Explore topic-wise InterviewSolutions in Current Affairs.

B. antinode.

A. fixed points

B. Barometer

B. Displacement and Velocity

The Correct option is C. 180°

D. The total gravitational force that Earth exerts on us - The total centripetal force required to keep us moving on Earth's axis

Explanation:

Divide 8 bags into three parts.

2,2 and 4 respectively.

If we compare the 2,2 bags on pans of a balance.

We can identify which side is the lighter bag placed.

And then we will need only one more weighing for identifying the faulty bag.

So only two weighings are required.

Alternative solution:

In the case of a weighing balance (i.e. beam balance) the following model can be observed.

1 - 3 → 1 weighing required

4 - 9 → 2 weighings required

10 - 27 → 3 weighings required

28 - 81 → 4 weighings required

And the process will go so on.

As there are eight objects. The minimum number of weighings required will be only 2.

Concept: -

1) As per IS 1893 (Part 1): 2016, cl. 3.1, closely spaced modes of a structure are those natural modes whose natural frequencies differ from each other by less than 10%.

2) As per IS 1893 (Part 1): 2016, cl. 7.7.5.3, If building does not have closely-spaced modes, then net peak response quantity can be calculated by Square Root of Sum of Squares method.

\(\lambda = \sqrt {\mathop \sum \limits_{k = 1}^{{N_m}} {{\left( {{\lambda _k}} \right)}^2}} \)

Here,

λ_k = peak response quantity in mode k,

B. the same as that of the lead ball

The two arrangements are (a) series arrangement, and (b) parallel arrangement. In a household circuit we will prefer the second circuit i.e., (b). In circuit (b) each appliance has same voltage of 220 V. Since all the appliances that we use have voltage rating of 220 V in our country, so each bulb works normally.

a. True. 

b. False. mA is in direct relation to the dose but noise is proportional to √mA, so the dose will be doubled and the noise will be √2. 

c. True. 

d. True. If the pixels are halved, then the incident photon will be (1/2)³ ; therefore, for the same signal-to-dose ratio, the dose needs to be eight times higher. 

e. True. mA and therefore dose is proportional to (SNR)² ; thus, (SNR x 2)² = 4 x dose.

a. True. 

b. True. CTDI: head CT > abdo-pelvis CT > chest CT. 

c. False. CTDI measures the dose efficiency. It is mainly for comparison between different models/protocols; it does not measure the dose for individual patients. 

d. False. Skin dose: prolonged fluoroscopy > CT > plain film. 

e. False.

a. False. 

b. False. A 16 cm diameter phantom is used for the head and a 32 cm diameter phantom is used for the body. 

c. False. E_DLP is measured in mSv mGy^-1 cm^-1 . The effective dose is derived from E_DLP x DLP. 

d. False. Effective doses in CT can be calculated using special computer programs. A special ionization chamber detector is used to measure CTDI. 

e. True. DLP: CTDI_w x 1/pitch x length; therefore, CTDI_w for 150 mAs is 15 mGy = 15 1/1.25 25 = 300 mGy cm. 

a. False. Fourth-generation CT uses rotate–stationary scanners. The advantages are: detector stability, simpler reconstruction and readjustment of calibration through the scanning cycle. The disadvantages are: an increase in the number of detectors by a factor of 6, it is prohibitive in multi-slice CT, and there is a higher dose because of the increased distance between the patient and the detectors. 

b. False.

c. True. 

d. True. 

e. False.

a. False. In first-generation scanners, the X-ray source and single detector both moved across the scanning plane at 180° rotation. 

b. True. 

c. False. In second-generation scanners, there are around 30 detectors, which are not enough for the whole cross-section. 

d. False. In fourth-generation scanners, the patient-to-detector distance is greater than in other generations. 

e. True. In third-generation scanners, data collection is continuous for the full 360°.

a. False. In multi-slice scanners, a larger change in sensitivity of the detector is needed to cause ring artefacts, so it is less susceptible. 

b. True. Blurry boundaries between high-contrast details results from the cone beam effect. 

c. True. 

d. False. 

e. True. 

a. True. If the repetition time (TR) is shorter than the T₂ of the tissue, residual transverse magnetization remains at the time of the next radio frequency (RF) excitation pulse. Spoiling eliminates this residual transverse magnetization and the image weighting becomes more T₁-like with increasing flip angle. 

