Refer to the circuit diagram given below. The heat generated across the $6 \Omega$ resistance in 100 second is $\frac{\alpha}{100} \mathrm{~J}$. The value of $\alpha$ is $\_\_\_\_$ . (Nearest integer)

An unpolarized light of intensity $I_0$ passes through polarizer and then through a certain optically active solution and finally it goes to analyser. If the angle between analyser and polariser is $0^{\circ}$ and intensity of light emerged from analyser is $\frac{3}{8} I_0$, the angle of rotation of the light by the solution with respect to analyser is $\_\_\_\_$ degrees.

The energy released when $\frac{7}{17.13} \mathrm{~kg}$ of ${ }_3^7 \mathrm{Li}$ is converted into ${ }_2^4 \mathrm{He}$ by proton bombardment is $\alpha \times 10^{32} \mathrm{eV}$. The value of $\alpha$ is $\_\_\_\_$ . (Nearest integer) (Mass of ${ }_3^7 \mathrm{Li}=7.0183 \mathrm{u}$, mass of ${ }_2^4 \mathrm{He}=4.004 \mathrm{u}$, mass of proton $=1.008 \mathrm{u}$ and $1 \mathrm{u}=931 \mathrm{MeV} / \mathrm{c}^2$ and Avogadro number $=6.0 \times 10^{23}$ )

A three coulomb charge moves from the point $(0,-2,-5)$ to the point $(5,1,2)$ in an electric field expressed as $\vec{E}=2 x \hat{\mathrm{i}}+3 \mathrm{y}^2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \mathrm{N} / \mathrm{C}$. The work done in moving the charge is $\_\_\_\_$ J.