{-2}T}{CL} = [MT^{-1}C^{-1}]$ .

A metal sample carrying a current along

$X-$ axis with density

$J_{x}$ is subjected to a magnetic field

$B_{z}$ (along

$z-$ axis). The electric field

$E_{y}$ developed along

$Y-$ axis is directly proportional to

$J_{x}$ as well as

$B_{z}$ . The constant of proportionality has

$SI$ unit.

0%
$\dfrac{m^{2}}{A}$
0%
$\dfrac{m^{3}}{As}$
0%
$\dfrac{m^{2}}{As}$
0%
$\dfrac{As}{m^{3}}$

Explanation

Given:

$E_{y} \alpha J_{x}$ and

$E_{y} \alpha B_{Z}$

$E_{y} \alpha J_{x}B_{Z}$

$\Rightarrow E_{y}=KJ_{x}B_{Z}$ (Where K = constant of proportionality)

By Dimensional analysis we have;

$\left [ E_{y} \right ] = [KJ_{x}B_{Z}]$

$\Rightarrow \left [ K \right ] = \dfrac{\left [ E_{y} \right ]}{\left [ J_{x}B_{Z} \right ]}=\dfrac{[]M^{1}L^{1}T^{3}A^{-1}]}{[L^{-2}A^{A}][M^{1}T^{-2}A^{-1}]}$

$\Rightarrow [K]=\dfrac{[L^{3}]}{[A^{1}T^{1}]}\rightarrow$ Dimensional formual of K

So, S.I unit of K will be :

$\dfrac{m^{3}}{As}$

Two particles of mass

${m}_{1}$ and

${m}_{2}$ are approaching towards each other under their mutual gravitational field. If the speed of the particle is

$v$ , the speed of the center of mass of the system is equal to:

0%
$\cfrac { { m }_{ 1 } }{ { m }_{ 2 } } v$
0%
zero
0%
$\cfrac { \left( { m }_{ 1 }-{ m }_{ 2 } \right) v }{ { m }_{ 1 }+{ m }_{ 2 } }$
0%
$\cfrac { { m }_{ 2 } }{ { m }_{ 1 } } v$

Explanation

Taking both particles as a system,

$F_{ext}=0$

as mutual gravitational force becomes internal force here.

$a_{cm}=\dfrac{F_{ext}}{m}=0$

$v_{cm}=u_{cm}+a_{cm}t$

$v_{cm}=0$

$u_{cm}$ of system is zero as system starts from rest.

Hence v_{cm}=0$$

Option

$\textbf B$ is the correct answer.

Give the MKS units for the following quantities.
Power of lens.

0%
hertz
0%
Dioptre.
0%
ohms
0%
unit less

Explanation

The dioptre is the unit of measure for the refractive power of a lens. The power of a lens is defined as the reciprocal of its focal length in meters, or

$D = \dfrac 1f$ , where

$D$ is the power in diopters and

$f$ is the focal length in meters.

In the formula

$X=3YZ^2$ , X and Z have dimensions of capacitance and magnetic induction respectively. The dimensions of Y in MKSQ system are ________.

0%
$-2$ in mass, $-2$ in length, $4$ in time and $5$ in charge.
0%
$-3$ in mass, $-2$ in length, $4$ in time and $4$ in charge.
0%
$-4$ in mass, $-5$ in length, $4$ in time and $4$ in charge.
0%
$-3$ in mass, $-2$ in length, $8$ in time and $7$ in charge.

Explanation

Given,

$X=3YZ^2$

Then,

$Y=\dfrac{X}{3Z^2}$

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

$[Z]=[B]=[MT^{-1}Q^{-1}]$

Then,

$[Y]=\dfrac{[X]}{[3Z^2]}$

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

That is,

$-3$ in mass,

$-2$ in length,

$4$ in time and

$4$ in charge.

Planck's constant has the dimensions ___________.