b. True. RF spoiling is the most common method for destroying transverse magnetization. Spoiling using various gradient ‘killer’ pulse schemes has been tried, but is not as effective. Examples of RF-spoiled sequences are SPGR (spoiled gradient recalled), FLASH (fast low-angle shot) and T₁-FFE (fast field echo). 

c. True. These sequences preserve residual magnetization (steady state) by using a short TR and rewinding the phase-encoding gradient after the echo has been acquired so that there is no residual phase shift dependent on the phase-encoding step. The resulting signal is proportional to the ratio of T₂/T₁; therefore, water is hyperintense on these sequences, making them perfectly suited to these types of imaging. Other uses include 3D and breath-hold imaging. Examples of such sequences are GRE, FISP (fast imaging with steady-state precession) and FFE. 

d. False. Motion and flow destroy the steady state on which these sequences depend. 

e. True. Due to the very short TRs, in order not to saturate the transverse magnetization, only very small excitation flip angles are used, resulting in poor T₁ contrast. This can be overcome by applying preparation pulses, such as a 180° inversion prior to acquisition, which improves the T₁ contrast between tissues.

a. True. This allows shorter repetition times (TRs) to be used as not all the longitudinal magnetization is tipped into the transverse plane with each excitation. The image contrast is therefore dependent not only on echo time (TE) and TR but also on the flip angle. 

b. True. 

c. False. The echo is formed by a reversal of the frequency-encoding gradient. A prephasing gradient is used initially to dephase the spins, followed by the standard frequency-encoding gradient of the opposite polarity, which rephases the spins. Rephasing (and therefore signal amplitude) is maximal in the middle of the frequency-encoding gradient. 

d. False. The Ernst angle is the excitation flip angle that maximizes the magnetic resonance signal for a given tissue T₁ and for a given TR. The Ernst angle (x_E) is given by cos(x_E) = exp(-TR/T₁). This angle may not necessarily provide the optimum contrast between any two tissues. 

e. True. This is because local field inhomogeneities are not eliminated due to the absence of a 180° refocusing pulse

a. False. Motion artefacts occur in the phase-encoding direction regardless of the actual direction of motion and are usually seen as either a smearing (induced by random motion) or ghosting (periodic motion). 

b. True. Aliasing is encountered in the phase-encoding direction when a part of the patient that lies beyond the chosen field of view appears on the opposite side of the image (known as a wrap-around artefact). A field of view smaller than the object in the frequency-encoding direction does not exhibit aliasing as the frequency-encoded signals are electronically filtered to remove the frequencies outside the receiver bandwidth. Note that in 3D scans, aliasing can also occur in the slice-select direction, as 3D slice encoding also relies on a phase shift. 

c. False. Magnetic susceptibility artefacts are worse on gradient-echo sequences, which do not provide compensation for local magnetic field inhomogeneities. 

d. False. Because the chemical shift (difference in resonance frequency) between fat and water is proportional to the static magnetic field strength, the resulting artefacts may be more pronounced at higher field strengths. 

e. True. Ringing appears as alternating dark and bright lines parallel to a high-contrast interface and is due to insufficient sampling of higher spatial frequencies. It can occur along the frequency- or phase-encoding direction, but is more commonly seen in the latter (due to the phase matrix being smaller in order to reduce scanning time).

a. False. The Ernst angle is the flip angle at which the signal is maximal for a given T₁ and repetition time (TR). Increasing the flip angle above this value would lead to a decrease in signal. 

b. False. SNR is inversely proportional to the square root of the bandwidth. A larger bandwidth increases the detected noise. 

c. True. This reduces the volume of tissue from which noise is detected. 

d. True. In general, spin echo gives higher signal than GRE. 

e. False. As the TE increases, more T₂ decay occurs resulting in a smaller signal.

a. True. The signal will increase at the expense of a reduction in spatial resolution. 

b. False. Decreasing the receiver bandwidth decreases the strength of the frequency-encoding gradient resulting in a larger chemical shift between water and fat. 

c. False. It is proportional to the square root of the number of repetitions. Thus, repeating the sequence four times would result in a twofold improvement in the SNR. 

d. True. As a first approximation, the SNR increases linearly with magnetic field strength. 

e. False. There are a number of methods to improve image contrast, including fat suppression, magnetization transfer and the use of contrast media. Image contrast will also depend on the chosen echo and repetition times, which will determine the image weighting.

