Class 11th

${\overrightarrow{v}}_{BW}=4\sqrt[]{2}\left(cos45°\widehat{i}+sin45°\widehat{j}\right)=4\widehat{i}+4\widehat{j},and{\overrightarrow{v}}_{Wg}=0\widehat{i}-\widehat{j}$              

${\overrightarrow{v}}_{Bg}={\overrightarrow{v}}_{BW}+{\overrightarrow{v}}_{Wg}=\left(4\widehat{i}+4\widehat{j}\right)+\left(0\widehat{i}-\widehat{j}\right)=4\widehat{i}+3\widehat{j}$      

$\left|{\overrightarrow{v}}_{Bg}\right|=5m/s\Rightarrow SpeedofButterfly$

Magnitude of displacement of Butterfly = $\left|{\overrightarrow{v}}_{Bg}\right|*t=5*3=15m$       

Since process is isothermal, so

$W=nRT\mathcal{l}n\left(\frac{V_f}{V_i}\right)=1*8.3*300*\mathcal{l}n\left(\frac{4}{2}\right)=1*8.3*300*\mathcal{l}n\left(2\right)$       

$\Rightarrow W=1*8.3*300*0.6931\approx 1725.82J\approx 17258*10^{-1}J$

Assuming Tension in metal wire suspended from roof varies linearly and 0 and T₀ ( developed due its own weight) are tensions are ends of wire of length $\mathcal{l}.$ So

$T_1\alpha \left({\mathcal{l}}_1-\mathcal{l}\right),and{T}_2\alpha \left({\mathcal{l}}_2-\mathcal{l}\right)$

$\frac{T_1}{T_2}=\frac{{\mathcal{l}}_1-\mathcal{l}}{{\mathcal{l}}_2-\mathcal{l}}\Rightarrow \mathcal{l}=\frac{T_1{\mathcal{l}}_2-T_2{\mathcal{l}}_1}{T_1-T_2}$

${?}^2={?}_0^2+2???{\left(\frac{2?}{60}*1800\right)}^2={\left(\frac{2?}{60}*600\right)}^2+2\left(\frac{2?}{60}*\frac{1200}{10}\right)?$

$?{\left(60?\right)}^2?{\left(20?\right)}^2=2\left(4?\right)???=\frac{80?*40?}{2*4?}=400?\text{?}\text{?}rad$

Number of rotations = $\frac{?}{2?}=\frac{400?}{2?}=200$

Consider translational motion of a body

mg sinq - f_S = ma             .(i)

Consider rotational motion of body about its centre of mass

$f_{S}R=l\alpha \Rightarrow \alpha =\frac{f_{S}R}{l}$ .(ii)

Using the condition of rolling without slipping at contact, we have

$\alpha R=a\Rightarrow \frac{f_S{R}^2}{l}=\frac{mgsin\theta -f_S}{m}\Rightarrow f_S=\frac{lmgsin\theta }{l+m{R}^2}$

Using the condition of static friction, we have

$f_S\le \mu N\Rightarrow \frac{lmgsin\theta }{l+m{R}^2}\le \mu mgcos\theta \Rightarrow \mu \ge \frac{ltan\theta }{l+m{R}^2}...........\left(iii\right)$

$\alpha =\frac{mgRsin\theta }{l+m{R}^2}\Rightarrow a=\alpha R=\frac{mg{R}^{2}sin\theta }{l+m{R}^2}=\frac{g{R}^{2}sin\theta }{k^2+R^2}...........\left(iv\right)$

$t=\sqrt[]{\frac{2\mathcal{l}}{a}}\Rightarrow$ Time required to reach the ground

$\Rightarrow \frac{v_{ring}}{v_{cylinder}}=\sqrt[]{\frac{2g\mathcal{l}{R}^{2}sin\theta }{k_{ring}^2+R^2}}*\sqrt[]{\frac{k_{cylinder}^2}{2g\mathcal{l}{R}^{2}sin\theta }}=\sqrt[]{\frac{\frac{R^2}{2}+R^2}{R^2+R^2}}=\frac{\sqrt[]{3}}{2}$

N_a -> Avogadro's Number

$J{\beta }^2\equiv \left[\mu kR\right]\Rightarrow {\beta }^2\equiv \left[kR\right]\Rightarrow \beta \equiv \left[k\right]\equiv \left[M{L}^2{T}^{-2}{K}^{-1}\right]$

, and

-> $S\equiv {\alpha }^2\beta \Rightarrow \left[M{L}^2{T}^{-2}{K}^{-1}\right]\equiv {\alpha }^2\left[M{L}^2{T}^{-2}{K}^{-1}\right]\Rightarrow \alpha \equiv \left[M^0{L}^2{T}^0{K}^0\right]$ Dimensionless

This is because acceleration depends on both force and mass (F = ma).

We know that from Newton's Third Law. While action-reaction forces are always equal, the objects they act on usually have very different masses.

If you consider a falling stone and the Earth as an action-reaction pair. The Third Law tells us,  

• Earth pulls stone down with force F

• Stone pulls Earth up with equal force F

• But Earth's mass is enormous, so its acceleration is tiny (F/huge mass = tiny acceleration)

• Stone's mass is small, so its acceleration is large (F/small mass = large acceleration)

• Result: Stone falls noticeably, Earth's motion is unnoticeable.

The third law of motion by Newton may confuse you into thinking that action causes reaction in sequence. This is incorrect.

In reality, both forces exist at the exact same moment. When you push a wall, your hand pushes the wall. At the same time, the wall pushes back simultaneously. Not one, then the other. Just note that there's no time delay between them.

Action-reaction forces act on different objects. That's why they don't or cannot cancel out.

For instance, when you push a wall, you can observe two things. 

• Your hand pushes the wall (action)

• The wall pushes your hand (reaction)

These are equal and opposite. But a close scientific examination will tell you that they are indeed acting on different things. To find your hand's motion, only consider the force on your hand (wall pushing back). The force from your hand affects the wall's motion, which is not yours.