Explore topic-wise InterviewSolutions in Current Affairs.

a. False. The area of a continuous spectrum graph represents the total output of X-ray photons emitted as a result of Bremsstrahlung radiation, not characteristic radiation. 

b. True. Increasing the tube voltage increases the frequency of photon production as a result of characteristic radiation; however, it does not influence photon energy, which is dependent on the target material. Photons produced by Bremsstrahlung radiation increase in both their frequency and energy (keV). 

c. False. Photons with an energy of less than 20 keV are absorbed by the glass tube and do not reach the patient. 

d. True. Increasing the filament voltage increases the number of electrons colliding with the target material causing an increase in the number of photons produced by both characteristic and Bremsstrahlung radiation. 

e. True. Filament voltage does not influence photon energy in either characteristic or Bremsstrahlung radiation.

a. False. Nucleon is a term used to describe protons and neutrons. Beta-positive decay results in the conversion of a proton to a neutron plus an electron. Hence, the number of nucleons is unchanged. 

b. False. The nucleons equal the atomic mass (i.e. protons þ neutrons). 

c. False. The difference between mass number and the number of protons is equal to the number of neutrons. 

d. False. Protons have a positive charge and neutrons have no charge. 

e. True. The number of protons (i.e. atomic number) influences the binding energy.

a. True. Increasing the tube voltage increases the kinetic energy of the electrons, the majority of which is lost as heat on collision with the target. 

b. False. This results in the heat being spread over a reduced area. 

c. False. The target angle is the angle between the target and the central photon beam. Reducing the target angle for a given spot size increases the area that the heat is spread over and therefore improves its heat rating. 

d. True. 

e. True. A rotating anode results in the heat being spread over a larger area and is one of the methods used to reduce target heating.

a. True. Nucleon is a collective term for neutrons and protons. Mass number is the number of neutrons and protons in an atom. 

b. False. A positron is a positively charged electron that is ejected following radioactive decay. 

c. False. The outer shell (valence shell) influences the chemical, thermal and optical properties of an atom. 

d. True.

e. True.

a. True. The filament is heated by passing an electrical current through it, known as the tube current, which subsequently emits electrons. 

b. False. The tube voltage affects the kinetic energy of each electron (keV), not the tube current, i.e. number of electrons. 

c. False. Electrons are released by heating the cathode filament. 

d. True. 

e. False. The collision of electrons with tungsten results mainly in the production of heat, due to interaction with outer electrons.

a. False. Electromagnetic rays originate from a point source and diverge out, travelling in a straight line, unless attenuated. 

b. True. 

c. True. The inverse square law states that the intensity of radiation emitted from a point source will reduce in intensity, proportional to the square of the distance from that point, i.e. the area covered by the beam increases as the rays are diverging from a point; however, the number of photons remains the same and hence their intensity reduces. 

d. False. 

e. False. Important points to remember regarding the inverse square law are: 

i. Radiation comes from a point source. 

ii. There is no absorption or scatter of radiation between source and point of measurement.

Given:

36 × 44 = x

Formula Used:

(a – b) × (a + b) = a² – b²

Calculation:

36 × 44 = x

⇒ (40 – 4) × (40 + 4)

⇒ (40)² – (4)²

⇒ 1600 – 16

⇒ 1584

∴ The value of x is 1584.

Given:

102 × 98 = x

Formula Used:

(a – b) × (a + b) = a² – b²

Calculation:

102 × 98 = x

⇒ (100 + 2) × (100 – 2)

⇒ (100)² – (2)²

⇒ 10000 – 4

⇒ 9996

∴ The value of x is 9996.

The maximum and minimum value  of cos(x) is  1 and -1.
Hence cos(cosx) = cos(cos90) at ,  x = 90 degree  is cos0 = 1 (maximum value )
And cos(cosx) = cos(cos0) at x = 0 is cos(1) , it is minimum value .
Here the cos function decreases from 0 to 90 degree.
Hence if one cos increases then other cos will decreases simultaneously.

As you know
The value of cos(x) lies between (-1 & 1).
So put any value of cos(x) between (-1) and (1) in cos(cos(x)).

cos(-1) = cos(1)   as [cos(-x) = cos(x)]

cos(0) = 1

It is clear from the above value that the value of cos(cos(x)) is between "cos(1)" and "1".