a. False. They are usually made from rare earth materials – typically neodymium-iron-boron (NIB) or samarium cobalt (SmCo). 

b. True. Old air-cored designs are no longer in use. 

c. False. The value for permanent magnets is correct; however, currently available iron-cored resistive magnets give field strengths of up to 0.6 T only. 

d. False. In permanent magnets, the fringe field is almost entirely contained within the magnet. 

e. True.

a. True. 

b. False. This contains the 3 mT (30 Gauss) field contour. 

c. True. However, they cannot be brought into the inner MR controlled area where a projectile hazard exists. Good practice would indicate that ferromagnetic objects are also kept out of the controlled area. 

d. False. Patients with pacemakers are not allowed to be exposed to a magnetic field stronger than 0.5 mT (MR controlled area; note this can extend beyond the actual MR room). This may change in the future given the availability of MRI-compatible pacemakers. 

e. False. Equipment marked as ‘MR Conditional’ can also be brought into the inner controlled area, providing the specified conditions are met.

a. True. This is due to eddy currents arising in conducting loops. Burns can also occur where the arms and legs are touching, forming a conducting loop pathway. 

b. False. The main risk is that of a ferromagnetic object entering the inner MR controlled area becoming a projectile. 

c. True. The Lenz effect is an induction of a magnetic field in a moving conductor that opposes the external magnetic field. It is of low significance at a 1.5 T field, but is a potential hazard at higher field strengths. 

d. False. This is caused by magnetic field gradients. 

e. True. In the first-level controlled mode with appropriate monitoring, the allowable temperature rise is 1°C.

(A) a_A= 20mm/s , a_B= 10mm/s

Centripetal force is the component of force acting on an object in curvilinear motion which is directed towards the axis of rotation or centre of curvature. Centrifugal force is a pseudo force in a circular motion which acts along the radius and is directed away from the centre of the circle.

Given here: Atomic mass of iron = 56
now, 1 mole of iron contains 56 g of iron.
no. of moles in 224 g iron = n= 224/56 = 4
Now,
Total number of electrons or protons in 4 moles of iron = n ×​ avagadro number = 4 ×​ 6.032 ×​ 10²³ 
0.01 % of  4 ×​ 6.032 ×​ 10²³ ​ =  a = (0.01/100) × 4 ×​ 6.032 ×​ 10²³ ​= 2.41×​10²⁰

Now, defecient negative charge = a ×​ charge on  electron  

= 2.41×​10²⁰  × 1.602 × 10^-19 = 38.6 C

Positive charge acquired by 224g iron = defecient negative charge​  = 38.6 C​

In this question there is no consideration of volume density change, we have only to use formulae σ=q/a
and also a= 4πr ²
r₁=10cm
r₂=40cm
According to equation 
q₁/q₂=r₂² /r₁2

Thus
q₁/q₂= (40x40)/(10x10)
q₁/q₂=16/1
 

Ratio of charges is 16:1

Procedure for doing real Lab:

Take two similar spring balances of different ranges, say A and B.

Note the least count of the spring balances.

Attach the ring of spring balance A on a hook fixed in the wall and the spring balance B is attached to the hook of spring balance A.

Hold the spring balances exactly horizontal to the table.

Pull the ring of spring balance B gently.

Observe and note the reading of both the spring balances. Repeat the experiment by applying different forces.

Procedure for doing simulator:

Change the ‘Applied force’ slider and observe the reading on the two spring balances.

Repeat the experiment by applying different forces.

Click on the reset button for reset the experiment.

Observations:

Least count of the spring balance = value of 1 small division

Least count of the spring balance A=………N

Least count of the spring balance B =…………………N

No. of observations

Reading of B (Second spring balance) when force applied (N)

Reading of A (First spring balance) when force applied (N)

Difference in reading of A and B (N)

Result:

The readings on both the balances are the same in each case. Thus action and reaction forces are equal and opposite and act on two different bodies.

Precautions:

Spring balance of different least count should be taken.

The spring balance should be brought in elastic mode before doing the experiment.

The second spring balance should not be pulled beyond its elastic limit.

The reading of the spring balance should be taken without any parallax error.

(c) Economic environment

D. 100 V in secondary coil 

Education and Health facility is not a factor of discouraging settlement.