Explore topic-wise InterviewSolutions in .

a. False. DICOM is a standard for the handling and transmission of data associated with medical imaging. One of the benefits is to keep data for the image filed alongside information relating to the imaging modality and patient-identifying information. 

b. True. 

c. False. This service retrieves the scheduled work list from the hospital information system (HIS) regarding examinations to be performed (hence reducing the need for repeated data entry) and may reduce data input errors. 

d. True. 

e. True.

a. True. This may be minimized by arranging the caesium iodide into narrow parallel columns, reducing the degree of reflection that may occur. 

b. False. The larger the detector element, the more light or X-ray photons are detected by that element so that structures smaller than the detector size cannot be distinguished separately. 

c. False. Increasing the thickness of the phosphor will increase the scattering of the laser light during image plate read-out. 

d. True. 

e. True.

a. False. It is still required.

b. False. The Royal College of Radiologists (RCR) IT guidance (Picture archiving and communication systems (PACS) and quality assurance, 2008) recommends an ideal screen resolution of 3 megapixels and a screen size of at least 50 cm with at least a 10-bit grey scale. 

c. True. See the RCR IT guidance (Picture archiving and communication systems (PACS) and quality assurance, 2008). 

d. True. 

e. True. This is a standard look-up table against which different display devices should be calibrated to allow for varying device performances.

a. True. Film-screen radiography has a narrower dynamic range. 

b. True. 

c. False. Separation of these stages is the characteristic of digital radiography, which allows flexibility in the post-acquisition processing that may be performed. 

d. False. 

e. True.

a. True. Computed radiography has a dynamic range in the order of 10,000:1. Digital radiography has a dynamic range in the order of 1000:1 to 10,000:1. For both, the low end of the system is determined by noise. In digital radiography, the high end of the range is determined by the charge-holding capacity of the detector elements.

b. False. They are both affected by this. This is noise associated with the image receptor, which typically cannot be corrected for (so-called electronic noise). 

c. False. This is related to an insufficient number of grey-scale values that can be demonstrated. 

d. True. 

e. True. Post-image acquisition analysis and processing can help with diagnosis of pathology.

a. True. The relationship differs from the characteristic curve of conventional radiography. Essentially, the response demonstrates wide latitude. 

b. True. Even though the computed radiography response demonstrates wide latitude, post-processing of the data may allow better contrast to be seen within the image. This may include rejection of signals outside the useful range and edge enhancement. 

c. False. 

d. False. The opposite is true. 

e. False. The laser light, which is used for processing of the latent image, may be spread out over the phosphor, increasing the area over which light is emitted. This effect is increased with increasing size of the phosphor crystals. The diameter of the laser beam also affects resolution.

The one whose speed is more will move a larger distance i.e. Q will move larger.

Let P moves x km before they meet so , Q will move (x + 60) km

Time taken by P = Time taken by Q

(Distance / speed )_P = (Distance / speed )_Q

x/15 = (x + 60)/45               

⇒ x = 30 km

Total distance between P and Q  = x + x + 60 = 30 + 30 + 60 = 120 km

Given:

Downstream speed = 4 km/h

Speed of current = 1 km/h

Formula used:

Speed of current = (Downstream speed – Upstream speed)/2

Calculation:

1 = (4 – Upstream speed)/2

⇒ Upstream speed = 4 - 2 = 2

⇒ Upstream speed = 2 km/h

Given: 

Distance (D1) = 120 km  (D2) = 160 km and (D3) = 140 km 

Time (T1) = 2 hours  (T2) = 3 hours and  (T3) = 2 hours.

Concept:

Average speed = (Total distance covered throughout the journey)/ (Total time taken for the journey).

Calculations:

Total DIstance = (120+ 160 +140) km = 420 km

Total Time = (2 + 3 +2) hours = 7 hours

Average speed = 420/7 = 60 km/hr

∴ The average speed of the train is 60 km/hr

a. False. It depends on photo-stimulated phosphorescence. Fluorescence describes the emission of light immediately after the phosphor is exposed to X-rays, while phosphorescence describes delayed emission, which is the property exploited in computed radiography. 

b. True. This is the name for the screen comprising barium fluorohalide crystals (halogen elements of bromine, iodine and chlorine) with europium ions, which are usually embedded in a light-reflective layer. 

c. True. High-resolution image plates are designed with a thinner layer of phosphor crystals without a light-reflective layer. The compromise with these plates is that they will require a higher dose to obtain an adequate image. 

d. False. Electrons located in ‘electron traps’ within the phosphor store the latent image. These represent energy levels, which differ from those in which the electrons usually exist (the valence bands) prior to irradiation. 

e. True.