0%
$ML^3T^{-1}$ .
0%
$ML^2T^{-1}$ .
0%
$ML^6T^{-1}$ .
0%
$ML^8T^{-1}$ .

Explanation

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

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

The dimensions of angular momentum, latent heat and capacitance are, respectively.

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

Explanation

For angular momentum:

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

The latent heat:
Latent Heat

$= \dfrac {Q}{m} = [L^{2}T^{-2}]$

The capacitance is given as:

$[C] = \dfrac {q}{V} =\dfrac{q^2}{W}$

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

The physical quantities which do not have same dimensions are

0%
Energy and Torque
0%
Pressure and Stress
0%
Work and Energy
0%
Strain and Youngs Modulus

Explanation

The energy and torque have the same dimensions as both can be represented as the product of force and distance.

The pressure and stress are the quantities that have the same dimensions. They both are defined as force per unit area.

The work and energy are the quantities that have the same dimensions. They are the product of force and distance.

The young's modulus is defined as the ratio of stress and strain. The strain is a dimensionless quantity. So, Young's modulus will have the same unit as of stress.

Strain =

$\dfrac { change\quad in\quad dimension }{ orginal\quad dimension }$

$={ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }$

Youngs Modulus

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

Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii

$R_1$ and

$R_2$ , respectively. The ratio of the masses of X to that of Y is

0%
$(R_1/R_2)^{1/2}$
0%
$R_2/R_1$
0%
$(R_1/R_2)^2$
0%
$R_1/R_2$

Explanation

We know,

$R=\dfrac{\sqrt{2mqV}}{qB}$

As q,B,V are same

$R\propto \sqrt m$

$\dfrac{R_1}{R_2}=\sqrt{\dfrac{m_1}{m_2}}$

$\dfrac{m_1}{m_2}=\dfrac{(R_1)^2}{(R_2)^2}$

Option

$\textbf C$ is the correct answer.

The equation of the state of some gases can be expressed as

$(P + \dfrac{a}{V^2} = \dfrac{R\theta}{V})$ , where P is the pressure. V the volume,

$\theta$ the absolute temperature and a and b are constant. The dimensional formula of a is

0%
$[ML^5T^{-2}]$
0%
$[M^{-1}L^5T^{-2}]$
0%
$[ML^{-1}T^{-2}]$
0%
$[ML^{-5}T^{-2}]$

Explanation

Dimensional formula of Pressure is

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

Dimensional formula of Volume is

$[M^{0} L^{3} T^{0}]$

From the given equation:

$P+\dfrac{a}{V^2}=\dfrac{R\theta}{V}$

Dimensional formula of

$\dfrac{a}{V^{2}}$ is same as pressure.

$\dfrac{a}{L^6}=[M^1L^{-1}T^{-2}]$

Therefore dimensional formula of a is

$[M^{1} L^{5} T^{-2}]$

$y$ represents pressure and

$x$ represents velocity gradient, then the dimensions of

$\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$ are

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

Explanation

Dimensions of

$\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$ = dimension of

$\dfrac { y }{ { x }^{ 2 } }$

Dimension of y

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

Dimension of x

${ T }^{ -1 }$

$\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \quad =\dfrac { M{ L }^{ -1 }{ T }^{ -2 } }{ { \left( { T }^{ -1 } \right) }^{ 2 } } \\ =M{ L }^{ -1 }{ T }^{ 0 }$

Dimensional equation of universal constant of gravitation is:

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

The dimensions of planck's constant equal to that of

0%
energy
0%
momentum
0%
angular momentum
0%
power

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.

In the formula : y = a sin ( at - b), where y - position, t - time and A, a, b are constants. What will be the dimensional formula of

$\frac{b}{a}$ ?