Also, cos(x) is a decreasing function in 0° and 90°. so as the maximum value of cos(x) will be at 0° and minimum at 1° in this question.

Given:

Ratio at the beginning of the year = 8 : 5

Ratio at the end of the year = 6 : 7

Number of boys reached the age of 16 years by the end of year = 40

Calculation:

Let the number of boys having age upto 16 years and those over the age of 16 years at the beginning of year be 8x and 5x respectively.

As by the end of year 40 boys had reached the age of 16 years, then

Number of boys having age upto 16 years = 8x - 40

Number of boys having age more than 16 years = 5x + 40

(8x - 40)/(5x + 40) = 6/7

⇒ 56x - 280 = 30x + 240

⇒ 56x - 30x = 240 + 280

⇒ 26x = 520

⇒ x = 20

Total number of boys in the school = 8x + 5x

⇒ 13x

⇒ 13 × 20

⇒ 260

∴The total number of boys in the school is 260.

Given:

A. Cost price = Rs. 60, Profit = Rs. 29

B. Cost price = Rs. 40, Profit = Rs. 17

C. Cost price = Rs 50, profit = Rs 21

D. Cost Price = Rs. 30, Profit = Rs. 14

Formula used:

Profit percentage = (P/CP) × 100

where,

P is Profit 

CP is Cost price

Calculations:

(1) When Cost Price is Rs. 60 and Profit is Rs. 29

Profit percentage = (P/CP) × 100

⇒ Profit percentage = (29/60) × 100

⇒ Profit percentage = 48.33%

(2) When Cost Price is Rs. 40 and Profit is Rs. 17

Profit percentage = (P/CP) × 100

⇒ Profit percentage = (17/40) × 100

⇒ Profit percentage = 42.5%

(3) When Cost Price is Rs. 50 and Profit is Rs. 21

Profit percentage = (P/CP) × 100

⇒ Profit percentage = (21/50) × 100

⇒ Profit percentage = 42%

(4) When Cost Price is Rs. 30 and Profit is Rs. 14

Profit percentage = (P/CP) × 100

⇒ Profit percentage = (14/30) × 100

⇒ Profit percentage = 46.66%

∴ The maximum profit percentage is 48.33%.

(3³ - 2²) × (2/3)^-2

= ^{\((3^3 - 2^2) × (\frac{3}{2})^2\)}

= \(23 × \frac{9}{4}\)

= \(\frac{207}{4}\)

Given:

HCF = 13

One number = 117

Concept Used:

Number 1 = HCF× (a) 

Number 2 = HCF× (b)  

Calculation:

Let the ratio of two numbers be a : b 

here 'a' and 'b' are co-prime numbers.

1st number = 117 and HCF = 13

⇒ 117 = 13 × a

⇒ a = 9

Now b will be a co-prime number of 9

Now comparing with the options,

Option 1 = 169 

⇒ 169 =  13 × b

⇒ b = 13

9 and 13 are co-prime numbers.

Hence the second number can be = 169

Now option 2 =143

⇒ 143 = 13 × b

⇒ b = 11

9 and 11 are co-prime numbers.

Hence the second number can be = 143

Now option 4 = 130

⇒ 130 = 13 × b

⇒ b = 10

9 and 10 are co-prime numbers.

Now option 3 = 156

⇒ 156 = 13 × b

⇒ b = 12

9 and 12 are not co-prime numbers.

∴ 156 cannot be a second number.

Given:

Four numbers are given 12346, 12348, 12344, and 12340

Concept Used:

We check the divisibility of 2 and 3. The number which is divisible by 2 and 3 simultaneously will be divisible by 6.

Divisibility of 2: The number must end with 0, 2, 4, 6, 8.

Divisibility of 3: Sum of all digits of the given number should be divisible by 3

Calculation:

12346 : Last digit is 6 so divisible by 2 but sum of digits (1 + 2 + 3 + 4 + 6 = 16) is not divisible by 3, so number is not divisible by 6.

12348 : Last digit is 8 so divisible by 2 but sum of digits (1 + 2 + 3 + 4 + 8 = 18) is divisible by 3, so number is divisible by 6.

