Units and Measurements Test

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Units and Measurements Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

Dimensions of $$\displaystyle \frac{1}{\mu_{0}{\varepsilon_{0}}}$$ , where symbols have their usual meaning, are :

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

B
$$[\mathrm{L}^{-2}\mathrm{T}^{2}]$$

C
$$[\mathrm{L}^{2}\mathrm{T}^{-2}]$$

D
$$[\mathrm{L}\mathrm{T}^{-1}]$$

Solution

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

Dimensions for $$\dfrac { 1 }{ { \mu }_{ 0 }{ \varepsilon }_{ 0 } } $$ will be $$\dfrac { 1 }{ [{ M }^{ -1 }{ L }^{ -3 }{ T }^{ 4 }{ I }^{ 2 }]\times [{ M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 }{ I }^{ -2 }] } =[L^2T^{-2}]$$
Question 2
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Let $$[\varepsilon _{0}]$$ denote the dimensional formula of the permittivity of vacuum. If $$M=$$ mass, $$L=$$ length, $$T=$$ time and $$A=$$ electric current, then:

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

B
$$[\varepsilon _{0}]=[M^{-1}L^{2}T^{-1}A^{-2}]$$

C
$$[\varepsilon _{0}]=[M^{-1}L^{2}T^{-1}A]$$

D
$$[\varepsilon _{0}]=[M^{-1}L^{-3}T^{2}A]$$

Solution

$$\displaystyle \frac{1}{4\pi\varepsilon _{0}}\frac{q^{2}}{r^{2}}=F$$
$$\displaystyle
\varepsilon_{0}=\frac{[A^{2}T^{2}]}{[MLT^{-2}L^{2}]}=[M^{-1}L^{-3}A^{2}T^{4}]
$$
Question 3
1 / -0

In the following $$I$$ refers to current and other symbols have their usual meaning. Choose the option that corresponds to the dimensions of electrical conductivity:

A
$$M{L}^{-3}{T}^{-3}{I}^{2}$$

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

C
$${M}^{-1}{L}^{3}{T}^{3}I$$

D
$${M}^{-1}{L}^{-3}{T}^{3}I$$

Solution

Conductivity has relation, $$\sigma =\dfrac { l }{ AR } =\dfrac { lI }{ AV } $$.

V has dimension $$[M{ L }^{ 2 }{ T }^{ -3 }{ I }^{ -1 }]$$. Hence dimension of $$\sigma$$ is $$[{ M }^{ -1 }{ L }^{ -3 }{ T }^{ 3 }{ I }^{ 2 }]$$.
Question 4
1 / -0

The physical quantities not having same dimensions are:

A
torque and work

B
momentum and Plancks constant

C
stress and Youngs modulus

D
speed and $$(\mu_{0}\varepsilon _{0})^{-1/2}$$

Solution

The and the work have the same dimensions because they both are defined as the product of the force and the distance.
The stress and the young's modulus have the same dimensions because the young's modulus is defined as the ratio of stress and strain and strain is a dimensionless quantity.
The speed of an electromagnetic wave is given by $$\dfrac{1}{\mu_0\epsilon_0}$$. so, both have the same dimension.
Dimensions of momentum $$=$$ kg $$\mathrm{m}/\sec=[\mathrm{M}\mathrm{L}\mathrm{T}^{-2}]$$
Dimensions of PIanck's constant $$=$$ joule $$\sec=[\mathrm{M}\mathrm{L}^{2}\mathrm{T}^{-1}]$$
$$\therefore $$Dimensions of momentum $$\neq$$ dimensions of PIanck's constant.
Question 5
1 / -0

The dimension of magnetic field in $$\mathrm{M},\ \mathrm{L},\ \mathrm{T}$$ and $$\mathrm{C}$$ (Coulomb) is given as :

