QJEEM25Phf7e0: JEE Main 2025 Physics - Class 11 JEEM25Phf7e0 Solution: Match List - I with List - II . List - I List - II (A) Young’s…

We know, Young's modulus, $E = {\sigma \over \varepsilon }$

$\Rightarrow E = {{{F \over A}} \over {{{\Delta L} \over L}}} = {F \over A}\left( {{L \over {\Delta L}}} \right)$

$\Rightarrow [E] = {{[F]} \over {[A]}} = {{\left[ {{M^1}{L^1}{T^{ - 2}}} \right]} \over {\left[ {{L^2}} \right]}}$

(A) $\Rightarrow [E] = [{M^1}{L^{ - 1}}{T^{ - 2}}]$

(B) we know, Torque = $r\times F$

$[\tau ] = [r][F]$

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

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

(C) We know, $F = \mu A{u \over y}$

where, F = force, $\mu$ = coefficient of viscosity

A = area $\frac{u}{y}$ = rate of shear deformation

$\Rightarrow \mu = {{Fy} \over {Au}}$

$\Rightarrow [\mu ] = {{[{M^1}{L^1}{T^{ - 2}}][L]} \over {[{L^2}][L{T^{ - 1}}]}}$

$\Rightarrow [\mu ] = [{M^1}{L^{ - 1}}{T^{ - 1}}]$

(D) We know, $F = {{G{m_1}{m_2}} \over {{r^2}}}$ (Newton's law of gravitation)

$\Rightarrow G = {{F{r^2}} \over {{m_1}{m_2}}}$ < $\Rightarrow G = {{F{r^2}} \over {{m_1}{m_2}}}$

$\Rightarrow [G] = {{[{M^1}{L^1}{T^{ - 2}}][{L^2}]} \over {[{M^2}]}}$

$\Rightarrow [G] = [{M^{ - 1}}{L^3}{T^{ - 2}}]$

Hence, option 4 is correct.