12344 : Last digit is 4 so divisible by 2 but sum of digits (1 + 2 + 3 + 4 + 4 = 14) is not divisible by 3, so number is not divisible by 6.

12340 : Last digit is 0 so divisible by 2 but sum of digits (1 + 2 + 3 + 4 + 0 = 10) is not divisible by 3, so number is not divisible by 6.

∴ 12348 is the number that is divisible by 6.

BOYCOTT 

“to refuse to take part in an event, buy things from a particular company, etc. because you strongly disapprove of it”.

Sin 30° x cos 15° + cos 30° x sin 15°

= Sin (30° + 15°) (∵ Sin A. cos B + cos A. sin B = sin (A+B))

= Sin 45°

= \(\frac {1}{\sqrt2}\) 

GIVEN:

Numbers b/w 6 and 34 which are divisible by 5 are 10, 15, 20, 25, 30

FORMULA USED:

Average = (Sum of all the observation)/ Numbers of observation

CALCULATION:

All the numbers b/w 6 and 34 which are divisible by 5 are 10, 15, 20, 25, 30

⇒ Average = (Sum of all the observation)/ Numbers of observation

⇒ (10 + 15 + 20 + 25 + 30)/5

⇒ 100/5 

⇒ 20

∴ The average of all the numbers b/w 6 and 34 which are divisible by 5 is 20

Correct answer is Carbohydrate

a₁ = 11/8, a₂ = 14/8, a₃ = 17/8,....

d = a₂ - a₁ = 14/8 - 11/8 = 3/8;

Also a₃ - a₂ = 17/8 - 14/8 = 3/8

Hence, they will form an A.P.

a_n = a + (n - 1)d = 11/8 + (n - 1)3/8

 = \(\frac{3n+8}8\) = 1 + \(\frac{3n}8\)

We get whole number, when

n \(\in\) {8, 16, 24....}

or n \(\in\) 8N

GIVEN:

L.C.M = 48 and the numbers are in the ratio 2 : 3

FORMULA USED:

Product of the two numbers = L.C.M × H.C.F

CALCULATION:

Let the H.C.F of the two numbers be x

⇒ The numbers are 2x and 3x and LC.M is 48

⇒ Product of the two numbers = L.C.M × H.C.F

⇒ 2x × 3x = 48 × x

⇒ x = 8

⇒ Numbers are 2 × 8 = 16 and 3 × 8 = 24

∴ Sum of the two numbers = 16 + 24 = 40

Correct option is D) robbed

a₁ = 16 = 4² = (4 x 1)²

a₂ = 64 = 8² = (4 x 2)²

∴ a_n = (4_n)²

∴ a₅ = (4 x 5)²

a₁ = 16 = 4² = (4 x 1)²

a₂ = 64 = 8² = (4 x 2)²

∴ a_n = (4_n)²

∴ a₅ = (4 x 5)²

= 20² = 400

Correct answer is Christmas

Correct option is B) adopt

Given that in A.P 4th term = 26 & 8th term = 72

= a + 3d = 26 and a + 7d = 72

∴ (a + 7d) - (a + 3d) = 72 - 26

= 4d = 46

= d = \(\frac {46}{4}\) = \(\frac {23}{2}\)

∴ a + 3 x \(\frac {23}{2}\)= 26

= a = 26 - \(\frac {69}{2} = \frac {52-69}{2} = \frac {-17}{2}\)

∴ 6th term = a+5d

= \( \frac {-17}{2} + 5\times \frac {23}{2}\)

= \( \frac {-17 + 115}{2} = \frac {98}{2} = 49\)

Hence, 6th term of A.P is 49

x = 90° (in angle formed in a enmicircle is right angled)

y° = 180° − (90° + 50°) [Angle curve property]

y = 40°

Correct answer is Kitten

Correct answer is Pork

Given system of equations are 2x – 3y = 5 and 6x – 9y = 15. 

Now, comparing given system of equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, 

We get a₁ = 2, b₁ = – 3 c₁ = – 5 and a₂ = 6, b₂ = – 9, c₂ = – 15. 

