Class 11th

Following are the dihedral angles of H₂O₂ in gas phase and solid (Reverse with respect to question).

Change of angle is the changing of intermolecular interaction in different phase

Given electronic configuration is for Ga and in 5^th period diagonally situated element is Sn with respect to Ga. Hence electronic configuration of Sn is [Kr]4d¹⁰5s²5p².

70 g – N = 70 a = 70 * 0.2 = 14

->N = 700 – 14 = 686 N

For ascent

$t_a=\sqrt[]{\frac{2\mathcal{l}}{a_{ascent}}}=\sqrt[]{\frac{2\mathcal{l}}{g\left(sin\theta +\mu cos\theta \right)}}$

For descent

$t_d=\sqrt[]{\frac{2\mathcal{l}}{a_{descent}}}=\sqrt[]{\frac{2\mathcal{l}}{g\left(sin\theta -\mu cos\theta \right)}}$

According question, we can write

$\Rightarrow \mu =\frac{3}{5}tan\theta =\frac{3}{5}tan30°=\frac{\sqrt[]{3}}{5}$

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

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

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

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