Units & Measurements — Page 18 | JEE Physics

Q171

Match List I with List II .tg .tg LIST I LIST II A. Torque I.

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

B. Magnetic field II.

\left[L^2 A^1\right]

C. Magnetic moment III.

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

D. Permeability of free space IV.

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

Choose the correct answer from the options given below:

A A-I, B-III, C-II, D-IV

B A-IV, B-II, C-III, D-I

C A-III, B-I, C-II, D-IV

D A-IV, B-III, C-II, D-I

Correct Answer

Option D

Solution

To determine the correct matches between List I and List II, we need to consider the dimensional formulas of the given physical quantities.

Torque (τ) is defined as the product of force (F) and distance (d): $\tau = F \cdot d$ The dimensional formula for torque is: $[\tau] = [M^1 L^2 T^{-2}]$ Thus, Torque corresponds to IV.

Magnetic field (B) is derived from the relation involving force (F), charge (q), and velocity (V): $F = qVB$ The dimensional formula for the magnetic field is: $[B] = [M^1 T^{-2} A^{-1}]$ Thus, Magnetic field corresponds to III.

Magnetic moment (M) is the product of current (I) and area (A): $M = I \cdot A$ The dimensional formula for the magnetic moment is: $[M] = [L^2 A^1]$ Thus, Magnetic moment corresponds to II.

Permeability of free space (μ₀) has the dimensional formula: $[μ₀] = [M^1 L^1 T^{-2} A^{-2}]$ Thus, Permeability of free space corresponds to I.

Based on the above analysis, the correct matches are: A.

Torque – IV B.

Magnetic Field – III C.

Magnetic Moment – II D.

Permeability of free space – I Therefore, the correct option is Option D: A-IV, B-III, C-II, D-I.

Q172

Match with \mathrm{L}^2 \mathrm{~T}^{-2}\right]$

List - I		List - II
(B)	Magnetic field	(II)	$\left[\mathrm{M} \mathrm{T}^{-2} \mathrm{~A}^{-1}\right]$
(C)	Magnetic moment	(III)	$\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2} \mathrm{~A}^{-2}\right]$
(D)	Torsional constant	(IV)	$\left[\mathrm{L}^2 \mathrm{~A}\right]$

A (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

B

(\mathrm{A})-(\mathrm{I}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{II}),(\mathrm{D})-(\mathrm{III})

C

(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{III}),(\mathrm{D})-(\mathrm{IV})

D (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

Correct Answer

Option A

Solution

The correct matching is: (A) Permeability of free space has dimensions

[MLT^{-2}A^{-2}]

which corresponds to entry (III). (B) Magnetic field has dimensions

[MT^{-2}A^{-1}]

which corresponds to entry (II). (C) Magnetic moment in SI units is expressed in Ampere-square meters, i.e.,

[L^2A]

which corresponds to entry (IV). (D) Torsional constant (torque per unit angular displacement) has dimensions

[ML^2T^{-2}]

which corresponds to entry (I). Thus, the correct answer is: Option A: (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

Q173

Match with . ^0 \mathrm{~T}^{-3}\right]$

List - I		List - II
(B)	Intensity of wave	(II)	$\left[\mathrm{ML}^{-2} \mathrm{~T}^{-2}\right]$
(C)	Pressure gradient	(III)	$\left[\mathrm{M}^{-1} \mathrm{LT}^2\right]$
(D)	Compressibility	(IV)	$\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]$

A (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

B (A)-(IV), (B)-(II), (C)-(I), (D)-(III)

C (A)-(IV), (B)-(I), (C)-(II), (D)-(III)

D (A)-(I), (B)-(IV), (C)-(III), (D)-(II)

Correct Answer

Option C

Solution

Here are the dimensional formulas: Coefficient of viscosity, μ

[\mu]=\frac{\text{pressure}\times\text{time}}{} =\frac{[M\,L^{-1}T^{-2}]\,[T]}{} =[M\,L^{-1}T^{-1}]\quad\longrightarrow\text{(IV)}

Intensity of a wave, I

[I]=\frac{\text{power}}{\text{area}} =\frac{[M\,L^2T^{-3}]}{[L^2]} =[M\,L^0T^{-3}]\quad\longrightarrow\text{(I)}

Pressure gradient, dP/dx

\left[\frac{dP}{dx}\right] =\frac{[M\,L^{-1}T^{-2}]}{[L]} =[M\,L^{-2}T^{-2}]\quad\longrightarrow\text{(II)}

Compressibility, κ

[\kappa]=\frac1{[\text{pressure}]} =[M^{-1}L\,T^2]\quad\longrightarrow\text{(III)}

Matching gives (A)–(IV), (B)–(I), (C)–(II), (D)–(III), i.e. Option C.

Q174

\text{ Match the LIST-I with LIST-II }

.tg .tg LIST-I LIST-II A.

