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Correct option: b) Radioactive waste

a. False. Protons have a positive charge. Electrons have a negative charge. 

b. True. 

c. True. Protons (atomic number (Z)) þ neutrons (A – Z) ¼ mass number (A)). 

d. False. Both protons and neutrons have a mass. Electrons have a negligible mass. 

e. False. Following electron capture the nucleus may increase its number of neutrons relative to protons by capturing an electron from the K-shell (p + þ → e n) 

a. False. Photon fluence is the number of photons through a cross-section of the beam (i.e. per unit area). Adding the particle energies gives the total amount of energy per unit area and is called energy fluence.

b. True. 

c. True. Wavelength is the distance between crests, when field strength is plotted against distance. Frequency multiplied by wavelength equals velocity (λ x  f V) 

d. False. Frequency is proportional to photon energy; the content of proportionality is called Planck’s constant. 

e. True.

a. True. This is known as beta-negative (b) decay and results in the conversion of a neutron to a proton, therefore increasing atomic number. 

b. False. The rate of decay is dependent on the characteristic of the radionuclide and is unaffected by factors such as heat or chemical reactions. 

c. False. Isomers have differing half-life and energy states, but the same atomic and mass numbers. 

d. True. This can occur either by beta-positive (b+) decay or K-electron capture, both of which result in the reduction of the number of protons and therefore atomic number. 

e. False. Positrons collide with negative beta particles resulting in the two masses being converted to energy.

a. True. When secondary electrons (e.g. photo electrons and recoil electrons) pass through tissue, they result in ionization and excitation of atoms resulting in tissue damage. 

b. False. Cells with higher mitotic levels are more prone to radiation damage. 

c. True. Ionization by secondary electrons, results in damage to biological tissue either by rupturing covalent bonds or by the production of free radicals, which result in oxidation of organic molecules. 

d. False. Secondary electrons have a tortuous path, as negative electrons easily deflect them, leaving a track of ionized atoms behind. 

e. False. X-rays and gamma rays result in ionization of atoms via secondary electrons and are therefore indirectly ionizing agents.

 a. False. Binding energy is the energy expended in completely removing the electron from the atom, against a positive force of the nucleus. 

b. True. The nucleus exerts a stronger pull on the inner electrons than the outer electrons. 

c. False. Neutrons have a charge of zero and hence do not affect the binding energy of electrons. 

d. False. Bremsstrahlung radiation is produced from filament electrons that penetrate the K-shell and approach the nucleus. Characteristic radiation is formed when an electron shifts from an outer to an inner shell, releasing a photon with energy equal to the difference in the binding energy of the two shells. 

e. True. In photoelectric absorption, when a photon collides with an electron from an inner shell, it ejects the electron, which is then termed a photo electron. Kinetic energy of photo electron (E_k for K-shell) = photon energy – binding energy.

Inductive Impedance V₁ I cos φ₁ = power; 230 × I₁ × 0.5 = 100 ; I₁ = 0.87 A 

Now, I₁² R₁ = power or 0.87² R₁ = 100; R₁ = 132 Ω ; Z₁ = 230/0.87 = 264 Ω 

X_L = √(Z²₁ - R²₁) = √(264² - 132²) = 229Ω

Capacitance Impedance I₂ = 60/230 × 0.6 = 0.434 A ; R₂ = 60/0.4342 = 318 Ω

Z₂ = 230/0.434 = 530 Ω ; X_C = √(530² - 318²) = 424Ω (capacitive)

When Z₁ and Z₂ are connected in series 

R = R₁ + R₂ = 132 + 318 = 450 Ω; X = 229 − 424 = − 195 Ω (capacitive)

Z = √(R² + X²) = √(450² + (-195)²) = 490 , 230 / 490 0.47 A

(i) Total power absorbed = I²R = 0.472 × 450 = 99 W, cos φ = R/Z = 450/490 = 0.92 (lead) 

(ii) Power factor will become unity when the net capacitive reactance is neutralised by an equal inductive reactance. The reactance of the required series pure inductive coil is 195 Ω.

a. True. Electromagnetic radiation is named according to how it is produced, e.g. X-rays (X-ray tube), gamma rays (radioactive nuclei). 

b. False. The different types of electromagnetic radiation differ in their properties and are made up of photons, which do not have a mass or electric charge. 

c. False. All forms of electromagnetic radiation travel with the velocity of light in a vacuum. 

d. True. Electromagnetic radiation produces a sinusoidal graph when electric or magnetic field strength is plotted against time or distance, travelling with velocity (C). The peak field strength is called the amplitude (A). 

e. False. Frequency (f ) is the number of crests passing a point in a second. The interval between successive crests is called the period.

