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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 94801. |
Open-chain saturated hydrocarbons are called___.(a). Alphatic(b). Cyclic (d). Acyclic(e) Paraffin |
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Answer» The correct option is (d) Paraffin An alkane, or paraffin (a historical name that also has other meanings), is a saturated hydrocarbon. Alkanes consist only of hydrogen and carbon atoms and all bonds are single bonds. Alkanes (technically, always acyclic or open-chain compounds) have the general chemical formula CnH2n+2. For example, methane is CH4, in which n = 1 (n being the number of carbon atoms). |
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| 94802. |
Which of the following is an amphoteric hydroxide?(1) Sr(OH)2 (2) Ca(OH)2 (3) Mg(OH)2 (4) Be(OH)2 |
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Answer» Correct option (4) Be(OH)2 Explanation: Be(OH)2 amphoteric in nature, since it can react both with acid and base Be(OH)2 + 2HCl → BeCl2 + 2H2O Be(OH)2 + 2NaOH → Na2 [Be(OH)4] |
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| 94803. |
Terry was jealous ______ Tom’s success. A) of B) on C) by D) at |
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Answer» Correct option is A) of |
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| 94804. |
I was very surprised ______ the news of your marriage. A) by B) with C) on D) at |
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Answer» Correct option is D) at |
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| 94805. |
He thinks he is superior ______ us because his father is very rich. A) to B) over C) on D) from |
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Answer» Correct option is A) to |
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| 94806. |
My father thinks that all heavy drug users should be put in jail. A) joints B) johnies C) junkies |
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Answer» Correct option is C) junkies |
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| 94807. |
When blood containing "Haemoglobin" as the colloidal partial is subjected to electro osmosis. The dispersion medium moves towards the anode. Hence, the incorrect statement will beA. `Fe(OH)_(3)` solution on mixing with haemoglobin in blood can cause coagulationB. `As_(2)S_(3)` solution can cause coagulation in blood on mixing.C. In electrophoresis of blood, movement of colloidal particle will be towards cathode.D. NaCl on mixing in blood can cause coagulation. |
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Answer» Correct Answer - A Dispersion medium moves towards anode `rArr` dispersion medium has -ve charge `rArr` dispersed phase has +v charge `rArr Fe(OH)_(3)` cannot cause coagulation |
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| 94808. |
He's in the toilet at the moment. A) can B) cupboard C) cold |
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Answer» Correct option is A) can |
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| 94809. |
I need to go to the toilet. Can you show me where it is? A) jump B) john C) jug |
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Answer» Correct option is B) john |
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| 94810. |
In this method approx. total length of walls is found in running metre and this total length multiplied by the rate per running metre of wall gives a fairly accurate cost.(a) Annual repair(b) Item rate estimate(c) Approximate quantity method estimate(d) Cubical content estimate |
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Answer» The correct option is (c) Approximate quantity method estimate The best I can explain: For this method the structure may be divided into two parts viz. Foundation including plinth and Superstructure. |
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| 94811. |
The relation between the direction of current and the direction of the force is ________(a) Same direction(b) Opposite direction(c) Perpendicular(d) Unrelated |
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Answer» The correct choice is (c) Perpendicular The explanation is: When a conductor carries a certain value of current, the force developed in the conductor, the current in the conductor and the magnetic field in the conductor are mutually perpendicular to each other. |
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| 94812. |
In long and short wall method of estimation, the length of long wall is the centre to centre distance between the walls and ________________(a) breadth of the wall(b) half breadth of wall on each side(c) one fourth breadth of wall on each side(d) length of the wall |
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Answer» The correct option is (b) half breadth of wall on each side To explain: This method is used for deriving quantities of different items of work in a load bearing structure. The wall that contains lengthiest length on plan is considered as “long wall” and all the walls parallel to the certain long wall (regardless of length) are also taken as long walls. The wall that is perpendicular to the direction of long wall is taken as “short wall” alias “cross wall”. |
