(i) Ajeet is intelligent, cooperative and has good command over Physics.
(ii) Rutherford used gold foil of 10^-8m thickness.
(iii) When Rutherford was performing  \(\alpha \) - particles scattering experiment, he observed that very less number of  \(\alpha \) - particles retrace their path. Thus, he concluded that there is a positive massive part at the centre of atom and named it nucleus.

(i) Here,
\(\lambda =600 \ nm=600\times { 10 }^{ -9 }m=6\times { 10 }^{ -7 }m\)
\(\theta ={ 0.1 }^{ \circ }=\frac { 0.1\pi }{ 180 } rad,d=?\)
From angular width,\(\theta =\frac { \lambda }{ d } \)
\(\Rightarrow \ d=\frac { \lambda }{ \theta } =\frac { 6\times { 10 }^{ -7 } }{ \frac { \pi }{ 180 } \times 0.1 } =3.44\times { 10 }^{ -4 }m\)
(ii) The frequency and wavelength of reflected wave will not change. The refracted wave will have same frequency. The velocity of light in water is given by \(v=f\lambda \)
where, v = velocity of light
f = frequency of light
​​​​​​​​​​​​\( ​​lambda =wavelength \ of \ light\)
If velocity will decrease, then wavelength (\(\lambda \)) will also decrease.

(ii) As, the number of point sources increases, their contribution towards intensity also increases. Intensity varies as square of the slit width. Thus, when the width of the slit is made double the original width, intensity will get four times of its original value.
Width of central maximum is given by \(\beta =\frac { 2D\lambda }{ b } \)
Where, D = distance between screen and slit,
 \(\lambda \) = wavelength of the light,b = size of slit.

So, with the increase in size of slit, the width of central maxima decreases. Hence, double the size of the slit would result as half the width of the central maxima.

A acts as the two inputs of the NAND gate and Y is the output, as shown in the following figure.

Hence, the output can be written as:
\(Y=\overline{A . A}=\bar{A}+\bar{A}=\bar{A}\)
The truth table for equation (i) can be drawn as:

 This circuit functions as a NOT gate. The symbol for this logic circuit is shown as:

Given, amplitude of an electromagnetic wave,
E₀ = 120 N/C
Frequency of wave, v = 50 MHz = 50 \(\times\) 10⁶ Hz
(i) Speed of light in vacuum, \(c=\frac{E_0}{B_0}\)
\(\begin{aligned} B_0=\frac{E_0}{c}= & \frac{120}{3 \times 10^8}=40 \times 10^{-8} \end{aligned}\)
= 400 \(\times\) 10^-9 T = 400 nT
Angular frequency of electromagnetic wave,
\(\omega=2 \pi \nu=2 \times 3.14 \times 50 \times 10^6\)
\(\omega=3.14 \times 10^8 \mathrm{rad} / \mathrm{s}\)
Wave number of electromagnetic wave,
\(k=\frac{\omega}{c}=\frac{3.14 \times 10^8}{3 \times 10^8}=1.05 \mathrm{rad} / \mathrm{m}\)
Wavelength of electromagnetic wave,
\(\lambda=\frac{c}{v}=\frac{3 \times 10^8}{50 \times 10^6}=6.00 \mathrm{~m}\)
(ii) Expression of electric field, E = E₀ sin (kx - \(\omega\)t)
E = 120 sin (1.05x - 3.14 \(\times\)10⁸ t)
Expression of magnetic field B,
B = B₀ sin (kx - \(\omega\)t)
E = 120 sin (kx - \(\omega\)t)
B = 4 \(\times\)10^-7 sin (1.05x - 3.14 \(\times\) 10⁸ t)