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

Explanation

Dimensions of at=Dimension of b

$\left[ a \right] \left[ t \right] =\left[ b \right] \\ \dfrac { b }{ a } =t\\ =T$

The radius of a sphere is

$(2.6 \pm 0.1)$ cm. The percentage error in its volume is

0%
$\dfrac{0.1}{2.6} \times 100%$
0%
$3 \times \dfrac{0.1}{2.6} \times 100%$
0%
$\dfrac{{8.49}}{{73.6}} \times 100$
0%
$\dfrac{0.1}{2.6} %$

Explanation

Given,

Radius

$= \left( {2.6 \pm 0.1} \right)cm$

$V = \frac{4}{3}\pi {r^3} = \frac{4}{3} \times 3.14 \times {\left( {2.6} \right)^3} = 73.6\,c{m^3}$

Now, in error

$\begin{array}{l} \frac { { \Delta V } }{ V } =3\left( { \frac { { \Delta r } }{ r } } \right) \\ \Delta V=v\propto 3\left( { \frac { { \Delta r } }{ r } } \right) \\ \Delta V=73.6\propto 3\left( { \frac { { 0.1 } }{ { 2.6 } } } \right) \\ \Delta V=8.49\, c{ m^{ 3 } } \\ \frac { { \Delta V } }{ V } \times 100=\frac { { 8.49 } }{ { 73.6 } } \times 100 \end{array}$

$=11.53$ %

$\therefore$ Option

$C$ is correct.

The dimensions of emf in MKS is:

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

Explanation

The emf across an inductor is given by:

$e = -L\dfrac {di}{dt}$

$e = \big[ {ML^{2}}{T^{-2}}{A^{-2}} \big] \big[ \dfrac {A}{T} \big]$

$e = \dfrac {{ML^{2}}{T^{-2}}}{A^{-1}{T}}$

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

The dimension of

$\left( \dfrac { 1 }{ 2 } \right) { \varepsilon }_{ 0 }{ E }^{ 2 }$ is(

${ \varepsilon }_{ 0 }$ : permittivity of free space, Electric field)

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

Explanation

As we know that

Energy Density

$= \dfrac{energy}{volume}= \left( \dfrac { 1 }{ 2 } \right) { \varepsilon }_{ 0 }{ E }^{ 2 }$

Hence ,

Dimesion of

$\left( \dfrac { 1 }{ 2 } \right) { \varepsilon }_{ 0 }{ E }^{ 2 }$ is, =

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

The velocity of a body moving viscous medium is given by

$V=\dfrac { A }{ B } [1-{ e }^{ { -t }/{ B } }]$ where

$t$ is time,

$A$ and

$B$ are constants. Then the dimensional formula of

$A$ is:

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

Explanation

$e^\frac{-t}{B}$ is the exponential term. So, the

$\dfrac{t}{B}$ is constant.

So, B has the dimensions as:

$B=[T]$

On the other hand, the dimension of

$\dfrac AB$ will be the same as that of the velocity.

$\dfrac{A}{B}$ has dimensions as velocity

$[LT^{-1}]=\dfrac{a}{[T]}$

dimensional formula of A is

$M^0L^1 T^0$

Write down the significant figures in the following

$5238$ N

$4200$ kg

0%
$4 , 4$
0%
$4 , 2$
0%
$4, 3$
0%
$5, 3$

Explanation

All non-zero digits are significant and last zeros are not considered as significant.

(a) : The given number

$5238$ has four significant figures.

(b) : The given number

$4200$ has two significant figures.

The dimension of the magnetic field intensity B is

0%
$(a)\,ML{T^{ - 2\,}}\,{A^{ - 1}}$
0%
$(b)\,M{T^{ - 2\,}}\,{A^{ - 1}}$
0%
$(c)\,M{L^{2\,}}\,T{A^{ - 2}}$
0%
$(d)\,{M^{2\,\,}}L{T^{ - 2}}\,{A^{ - 1}}$