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| 94813. |
Magnetic field is strong when____________(a) magnetic field lines are closer(b) magnetic field lines are farther(c) magnetic field lines are longer(d) magnetic field lines are thicker |
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Answer» Correct option is (a) magnetic field lines are closer To explain: Magnetic field is strong where magnetic field lines are closer and weak where magnetic field lines are farther. |
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| 94814. |
The expected out turn for earth work in excavation in ordinary soil per labourer per day is _____(a) 1 cum(b) 2 cum(c) 3 cum(d) 4 cum |
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Answer» Right answer is (c) 3 cum Explanation: Sand = 4 cum Ordinary soil = 3 cum Compacted soil = 2 cum Rock = 1 cum. |
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| 94815. |
Excavation may be classified by type of material- Earth excavation ________ excavation, rock excavation, muck excavation, unclassified excavation.(a) bottom soil(b) sub-soil(c) topsoil(d) surface soil |
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Answer» Right answer is (c) topsoil Explanation: Topsoil is the upper, outermost layer of soil, usually the top 2 inches (5.1 cm) to 8 inches (20 cm). It has the highest concentration of organic matter and microorganisms and is where most of the Earth’s biological soil activity occurs. Four elements constitute the composition of soil. Those elements are mineral particles, organic matter, water, and air. The volume of top soil consists of 50 to 80% of these particles which form the skeletal structure of most soils. |
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| 94816. |
If the volume of a sphere increases at constant rate `((dv)/(dt)=4)`. If radius of the sphere is denoted by `r` then surface area of the sphere inceases at the rate`:-`A. `(4)/(r )`B. `(8)/(r )`C. `(12)/(r )`D. `(16)/(r )` |
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Answer» Correct Answer - B `V=(4)/(3)pir^(2)rArr(dV)/(dt)=(dV)/(dt)xx(dr)/(dt)=[(4)/(3)pi(2r)](dr)/(dt)=4` `rArr(dr)/(dt)=(4)/(4pir^(2))` `5=4pir^(2)rArr(dS)/(dt)=(dS)/(dr)xx(dr)/(dt)=[4pi(2r)](dr)/(dt)` `(dS)/(dt)=(8pir)((4)/(4pir^(2)))=(8)/(r)` |
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| 94817. |
Potential Energy `U(x)` and associated force `F(x)` bears the relation `F(x)=-(d)/(dx)U(x)`. Dependence of potential energy of a two`-`particle system separation `x` between them is shown in the following figure. Which of the following graphs shows correct variation in assoicated force `F(x)` with `x`?B. C. D. |
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Answer» Correct Answer - B `F(x)=-(dU(x))/(dt)=-`slope of `U (v)//(s)x` curve Intially slope is constant and then it becomes zero. |
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| 94818. |
Electric current in a wire is time rate of flow of charge. The charge in coulombs that passes through a wire after `t` seconds is given by the function `q(t)=t^(2)-2t^(2)+5t+2`. Determine the average current (in coulmbe per second) during the first two seconds.A. `4A`B. `5A`C. `6A`D. `12A` |
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Answer» Correct Answer - B `q(0)-=2,q(2)=12` `i_(a^(1***))=(Deltaq)/(Deltat)=(12-2)/(2)=(10)/(2)=5` |
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| 94819. |
A metal rod extends on the `x-`axis from `x=0m` to `x=4m`. Its linear density `rho` (mass`//`length) id not uniform and varies with the distance `x` from the origin according to equation `rho=(1.5+2x)kg//m` What is the mass of rod`:-`A. `16kg`B. `22kg`C. `28kg`D. `32kg` |
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Answer» Correct Answer - B `(dm)/(dx)=(1.5+2x)rArrm=int_(0)^(4)(1.5+2x)dx` `implies m={:(4),(0):}[1.5x+(cancel(2)x^(2))/(cancel(2))]`=1.5(4)+16=22kg` |
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| 94820. |
For the given cyclic process CAB as shown for a gas, the work done is :(1) 1J (2) 5J (3) 10J (4) 30J |
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Answer» The correct option (3) 10 J Explanation: ∫pdV = 1/2(4)(5) = 10J |
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| 94821. |
The figure shows a network of currents. The magnitude of currents is shown here. The current I will be (a) -3A(b) 3A(c) 13A(d) 20A |
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Answer» Correct option(d) Explanation: According to Kirchhoff’s Ist law I = 7A + 13 A I = 20 A |
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| 94822. |
The approximate protection requirements for the walls/screens of a typical diagnostic X-ray room would be: a. 2 mm of lead b. 2 cm of glass c. 2 cm of barium plaster d. 2 cm of concretee. 24 cm of brick |
