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Mechanical Properties of Fluids formulas

Class 11 physics formula sheet for NEET and JEE: the key equations of NCERT chapter 9, the special cases questions are built on, and diagrams where they help.

21 formulas4 sectionsClass 11 · Chapter 91 of 4 sections free

By Sreeraj P, M.Sc Physics · 10+ years teaching NEET and JEE

Most used formulasOther formulas and cases

1Pressure and hydrostatics

$$\begin{array}{l}\displaystyle P=P_0+\rho gh,\quad P_{\text{gauge}}=\rho gh\\[5pt]\displaystyle 1\ \text{atm}=1.013\times10^5\ \text{Pa}=76\ \text{cm Hg}\end{array}$$
$$\begin{array}{l}\displaystyle P=P_0+\rho gh\\[6pt]\displaystyle P_{\text{gauge}}=\rho gh\\[6pt]\displaystyle 1\ \text{atm}=1.013\times10^5\ \text{Pa}=76\ \text{cm Hg}\end{array}$$

Same at the same level in a connected liquid at rest; independent of vessel shape (hydrostatic paradox). Force on the base $=\rho ghA$, may differ from liquid weight. Barometer: $h=\dfrac{P_0}{\rho g}$ (10.3 m of water).

CaseResult
U-tube with two immiscible liquids$\rho_1h_1=\rho_2h_2$ (heights above the interface level)
Force on a vertical wall (width $b$, depth $H$)$F=\tfrac12\rho gbH^2$, acting at $H/3$ from the bottom; torque about base $\tfrac16\rho gbH^3$
Container accelerating horizontally ($a$)surface tilts: $\tan\theta=a/g$
Container accelerating up / down$P=P_0+\rho(g\pm a)h$; free fall: no pressure difference
Rotating about vertical axisparaboloid: $h=\dfrac{\omega^2r^2}{2g}$
U-tube accelerating horizontally (arms $L$ apart)level difference $\dfrac{aL}g$
Rotating U-tube (one arm on axis)$\Delta h=\dfrac{\omega^2L^2}{2g}$
$$\frac{F_1}{A_1}=\frac{F_2}{A_2}\ (\text{Pascal}),\qquad F_B=\rho_lVg\ (\text{Archimedes})$$
$$\begin{array}{l}\displaystyle \frac{F_1}{A_1}=\frac{F_2}{A_2}\ (\text{Pascal})\\[6pt]\displaystyle F_B=\rho_lVg\ (\text{Archimedes})\end{array}$$

Hydraulic lift, brakes: force multiplied, work not ($F_1d_1=F_2d_2$). Floating: $\dfrac{V_{\text{in}}}V=\dfrac{\rho}{\rho_l}$. Apparent weight $W\left(1-\dfrac{\rho_l}\rho\right)$; relative density $=\dfrac{W}{W-W_{\text{in water}}}$.

Floating caseResult
Body between two liquids$\rho V=\rho_1V_1+\rho_2V_2$
Ice floating meltswater level unchanged (stone inside ice: level falls; wood/air: unchanged)
Lift acceleratingfraction immersed unchanged
Hollow sphere just floats (radii $r$, $R$)$\rho\left(R^3-r^3\right)=\rho_lR^3$
Body released at depth $h$ ($\rho<\rho_l$)$a=g\left(\dfrac{\rho_l}\rho-1\right)$; rises above the surface to $h\left(\dfrac{\rho_l}{\rho}-1\right)$ (no drag)
Beaker on a balance, body hung insidereading increases by the buoyant force
Floating block pushed down and releasedSHM, $T=2\pi\sqrt{h/g}$

3 more sections and 16 formulas in the full chapter

  1. 2Fluid flow2 formulas · 1 case table · 1 diagram
  2. 3Viscosity8 formulas
  3. 4Surface tension6 formulas · 1 case table

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