If energy $(E)$, velocity $(v)$and force $(F)$ be taken as fundamental quantity, then what are the dimensions of mass
If energy $(E)$, velocity $(v)$and force $(F)$ be taken as fundamental quantity, then what are the dimensions of mass
- A
$E{v^2}$
- B
$E{v^{ - 2}}$
- C
$F{v^{ - 1}}$
- D
$F{v^{ - 2}}$
Similar Questions
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
Sometimes it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity. In one such system, dimensions of different quantities are given in terms of a quantity $X$ as follows: [position $]=\left[X^\alpha\right] ;[$ speed $]=\left[X^\beta\right]$; [acceleration $]=\left[X^{ p }\right]$; [linear momentum $]=\left[X^{ q }\right]$; [force $]=\left[X^{ I }\right]$. Then -
$(A)$ $\alpha+p=2 \beta$
$(B)$ $p+q-r=\beta$
$(C)$ $p-q+r=\alpha$
$(D)$ $p+q+r=\beta$
Sometimes it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity. In one such system, dimensions of different quantities are given in terms of a quantity $X$ as follows: [position $]=\left[X^\alpha\right] ;[$ speed $]=\left[X^\beta\right]$; [acceleration $]=\left[X^{ p }\right]$; [linear momentum $]=\left[X^{ q }\right]$; [force $]=\left[X^{ I }\right]$. Then -
$(A)$ $\alpha+p=2 \beta$
$(B)$ $p+q-r=\beta$
$(C)$ $p-q+r=\alpha$
$(D)$ $p+q+r=\beta$
- [IIT 2020]
Match the following two coloumns
Column $-I$
Column $-II$
$(A)$ Electrical resistance
$(p)$ $M{L^3}{T^{ - 3}}{A^{ - 2}}$
$(B)$ Electrical potential
$(q)$ $M{L^2}{T^{ - 3}}{A^{ - 2}}$
$(C)$ Specific resistance
$(r)$ $M{L^2}{T^{ - 3}}{A^{ - 1}}$
$(D)$ Specific conductance
$(s)$ None of these
Match the following two coloumns
| Column $-I$ | Column $-II$ |
| $(A)$ Electrical resistance | $(p)$ $M{L^3}{T^{ - 3}}{A^{ - 2}}$ |
| $(B)$ Electrical potential | $(q)$ $M{L^2}{T^{ - 3}}{A^{ - 2}}$ |
| $(C)$ Specific resistance | $(r)$ $M{L^2}{T^{ - 3}}{A^{ - 1}}$ |
| $(D)$ Specific conductance | $(s)$ None of these |
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 .............. $10^{-1}\mathrm{N/m}^{2}$
In a system of units if force $(F)$, acceleration $(A) $ and time $(T)$ are taken as fundamental units then the dimensional formula of energy is
In a system of units if force $(F)$, acceleration $(A) $ and time $(T)$ are taken as fundamental units then the dimensional formula of energy is