In the relation : $\frac{d y}{d x}=2 \omega \sin \left(\omega t+\phi_0\right)$ the dimensional formula for $\left(\omega t+\phi_0\right)$ is :
In the relation : $\frac{d y}{d x}=2 \omega \sin \left(\omega t+\phi_0\right)$ the dimensional formula for $\left(\omega t+\phi_0\right)$ is :
- A
$MLT$
- B
$MLT ^0$
- C
$ML ^0 T ^0$
- D
$M ^0 L ^0 T ^0$
Similar Questions
To determine the Young's modulus of a wire, the formula is $Y = \frac{FL}{A\Delta L};$ where $L$ = length, $A = $area of cross-section of the wire, $\Delta L = $change in length of the wire when stretched with a force $F$. The conversion factor to change it from $CGS$ to $MKS$ system is .............. $10^{-1}\mathrm{N/m}^{2}$
Density of wood is $0.5 \;gm / cc$ in the $CGS$ system of units. The corresponding value in $MKS$ units is.?
In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, $[E]$ and $[B]$ stand for dimensions of electric and magnetic fields respectively, while $\left[\varepsilon_0\right]$ and $\left[\mu_0\right]$ stand for dimensions of the permittivity and permeability of free space respectively. $[L]$ and $[T]$ are dimensions of length and time respectively. All the quantities are given in $SI$ units.
($1$) The relation between $[E]$ and $[B]$ is
$(A)$ $[ E ]=[ B ][ L ][ T ]$ $(B)$ $[ E ]=[ B ][ L ]^{-1}[ T ]$ $(C)$ $[ E ]=[ B ][ L ][ T ]^{-1}$ $(D)$ $[ E ]=[ B ][ L ]^{-1}[ T ]^{-1}$
($2$) The relation between $\left[\varepsilon_0\right]$ and $\left[\mu_0\right]$ is
$(A)$ $\left[\mu_0\right]=\left[\varepsilon_0\right][ L ]^2[ T ]^{-2}$ $(B)$ $\left[\mu_0\right]=\left[\varepsilon_0\right][ L ]^{-2}[ T ]^2$ $(C)$ $\left[\mu_0\right]=\left[\varepsilon_0\right]^{-1}[ L ]^2[ T ]^{-2}$ $(D)$ $\left[\mu_0\right]=\left[\varepsilon_0\right]^{-1}[ L ]^{-2}[ T ]^2$
Give the answer or quetion ($1$) and ($2$)
In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, $[E]$ and $[B]$ stand for dimensions of electric and magnetic fields respectively, while $\left[\varepsilon_0\right]$ and $\left[\mu_0\right]$ stand for dimensions of the permittivity and permeability of free space respectively. $[L]$ and $[T]$ are dimensions of length and time respectively. All the quantities are given in $SI$ units.
($1$) The relation between $[E]$ and $[B]$ is
$(A)$ $[ E ]=[ B ][ L ][ T ]$ $(B)$ $[ E ]=[ B ][ L ]^{-1}[ T ]$ $(C)$ $[ E ]=[ B ][ L ][ T ]^{-1}$ $(D)$ $[ E ]=[ B ][ L ]^{-1}[ T ]^{-1}$
($2$) The relation between $\left[\varepsilon_0\right]$ and $\left[\mu_0\right]$ is
$(A)$ $\left[\mu_0\right]=\left[\varepsilon_0\right][ L ]^2[ T ]^{-2}$ $(B)$ $\left[\mu_0\right]=\left[\varepsilon_0\right][ L ]^{-2}[ T ]^2$ $(C)$ $\left[\mu_0\right]=\left[\varepsilon_0\right]^{-1}[ L ]^2[ T ]^{-2}$ $(D)$ $\left[\mu_0\right]=\left[\varepsilon_0\right]^{-1}[ L ]^{-2}[ T ]^2$
Give the answer or quetion ($1$) and ($2$)
- [IIT 2018]
The potential energy of a point particle is given by the expression $V(x)=-\alpha x+\beta \sin (x / \gamma)$. A dimensionless combination of the constants $\alpha, \beta$ and $\gamma$ is
The potential energy of a point particle is given by the expression $V(x)=-\alpha x+\beta \sin (x / \gamma)$. A dimensionless combination of the constants $\alpha, \beta$ and $\gamma$ is
- [KVPY 2012]
If velocity $v$, acceleration $A$ and force $F$ are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of $v,\,A$ and $F$ would be
If velocity $v$, acceleration $A$ and force $F$ are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of $v,\,A$ and $F$ would be