A
$$\mathrm{M}\mathrm{T}^{-1}\mathrm{C}^{-1}$$

B
$$\mathrm{M}\mathrm{T}^{-2}\mathrm{C}^{-1}$$

C
$$\mathrm{M}\mathrm{L}\mathrm{T}^{-1}\mathrm{C}^{-1}$$

D
$$\mathrm{M}\mathrm{T}^{2}\mathrm{C}^{-2}$$

Solution

Magnetic force $$\mathrm{F}=\mathrm{i}\mathrm{B}l$$
$$\implies $$ Magnetic field $$B = \dfrac{F}{il}$$
Dimensions of magentic field $$[B] = \dfrac{MLT^{-2}}{[CT^{-1}][L]}$$
$$\Rightarrow [\mathrm{B}]=\lfloor \mathrm{M}\mathrm{T}^{-1}\mathrm{C}^{-1}\rfloor$$
Question 6
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Which of the following does not have the same dimension?

A
Electric flux, Electric field, Electric dipole moment

B
Pressure, stress, Youngs modulus

C
Electromotive force, Potential difference, Electric voltage

D
Heat, Potential energy, Work done

Solution

B) Pressure: $$\dfrac { force }{ Area } $$
Stress : $$\dfrac { force }{ Area } $$
Youngs modulus : $$\dfrac { stress }{ strain } $$, here strain is dimension less So, all three has same dimension.

C) Electromotive force: $$\dfrac { Energy }{ charge } $$
Potential difference: $$\dfrac { W}{ Q } $$, with is same as Electromotive
Electric voltage: If you see in a capacitor, then Voltage is given by $$\dfrac { Energy }{ charge } $$

D) Heat: $$ML^2T^{-2} $$
Potential energy: $$mgh \Rightarrow ML^2T^{-2}$$
Work done $$=f.d\Rightarrow ML^2T^{-2}$$

A) Electric flux: $$ M^{-2}SA $$
Electric field: $$ M^2S^{-3}A^{-1} $$

Hence, it has quantities with different units.
Question 7
1 / -0

A cube has a side of length $$1.2\times 10^{-2}\ \mathrm{m}$$. Calculate its volume ?

A
$$1.7\times 10^{-6}\ \mathrm{m}^{3}$$

B
$$1.73\times 10^{-6}\ \mathrm{m}^{3}$$

C
$$1.70\times 10^{-6}\ \mathrm{m}^{3}$$

D
$$1.732\times 10^{-6}\ \mathrm{m}^{3}$$

Solution

Given, side of cube $$=1.2 \times 10 ^{-2}\ m$$
$$V={ t }^{ 3 }={ \left( 1.2\times { 10 }^{ -2 }\ m \right) }^{ 3 }=1.728\times { 10 }^{ -6 }\ { m }^{ 3 }$$
Since length is expressed in two significant figures, the volume will also be expressed in two significant figures.
So, volume $$=1.7 \times 10^{-6}\ m^3$$
Question 8
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Directions For Questions

In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, $$[E]$$ and $$[B]$$ stand for dimensions of electric and magnetic fields respectively, while $$[\in_0]$$ and $$[\mu_0]$$ stand for dimensions of the permittivity and permeability of free space respectively. $$[L]$$ and $$[T]$$ are dimensions of length and time respectively. All the quantities are given in $$SI$$ units.

...view full instructions

The relation between $$[\in_0]$$ and $$[\mu_0]$$ is

A
$$[\mu_0] = [\in_0] [L]^2[T]^{-2}$$

B
$$[\mu_0] = [\in_0] [L]^{-2}[T]^{2}$$

C
$$[\mu_0] = [\in_0]^{-1} [L]^2[T]^{-2}$$

D
$$[\mu_0] = [\in_0]^{-1} [L]^{-2}[T]^{2}$$

Solution

We have,
$$C = \dfrac{1}{\sqrt{\mu_0 \in_0}}$$

$$\therefore [C^2] = \left[\dfrac{1}{\mu_0 \in_0}\right]$$

$$\Rightarrow L^2T^{-2} = \dfrac{1}{[\mu_0][\in_0]}$$

$$[\mu_0] = [\in_0]^{-1} [L]^{-2} [T]^2$$
Question 9
1 / -0

The dimensions of $${ \left( { \mu }_{ 0 }{ \varepsilon }_{ 0 } \right) }^{ -1/2 }$$ are :