\text{ Boltzmann constant }

I

\mathrm{ML}^2 \mathrm{~T}^{-1}

B

\text{ Coefficient of viscosity }

II

\mathrm{MLT}^{-3} \mathrm{~K}^{-1}

C

\text{ Planck's constant }

III

\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}

D

\text{ Thermal conductivity }

IV

\mathrm{ML}^{-1} \mathrm{~T}^{-1}

Choose the correct answer from the options given below:

A A - III, B - IV, C - I, D - II

B A - III, B - IV, C - II, D - I

C A - III, B - II, C - I, D - IV

D A - II, B - III, C - IV, D - I

Correct Answer

Option A

Solution

Explanation of Dimensional Analysis In order to match the quantities from LIST-I with their respective dimensions in LIST-II, we need to analyze their dimensional formulas: (A) Boltzmann Constant [k]: The equation relating pressure (P), volume (V), number of moles (N), and temperature (T) gives us: $[\mathrm{k}] = \dfrac{\mathrm{PV}}{\mathrm{NT}} = \dfrac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~K}} = \mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}$ This corresponds to option III.

(B) Coefficient of Viscosity [η]: The coefficient of viscosity can be defined using the relation: $[\eta] = \dfrac{\mathrm{F}}{6 \pi \mathrm{rv}} = \dfrac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2 \mathrm{~T}^{-1}} = \mathrm{ML}^{-1} \mathrm{~T}^{-1}$ This corresponds to option IV.

(C) Planck's Constant [h]: The relationship between energy (E) and frequency (f) gives the dimensional formula: $[\mathrm{h}] = \dfrac{\mathrm{E}}{\mathrm{f}} = \dfrac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~T}^{-1}} = \mathrm{ML}^2 \mathrm{~T}^{-1}$ This corresponds to option I.

(D) Thermal Conductivity [k]: From the heat conduction equation, we get: $\mathrm{k} = \dfrac{\left(\mathrm{ML}^2 \mathrm{~T}^{-3}\right) \mathrm{L}}{\mathrm{~L}^2 \cdot \mathrm{~K}} = \mathrm{MLT}^{-3} \mathrm{~K}^{-1}$ This corresponds to option II.

To conclude, the correct matches are: A - III B - IV C - I D - II This analysis ensures each physical quantity is aligned with its correct dimensional representation.

Q175

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 ML

-

3 T

-

3 I2

B M

-

1 L3 T3 I

C M

-

1 L

-

3 T3 I2

D M

-

1 L

-

3 T3 I

Correct Answer

Option C

Solution

We know. resistivity ( $\rho$ ) =

{{RA} \over L}

and conductivity =

{1 \over \rho }

=

{1 \over {RA}}

As R =

{V \over {\rm I}}

$\therefore$ conductivity =

{{L{\rm I}} \over {VA}}

Also V =

{\omega \over q}

=

{\omega \over {it}}

=

{{\left[ {M{L^2}{T^{ - 2}}} \right]} \over {\left[ {\rm I} \right]\left[ T \right]}}

=

\left[ {M{L^2}{T^{ - 3}}{{\rm I}^{ - 1}}} \right]

$\therefore$ Conductivity =

{{\left[ {\rm{L}} \right]\left[ {\rm I} \right]} \over {\left[ {M{L^2}{T^{ - 3}}{{\rm{I}}^{ - 1}}} \right]\left[ {{L^2}} \right]}}

=

\left[ {{M^{ - 1}}{L^{ - 3}}{T^3}{{\rm I}^2}} \right]

Q176

Time (T), velocity (C) and angular momentum (h) are chosen as fundamentalquantities instead of mass, length and time. In terms of these, the dimensions of mass would be :

A [M] = [T

-

1 C

-

2 h]

B [M] = [T

-

1 C2 h]

C [M] = [T

-

1 C

-

2 h

-

1]

D [M] = [T C

-

2 h]

Correct Answer

Option A

Solution

Let, M $\propto$ Tx Cy hz

\therefore\,\,\,

[M1LoTo] = [T1]x [L1 T $-$ 1]y [M1L2T $-$ 1]z [M1Lo To] = [Mz Ly + 2z Tx $-$ y $-$ z] By comparing both sides we get, z = 1 y + 2z = 0 x $-$ y $-$ z = 0

\therefore\,\,\,

y = $-$ 2z = $-$ 2 x = y + z = $-$ 2 + 1 = $-$ 1

\therefore\,\,\,

[M] = [M $-$ 1 C $-$ 2 h1]

Q177

The force of interaction between two atoms is given by F =

\alpha

\beta

exp

\left( { - {{{x^2}} \over {\alpha kt}}} \right)

; where x is the distance, k is the Boltzmann constant and T is temperature and

\alpha

and

\beta

are two constants. The dimension of

\beta

is :

