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The chemical bonds can be primary or strong bonds such as ionic, covalent, and metallic bonds. It can also be the secondary or weak bonds like hydrogen bonding, London dispersion force, and dipole interactions.

Rate of burning of fuel $\left(\frac{dm}{dt}\right)=1kg/s$  and velocity of ejected fuel

$\left(v\right)=60km/s$

$=60\times 10^{3}m/s.$  

Forcev =  Rate of change of momentum

$=\frac{dp}{dt}=\frac{d\left(m\nu \right)}{dt}=v\frac{dm}{dt}=\left(60\times 10^3\right)\times 1$

  $=60000$ N.

It is the bond that holds the ions, atoms or molecules to form stable substances, compounds and larger structures.

$\stackrel{?}{F}=e{E}_0\stackrel{?}{j}$

${\stackrel{?}{a}}_y=\frac{e{E}_0}{m}\stackrel{?}{j}$

$=v_0\stackrel{?}{i}+\left(\frac{e{E}_{0}t}{m}\right)\stackrel{?}{j}$

$\left|\stackrel{?}{v}\right|=\sqrt[]{v_0^2+{\left(\frac{e{E}_{0}t}{m}\right)}^2}=v_0\sqrt[]{1+{\left(\frac{e{E}_{0}t}{m}\right)}^2}$

$\lambda =\frac{{\lambda }_0}{\sqrt[]{1+{\left(\frac{e{E}_{0}t}{m}\right)}^2}}$

In Chemistry, the bonds are of four types - Covalent bond, Ionic bond, Hydrogen bond, and Metallic bond.

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Force $\left(F\right)=10N$  and acceleration

From Lenz law, current in the square loop will be in anticlockwise sense.

Direction of  is perpendicular to equipotential surface.

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For diamagnetic ; $\chi <0andl=\chi H$

At resonance V_L = V_C

_{$\Rightarrow cos?=\frac{V_R}{\sqrt[]{V_R^2+{\left(V_L-V_C\right)}^2}}=1$}

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Density of nuclei is order of 10¹⁷ kg m^-3.

Photocell works on photoelectric effect.

$U_i=-\frac{GMm}{R}$

${\left(T.E.\right)}_f=-\frac{GMm}{8R}$

$\Rightarrow Energy={\left(T.E\right)}_f-\left(U_i\right)$

$=\frac{7GMm}{8R}$

$\Rightarrow \frac{R}{100}=\frac{40}{60}$

$\Rightarrow \Delta R=\left(\frac{\Delta l}{l}+\frac{\Delta l}{\left(100-l\right)}\right)R$

$=\left(\frac{0.1}{40}+\frac{0.1}{60}\right)\times 66.7=0.27=0.3$

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$C_1=1\mu F{V}_1=100V?U_1=\frac{1}{2}\times 1\times {10}^{-6}\times {\left(100\right)}^2=5\times {10}^{-3}J$  

$C_2=1\mu F$ , common potential $V=\frac{C_1{V}_1}{C_1{C}_2}=\frac{100}{2}=50V$

Final electrostatic energy stored in both the capacitors.

$=\frac{1}{2}\left(C_1+C_2\right)V^2=\frac{1}{2}\times 2\times {\left(50\right)}^2\times {10}^{-6}$

$=2.5\times {10}^{-3}J$

$Energylost=2.5\times {10}^{-3}$

$\%lossofenergy=\frac{2.5\times {10}^{-3}}{5\times {10}^{-3}}\times 100$

$=50\%$