Explanation

$F=qVB\sin { \theta }$

$\left[ F \right] =\left[ ML{ T }^{ -2 } \right]$

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

$\left[ q \right] =\left[ AT \right]$

$\sin \theta =$ dimensionless

$\left[ B \right] =\cfrac { \left[ F \right] }{ \left[ q \right] \left[ V \right] }$

$=\cfrac { \left[ ML{ T }^{ -2 } \right] }{ \left[ AT \right] \left[ L{ T }^{ -1 } \right] }$

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

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

In the context of accuracy to measurement and significant figures in expressing results of experiment, which of the following is/are correct?
(1) Out of the measurements 50.14 cm and 0.00025 ampere, the first one has greater accuracy.
(2) If one travels 478 km by rail and 397 m. by road, the total distance traveled is 478 km.

0%
Only (1) is correct
0%
Only (2) is correct
0%
Both are correct
0%
None of these is correct.

Explanation

Significant figures are the number of digits in a value, often a measurement, that contribute to the degree of accuracy of the value. We start counting significant figures at the first non-zero digit.

Since for

$50.14cm$

Now, the significant number

$=4$

and for

$0.00025$

Significant number

$=2$

Both are correct

Dimension of

$\epsilon_0$ are

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

Explanation

The electrostatic force is given as,

$F = \dfrac{{{Q_1}{Q_2}}}{{4\pi {\varepsilon _0}{R^2}}}$

${\varepsilon _0} = \dfrac{{{Q_1}{Q_2}}}{{4\pi F{R^2}}}$

Dimensions of ${\varepsilon _0}$ are,

$= \dfrac{{{A^2}{T^2}}}{{ML{T^{ - 2}} \times {L^2}}}$

$= \dfrac{{{A^2}{T^4}}}{{M{L^3}}}$

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

Which of the following is dimensionally incorrect ?

0%
$u=v-at$
0%
$s-ut=\dfrac{1}{2}at^2$
0%
$u^2=2a(gt-1)$
0%
$v^2-u^2=2as$

Explanation

A> We know acceleration a=rate of change of velocity

$a=\dfrac{v-u}{t}$ where v=final velocity and u = initial velocity and t= time

Thus we can rewrite the above equation as,

$u=v-at$

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

$[LT^{-1}]=[LT^{-1}]$ (Dimensionally correct)

B> From equation of motion we get displacement

$s=ut+\dfrac{1}{2}a{t}^{2}$

$s-ut=\dfrac{1}{2}a{t}^{2}$

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

Here also, LHS =RHS

C>The equation

$u^2=2a(gt-1)$

Substituting the dimensions:

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

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

LHS is not equal to RHS.

D> From equation of motion,

${v}^{2}={u}^{2}+2as$

${v}^{2}-{u}^{2}=2as$

$[LT^{-1}]^2-[LT^{-1}]^2=[LT^{-2}][L]$

So, LHS=RHS

Whereas the dimensions of the equation are not the same on the LHS and RHS.

Thus option C is incorrect

The dimensions of energy per unit volumes are the same as those of :

0%
Work
0%
Stress
0%
Pressure
0%
Modulus

Explanation

The dimension of energy per unit volume is equal to the dimension of pressure.

Firstly, the dimension of energy is :

$[ML^2T^{−2}]$

The dimension of volume:

$[L^3]$

The dimension of pressure :

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

Now,

$\dfrac{energy}{volume} = \dfrac{[ML^2T^{−2}]}{[L^3]} = [L^{-1}M T^{-2}]$

Pressure P varies as

$P=\cfrac { \alpha }{ \beta } exp\left( \cfrac { -\alpha }{ { K }_{ B }\theta } Z \right)$ where Z denotes distance.

${ K }_{ B }$ Boltzman's constant.

