If the buoyant force F acting on an object de | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(a)

$F \propto V^a \rho^b g^c$

$F=\left[L^3\right]^a\left[ ML ^{-3}\right]^b\left[ LT ^{-2}\right]^c$

$\left[ ML ^{-2}\right]=F=\left[ M ^{ b

Vedclass · Accepted Answer

$V \rho g$

Vedclass · Answer

(a)

$F \propto V^a \rho^b g^c$

$F=\left[L^3\right]^a\left[ ML ^{-3}\right]^b\left[ LT ^{-2}\right]^c$

$\left[ ML ^{-2}\right]=F=\left[ M ^{ b } L ^{3 a -3 b +c} T ^{-2 c}\right]$

On comparing.

$b=1,$

$-2 c=-2$

$\Rightarrow c=1$

$3 a-3 b+c=1$

$\Rightarrow 3 a-3+1=1$

$\Rightarrow 3 a-2=1$

$\Rightarrow 3 a=3 \Rightarrow a=1$

So, on putting all these values,

$F=V \rho g$

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

Similar Questions

If the present units of length. time and mass $(m, s, k g)$ are changed to $100\; m, 100\; s$. $\frac{1}{10} \;k g$ then

If force $(F)$, length $(L) $ and time $(T)$ are assumed to be fundamental units, then the dimensional formula of the mass will be

If ${E}, {L}, {m}$ and ${G}$ denote the quantities as energy, angular momentum, mass and constant of gravitation respectively, then the dimensions of ${P}$ in the formula ${P}={EL}^{2} {m}^{-5} {G}^{-2}$ are

The time dependence of a physical quantity $P$ is given by $ P = P_0 exp^{(-\alpha t^{2})} $ where $\alpha$ is a constant and $t$ is time. The constant $\alpha$

If $P$ represents radiation pressure, $c$ represents speed of light and $Q$ represents radiation energy striking a unit area per second, then non-zero integers $x,\,y$ and $z$ such that ${P^x}{Q^y}{c^z}$ is dimensionless, are