Class 11th

Rotational motion in Class 11 Physics is regarded as a challenging chapter, particularly due to the transition from point particles to extended bodies. Additionally, the concepts covered here are analogous to, yet distinct from, linear motion. Even vector calculations here can be difficult. But the key to mastering this chapter is to be conceptually clear with your previous chapters in Physics. You also need to be able to visualise rotational motion instead of memorising definitions, to be able to master it.  

Yes, rotational motion is important for the NEET exam. Expect and prepare for, at least, two to three questions on rotational motion as per the latest NEET Physics syllabus for Mechanics. Questions from this chapter are clubbed into the application of principles from other areas of mechanics, such as Laws of Motion, Work, Energy, and Power, and Conservation of Momentum. 

${\overrightarrow{r}}_1=2\widehat{i}+4\widehat{j}-2\widehat{k},{\overrightarrow{r}}_2=\widehat{i}+2\widehat{j}+\alpha \widehat{k}$

Projection of ${\overrightarrow{r}}_1$ ${\stackrel{}{}}_{}$ on ${\overrightarrow{r}}_2$ ${\stackrel{}{}}_{}$ is zero $\Rightarrow {\overrightarrow{r}}_1.{\overrightarrow{r}}_2$ ${\stackrel{}{}}_{}{\stackrel{}{}}_{}$ = 0

2 + 8 - 2 = 0=> α= 5

$N_2{O}_4?2N{O}_2$

$\underset{=0.5mol}{\left(1-0.5\right)}$ $2*0.5=1mol$

$k_P=\frac{{\left(\frac{1}{1.5}*1\right)}^2}{\left(\frac{0.5}{1.5}*1\right)}$

$=\frac{4}{3}=1.33$

$=-710.15J/mol$

1000 ml solution contains 0.02 milli mole (mm)

$\therefore$ 500 ml solution contains 0.02 m.m

$\therefore$ Solution made 1000 ml with H₂O

$\therefore$ m.m in final solution = 0.01 mm

Solution (A) + 0.01 m. m H₂SO₄

= 0.01 + 0.01

= 0.02 m.m

= 0.00002 * 10³ mm

$\Delta H=-165kJ/mole$

T =?

$\Delta S=-550J{K}^{-1}$

At equilibrium ; $\Delta G=0$

$\therefore T=\frac{\Delta H}{\Delta S}=\frac{-165*1000}{-550}K$

$\begin{array}{l}=3*100K\\ =300K\end{array}$

$E_{1_H}=-2.2*10^{-18}J$

$Li\to L{i}^{+2}+2e^{-},n=2$

$E_{L{i}^{+2}}=E_{1H}*\frac{Z^2}{n^2}=-2.2*10^{-18}*\frac{3^2}{2^2}$

$\lambda =4*10^{-8}$ m

Molar mass of C₇H₅N₃O₆ = 12 * 7 + 1 * 5 + 14 * 3 + 16 * 6 = 84 + 42 + 96 = 227 g/mol

Number of moles = $\frac{681}{227}$  = 3 mol

$\therefore$ number of N-atoms

$=3*6.02*10^{23}*3=9*6.02*10^{23}$

$=5418*10^{21}$

$\therefore x=5418$ 

$\frac{2}{?4}=\frac{?1}{2}$

y – 4 = 2 (x – 3)

y = 2x – 2

x² + (2x – 2)² = 25

$5x^2?8x?21=0$

$z\left(\frac{?7}{5},\frac{?24}{5}\right)$

$t_n=\frac{3^n-2^n}{3^n}=1-{\left(\frac{2}{3}\right)}^n$

$S_{100}=100-\frac{\frac{2}{3}\left({\left(\frac{2}{3}\right)}^{100}-1\right)}{\frac{2}{3}-1}=100+2.\frac{\left(2^{100}-3^{100}\right)}{3^{100}}$

$=100-2+\frac{2^{101}}{3^{100}}=98+{\left(\frac{2}{3}\right)}^{100}.2<98+1$