Given:

Two speed are 3 km/h and 9 km/h

Formula used:

If time₁ = time₂, and two different speed are given, then formula for calculation average speed is as follows:

Average speed = Sum of speed/Number of different speed:

Calculation:

Average speed = (3 + 9)/2

⇒ 6 km/h

Alternative method

Total distance = (3 × 5) + (9 × 5) = 60 km

Total time = (5 + 5) hours = 10 hours

a. False. This is when the image plate has been incompletely erased since the previous exposure. 

b. False. Grids may be used in computed radiography as in film-screen radiography with the same effect on resultant image quality. 

c. True. For example, applying incorrect parameters involved in analysis of the intensity values to produce a specific gradation curve for a particular image. 

d. True. This may produce excessive shading across the image if not functioning correctly. 

e. False. If the problem is related to a faulty image plate, then data processing techniques will not be able to improve the diagnostic quality of the final image if the artefact is significant.

Given : 

Time taken to walk around field 1 = 11 s

Time taken to walk around field 2 = 33 s

Formula used :

Speed = Distance/ Time 

Calculations :

Let the radius of field 1 and field 2 be r₁ and r₂ respectively.

And the speeds while traveling in field 1 and field 2 are 2x and x respectively.

Distance traveled = Perimeter of the fields 

⇒ Distance 1/Distance 2 = (speed× time)₁ / (speed× time)₂

⇒ Distance 1/Distance 2 = (2x × 11)/(x × 33)

⇒ 2/3 = r₁/r₂

⇒ r₂ = 3/2 × r₁

⇒ r₂ = 3/2 × 14

⇒ r₂ = 21 m

∴ The radius of field two is 21 m.

Given:

Total time taken = 10 hours

Speed in first half = 30 km/hr

Speed in second half = 20 km/hr

Formula used:

Time = Distance/Speed

Calculation:

Let the total distance covered be 2x km

As, he covered the first half (i.e. x km) with 30 km/hr

Time taken to cover first-half = Distance/Speed = x/30 hours

He covered the second half (i.e. x km) with 20 km/hr

Time taken to cover second half = x/20 hours

Total time taken = 10 hours

⇒ x/30 + x/20 = 10

⇒ 5x/60 = 10

⇒ x = 120

Total distance covered by him = 2x = 2 × 120 = 240 km

∴ Total distance covered by him is 240 km

Given:

The ratio of three number a, b, and c = 2 ∶ 3 ∶ 1

Sum of three numbers = 120

Calculation:

Let the values be

a = 2x, b = 3x, and c = x

So, a + b + c = 2x + 3x + x = 6x

Given that the sum of the numbers is 120

So, 6x = 120

⇒ x = 20

So, b = 3 × 20 = 60

∴ The required value of b is 60

Given:

A can-do 1 unit work = 1/12 minutes                                   

B can do 1 unit work = 1/15 minutes 

C can do 1 unit work = - 1/x minutes

Together they can fill the tank = 1/10 minutes

Formula used:

Total work = Efficiency × days

Solutions:

⇒ 1/total days = 1/A + 1/B - 1/C

⇒ 1/10 = 1/12 + 1/15 - 1/x

⇒ 1/x = 1/12 + 1/15 - 1/10

⇒ 1/x = (5 + 4 - 6)/60

⇒ 1/x = 3/60

⇒ x = 20

∴  The value of x is 20.

Correct option is B) ten-kilometer

Correct option is A) twenty-kilometer

1. Tagore was awarded Nobel prize.

2. IIM Ahmedabad was established in 1961.

3. Chhattisgarh was formed in 2000.

4. The first passenger train was inaugurated in 1853.

5. Indian Airlines was set up in 1953.

1. C₃ cycle 

2. Chloroplast

It's auxiliary equation is

m² +2 = 0

m = \(\pm\sqrt2i\) 

∴ C.F. = C₁ cos\(\sqrt2\)x + C₂sin √2x

P. I. = \(\frac1{D^2+2}x^2e^{3x}+\frac1{D^2+2}e^xcos2x\)

  = e^3x \(\frac1{(D+3)^2+2}x^2+e^x\frac1{(D+1)^2+2}cos 2x\)

 = e^3x\(\frac1{11+6D+D^2}x^2+e^x\frac{1}{D^2+2D+3}cos 2x\) 

 = \(\frac{e^{3x}}{11}(1+\frac{6D+D^2}{11})^{-1}\)x² + e^x \(\frac1{-4+2d+3}cos 2x\)

 = \(\frac{e^{3x}}{11}(1-\frac{6D+D^2}{11}+\frac{36}{121}D^2+....)x^2\) + e^x\(\frac{2D+1}{4D^2-1}cos 2x\)

 = \(\frac{e^{3x}}{1331}(121x^2-132x + 50)-\frac{e^x}{17}(cos 2x - 4sin 2x)\)

∴ y = C. F. + P. I.

 = C₁cos(√2x) + C₂sin(√2x) + \(\frac{e^{3x}}{1331}\)(121x² - 132x + 50) - \(\frac{e^x}{17}\)(cos 2x - 4 sin 2x)

A statement is a declarative sentence if it is either true or false but not both.