Now, \(\frac{a_1}{a_2}\) = \(\frac{2}{6} = \frac{1}{3}\) , \(\frac{b_1}{b_2} = \frac{-3}{-9}\) = \(\frac{1}{3}\) and \(\frac{c_1}{c_2} = \frac{-5}{-15} = \frac{1}{3}\). 

Hence, \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\)= \(\frac{c_1}{c_2}\) which is condition of infinite many solutions . 

Hence, the given system of equations has an infinite number of solutions.

Direction ratios of vectors (\(2\hat i-3\hat j+4\hat k\)) and (\(a\hat i+6\hat j-8\hat k\)) are 2, −3, 4 and a, 6, −8, respectively.

The vectors \((2\hat i-3\hat j+4\hat k)\) and \((a\hat i+6\hat j-8\hat k)\) are collinear if \(\cfrac2a=\cfrac{-3}6=\cfrac{4}{-8}\)

⇒ \(\cfrac2a=-\cfrac12\)

⇒ a = -4.

If a = −4 then the vectors \((2\hat i-3\hat j+4\hat k)\) and \((a\hat i+6\hat j-8\hat k)\) are collinear.

Given system of equations are – x + 2y + 2 = 0 and \(\frac{1}{2}\) x – \(\frac{1}{2}\) y – 1 = 0. 

By comparing given system of equations with a₁x + b₁y + c₁ = 0 

and a₂x + b₂y + c₂ = 0, 

We get a₁ = –1, b₁ = 2, c₁ = 2 and a₂ = 1 2 , b₂ = – 1 2 , c₂ = – 1. 

Now,\(\frac{a_2}{a_2}\)= \(\frac{-1}{\frac{1}{2}}\) = −2, 

And \(\frac{b_1}{b_2}\) = \(\frac{2}{\frac{-1}{2}}\) = –4. 

Hence, \(\frac{a_2}{a_2}\) ≠ \(\frac{b_1}{b_2}\) which is condition of unique solution . 

Therefore the given system of equations has a unique solution.

Since the highest order derivative present in the given differential equation is \(\cfrac{d^2y}{dx^2}\) . So, its order is 2. It is a polynomial equation in \(\cfrac{d^2y}{dx^2}\) and \(\cfrac{dy}{dx}\) . The highest power of the highest order derivative \(\cfrac{d^2y}{dx^2}\) in the differential equation is 1. So, its degree is 3.

Hence, the order and the degree of the differential equation \(\cfrac{d^2y}{dx^2}\) + 5x ( \(\cfrac{dy}{dx}\))² − 6y = log x is 2  and 1, respectively.

Let the radius of the sphere is r. 

Given that the diameter of the sphere is 14 cm. 

∴ 2r = 14 cm. (∵ diameter is double of the radius)

⇒ r = \(\frac{14}{2}\) = 7 cm.

Hence the radius of the sphere is 7 cm. 

We know the volume of the sphere whose radius is r is given by V = \(\frac{4}{3} \pi r^3\)

Therefore the volume of given sphere is V = \(\frac{4}{3} \times \frac{22}{7} \times7^3\) (\(\because\)\(\pi\) = \(\frac{22}{7}\))

\(\frac{4}{3} \times \frac{22}{7} \times49\) = \(\frac{4312}{3}\)  = \({1437.33} \,cm^3\).

Hence, the volume of the sphere is 1437.33 cm³.

The a₃₂ element of \(\begin{bmatrix}2&-3&5\\6&0&4\\1&5&-7\end{bmatrix}\)is a₃₂ = 5.

We know that the cofactor of the element a_ij of a matrix is given by A_ij = (−1)^{i + j} M_ij,

where M_ij is minor of the element a_ij of that matrix.

Therefore, the cofactor of the element a₃₂ of \(\begin{vmatrix}2&-3&5\\6&0&4\\1&5&-7\end{vmatrix}\) is A₃₂ = (-1)^{(3 + 2)} M₃₂ = \(-\begin{vmatrix}2&5\\6&4\end{vmatrix}\)

= −(8 − 30) = −(−22) = 22.

Therefore, a₃₂A₃₂ = 5 × 22 = 110.

Hence, the value of a₃₂ A₃₂ of \(\begin{vmatrix}2&-3&5\\6&0&4\\1&5&-7\end{vmatrix}\) is 110.