Physics

The principle of circular motion follows Newtonian mechanics. It can be stated that an object will move in a circle only when there is a centripetal force that continuously acts perpendicular to its tangential velocity. That changes the direction but not its speed. The inward acceleration a_c = v squared/r obeys Newton's laws and ensures circular path motion. 

Circular motion has the following characteristics.

• Fixed radius with the same and constant distance from the centre
• Constant speed, when the motion is uniform
• A changing velocity vector with continuous direction of change
• Centripetal acceleration that is directed radially inwards, while the force that causes it is the centripetal force that maintains the curved path

$\mathrm{W}={\mathrm{Q}}_1-{\mathrm{Q}}_2={\mathrm{Q}}_2-{\mathrm{Q}}_3$

${\mathrm{Q}}_2=\frac{{\mathrm{Q}}_1+{\mathrm{Q}}_3}{2}$

$2=\frac{{\mathrm{Q}}_1}{{\mathrm{Q}}_2}+\frac{{\mathrm{Q}}_3}{{\mathrm{Q}}_2}$

$2=\frac{T_1}{T}+\frac{T_2}{T}$

$\mathrm{T}=\frac{{\mathrm{T}}_1+{\mathrm{T}}_2}{2}$

At any time  $t$

$?=\mathrm{B}\mathrm{A}\mathrm{c}\mathrm{o}\mathrm{s}?\frac{\pi \mathrm{t}}{5}\mathrm{}\left(?\omega =\frac{2\pi }{\mathrm{T}}=\frac{2\pi }{10}=\frac{\pi }{5}\right)$

$\frac{\mathrm{d}?}{\mathrm{d}\mathrm{t}}=-\frac{\pi }{5}\mathrm{B}\mathrm{A}\mathrm{s}\mathrm{i}\mathrm{n}?\left(\frac{\pi \mathrm{t}}{5}\right)$

$\mathrm{e}=\frac{\pi }{5}\mathrm{B}\mathrm{A}\mathrm{s}\mathrm{i}\mathrm{n}?\left(\frac{\pi \mathrm{t}}{5}\right)$

e will be maximum when $\frac{\pi \mathrm{t}}{5}$  is $\frac{\pi }{2}$ .

$\mathrm{t}=2.5\mathrm{s}\mathrm{e}\mathrm{c}$
e will be minimum when $\frac{\pi \mathrm{t}}{5}$ is $\pi ,0$

$\mathrm{t}=\mathrm{0,5}\mathrm{s}\mathrm{e}\mathrm{c}$

So, answer will be (1).

Moment of inertia of ring

$\mathrm{d}\mathrm{l}={\mathrm{d}\mathrm{m}\mathrm{r}}^2$

$\mathrm{I}={\int }_0^{\mathrm{a}}?(\mathrm{A}+\mathrm{B}\mathrm{r})2\pi \mathrm{r}\mathrm{d}\mathrm{r}\pi {\mathrm{r}}^2$

$=2\pi A{\int }_0^a?r^{3}dr+2\pi B{\int }_0^a?r^{4}dr$

 

$=2\pi \left. [A\frac{a^4}{4}+B\frac{a^5}{5}\right]$

$=2\pi {\mathrm{a}}^4\left. [\frac{\mathrm{A}}{4}+\frac{\mathrm{B}\mathrm{a}}{5}\right]$

Kindly consider the following answer

$\overrightarrow{\mu }=-\frac{e}{2m}\overrightarrow{L}$

$f_1=f_o\left(\frac{v}{v-v_s}\right)\mathrm{}{f}_1-f_2=2$

$f_2=f_o\left(\frac{v}{v+v_s}\right)\mathrm{}{f}_{0}v\left[\frac{v+v_s-\left(v-v_s\right)}{v^2-v_s^2}\right]=2;\mathrm{}\frac{2f_{0}v{v}_s}{v^2-v_s^2}=2$

$?{\mathrm{v}}_{\mathrm{s}}?\mathrm{v}\mathrm{}\therefore {\mathrm{v}}_{\mathrm{s}}=\frac{1}{4}\mathrm{m}/\mathrm{s}$

mvr = $\frac{nh}{2\pi }$ $\frac{}{}$

So momentum (mv) = $\frac{nh}{2\pi r}$

 ${}_{}\frac{}{}$ $K.E_1=\frac{1230}{800}-?$ - (i)

${}_{}{}_{}\frac{}{}$  $K.E_2=2K.E_1=\frac{1230}{500}-?$ - (ii)

from (i) & (ii)

$$ $?=0.615$  ev