A M2L2T

-

2

B M2LT

-

4

C MLT

-

4

D M0L2LT

-

4

Correct Answer

Option B

Solution

F = \alpha \beta {e^{\left( {{{ - {x^2}} \over {\alpha KT}}} \right)}}

\left[ {{{{x^2}} \over {\alpha KT}}} \right] = {M^o}{L^o}{T^o}

{{{L^2}} \over {\left[ \alpha \right]M{L^2}{T^{ - 2}}}}

$=$

{M^o}{L^o}{T^o}

$\Rightarrow$

\left[ \alpha \right] = {M^{ - 1}}{T^2}

\left[ F \right] = \left[ \alpha \right]\left[ \beta \right]

MLT $-$ 2 = M $-$ 1T2[ $\beta$ ] $\Rightarrow$ [ $\beta$ ] = M2LT $-$ 4

Q178

In SI units, the dimensions of

\sqrt {{{{ \in _0}} \over {{\mu _0}}}}

is :

A A–1 TML3

B A2T3M–1L–2

C AT–3ML3/2

D AT2M–1L–1

Correct Answer

Option B

Solution

\sqrt {{{{ \in _0}} \over {{\mu _0}}}}

=

{{{ \in _0}} \over {\sqrt {{\mu _0}{ \in _0}} }}

= c $\times$

{{ \in _0}}

$\therefore$

\left[ {\sqrt {{{{ \in _0}} \over {{\mu _0}}}} } \right]

=

\left[ {L{T^{ - 1}}} \right] \times \left[ {{ \in _0}} \right]

We know, F =

{1 \over {4\pi { \in _0}}}{{{q^2}} \over {{r^2}}}

$\therefore$

{ \in _0} = {{{q^2}} \over {4\pi {r^2}F}}

$\Rightarrow$

\left[ {{ \in _0}} \right] = {{{{\left[ {AT} \right]}^2}} \over {\left[ {ML{T^{ - 2}}} \right] \times \left[ {{L^2}} \right]}}

=

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

$\therefore$

\left[ {\sqrt {{{{ \in _0}} \over {{\mu _0}}}} } \right]

=

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

$\times$

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

= [A2T3M–1L–2]

Q179

Expression for time in terms of G(universal gravitional constant), h (Planck constant) and c (speed of light) is proportional to :

A

\sqrt {{{h{c^5}} \over G}}

B

\sqrt {{{{c^3}} \over {Gh}}}

C

\sqrt {{{Gh} \over {{c^5}}}}

D

\sqrt {{{Gh} \over {{c^3}}}}

Correct Answer

Option C

Solution

Let t $\propto$ Gx hy cz $\therefore$ [t] = [G]x [h]y [c]z . . . . . (1) We know, F =

{{G{M^2}} \over {{R^2}}}

$\Rightarrow$ G =

{{F{R^2}} \over {{M^2}}}

$\therefore$ [G] =

{{\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]} \over {\left[ {{M^2}} \right]}}

[G] =

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

Also, E = hf $\therefore$ [h] =

{{[E]} \over {[F]}}

=

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

=

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

[C] =

\left[ {{M^o}L\,{T^{ - 1}}} \right]

From equation (1) we get,

\left[ {{M^o}\,{L^o}\,{T^1}} \right] = {\left[ {{M^{ - 1}}\,{L^3}\,{T^{ - 2}}} \right]^x}{\left[ {M\,{L^2}\,{T^{ - 1}}} \right]^y}{\left[ {{M^o}L{T^{ - 1}}} \right]^z}

\left[ {{M^o}\,{L^o}\,{T^1}} \right] = \left[ {{M^{ - x + y}}\,{L^{3x + 2yz}}\,{T^{ - 2x - y - z}}} \right]

By comparing the power of M, L, T $-$ x + y = 0 $\Rightarrow$ x = y 3x + 2y + z = 0 $\Rightarrow$ 5x + z = 0 . . . . . (2) $-$ 2x $-$ y $-$ z = 1 $\Rightarrow$ $-$ 3x $-$ z = 1 . . . .

(3) By solving (2) and (3), we get, x =

{1 \over 2}

= y and z = $-$

{5 \over 2}

$\therefore$ t $\propto$

\sqrt {{{Gh} \over {{C^5}}}}

Q180

A copper wire is stretched to make it 0.5% longer. The percentage change in its electrical resistance if its volume remains unchanged is :

A 2.0 %

B 2.5 %

C 1.0 %

D 0.5 %

Correct Answer

Option C

Solution

We know,

R = {{\rho l} \over A}

and Volume (V) = A

l

$\Rightarrow$ A $=$

{V \over l}

$\therefore$

R = {{\rho {l^2}} \over v}

$\therefore$

{{\Delta R} \over R} = 2{{\Delta l} \over l}

$=$ 2 $\times$ 0.5 $=$ 1%