Match List I with List II .tg border-collapse:collapse;border-spacing:0; .tg tdborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal; .tg thborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal; .tg .tg-c3owborder-color:inherit;text-align:center;vertical-align:top .tg .tg-7bttborder-color:inherit;font-weight:bold;text-align:center;vertical-align:top .tg .tg-0pkyborder-color:inherit;text-align:left;vertical-align:top LIST I LIST II A. Torque I. [M^1 L^1 T^-2 A^-2] B. Magnetic field II. [L^2 A^1] C. Magnetic moment III. [M^1 T^-2 A^-1] D. Permeability of free space IV. [M^1 L^2 T^-2] Choose the correct answer from the options given below:

Units and Measurements PYQ — JEE Main 2024 · 6 Apr · Shift 1 · Q69 with Solution

Match List I with List II

	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

View written solutionFree

Correct answer: D

Find dimensions of each quantity in List I

We match each physical quantity with its dimensional formula.

A. Torque

Torque $= \text{Force} \times \text{distance}$

Since, $[\text{Force}] = [M^1L^1T^{-2}]$

Therefore, $[\tau] = [M^1L^1T^{-2}]\cdot [L] = [M^1L^2T^{-2}]$

So, $A \to IV$

B. Magnetic field

Using Lorentz force, $F = qvB$

So, $B = \frac{F}{qv}$

Now,

$[F] = [M^1L^1T^{-2}]$
$[q] = [AT]$
$[v] = [LT^{-1}]$

Hence, $[B] = \frac{[M^1L^1T^{-2}]}{[AT][LT^{-1}]}$

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

So, $B \to III$

C. Magnetic moment

Magnetic moment of a current loop, $\mu = I \times \text{area}$

Thus, $[\mu] = [A]\cdot [L^2] = [L^2A^1]$

So, $C \to II$

D. Permeability of free space

From magnetic force between two long parallel wires, $\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}$

So, $\mu_0 = \frac{F}{L}\cdot \frac{r}{I^2}$

Now, $\left[\frac{F}{L}\right] = [M^1T^{-2}]$

Multiplying by $r$ : $[M^1T^{-2}]\cdot [L] = [M^1L^1T^{-2}]$

Dividing by $[A^2]$ : $[\mu_0] = [M^1L^1T^{-2}A^{-2}]$

So, $D \to I$

Final matching

Thus the correct match is:

$A \to IV$
$B \to III$
$C \to II$
$D \to I$

This corresponds to Option D.

Comparison with stored answer

Stored correct answer = D

Our derived answer = D

So, the answer agrees with the stored answer.

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Match List I with List II

	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

View written solutionFree

Correct answer: D

Find dimensions of each quantity in List I

We match each physical quantity with its dimensional formula.

A. Torque

Torque $= \text{Force} \times \text{distance}$

Since, $[\text{Force}] = [M^1L^1T^{-2}]$

Therefore, $[\tau] = [M^1L^1T^{-2}]\cdot [L] = [M^1L^2T^{-2}]$

So, $A \to IV$

B. Magnetic field

Using Lorentz force, $F = qvB$

So, $B = \frac{F}{qv}$

Now,

$[F] = [M^1L^1T^{-2}]$
$[q] = [AT]$
$[v] = [LT^{-1}]$

Hence, $[B] = \frac{[M^1L^1T^{-2}]}{[AT][LT^{-1}]}$

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

So, $B \to III$

C. Magnetic moment

Magnetic moment of a current loop, $\mu = I \times \text{area}$

Thus, $[\mu] = [A]\cdot [L^2] = [L^2A^1]$

So, $C \to II$

D. Permeability of free space

From magnetic force between two long parallel wires, $\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}$

So, $\mu_0 = \frac{F}{L}\cdot \frac{r}{I^2}$

Now, $\left[\frac{F}{L}\right] = [M^1T^{-2}]$

Multiplying by $r$ : $[M^1T^{-2}]\cdot [L] = [M^1L^1T^{-2}]$

Dividing by $[A^2]$ : $[\mu_0] = [M^1L^1T^{-2}A^{-2}]$

So, $D \to I$

Final matching

Thus the correct match is:

$A \to IV$
$B \to III$
$C \to II$
$D \to I$

This corresponds to Option D.

Comparison with stored answer

Stored correct answer = D

Our derived answer = D

So, the answer agrees with the stored answer.

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