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Answer» a. True. b. False. c. True. d. False. Generally, 150 mm solid concrete provides sufficient shielding. e. True. 120 mm solid brick = 1 mm lead. |
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| 94823. |
Which of the following are correct regarding electronic dosimeters? a. They are not highly energy dependent b. Their sensitivity can be 100 times that of a film badge c. They can be gas-filled tubes d. They do not need a filter e. They are useful in identifying methods of dose reduction for procedures with a potential high dose to staff |
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Answer» a. False. Their response is highly energy dependent. b. True. They are able to measure down to 1 mSv, while film sensitivity is not better than 0.1 mSv. c. True. They can be based on Geiger-Müller tubes (gas-filled tubes). d. False. They are placed behind a filter to give an accurate reading. e. True. Because they provide a direct reading, they are useful for dose reduction for high-dose procedures. |
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| 94824. |
Magnetic force on a charged particle is given by `vec F_(m) = q(vec(v) xx vec(B))` and electrostatic force `vec F_(e) = q vec (E)`. A particle having charge q = 1C and mass 1 kg is released from rest at origin. There are electric and magnetic field given by `vec(E) = (10 hat(i)) N//C for x = 1.8 m` and `vec(B) = -(5 hat(k)) T` for `1.8 m le x le 2.4 m` A screen is placed parallel to y-z plane at `x = 3 m`. Neglect gravity forces. y-coordinate of particle where it collides with screen (in meters) isA. `(0.6(sqrt(3)-1))/(sqrt(3))`B. `(0.6(sqrt(3)+1))/(sqrt(3))`C. `1.2(sqrt(3)+1)`D. `(1.2(sqrt(3)-1))/(sqrt(3))` |
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Answer» Correct Answer - D In magnetic field path of the particle is circle. Radius of circular path is `r=(mv)/(qB)=((1)(6))/((1)(5))=1.2m` `d=2.4m-1.8m=0.6m` Since `d lt r sin theta=d/r=0.6/1/2 = 1/2` `implies theta=30^(@)` `AE=AD-DE=r-r cos theta` `=r(1-cos theta) = 1.2 (1-(sqrt(3))/(2))=0.6(2-sqrt(3))` `FC=BF tan theta = 0.6/(sqrt(3))` `:.` y-co-ordinate `=AE+FC` `=0.6 (2-sqrt(3))+(0.6)/(sqrt(3))` `0.6[2-sqrt(3)+(1)/(sqer(3))]=(1.2(sqrt(3)-1))/(sqrt(3))m`. Hence choice (d) is correct and other choices are wrong. |
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| 94825. |
Which of the following are correct regarding the stochastic effects of radiation? a. They can be hereditary in nature b. They arise as a result of chance c. The probability of the effect occurring increases with dose d. Increasing the dose results in an increase in the severity of the disease e. There is no threshold dose for stochastic effects |
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Answer» a. True. They may occur in descendants of individuals exposed as a result of lesions in the germinal cells. b. True. c. True. d. False. This is true of deterministic effects. Stochastic effects either occur or do not occur. Their severity is not affected by the dose. e. True. |
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| 94826. |
Which of the following are correct regarding thermoluminescent dosimeters (TLDs)? a. A TLD is not susceptible to environmental effects, especially high temperature b. A TLD shows a linear response to illumination and can be read only once c. TLD sensitivity is not significantly better than film d. A TLD needs a filter e. They can give readings down to 0.01 mSv |
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Answer» a. False. High temperatures can remove all the information from the TLD. b. True. c. True. d. True. They are used in conjunction with filters set in the badge holder. e. False. Optically simulated luminescent dosimeters can give readings down to 0.01 mSv. TLD sensitivity is similar to that of films (0.1 mSv). |
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| 94827. |
Which of the following are true regarding dosimetry? a. Kerma takes into account the type of tissue being irradiated b. The absorbed dose is measured in grays (Gy) c. The equivalent dose takes into account the radiation weighting factor (wR) d. The effective dose is measured in Gy e. 1 Gy = 1J g-1 |
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Answer» a. False. Kerma is the kinetic energy of the secondary electrons released per unit mass of the irradiated material. It does not factor in the type of material being irradiated. b. True. c. True. Equivalent dose = absorbed dose radiation weighting factor (wR). It takes into account the absorbed dose and the type of radiation, e.g. X-ray, gamma ray or beta particle. d. False. The effective dose is measured in sieverts (Sv) e. False. 1 Gy = 1 J kg-1 |
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| 94828. |