(a) Z = 100/5 = 20 Ω 

(b) P = VI cos φ or 250 = 100 × 5 × cos φ 

∴ cos φ = 250/500 = 0.5 

(c) Total loss = loss in resistance + iron loss 

∴ 250 = 20 × 5 + Pi 

∴ P_i = 250 − 100 = 150 W

(d) Effective resistance of the choke is P/I² = 250/25 = 10Ω

∴ X_L = √(Z² - R²) = √(400 - 100) = 17.32Ω

X_L = 2 π fL = 2π × 50 × 0.2 = 62.8Ω ; X_C = 1/2π fC 

= 10⁻⁶ 2π × 50 × 150 = 21.2 Ω ; X = (X_L − X_C) = 41.6 Ω ;

Z = √(R² + X²) = √(20² + 41.6²)

= 46.2Ω; I = V/Z = 230/46.2 = 4.98A

Also, Z = R + jX = 20 + j 41.6 = 46.2 ∠ 64.3° ohm 

∴ Y = 1/Z = 1/46.2 ∠ 64.3° = 0.0216 ∠ − 64.3° siemens 

p.f. = cos 64.3° = 0.4336 (lag) 

Active power = VI cos φ = 230 × 4.98 × 0.4336 = 497 W 

Reactive power = VI sin φ = 230 × 4.98 × sin 64.3° = 1031 VAR

X_C = 0.2 × 314 = 63Ω , C = 10μF = 100 × 10⁻⁶ = 10⁻⁴ farad

X_C = 1/ωC = 1/(314 x 10^-4) = 32Ω, X = 63 − 32 = 31 Ω (inductive)

(a) Z = √(20² + 31)² = 37 Ω 

(b) I = 220/37 = 6 A (approx)

(c) V_R = I × R = 6 × 20 = 120 V; V_L = 6 × 63 = 278 V, 

V_C = 6 × 32 = 192 V 

(d) Power in V_A = 6 × 220 = 1320 

Power in watts = 6 × 220 × 0.54 = 713 W 

(e) p.f. = cos φ = R/Z = 20/37 = 0.54; φ = cos⁻¹(0.54) = 57°18′

It should be kept in mind the coil offers only resistance to direct voltage whereas it offers impedance to an alternating voltage.

∴ R = 24/6 = 4 Ω; Z = 30/6 = 5Ω

(i) ∴  X_L = √(Z² - R²) = √(5² - 4²) = 3Ω 

(ii) p.f. = cos φ = R/Z = 4/5 = 0.8 (lag)

X_C = 10⁶ π /2π × 50 × 100 = 32 Ω ; R = 50 Ω

(a) Z = √(50² + 32²) + 59.4 Ω 

(b) I = V/Z = 100/59.4 = 1.684 A 

(c) p.f. = R/Z = 50 = 59.4 = 0.842 (lead) 

(d) φ = cos⁻¹ (0.842)= 32°36′ 

(e) V_R = I_R = 50 × 1.684 = 84.2V 

(f) V_C = IX_C = 32 × 1.684 = 53.9V

X_C = 1/(3.14 × C) = 28.952 ohms

∴ Coil Impedance, Z = 28.952Ω 

Coil resistance = 28.952 × 0.8 = 23.162 Ω 

Coil reactance = 17.37 ohms 

Coil-inductance = 17.37/314 = 55.32 milli-henrys 

Total impedance, Z_T = 23.16 + j 17.37 − j 28.952 

= 23.162 − j 11.582 = 25.9 ohms 

Net power-factor = 23.162/25.9 = 0.8943 leading

The waves associated with material particles in motion is called matter waves. The expression for de-Broglie wavelength is

\(\lambda = \frac{h}{mv}\)

Where

h → is Planck’s constant 

m → is mass of moving particle 

v → is the velocity of moving particle.

