Question

JEE MAINS PHYSICS QUESTION SEPTEMBER 2020 Dimensional formula of thermal conductivity will be?

Answer 1

Answer.

$\sf \to \: dimensional \: formula \: of \: thermal \: conductivity = M {}^{ - 1} L {}^{ - 1} T {}^{ - 3} \theta {}^{ - 1}$

Option [ 1 ] is correct answer.

EXPLANATION.

$\sf \to \: formula \: of \: thermal \: conductivity \: \\ \\ \sf \to \: \theta \: = \frac{Q \Delta \: x}{A \Delta T t} \\ \\ \sf \to \: \theta \: = \frac{(heat)(length)}{(area)(temperature)(time)}$

$\sf \to \: dimensional \: formula \\ \\ \sf \to \: heat \: = q \: = (ml {}^{2}t {}^{ - 2} ) \\ \\ \sf \to \: length \: = l \\ \\ \sf \to \: area \: = {l}^{2} \\ \\ \sf \to \: temperature \: = \theta\\ \\ \sf \to \: time \: = t$

$\sf \to \: \theta \: = \dfrac{(ml {}^{2}t {}^{ - 2})(l) }{( {l}^{2})( \theta)(t) } \\ \\ \sf \to \: \theta \: = ml {t}^{ - 3} { \theta}^{ - 1}$

Answer 2

Answer.

$\sf \to \: dimensional \: formula \: of \: thermal \: conductivity = M {}^{ - 1} L {}^{ - 1} T {}^{ - 3} \theta {}^{ - 1}$

Option [ 1 ] is correct answer.

EXPLANATION.

$\sf \to \: formula \: of \: thermal \: conductivity \: \\ \\ \sf \to \: \theta \: = \frac{Q \Delta \: x}{A \Delta T t} \\ \\ \sf \to \: \theta \: = \frac{(heat)(length)}{(area)(temperature)(time)}$

$\sf \to \: dimensional \: formula \\ \\ \sf \to \: heat \: = q \: = (ml {}^{2}t {}^{ - 2} ) \\ \\ \sf \to \: length \: = l \\ \\ \sf \to \: area \: = {l}^{2} \\ \\ \sf \to \: temperature \: = \theta\\ \\ \sf \to \: time \: = t$

$\sf \to \: \theta \: = \dfrac{(ml {}^{2}t {}^{ - 2})(l) }{( {l}^{2})( \theta)(t) } \\ \\ \sf \to \: \theta \: = ml {t}^{ - 3} { \theta}^{ - 1}$