Which of the following statements are correct regarding a patient’s dose and the dose–area product (DAP)? a. The DAP can easily be measured and is directly related to the radiation risk b. The DAP can be converted to the absorbed dose using a conversion factor c. The conversion factors depend only on the region of the body, not the kV or filtration d. For the same region of the body, the conversion factor is the same for anterior–posterior (AP) and posterior–anterior (PA) views e. The entrance surface dose in the lateral spine view is more than that of the AP view |
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Answer» a. False. The DAP can easily be measured, but it is not directly related to the radiation risk. b. False. The DAP can be converted to the effective dose using the conversion factor. c. False. Conversion factors depend on the region of the body and to a lesser extent on kV and beam filtration. d. False. Conversion factors for PA examinations are less than AP examinations of the same region, because generally the organs and tissues with higher weighting factors are located anteriorly. e. True. The entrance surface dose in the lateral spine view could be around 10 mGy, while it is only around 4.3 mGy in the AP view. |
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| 94829. |
Which of the following are true regarding film badges in personal dosimetry? a. The film is highly energy dependent because of the intensifier b. The film has two different emulsions on each side c. The film can detect beta radiation and radioactive splash d. The aluminium filter will stop beta particles and the cadmium filter stops alpha particles e. As the film provides a permanent record of exposure, it can be used for more than 6 months |
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Answer» a. False. The film is highly energy dependent because of the high atomic number of silver and bromide. There is no intensifier in the film badge dosimeter. b. True. c. True. d. True. e. False. The film is subject to environmental effe |
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| 94830. |
Which of the following are correct regarding deterministic effects of radiation? a. It has a minimum threshold below which it does not occur b. The severity of the effect increases with dose c. The probability of the effect occurring increases with dose d. Breast cancer is a type of deterministic effect e. Skin erythema is a type of deterministic effect |
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Answer» a. True. Deterministic effect is characterized as having a threshold dose below which the effect will not occur. b. True. Once the threshold has been exceeded, increasing the dose results in the severity of the disease increasing. c. False. The effect occurs once the threshold is exceeded. d. False. This is a type of stochastic effect. e. True. |
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| 94831. |
With regard to diagnostic reference levels (DRLs), which of the following are true? a. They are defined as the maximum dose for a particular examination b. They are adjusted according to the patient’s body habitus c. They are set at the local level by the employer d. They can be assessed in terms of screening time e. If the DRL is exceeded, this must be reported to the Healthcare Commission |
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Answer» a. False. DRLs are defined as doses for typical examinations of that type. They can be thought of as performance standards against which individual patient doses can be judged. b. False. They are set for standard-sized patients. c. True. They are set locally by performing patient dose audits. d. True. DRLs are set in terms of measurable quantities such as screening time, dose–area product and entrance surface dose. e. False. The patient dose may exceed the DRL for that examination, especially if the patient is overweight. |
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| 94832. |
Regarding radiation dose: a. Linear energy transfer (LET) is the total energy deposited by a particle along its entire path b. An alpha particle has a lower LET than an electron c. Particles with high LET are more hazardous than low-LET particles d. The effective dose takes into account the radio sensitivity of the tissues e. 1 Sv = 1 J kg-1 |
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Answer» a. False. LET is the sum of the energy deposited in tissue per unit path length it travels. b. False. Alpha particles are much heavier than electrons and hence for the same initial energy as an electron, they travel a much shorter distance and therefore disperse more energy per unit area travelled. c. True. High-LET particles disperse more energy per unit area travelled, resulting in more non-repairable damage. d. True. The effective dose incorporates factors to account for the variable radio sensitivity of organs and tissues. e. True. |
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| 94833. |
Factors that reduce the dose to the patient include: a. Increased filtration b. Reduced collimation c. Using a grid with paediatric patients d. Using an intensifying screen e. Increasing the X-ray focus-to-patient distance |
