Class 11th

Use formula for M.I.

$x=4sin\left(\frac{\pi }{2}-\omega t\right)$  - (i)

$y=4sin\left(\omega t\right)$   - (ii)

From (i) and (ii) cos^{2 $\omega t+si{n}^2\omega t={\left(\frac{x}{4}\right)}^2+{\left(\frac{y}{4}\right)}^2=1$}

$\Rightarrow x^2+y^2={\left(4\right)}^2$  

For 2S, number of radial moles = $n-\mathcal{l}-1=2-0-1=1$ and ${\Psi }^2\left(r\right)$  will always be positive

$\therefore option\left(B\right)iscorrect?$ it has one radial node and it is positive.

a = k2rt2

$\Rightarrow \frac{v^2}{r}=k^{2}r{t}^2$

$\Rightarrow v=krt$

$a_t=\frac{dv}{dt}=kr$

? tangential force, F_t = ma_t = mkr

$\therefore Powerdelivered,P=F_{t}v=\left(mkr\right)\left(krt\right)=m{k}^2{r}^{2}t$  

 Note ® Power delivered by centripetal force will be zero.

Kindly go through the solution 

 $\frac{0.02858*0.112}{0.5702}=0.0561$

According to question, we can write

$\mathcal{l}=2\pi r\Rightarrow r=\frac{\mathcal{l}}{2\pi }$  

$I_1=\frac{m{\mathcal{l}}^2}{3},and{I}_2=\frac{m{r}^2}{2}=\frac{m{\mathcal{l}}^2}{8{\pi }^2}$

$\Rightarrow \frac{I_1}{I_2}=\frac{m{\mathcal{l}}^2}{3}*\frac{8{\pi }^2}{m{\mathcal{l}}^2}=\frac{8{\pi }^2}{3}$

According to Concept of resonance tube, we can write

$\frac{\lambda }{4}+e={\mathcal{l}}_{1}and\frac{3\lambda }{4}+e={\mathcal{l}}_2\Rightarrow \frac{\lambda }{2}={\mathcal{l}}_2-{\mathcal{l}}_1\Rightarrow \frac{V}{2\nu }={\mathcal{l}}_2-{\mathcal{l}}_1$  

$\Rightarrow {\mathcal{l}}_2=\frac{v}{2\nu }+{\mathcal{l}}_1=\frac{336}{400}+0.20=0.84+0.20=1.04m=104cm$  

Since velocity does not change, so acceleration will be zero.

mg = F_B + F_v $\Rightarrow \frac{4\pi }{3}{r}^3\rho g=\frac{4\pi }{3}{r}^3\sigma g+6\pi \eta rv$  

$\Rightarrow v=\frac{2r^2\left(\rho -\sigma \right)g}{9\eta }=\frac{2*0.1*0.1*10^{-6}*\left(10^4-10^3\right)*10}{9*1.0*10^{-5}}$  

$\Rightarrow h=\frac{400}{2g}=20m$  

Least  count of Vernier = 0.1mm

$\Rightarrow$ Reading of Vernier Scale = 5 * 0.1 = 0.5mm

The corrected diameter of sphere = Main Scale Reading + Vernier Scale reading + Zero correction = 1.7 + 0.05 + 0.05 = 1.8cm = 180 * 10^-2 cm.

According to question, we can write

$R=\frac{u^{2}sin2\theta }{g}\Rightarrow u^2=g{R}_{max},where\theta =45°$

$\Rightarrow H_{max}=\frac{u^2}{2g}=\frac{g{R}_{max}}{2g}=\frac{R_{max}}{2}=\frac{100}{2}=50m$