ncert solutions physics class 11th

Yes, there are many numerical problems in class 11 Physics. All important formulas must be on figure tips.

Class 11 Physics Chapter 11 is Thermodynamics. It is one of the most important topic in Physics. 

There are three main processes Isothermal, adiabatic and cyclic process. In isothermal, the system is thermally conductive and the temperature remains constant. In adiabatic process, the system is thermally isolated and there is no change in heat temperature. The system returns to its initial stage

A frame of reference is a system that helps an observer measure the position, speed, and time of an object. It is important because motion is always relative to something else. Without a frame of reference, we cannot describe how an object is moving or if it is moving at all. It provides a fixed vie

For students using the Physics NCERT class 11 solution popular and helpful books for similar content and problem solving practice include concepts of Physics by HC Verma understanding Physics by DC Pandey and Fundamentals of Physics by halliday Resnick and Walker additionally NCERT exemplar problems

4.22

Let us consider a vector $\stackrel{?}{P}$ . The equation can be written as

Px = Py = 1 $\left|\stackrel{?}{P}\right|$ = $\surd (P_x^2+P_y^2)$ $\left|\stackrel{?}{P}\right|$ = $\surd (1^2+1^2)\left|\stackrel{?}{P}\right|$ = $\sqrt{2}$ …….(i)

So the magnitude of vector $\stackrel{?}{i}$ + $\stackrel{?}{j}$ = $\sqrt{2}$

Let $\theta$ be the angle made by vector $\stackrel{?}{P}$ , with the x axis as given in the above fi

4.32

(a) Let $\mathit{\theta }\left(\mathit{t}\right)=\mathit{}{\mathbf{tan}}^{-1}?\left(\frac{{\mathit{v}}_{\mathit{o}\mathit{y}}-\mathit{}\mathit{g}\mathit{t}}{{\mathit{v}}_{\mathit{o}\mathit{x}}}\right)$ be the angle at which the projectile is fired w.r.t. the x-axis, since ${\mathit{\theta }}_0=\mathit{}{\mathbf{tan}}^{-1}?\left(\frac{4{\mathit{h}}_{\mathit{m}}}{\mathit{R}}\right)$ depends on t

Therefore tan $\theta$ since (Vy=Voy -gt and Vx = Vox)

(b) Since $\theta$ = $\theta \left(t\right)=\frac{Vx}{Vy}=\frac{Voy-gt}{Vox}$ sin2 $\theta \left(t\right)={\tan }^{-1}?\left(\frac{Voy-gt}{Vox}\right)$ /2g…….(1)

R = $h_{max}$ sin2 $u^2$ /g …….(2)

Dividing (1) by (2)

( $\theta$ /R) =