$\theta$ temperature and

$\alpha ,\beta$ are constants.Derive the dimensions of

$\beta$

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

Explanation

$\dfrac { \alpha Z }{ { K }_{ B }\theta } ={ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }{ K }^{ 0 }\\ \alpha =\dfrac { { K }_{ B }\theta }{ Z } =\dfrac { \left[ M{ L }^{ 2 }{ T }^{ -2 }{ K }^{ -1 } \right] \left[ K \right] }{ \left[ L \right] } \\ \alpha =ML{ T }^{ -2 }\\ \left[ P \right] =\dfrac { \left[ \alpha \right] }{ \left[ \beta \right] } \\ \beta =\dfrac { \alpha }{ P } =\dfrac { ML{ T }^{ -2 } }{ M{ L }^{ -1 }{ T }^{ -2 } } ={ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }\\$

Ans:

${ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }$

The dimensional formula of angular velocity

$\left(\omega=\cfrac{v}{r}\right)$ is

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

Explanation

The angular velocity of a body is given by:

$\omega=\cfrac{V}{r}$

The dimension of

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

The dimension of

$r=[L]$

$\omega=\cfrac{LT^{-1}}{L}=T^{-1}$

$\omega=M^0L^0T^{-1}$

The dimension of combination

$\dfrac{L}{CVR}$ are same as dimensions of

0%
Change
0%
Current
0%
Charge $^{-1}$
0%
Current $^{-1}$

Explanation

Dimension of

$Resistance = \dfrac{Potential \,Difference}{Current} = \dfrac{ML^2T^{-3}A^{-1}}{ A} = [ML^2T^{-3}A^{-2}[$

Dimension of

$Capacitance = \dfrac{Charge}{ Potential\, Difference} = \dfrac{AT}{ML^2T^{-3}A^{-1}} = [M^{-1}L^{-2}T^4A^2[$

Dimension of

$Self Inductance [L]=\dfrac {[\phi]}{[I]}= \dfrac{MT^{−2}L^2A^{−1}}{ A} = [ML^2T^{-2}A^{-2}[$

Dimension of

$Voltage=\dfrac{Energy}{Charge}=\dfrac{[ML^2T^{-2}]}{AT}=[ML^2T^{-3}A^{-1}]$

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

Pressure P varies as

$P=\dfrac{\alpha}{\beta}$ exp

$(\dfrac{-\alpha}{K_B \theta}Z)$ where Z denotes distance,

$K_B$ Boltzman's constant,

$\theta$ absolute temperature and

$\alpha,\beta$ are constants.Derive the dimensions of

$\beta$

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

Explanation

$P=\dfrac{\alpha}{\beta}$ exp

$(\dfrac{-\alpha}{k_B \theta}Z)$

Since the terms inside the exponential do not have any dimension,we can find out the dimension of

$\alpha$

$\dfrac{\alpha \times L' \times k}{J \times k}$

$\alpha$ has dimension of

$\dfrac{J}{L'}=\dfrac{ML^2 T^{-2}}{L'}=ML T^{-2}$

For

$\dfrac{\alpha}{\beta}$ to have dimension of pressure ,i.e

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

$\beta$ should have the dimension

$L^2$ i.e,

$M^0L^2T^0$

If temperature of sun is decreased by

$1\%$ then the value of solar constant will change by

0%
$2\%$
0%
$-4\%$
0%
$-2\%$
0%
$4\%$

Explanation

Solar constant

$=G=\sigma T^4\left[\cfrac{4\pi R^2}{4\pi O^2}\right]$

$\therefore$ change of G w.r.t

$O$ on both sides

$\triangle G=4\sigma T^3 \triangle T \left[\cfrac{4\pi R^2}{4\pi O^2}\right]$

$\cfrac{\triangle G}{G}=\cfrac{4\triangle T}{T}=-4\%$

A dimensionless quantity :

0%
may have a unit
0%
never has a unit
0%
always has a unit
0%
does not exist

If the dimension of a physical quantity is given by

$[M^aL^bT^c]$ then the physical quantity will be

0%
acceleration, if a = 1, b =-1, c =-2
0%
velocity, if a = 1, b =0, c =-1
0%
pressure, if a = 1, b =-1, c =-2
0%
force, if a = 0, b =-1, c =-2

