The dimension of the ratio of magnetic flux and the resistance is equal to that of :
The dimension of the ratio of magnetic flux and the resistance is equal to that of :
- A
induced $emf$
- B
charge
- C
inductance
- D
current
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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 .............. $10^{-1}\mathrm{N/m}^{2}$
Let $[{\varepsilon _0}]$ denotes the dimensional formula of the permittivity of the vacuum and $[{\mu _0}]$ that of the permeability of the vacuum. If $M = {\rm{mass}}$, $L = {\rm{length}}$, $T = {\rm{Time}}$ and $I = {\rm{electric current}}$, then
Let $[{\varepsilon _0}]$ denotes the dimensional formula of the permittivity of the vacuum and $[{\mu _0}]$ that of the permeability of the vacuum. If $M = {\rm{mass}}$, $L = {\rm{length}}$, $T = {\rm{Time}}$ and $I = {\rm{electric current}}$, then
- [IIT 1998]
The volume of a liquid flowing out per second of a pipe of length $l$ and radius $r$ is written by a student as $V\, = \,\frac{{\pi p{r^4}}}{{8\eta l}}$ where $p$ is the pressure difference between the two ends of the pipe and $\eta $ is coefficent of viscosity of the liquid having dimensional formula $[M^1L^{-1}T^{-1}] $. Check whether the equation is dimensionally correct.
The volume of a liquid flowing out per second of a pipe of length $l$ and radius $r$ is written by a student as $V\, = \,\frac{{\pi p{r^4}}}{{8\eta l}}$ where $p$ is the pressure difference between the two ends of the pipe and $\eta $ is coefficent of viscosity of the liquid having dimensional formula $[M^1L^{-1}T^{-1}] $. Check whether the equation is dimensionally correct.
$A$ and $B$ possess unequal dimensional formula then following operation is not possible in any case:-
$A$ and $B$ possess unequal dimensional formula then following operation is not possible in any case:-