A
$$[{L}^{-1}T]$$

B
$$[L{T}^{-1}]$$

C
$$[{L}^{1/2}{T}^{1/2}]$$

D
$$[{L}{T}^{1/2}]$$

Solution

Since expression for speed of light $$c=\dfrac { 1 }{ \sqrt { { \mu }_{ 0 }{ \varepsilon }_{ 0 } } } $$, the given expression has the dimensions of velocity i.e. $$L{ T }^{ -1 }$$.
Hence, option B is correct.
Question 10
1 / -0

The work done to raise a mass m from the surface of the earth to a height h, which is equal to the radius of the earth, is :

A
mgR

B
2 mgR

C
$$\dfrac{1}{2} mgR$$

D
$$\dfrac{3}{2} mg R$$

Solution

Initial potential energy at earths surface is
$$U_i = \dfrac{-GMm}{R}$$
Final potential energy at height h = R
$$U_f = \dfrac{-GMm}{2R}$$
As work done = change in PE
$$\therefore W = U_f - U_i$$
$$= \dfrac{GMm}{2R} $$
$$(\because GM = gR^2)$$
$$= \dfrac{gR^2m}{2R} = \dfrac{mgR}{2}$$

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Re-Attempt Weekly Quiz Competition

Units and Measurements Test - 17

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Units and Measurements Test - 17

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Question 1

$$[\mathrm{L}^{-1}\mathrm{T}]$$

$$[\mathrm{L}^{-2}\mathrm{T}^{2}]$$

$$[\mathrm{L}^{2}\mathrm{T}^{-2}]$$

$$[\mathrm{L}\mathrm{T}^{-1}]$$

Solution

Question 2

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

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

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

$$[\varepsilon _{0}]=[M^{-1}L^{-3}T^{2}A]$$

Solution

Question 3

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

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

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

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

Solution

Question 4

torque and work

momentum and Plancks constant

stress and Youngs modulus

speed and $$(\mu_{0}\varepsilon _{0})^{-1/2}$$

Solution

Question 5

$$\mathrm{M}\mathrm{T}^{-1}\mathrm{C}^{-1}$$

$$\mathrm{M}\mathrm{T}^{-2}\mathrm{C}^{-1}$$

$$\mathrm{M}\mathrm{L}\mathrm{T}^{-1}\mathrm{C}^{-1}$$

$$\mathrm{M}\mathrm{T}^{2}\mathrm{C}^{-2}$$

Solution

Question 6

Electric flux, Electric field, Electric dipole moment

Pressure, stress, Youngs modulus

Electromotive force, Potential difference, Electric voltage

Heat, Potential energy, Work done

Solution

Question 7

$$1.7\times 10^{-6}\ \mathrm{m}^{3}$$

$$1.73\times 10^{-6}\ \mathrm{m}^{3}$$

$$1.70\times 10^{-6}\ \mathrm{m}^{3}$$

$$1.732\times 10^{-6}\ \mathrm{m}^{3}$$

Solution

Question 8

$$[\mu_0] = [\in_0] [L]^2[T]^{-2}$$

$$[\mu_0] = [\in_0] [L]^{-2}[T]^{2}$$

$$[\mu_0] = [\in_0]^{-1} [L]^2[T]^{-2}$$

$$[\mu_0] = [\in_0]^{-1} [L]^{-2}[T]^{2}$$

Solution

Question 9

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

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

$$[{L}^{1/2}{T}^{1/2}]$$

$$[{L}{T}^{1/2}]$$

Solution

Question 10

mgR

2 mgR

$$\dfrac{1}{2} mgR$$

$$\dfrac{3}{2} mg R$$

Solution

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