Physics

$F=qE\left[E=\frac{\sigma }{{\epsilon }_0}\right]$

$F=\frac{q\sigma }{{\in }_0}=10N$

Final case – Now one plate is removed, Electric field will becomes half; So, force will become half.

$F\text{'}=\frac{10}{2}=5N$

$\lambda =\frac{h}{mv}=\frac{h}{\sqrt[]{2MK.E}}$

$\lambda \alpha \frac{1}{\sqrt[]{m}}$

$\Rightarrow {\lambda }_{\alpha }\sqrt[]{m_{\alpha }}={\lambda }_{C_{12}}\sqrt[]{m_{C_{12}}}$

$\Rightarrow \frac{{\lambda }_{\alpha }}{{\lambda }_{C_{12}}}=\sqrt[]{\frac{m_{C_{12}}}{m_{\alpha }}}=\sqrt[]{\frac{12}{4}}=\sqrt[]{3}:1$

Ultraviolet rays – sterilizing surgical instrument

Microwaves – Radar system

Infrared waves – Greenhouse effect

X-rays – Study crystal structure

Given, I = 5 sin (120πt), $\omega$ $$ = 120

$T=\frac{2\pi }{\omega }$

$T=\frac{2\pi }{120\pi }=\frac{1}{60}sec,\frac{T}{4}=\frac{1}{60*4}=\frac{1}{240}sec$

Given, x = 99

${\mu }_0=4\pi *10^{-7}Wb/A-m$

${}_{}$ ${\mu }_m=?$   (Permeability of the material)

${\mu }_m=4\pi *10^{-5}$

mg = pvg

mg = d₁vg           ……. (i)

F_b = weight of fluid displaced = d₂vg         …… (ii)

by NLM 1            F_b + F_v = mg

F_V = mg - F_b

$F_v=mg\left(1-\frac{d_2}{d_1}\right)$

Com (Conservation of Momentum)

Pi = Pf

$\Rightarrow m{u}_1+5m{u}_2=m{v}_1+5m{v}_2$

$\Rightarrow m{u}_1=m{v}_1+5m{v}_2$

${}_{}$ $\Rightarrow u_1+v_1+5v_2$  …….(i)

& $e=1=\frac{v_2-v_1}{U_1-U_2}=\frac{v_2-v_1}{u_1}$ $\frac{{}_{}{}_{}}{{}_{}{}_{}}\frac{{}_{}{}_{}}{{}_{}}$

${}_{}{}_{}{}_{}$  $\Rightarrow u_1=v_2-v_1$ …….(ii)

Equation (i) + (ii)

$k.E_T=\frac{5}{9}{\left(K.E\right)}_i=55.6\mathrm{\%}$

The law of gravitation holds good for any pair of bodies in the universe.

This is correct statement.

The weight of any person becomes zero when the person is at the centre of the Earth.

Weight = mg

= m * 0 [g = 0 at the centre of the Earth]

= 0

$\Rightarrow I_{g}G=\left(l-l_g\right)8\Rightarrow l_g*72=\left(l-l_g\right)8\Rightarrow 9l_g=l-l_g$

$10l_g=l{l}_g=\frac{1}{10}l$

% of I which passes through galvanometer

$$ $l_g=10\mathrm{\%}$  of I

To keep,   $\Delta {\alpha }_{cm}=0$

      $M_1\Delta x_1+M_2\Delta X_2=0$          

$\Rightarrow 10*6+30\Delta X_2=0$      

$\Delta X_2=\frac{-10*6}{30}$

$\Delta X_2=-2cm$        

i.e. 2 cm towards the 10 kg block.