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Answer» a. True. b. False. Better collimation prevents unnecessary exposure to tissues and also acts to reduce scatter. c. False. As younger patients are smaller, a lower kV may be used and so scatter will be less and grids can be avoided. Because grids reduce the intensity of the main radiation beam, a grid will result in an increased dose to the patient by increasing the required mA. d. True. e. True. However, this can lead to a reduction in magnification. |
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| 94834. |
Regarding dose and hazards from X-rays: a. The dose from radiation scattered from the patient reduces roughly in proportion to the reduction in the area of the irradiated field b. A 0.25 mm lead apron will attenuate the scatter radiation by a factor of about 100 c. Doubling the area of the X-ray beam approximately doubles the skin dose rate d. The effective dose to the patient can be measured directly using a thermoluminescent dosimeter (TLD) e. Someone in the primary beam at 2 m from the X-ray set could receive the effective dose limit in a few seconds |
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Answer» a. True. b. False. A 0.25 mm apron transmits 5% of the scatter radiation. c. False. The skin dose rate will remain the same. However, the skin dose itself will increase. d. False. e. True. A few seconds is long enough for the primary X-ray beam to reach the dose limit. |
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| 94835. |
Regarding the high-voltage supply to the X-ray tube: a. In a self-rectified tube, X-rays are produced during both the positive and the negative halves of the supplying voltage waveform b. In a tube connected to a half-wave rectifying circuit, X-rays are produced during both halves of the supplying voltage waveform c. In a fully rectified one-phase circuit, the voltage across the tube is almost constant d. A fully rectified three-phase supply provides more constant voltage throughout the cycle than a one-phase supply e. A tube supplied from a high-frequency generator requires less filtration in comparison with that supplied from rectified circuits |
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Answer» a. False. In a self-rectified tube, electrons travel towards the anode only during the positive half of the waveform. During the negative half of the cycle, the filament has positive potential and does not emit electrons; at high tube currents, the anode may reach temperatures sufficient for thermionic emission and electrons may flow towards the filament, destroying the tube. b. False. In a half-rectified circuit, the negative half of the voltage waveform is eliminated. As potential across the tube during this period of the cycle is zero, there is no electron flow and no X-ray production. c. False. In a fully rectified circuit, the negative half of the cycle is ‘flipped’ into positive. Therefore, in both halves of the cycle, the filament and the target have the correct polarity to produce X-rays. However, the voltage across the tube fluctuates from zero to maximum, and the mean energy of the photons produced is relatively low. d. True. With three overlapping fully rectified wave forms, the fluctuations in the tube voltage (ripple) are greatly diminished. The voltage across the tube never falls to zero and the efficiency of X-ray production is higher, as is the mean photon energy. High-frequency generators provide yet further improvements with a practically constant potential. e. True. More filtration is needed to eliminate photons with lower energies produced by the latter. |
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| 94836. |
Increasing the tube potential: a. Improves film contrast b. Increases the dose to the patient c. Reduces the exposure time d. Increases geometric unsharpness e. Has no effect on motion unsharpness |
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Answer» a. False. It decreases film contrast by reducing the probability of photoelectric absorption in tissues and increasing the scattered radiation contributing to the final image. b. False. The beam is more penetrating; therefore, less radiation is required to produce an adequate image. c. True. d. False. It has no effect. e. False. Using a higher kV reduces the required exposure time so it can indirectly reduce motion unsharpness. |
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| 94837. |
The amount of scattered radiation leaving the patient is decreased by: a. Decreasing the tube kV b. Reducing the field area by collimation c. Compression of the patient d. Using a radiation grid e. Using the air-gap technique |
