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(A) The dimension of impulse is $J = F 	imes \Delta t = [MLT^{-2}] 	imes [T] = [MLT^{-1}]$.Therefore,the dimension of "impulse per unit area" is $\f

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Viscosity

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(A) The dimension of impulse is $J = F 	imes \Delta t = [MLT^{-2}] 	imes [T] = [MLT^{-1}]$.Therefore,the dimension of "impulse per unit area" is $\frac{[MLT^{-1}]}{[L^2]} = [ML^{-1}T^{-1}]$.Now,let us check the dimensions of the given options:$A$. Viscosity (Coefficient of viscosity $\eta$): The force $F = \eta A \frac{dv}{dx}$,so $\eta = \frac{F}{A (dv/dx)} = \frac{[MLT^{-2}]}{[L^2] [LT^{-1}/L]} = [ML^{-1}T^{-1}]$.$B$. Surface tension: $[MT^{-2}]$.$C$. Bulk modulus: $[ML^{-1}T^{-2}]$.$D$. Force: $[MLT^{-2}]$.Thus,the unit of "impulse per unit area" is the same as that of viscosity.

List-$I$	List-$II$
$A$. Spring constant	$I$. $ML^2 T^{-2} K^{-1}$
$B$. Thermal conductivity	$II$. $ML^0 T^{-2}$
$C$. Boltzmann constant	$III$. $ML^2 T^{-3} A^{-2}$
$D$. Inductive reactance	$IV$. $MLT^{-3} K^{-1}$

List-$I$	List-$II$
$(i)$ Curie	$(A)$ $MLT^{-2}$
$(ii)$ Light year	$(B)$ $M$
$(iii)$ Dielectric strength	$(C)$ Dimensionless
$(iv)$ Atomic weight	$(D)$ $T$
$(v)$ Decibel	$(E)$ $ML^2T^{-2}$
	$(F)$ $MT^{-3}$
	$(G)$ $T^{-1}$
	$(H)$ $L$
	$(I)$ $MLT^{-3}I^{-1}$
	$(J)$ $LT^{-1}$

The unit of "impulse per unit area" is the same as that of

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