If the buoyant force $F$ acting on an object depends on its volume $V$ immersed in a liquid, the density $\rho$ of the liquid and the acceleration due to gravity $g$. The correct expression for $F$ can be
If the buoyant force $F$ acting on an object depends on its volume $V$ immersed in a liquid, the density $\rho$ of the liquid and the acceleration due to gravity $g$. The correct expression for $F$ can be
- A
$V \rho g$
- B
$\frac{\rho g}{V}$
- C
$\rho g V^2$
- D
$\sqrt{\rho g V}$
Similar Questions
$A$ 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:-
If velocity of light $c$, Planck’s constant $h$ and gravitational constant $G$ are taken as fundamental quantities, then express length in terms of dimensions of these quantities.
If velocity of light $c$, Planck’s constant $h$ and gravitational constant $G$ are taken as fundamental quantities, then express length in terms of dimensions of these quantities.
Time period $T\,\propto \,{P^a}\,{d^b}\,{E^c}$ then value of $c$ is given $p$ is pressure, $d$ is density and $E$ is energy
Time period $T\,\propto \,{P^a}\,{d^b}\,{E^c}$ then value of $c$ is given $p$ is pressure, $d$ is density and $E$ is energy
Given below are two statements: One is labelled as Assertion $(A)$ and other is labelled as Reason $(R)$.
Assertion $(A)$ : Time period of oscillation of a liquid drop depends on surface tension $(S)$, if density of the liquid is $p$ and radius of the drop is $r$, then $T = k \sqrt{ pr ^{3} / s ^{3 / 2}}$ is dimensionally correct, where $K$ is dimensionless.
Reason $(R)$: Using dimensional analysis we get $R.H.S.$ having different dimension than that of time period.
In the light of above statements, choose the correct answer from the options given below.
Given below are two statements: One is labelled as Assertion $(A)$ and other is labelled as Reason $(R)$.
Assertion $(A)$ : Time period of oscillation of a liquid drop depends on surface tension $(S)$, if density of the liquid is $p$ and radius of the drop is $r$, then $T = k \sqrt{ pr ^{3} / s ^{3 / 2}}$ is dimensionally correct, where $K$ is dimensionless.
Reason $(R)$: Using dimensional analysis we get $R.H.S.$ having different dimension than that of time period.
In the light of above statements, choose the correct answer from the options given below.
- [JEE MAIN 2022]
The dimensions of solar constant (energy falling on earth per second per unit area) are
The dimensions of solar constant (energy falling on earth per second per unit area) are