Explanation

Pressure,

$P=\dfrac{Force}{Area}=\dfrac{Mass\times Acceleration}{Area}$

$[P]=\dfrac{ML^1T^{-2}}{L^2}=[M^1L^{-1}T^{-2}]=[M^aL^bT^c]$

$a=1\,,b=-1\,,c=-2$

The equation of state of some gases can be expressed as

$\left(P+\dfrac{a}{V^{2}}\right)\left(V-b\right)=RT$ . Here P is the pressure,

$V$ is the volume,

$T$ is the absolute temperature and

$a,b,R$ are constant. The dimensions of

$a$ are:

0%
$ML^{5}T^{-2}$
0%
$ML^{-1}T^{-2}$
0%
$M^{0}L^{3}T^{0}$
0%
$M^{0}L^{6}T^{0}$

Explanation

Dimension, $\dfrac{a}{{{V}^{2}}}$ is the same as pressure dimension is $N{{m}^{-2}}$

$\dfrac{a}{{{V}^{2}}}\,=\dfrac{a}{{{\left( {{m}^{3}} \right)}^{2}}}=N{{m}^{-2}}$

$a=N{{m}^{4}}=\,\left[ M{{L}^{1}}{{T}^{-2}} \right]\left[ {{M}^{o}}{{L}^{4}}{{T}^{o}} \right]=\left[ M{{L}^{5}}{{T}^{-2}} \right]$

Hence, unit of $a$ is $\left[ M{{L}^{5}}{{T}^{-2}} \right]$

If there is a positive error of

$50\%$ in the measurement of velocity of a body, then the error in the measurement of kinetic energy is :

0%
$25\%$
0%
$50\%$
0%
$100\%$
0%
$125\%$

Explanation

Given that,

$\dfrac{\Delta v}{v}\times 100$ is the % error in velocity = 50%

Kinetic energy $K.E=\dfrac{1}{2}m{{v}^{2}}$

Error in the kinetic energy

$\dfrac{\Delta K.E}{K.E}\times 100=m\times 2\dfrac{\Delta v}{v}\times 100$

$m$ is as a constant

Now, percentage error

$\dfrac{\Delta K.E}{K.E}\times 100=2\dfrac{\Delta v}{v}\times 100$

$\dfrac{\Delta K.E}{K.E}\times 100=2\times 50$ %

$\dfrac{\Delta K.E}{K.E}\times 100=100$ %

Hence, the error in the measurement of kinetic energy is $100$ %

Choose the incorrect statements out of the following:-

0%
Every measurement by any measuring instrument has some errors
0%
Every calculated physical quantity that is based on measured values has some error
0%
A measurement can have more accuracy but less precision and vice-versa
0%
The percentage error is different from relative error.

Explanation

$\text { We have relative error as }$

$\text { Relative error = } \frac{\text { measured value - real value }}{\text { real value }}$

$\text { We have persentage error as }$

$\text { persentage error = } \frac{\text { measured value - real value }}{\text { real value }}$

$\begin{array}{l} \text { So we get a Relation } \\ \qquad\text { Relative error }=\frac{\text { Percent error }}{100} \end{array}$

A student uses a metre scale, measuring upto

$1\ mm$ , to measure the length an breadth of a retangular plate. He finds that length

$=5.7\ cm$ and breadth

$=3.4\ cm$ . What is the percentage error in the area of the plate?