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Answer» a. True. There is less scatter in the forward direction and it is less penetrating. b. True. This reduces the amount of scatter produced. c. True. This also reduces the amount of scatter produced. d. False. A grid reduces the amount of scattered radiation reaching the image, but not the amount of scatter leaving the patient. As it necessitates an increase in tube output, the grid may actually increase the amount of scatter produced (and the patient’s dose). e. False. This is based on the same principle as (d). |
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| 94838. |
Regarding factors affecting X-ray emission: a. The tube voltage affects the maximum energy of Bremsstrahlung radiation b. Due to inherent tube filtration, the maximum energy of photons measured in keV is slightly lower than the tube potential in kV c. Beam filtration affects both the quantity and quality of the X-ray beam d. Increasing peak kV (kVp) increases the quantity and quality of the emitted X-ray beam e. Increasing the tube current decreases the half-value layer (HVL) of the beam |
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Answer» a. True. b. False. Filtration does not affect the maximum energy of the beam spectrum, which is always equal in keV to the tube potential in kV. c. True. Filtration selectively removes the lower energy photons so it reduces the quantity. It shifts the mean energy of the remaining photons to a higher level, so it increases the beam quality. d. True. e. False. The HVL depends on the quality of radiation (energy spectrum), which is not changed by altering mA. |
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| 94839. |
Patient dose in radiography can be reduced by: a. Increasing the focus-to-film distance b. Compression c. Collimation d. Using a secondary radiation grid e. Using the air-gap technique |
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Answer» a. True. Even though increasing the distance necessitates the increase in mAs, it decreases the skin dose as the beam entering the patient is spread over a larger surface area. It also decreases the dose to deeper tissues to a lesser extent. b. True. This is due to a reduction of the required exposure factors (less tissue thickness to penetrate) and the amount of scattered radiation. c. True. d. False. The exit dose needs to be increased by a factor equal to the Bucky factor of the grid (the ratio of radiation incident on the grid to the transmitted radiation) to obtain the same film exposure. e. False. Use of an air gap requires a higher tube output and increases the patient dose. |
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| 94840. |
Regarding patient dose in radiography: a. The ratio of entrance surface dose (ESD) to exit dose is similar in most radio graphic examinations b. The film dose required to produce an image in film-screen radiography is in the range of 0.2–0.5 μGy c. Increasing mAs increases the exit dose and ESD proportionately d. Increasing the tube kV increases the ratio of ESD to the exit dose e. Increasing filtration reduces the ratio of ESD to the exit dose |
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Answer» a. False. Depending on examination type, this ratio may vary from 10 (for a PA chest X-ray) to over 5000 (in the case of a lateral lumbar spine radiograph). b. False. A dose of approximately 3 μGy is required. c. True. d. False. With increased kV, the X-ray beam is more penetrating and a lower entrance dose is required to produce the same exit dose. In general, the highest peak kV (kVp) producing acceptable image contrast should be used. e. True. This is because increasing filtration makes the beam more penetrating. |
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| 94841. |
X-ray exposure ratings: a. Determine the operational limits of the X-ray equipment b. May be illustrated graphically c. Are independent of focal spot size d. Are dependent on the anode rotation speed e. Are the same for single and repeated exposures |
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Answer» a. True. They describe the maximal operational settings that are permissible without damage to the anode and tube housing. b. True. There are rating charts that display curves showing the maximum permissible mA depending on exposure time and peak kV (kVp) for a particular X-ray tube and focal spot size. c. False. The size of the focal spot has an influence on the tube current that can safely be tolerated. d. True. e. False. Ratings for repeated exposures are lower as heat accumulation needs to be taken into account. |
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| 94842. |
Regarding quality assurance in radiography: a. The tube kV is tested using a dosimeter b. Tube output is assessed with an ionization chamber or a solid-state detector c. For a given mAs, the output of the tube should be constant over the entire voltage range d. Tube filtration is assessed by measuring the half-value layer (HVL) e. Light beam diaphragm alignment should be checked yearly. |
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Answer» a. False. It is tested using a digital kV meter. b. True. c. False. In the range of 60–120 kV, tube output is proportional to the square of the peak kV (kVp2 ). For lower kV (e.g. in mammography), output is proportional to kVp3 . d. True. e. False. It is usually tested every 1–2 months. |
