This can be solve by using Boolean identity but using k map it will be more easy

(A + B + C) (A + B + C) (A + B + C).

? V_P = 3.46 x 10⁸ m/s.

Consider

Number of samples per period = 10p

This is not a rational number. Hence signal is non periodic.

The area of the shaded region

.

Statement (ii) is false because maximum energy of photons varies linearly with frequency of incident light. It can be shown as follows :

F₁ > F₂ > F₃

j = light intensity = constant

Photocurrent V_s Anode voltage with frequency and incident light as a parameter.

The light intensity is constant.

Since number of comparators = 2ⁿ - 1 = 15;

Then number of bits are '4'

Resolution

Use the condition of controllability.

2^{n - 1} R = 32 K&Omeg;

n = 4

? 8R = 32

R = 4 K?.

V₁ = i₁R₁ + i₂R₂

i₁ + ai₁ = i₂

V₂ = i₂R₂

Putting a = - 1

= 0 V.

A quantity of any substance equal to its molecular weight in grams is a mole of that substance contains the same number of molecules as one mole of any other substance. This Avogadro's number and equals 6.02 x 10²³ molecules per mole. Thus

= 9.03 x 10²⁸ electrons/m³

Since for tungsten the atomic and the molecular weights are the same. Therefore for tungsten

E_F = 3.64 x 10^-19(?)^2/3

= 3.64 x 10^{- 19}(9.03 x 10²⁸)^2/3

E_F = 7.326 eV.