A. Glucuronidation

Correct answer is (c) four times

(i)  True

(ii)  True

(iii)  True

(iv)  True

(v)  False

\(\frac {8^x-2^x}{6^x-3^x} = \frac {2^x(2^{2x}-1)}{3^x (2^x-1)}\)

= \(\frac {2^x (2^x-1)(2^x+1)}{3^x (2^x-1)}\) 

= \(\frac {2^x (2^x+1)}{3^x}\)

\(\therefore\) \(\frac {2^x (2^x+1)}{3^x}\) = 2 (Given)

\(\Rightarrow\) \(2^{2x} +2^x = 2.3^x\)

For x = 0, \(2^{2x} + 2^x = 2^0 + 2^0 = 1+1 = 2\) 

\(2.3^x = 2.3^0 = 2 \times 1 = 2\)

Hence, \(2^{2x}+2^x = 2.3^x \) then x = 0

6^x + 9^x = 2^{2x + 1}

⇒ 3^2x + 3^x.2^x = 2^{2x + 1}

⇒ 3^2x + 2.3^x .\((\frac{2^x}2)+\frac{2^{2x}}4\) = 2^2x+1 + \(\frac{2^{2x}}4\)

⇒ (3^x + \(\frac{2^x}2\))² = 2^2x\((2+\frac14)\) (\(\because\) a² + 2ab + b² = (a + b)²)

⇒ (3^x + 2^x-1) = 3/2 . 2^x = 3.2^x-1

⇒ 3^x = 3.2^x-1- 2^x-1 = 2.2^x-1

⇒ 3^x = 2^x

which is possible only when x = 0

hence, if 6^x + 9^x = 2^2x+1 then x = 0

(b) 2(ΔP / P)

Explanation:

k = (p² / 2m)

(Δk / k) = 2(Δp / p)  

(a) M^–1L^–2T⁴A²  

Explanation:

 Q = CV

Q – charge,   C – capacitance, V – voltage

C = (Q/V)

Q = It where I is current and t is time

 V = (W/Q) where W is work done and Q is charge

 W = F × S where F is force and S is displacement

  ∴ V = (W/Q) = [(F × S) / Q] = [(m × a × s) / Q]

    ∴ Q = [A¹T¹]

   V = ([M¹L¹T^–2L¹] / [A¹T¹]) = [M¹L²T^–3A^–1]

  ∴ C = (Q/V) = [A¹T¹] / [M¹L²T^–3A^–1] = [M^–1L^–2T⁴A²] 

The correct answer is c (i:e) Dipole-Dipole interaction and London forces.

• The predominant intermolecular forces present in ethyl acetate liquid are London dispersion and dipole-dipole interaction.
•  Ethyl acetate is a polar molecule, therefore, dipole-dipole interaction will be present there.
   

• Fe³⁺ gives blood red color with ammonium thiocyanate forming ferric thiocyanate.
• Fe(SCN)₃ is a red salt soluble in water.

Fe³⁺ + SCN → [Fe(SCN)]²⁺

Correct option is D) intend

Correct option is B) convince

Correct option is B) pursue

Correct option is D) devising

Answer is (4) Cs

Cs used in photoelectric cell as it has least ionisation energy.

Concept used :

The given numbers are exactly divisible by their H.C.F.

Solution :

∵ 12 is H.C.F.

⇒ The number in the ratio will exactly be divisible by 12.

Let the numbers will be x : 2x : 3x

As, HCF is 12

⇒ x = 12

The numbers will be  12, 12 × 2 = 24 and 12 × 3 = 36

∴ The required number is 12, 24, 36.

Concept used :

Factorize the given number and make a pair of unpaired factors.

Solution :

Factorization of 60

⇒ 60 = 2 × 2 × 3 × 5

Make a pair of unpaired factors

⇒ Required number = (2 × 2) × (3 × 3) × (5 × 5)

∴ The required number is 900.