So, the given sentence is true.

Hence, it is a true statement

Correct option is: (B) \(\frac{1}{2}\)

Correct option is: (C) \(\frac{\sqrt{3}}{2}\)

Correct answer is (A) Thales

Answer is (b) [(cosθ,sinθ),(-sinθ,cosθ)]

Correct option:

(B) 2 

(D) None of these

Correct option:

(B) ±6 

Correct option:

(D) 2 sin√y

The correct answer is : (a) [1, 2]

Correct option:

(D) 5/2

The correct answer is : (c) -1

The correct answer is: (c) -1

u = tan^-1\((\frac{x^4+y^4}{x^2+y^2})\)

⇒ tan u = \((\frac{x^4+y^4}{x^2+y^2})\).....(i)

Partial differentiate (i) w. r. t x and y, we get

sec²y = \(\frac{∂y}{∂x}=\frac{4x^3(x^2+y^2)-(x^4+y^4)2x}{(x^2+y^2)^2}\)\(=\frac{2x^5+4x^3y^2-2xy^4}{(x^2+y^2)^2}\)

sec²y = \(\frac{∂y}{∂x}=\frac{4y^3(x^2+y^2)-(x^4+y^4)2y}{(x^2+y^2)^2}\)\(=\frac{2y^5+4y^3x^2-2x^4y}{(x^2+y^2)^2}\)

sec²y \((x\frac{∂y}{∂x}+y\frac{∂y}{∂y})\)\(=\frac{2x^6+4x^4y^2-2x^2y^4+2y^6+4y^4x^2-2x^4y^2}{(x^2+y^2)^2}\)

\(=\frac{2x^6+2x^4y^2+2x^2y^4+2y^6}{(x^2+y^2)^2}\)

\(=\frac{2x^4(x^2+y^2)+2y^4(x^2+y^2)}{(x^2+y^2)^2}\)\(=2\frac{x^4+y^4}{x^2+y^2}\) 

= 2 tan u (by (i))

⇒ \(x\frac{∂y}{∂x}+y\frac{∂y}{∂y}\) = \(\frac{2tan u}{sec^2y}\)

\(=\frac{2sin u}{cos u}\times cos^2y\)

 = 2 sin u cos u = sin 2u

Hence Proved

Correct option:

(A) B ⊂ A

(C) (x – y)dy + y²dx = 0

a, b \(\in\) Q

a * b = ab

∵ If a & b are rational numbers then their multiple is also a rational number.

∴ If a.b \(\in\) Q

⇒ ab \(\in\) Q

⇒ a * b \(\in\) Q  \(\forall\) a,b \(\in\) Q

∴ operation * is binary.

u = x² tan^-1\(\frac xy\)

\(\frac{\partial u}{\partial y}=\cfrac{x^2}{1+\frac{x^2}{y^2}}\times\frac{-x}{y^2}\) = \(=\frac{-x^3}{x^2+y^2}\)

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

 = \(-\left(\frac{3x^4+3x^2y^2-2x^4}{(x^2+y^2)^2}\right)\) = \(-\frac{x^2(x^2+3y^2)}{(x^2+y^2)^2}\)

\(\frac{\partial u}{\partial y}=\cfrac{x^2}{1+\frac{x^2}{y^2}}\times\frac 1y\) + 2x tan^-1\(\frac xy\) = \(\frac{x^2y}{x^2+y^2}+2xtan^{-1}\frac xy\)

\(\frac{\partial^2y}{\partial x\partial y}\) = \(\frac{(x^2+y^2)x^2-x^2y(2y)}{(x^2+y^2)^2}\) + \(\cfrac{2x}{1+\frac{x^2}{y^2}}\times\frac{-x}{y^2}\) 

 = \(\frac{x^4+x^2y^2-2x^2y^2}{(x^2+y^2)^2}-\frac{2x^2}{x^2+y^2}\) 

 = \(\frac{x^4-x^2y^2-2x^2(x^2+y^2)}{(x^2+y^2)^2}\) 

 = \(\frac{-x^4-3x^2y^2}{(x^2+y^2)^2}=\frac{-x^2(x^2+3y^2)}{(x^2+y^2)^2}\)

\(\therefore\) \(\frac{\partial^2 y}{\partial x\partial y} = \frac{\partial y}{\partial y\partial x}\) Hence Proved.

Correct option:

(C) 0

Correct option:

(C) 61 

Correct option:

(A) 2 

(B) (x - 1)/2 = (y - 1)/3 = (z - 1)/-1 

Major axes and minor axes of an ellipse are conjugate diameters.

The tangents at the ends of a pair of conjugate diameters of an ellipse which forms a parallelogram and the area of the parallelogram are constant and are equal to the product of the axis.

If ellipse has an equation x²/a² + y²/b² = 1 then,
Area of parallelogram =  4ab

= 4 x 3 x 2 = 24cm²