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| 94843. |
Find the current I in the j10 Ω branch of the given circuit shown in Fig. using the Nodal Method. |
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Answer» There are two principal nodes out of which node No. 2 has been taken as the reference node. V1(1/(6 + j8) + 1/(6 - j8) + 1/j10) - (100 ∠ 0º)/(6 + j8) (100 ∠- 60º)/(6 - j8) = 0 V1(0.06 −j0.08 + 0.06 + j0.08 −j0.1) = 6 −j8 + 9.93 − j1.2 = 18.4 ∠−30º ∴ V1(0.12 −j0.1) = 18.4∠30º or V1 × 0.156 ∠−85.6º = 18.4 ∠− 30º ∴ V1 = 18.4 ∠− 30º/0.156∠−85.6º = 118∠ 55.6ºV ∴ V = V1/j10 = 118∠55.6º/j10 = 11.8∠−34.4ºA |
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| 94844. |
In the network of Fig. determine the current flowing through the branch of 4 Ω resistance using nodal analysis. |
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Answer» We will find voltages VA and VB by using Nodal analysis and then find the current through 4Ω resistor by dividing their difference by 4. V2(1/5 + 1/4 + 1/j2) - VB/4 - (50∠30º)/5 = 0 ∴ VA(9 −j10) − 5VB = 200 ∠30º ...(i) Similarly, from node B, we have VB(1/4 + 1/2 + 1/-j2) - VA/4 - (50 ∠ 90º)/2 = 0 ∴ VB (3 + j2) −VA = 100 ∠90º = j 100 ...(ii) VA can be eliminated by multiplying. Eq. (ii) by (9−j10) and adding the result. ∴ VB(42 −j12) = 1173 + j1000 or VB = (1541.4 ∠ 40.40º)/(43.68 ∠-15.9º) = 35.29∠56.3º = 19.58 + j29.36 Substituting this value of VB in Eq. (ii), we get VA = VB(3 + j2) −j100 = (19.58 + j29.36) (3 + j2) −j100 = j27.26 ∴ VA −VB = j 27.26 − 19.58 −j29.36 = − 19.58 −j2.1 = 19.69∠186.12º ∴ I2 = (VA −VB)/4 = 19.69∠186.12º/4 = 4.92∠186.12º |
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| 94845. |
The molar heat capacity of an ideal gas in a process varies as `C=C_(V)+alphaT^(2)` (where `C_(V)` is mola heat capacity at constant volume and `alpha` is a constant). Then the equation of the process isA. `Ve^(-((alphaT^(2))/(2R)))=` ConstantB. `Ve^(-((alphaT^(2))/(R)))=` constantC. `Ve^(-((2alphaT^(2))/(R)))=` constantD. `Ve^(-((3alphaT^(2))/(2R)))=` constant |
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Answer» Correct Answer - A Given `C=C_(V)+alphaT^(2)` `C_(V)+(RT)/(V)(dV)/(dT)=C_(V)+alphaT^(2)` `int(alphaT)/(R)dT=int(dV)/(V)+lnk` `(alphaT^(2))/(2R)=ln(kV)` `kV=e^((alphaT^(2))/(2R))` `thereforeVe^(-((alphaT^(2))/(2R)))`=constant |
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| 94846. |
Correct order is :A. X-rays `lt` UV rays `lt` Radio waves (energy)B. Radio waves `lt` Micro waves `lt` Infra red rays (wave number)C. Cosmic rays `lt gamma`- rays `lt` X-rays (frequwncy)D. All are correct |
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Answer» Correct Answer - B The maximum volume at STP is …………… moles of `SO_(2) = 12.8//64 = 0.2 , V = 0.2xx22.4L` moles of `NO_(2) = 0.5 , V = 0.5xx22.4L` moles of `CH_(4) = (6.02xx10^(22))/(6.023xx10^(23)) = 0.1 V= 0.1xx22.4L` moles of `CO_(2) = 1` mole , `V= 1xx22.4L` |
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| 94847. |
Identify the correct option for the order of specific charge of the given particles.A. `n lt alpha lt p lt e` (considering sign of charged particle)B. `n lt alpha lt p lt e` (not considering sign of charged particle)C. `e lt n lt p lt alpha` (considering sign of charged particle)D. `n lt e lt alpha lt p` (considering sign of chatged particle) |
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Answer» Correct Answer - B Identify the correct option for the order ………….. `{:("e/m ratio","e/m ratio"),(("With sign"),"Without considering sign"),(eltnltalphaltp,nltalphaltplte):}` |
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| 94848. |
From a container having 64 g Oxygen, `11.2L` Oxygen gas at STP and `6.022xx10^(23)` Oxygen atoms are removed. Find the mass of the oxygen gas left :A. zeroB. 32 gC. 16 gD. none |
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Answer» Correct Answer - B From a container having 64 g Oxygen …………. Removed mass `= (11.2)/(22.4)xx32+(6.02xx10^(23))/(6.02xx10^(23))xx16=32g` mass left `= 64-32 = 32g`. |
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| 94849. |
Which of the following does not contain 32 g of oxygen ?A. 2 mol of `H_(2)O`B. 60 g of `C_(6)H_(12)O_(6)`C. `N_(A)` molecules of `CO_(2)`D. 60 g of urea, `NH_(2)-underset(O)underset(||)(C )-NH_(2)` |
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Answer» Correct Answer - D Which of the following does not ………….. (1) `n_(0) = 2mol` (2) `n_(C_(6)H_(12)O_(6)) = (60)/(180) = (1)/(3)mol` `n_(0) = (1)/(3)xx6 = 2mol` (3) `n_(0)=1xx2=2mol` |
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| 94850. |
His work was NOT INTERESTING. A) dull B) bright C) troublesome D) full E) difficult |
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Answer» Correct option is A) dull |
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