Given equation is,

\(x\,+\sqrt{x}\,+\sqrt{x+2}\,+\sqrt{x^2+2x}\) = 3

\(\Rightarrow\) \(x\,+\sqrt{x}\,+\sqrt{x+2}\,+\sqrt{x^2+2x}-3=0\)

Let f(x) = \(x\,+\sqrt{x}\,+\sqrt{x+2}\,+\sqrt{x^2\,+\,2x}-3\)

Therefore, the domain of function f(x) is [0,∞).

f(o) = - 3 < 0

f(1) = 1 + 1 + \(\sqrt{3}\,+\sqrt{3}\)

= 2 + \(2\sqrt{3}\) > 0

Hence, at least one solution lies between 0 and 1.

f¹(x) = 1 + \(\frac{1}{2\sqrt{x}}\) + \(\frac{1}{2\sqrt{x\,+\,2}}\) + \(\frac{2(x\,+\,1)}{2\sqrt{x^2\,+\,2x}}\)> 0 (\(\because\) \(x\in\) [0,∞)]

Hence, f(x) is increasing in its domain 

therefore, only one positive solution of given equation exist which lies between 0 and 1.

Hence, option (C) is correct

Correct option is (D) neither R₁ nor R₂ is an equivalence relation

\(R_1 = \{xy\ge 0 , x, y\in R\}\)

For reflexive x \(\times\) x ≥ 0 which is true.

For symmetric 

If xy ≥ 0 ⇒ yx ≥ 0

If x = 2, y = 0 and z = -2

Then x.y ≥ 0 & y.z ≥ 0 but x.z ≥ 0 is not true 

⇒ not transitive relation. 

⇒ R₁ is not equivalence 

R₂ if a ≥ b it does not implies b ≥ a 

⇒ R₂ is not equivalence relation 

⇒ D

If we know the normal vector of a plane and a point passing through the plane, the equation of the plane is established. a ( x − x₁) + b (y − y₁) + c (z − z₁) = 0.

3, \(\frac{3}{2}\), \(\frac{3}{4}\)

Here r = \(\frac{1}{2}\)< 1

sum of G.P = \(\frac{a(1-r^n)}{1-r}\)

⇒ \(\frac{255}{128} = \frac{3(1-(\frac{1}{2})^n)}{1- (\frac{1}{2})}\) = \(\frac{3(1- \frac{1}{2^n})}{\frac{1}{2}} = 6({1- \frac{1}{2^n}})\)

⇒ \(\frac{255}{256}\) = \(1- \frac{1}{2^n}\)

⇒ \(\frac{1}{2^n} = 1 - \frac{255}{256}\) = \(\frac{1}{256} = \frac{1}{2^8}\)

⇒ n = 8

Solution:

Given pth term = 1/q
That is ap = a + (p - 1)d = 1/q
aq + (pq - q)d = 1  --- (1)
Similarly, we get ap + (pq - p)d = 1  --- (2)
From (1) and (2), we get
aq + (pq - q)d = ap + (pq - p)d 
aq - ap = d[pq - p - pq + q]
a(q - p) = d(q - p)
Therefore, a = d
Equation (1) becomes,
dq + pqd - dq = 1  
d = 1/pq
Hence a = 1/pq
Consider, Spq = (pq/2)[2a + (pq - 1)d]
                    = (pq/2)[2(1/pq) + (pq - 1)(1/pq)]
                    = (1/2)[2 + pq - 1]
                    = (1/2)[pq + 1]

Answer : Given :if pth term of an ap is 1/q and the qth term of an ap is 1/p

To prove : the sum of pq terms is pq+1/2  

Let first term and common difference be a and d respectively

Now according to the question

T_p=a+(p-1)d=1/q ..................eq (1)

T_q=a+(q-1)d=1/p ...................eq(2)

Subtracting eq 2 by eq1, we get

=> (p-q)d = (1/q )- (1/p )

 => (p-q)d = [ (p-q) / pq ]

 => d =1/pq......................eq 3

Substituting the value of d in eq 1

=> 1/q=a+[(p-1) (1/(pq)) ]

=> 1/q = a + [1/q] - [1/(pq)]

=> a=1/pq.................................eq 4

Now the sum of pq terms using the formula S_n = n/2 ( 2a +(n-1) d ) is given by

=> S_pq=(pq/2 ) [ (2/(pq)) + (pq-1) (1/pq)]   {using the value of eq 3 and eq 4}

=> S_pq= (pq/2 ) [ (2/(pq)) + 1 -(1/pq)]     

=> S_pq= (pq/2 ) [ (1/(pq)) + 1]     

=> S_pq = ( 1/2 ) + (pq /2)

=> S_pq=(pq+1)/2