Match List - I with List - II : List I List II (a) h (Planck's constant) (i) [MLT^ - 1] (b) E (kinetic energy) (ii) [ML^2T^ - 1] (c) V (electric potential) (iii) [ML^2T^ - 2] (d) P (linear momentum) (iv) [ML^2I^ - 1T^ - 3] Choose the correct answer from the options given below :

(a) \to (ii), (b) \to (iii), (c) \to (iv), (d) \to (i)

(a) \to (i), (b) \to (ii), (c) \to (iv), (d) \to (iii)

(a) \to (iii), (b) \to (ii), (c) \to (iv), (d) \to (i)

(a) \to (iii), (b) \to (iv), (c) \to (ii), (d) \to (i)

Units and Measurements PYQ — JEE Main 2021 · 25 Feb · Shift 1 · Q51 with Solution

Match List - I with List - II :

Choose the correct answer from the options given below :

View written solutionFree

Correct answer: A

Find dimensions of each physical quantity

(a) Planck's constant, $h$

Using $E = h\nu$ , $h = \frac{E}{\nu}$ Dimensions of energy: $[E] = [ML^2T^{-2}]$ Dimensions of frequency: $[\nu] = [T^{-1}]$ Therefore, $[h] = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$ So, $(a) \to (ii)$ .
(b) Kinetic energy, $E$

Kinetic energy has dimensions of energy: $[E] = [ML^2T^{-2}]$ So, $(b) \to (iii)$ .
(c) Electric potential, $V$

Electric potential is work done per unit charge: $V = \frac{W}{q}$ Dimensions of work: $[W] = [ML^2T^{-2}]$ Dimensions of charge: $[q] = [IT]$ Therefore, $[V] = \frac{[ML^2T^{-2}]}{[IT]} = [ML^2I^{-1}T^{-3}]$ So, $(c) \to (iv)$ .
(d) Linear momentum, $P$

$P = mv$ Therefore, $[P] = [M][LT^{-1}] = [MLT^{-1}]$ So, $(d) \to (i)$ .
Final matching
$(a) \to (ii),\quad (b) \to (iii),\quad (c) \to (iv),\quad (d) \to (i)$
Compare with options

This corresponds to Option A.
Comparison with stored correct answer

Stored correct answer = A

Our derived answer = A

Hence, the stored answer is correct.

Match List - I with List - II :

Choose the correct answer from the options given below :

View written solutionFree

Correct answer: A

Find dimensions of each physical quantity

(a) Planck's constant, $h$

Using $E = h\nu$ , $h = \frac{E}{\nu}$ Dimensions of energy: $[E] = [ML^2T^{-2}]$ Dimensions of frequency: $[\nu] = [T^{-1}]$ Therefore, $[h] = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$ So, $(a) \to (ii)$ .
(b) Kinetic energy, $E$

Kinetic energy has dimensions of energy: $[E] = [ML^2T^{-2}]$ So, $(b) \to (iii)$ .
(c) Electric potential, $V$

Electric potential is work done per unit charge: $V = \frac{W}{q}$ Dimensions of work: $[W] = [ML^2T^{-2}]$ Dimensions of charge: $[q] = [IT]$ Therefore, $[V] = \frac{[ML^2T^{-2}]}{[IT]} = [ML^2I^{-1}T^{-3}]$ So, $(c) \to (iv)$ .
(d) Linear momentum, $P$

$P = mv$ Therefore, $[P] = [M][LT^{-1}] = [MLT^{-1}]$ So, $(d) \to (i)$ .
Final matching
$(a) \to (ii),\quad (b) \to (iii),\quad (c) \to (iv),\quad (d) \to (i)$
Compare with options

This corresponds to Option A.
Comparison with stored correct answer

Stored correct answer = A

Our derived answer = A

Hence, the stored answer is correct.