(A) $1$. Spring constant $(k)$: $F = kx \implies [k] = [F]/[x] = (MLT^{-2}) / L = MT^{-2}$. Thus,$A-II$.$2$. Angular speed $(\omega)$: $\omega = \Delt

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$(A)-(II), (B)-(I), (C)-(IV), (D)-(III)$

(A) $1$. Spring constant $(k)$: $F = kx \implies [k] = [F]/[x] = (MLT^{-2}) / L = MT^{-2}$. Thus,$A-II$.$2$. Angular speed $(\omega)$: $\omega = \Delta\theta / \Delta t \implies [\omega] = [1] / T = T^{-1}$. Thus,$B-I$.$3$. Angular momentum $(L)$: $L = mvr \implies [L] = M(LT^{-1})L = ML^2T^{-1}$. Thus,$C-IV$.$4$. Moment of inertia $(I)$: $I = mr^2 \implies [I] = ML^2$. Thus,$D-III$.Therefore,the correct matching is $(A)-(II), (B)-(I), (C)-(IV), (D)-(III)$.

Class 11 Physics Units and Measurements Dimensions and Dimensional Formula

Medium

Match List $I$ with List $II$:
List $I$ List $II$
$A$. Spring constant $I$. $(T^{-1})$
$B$. Angular speed $II$. $(MT^{-2})$
$C$. Angular momentum $III$. $(ML^2)$
$D$. Moment of inertia $IV$. $(ML^2T^{-1})$

Choose the correct answer from the options given below:

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A
$(A)-(II), (B)-(I), (C)-(IV), (D)-(III)$
B
$(A)-(IV), (B)-(I), (C)-(III), (D)-(II)$
C
$(A)-(II), (B)-(III), (C)-(I), (D)-(IV)$
D
$(A)-(I), (B)-(III), (C)-(II), (D)-(IV)$

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Different physical quantities are given in Column-$I$ and their dimensional formulas are given in Column-$II$. Match them appropriately.

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According to Newton,the viscous force acting between liquid layers of area $A$ and velocity gradient $\Delta v/\Delta z$ is given by $F = - \eta A \frac{\Delta v}{\Delta z}$,where $\eta$ is a constant called the coefficient of viscosity. The dimensions of $\eta$ are:

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List $I$	List $II$
$A$. Spring constant	$I$. $(T^{-1})$
$B$. Angular speed	$II$. $(MT^{-2})$
$C$. Angular momentum	$III$. $(ML^2)$
$D$. Moment of inertia	$IV$. $(ML^2T^{-1})$

Match List $I$ with List $II$:List $I$List $II$$A$. Spring constant$I$. $(T^{-1})$$B$. Angular speed$II$. $(MT^{-2})$$C$. Angular momentum$III$. $(ML^2)$$D$. Moment of inertia$IV$. $(ML^2T^{-1})$Choose the correct answer from the options given below:

Similar Questions

If $E$ and $G$ respectively denote energy and gravitational constant,then $\frac{E}{G}$ has the dimensions of :

The focal power of a lens has the dimensions

Different physical quantities are given in Column-$I$ and their dimensional formulas are given in Column-$II$. Match them appropriately. Column-$I$ Column-$II$ $(a)$ Viscous force $(i)$ $[M^1 L^1 T^{-2}]$ $(b)$ Coefficient of viscosity $(ii)$ $[M^1 L^{-1} T^{-1}]$ $(iii)$ $[M^1 L^{-1} T^{-2}]$

The dimensional formula of relative density is

According to Newton,the viscous force acting between liquid layers of area $A$ and velocity gradient $\Delta v/\Delta z$ is given by $F = - \eta A \frac{\Delta v}{\Delta z}$,where $\eta$ is a constant called the coefficient of viscosity. The dimensions of $\eta$ are:

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Match List $I$ with List $II$:
List $I$ List $II$
$A$. Spring constant $I$. $(T^{-1})$
$B$. Angular speed $II$. $(MT^{-2})$
$C$. Angular momentum $III$. $(ML^2)$
$D$. Moment of inertia $IV$. $(ML^2T^{-1})$

Choose the correct answer from the options given below:

Different physical quantities are given in Column-$I$ and their dimensional formulas are given in Column-$II$. Match them appropriately.

Column-$I$ Column-$II$

$(a)$ Viscous force $(i)$ $[M^1 L^1 T^{-2}]$

$(b)$ Coefficient of viscosity $(ii)$ $[M^1 L^{-1} T^{-1}]$

$(iii)$ $[M^1 L^{-1} T^{-2}]$