0%
$3.58\%$
0%
$4.7\%$
0%
$5.3\%$
0%
$6.2\%$

Explanation

$\begin{array} \text { Given } \rightarrow \text { Measuring scale with least count } 1 \mathrm{~mm} \\ \text { length }=5.7 \mathrm{~cm}=57 \mathrm{~mm} \\ \text { Breadth }=3.4 \mathrm{~cm}=34 \mathrm{~mm} \end{array}$

$\begin{aligned} \text { Now } \rightarrow \text { Area } &=\text { Length } \times \text { Bredth. } \\ A &=l \times b=57 \times 34=1938 \mathrm{~mm}^{2} \end{aligned}$

$\begin{array}{l} \text { Using error analysis } \rightarrow \\ \text { Take logarithm both sides } \\ \qquad \log A=\log (l)+\log (b) \end{array}$

$\begin{array}{l} \text { difterentiate both sides - } \\ \frac{d A}{A}=\frac{d l}{l}+\frac{d b}{b} \quad \Rightarrow d A=A\left[\frac{d l}{l}+\frac{d b}{b}\right] \end{array}$

$\frac{d A}{A}=1938\left[\frac{1}{57}+\frac{1}{34}\right]=0.047=4.7 \%$

The dimensional formula for thermal resistance is

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

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).

$Thermal\, Resisatnce$

$=\dfrac{Temperature\,difference}{Thermal\,current}=\dfrac{\Delta T}{Rate\,of\,flow\,of\,heat}$

$\implies \dfrac{\Delta T}{\dfrac{\Delta Q}{\Delta t}}=\dfrac{[K]}{[ML^2T^{-3}]}=[M^{-1}L^{-2}T^3K]$

The dimensional formula for thermal resistance is

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

Find the dimensions of inductance

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

The dimension formula of torque is

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

Explanation

Torque,

$\tau=r\times F=rma$

The dimensional formula of torque,

$[\tau]=[r][m][a]$

$[\tau]=[L][M][LT^{-2}]$

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

The correct option is D.

Find the maximum possible percentage error in the measurement of force on an object(on mass m) travelling at velocity v in a circle of radius r, if m=(4.0 plus minus 0.1)kg, v=(10 plus minus 0.1)m/s and r=(8.0 plus minus 0.2)m

0%
( 50 ± 1) N
0%
( 50 ± 2.5) N
0%
( 50 ± 1.5) N
0%
( 50 ± 3.5) N

Explanation

Force

$(f)=\cfrac{mv^2}{r}=\cfrac{4\times(10)^2}{8}=50\\ \therefore\cfrac{\Delta F}{F}=\cfrac{\Delta m}{m}+2(\cfrac{\Delta V}{V})+(\cfrac{\Delta r}{r})\\ \quad=\cfrac{0.1}{4}+2\cfrac{(0.1)}{10}+\cfrac{0.8}{8}\\ \therefore \cfrac{\Delta F}{F}=0.07$

Percentage error in

$F=25\%+2+2.5\%$

$\therefore$ Percentage error in

$F=7\%$

Eror in

$\Delta F$ in

$F=F\times0.07\\ \quad=0\times0.07$

$\therefore$ Error in

$\Delta F$ in

$F=3.5N$

$\therefore F=(50\pm3.5)N$

A physical quantity

$P$ is given by the relation,

$P={P}_{0}{e}^{(-\alpha {t}^{2})}$ . If

$t$ denotes the time, the dimensions of constant

$\alpha$ are

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

Explanation

Given,

$P=P_0e^{(-\alpha t^2)}$

As we know, both

$P$ and

$P_0$ are pressure, it have the same units. Therefore,

$\alpha t^2$ must be dimensionless for which,

$\alpha=\dfrac{1}{T^2}=T^{-2}$

So the dimension of

$\alpha$ is

$[T^{-2}]$ .

$L=2.331 cm$ ,

$B = 2.1 cm$ , then L+B=

0%
$4.431cm$
0%
$4.43cm$
0%
$4.5cm$
0%
$4.2cm$

Explanation

Step 1

$L = 2.331 cm = 2.33$ (corrected to two decimal places) and

$B = 2.1 cm$

Step 2

$L + B = 2.33 + 2.1 = 4.43cm. = 4.4cm$

(by the rule of addition the sum is expressed in minimum decimal places of terms in addition).

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

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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