InterviewSolution
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.
| 4051. |
depend ?e.f.Sn11. (i) Acurrent of 1 ) Arnpere is nowingin long straightconductor. Calculate magnetic flux density (B) at a distance[Ans. 2 ×10-4 NAA-m)14of 1 em from conductorOx |
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| 4052. |
5 MARKS THEORY QUESTIONS1)Explain ohm's law applied to complete circuit.A conductor having diameter 0.4 mm has resistivity 5 x 10-7 Ωm and its length is 10.05 m. What isthe resistance of conductor?(R = 40Ω) |
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Answer» Ohm'slawstates that the current through aconductorbetween two points is directly proportional to the voltage across the two points. Thislawis one of the most basiclawsof electricity. It helps to calculate power, efficiency, current, voltage, and resistance of an element of an electricalcircuit. Resistance is given by: R=ρL/A ρ=resistivity L=length of the conductor(in m) A=area of cross section of the conductor (in m2) diameter is 0.4mm=0.4×10^-3m Area =πd^2/4=1.25×10^-7m2 ρ=RA/L=2.5×1.25×10^-7/1.2=2.6×10^-7ohm m |
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| 4053. |
38.Apersonaiming to reach the exactly opposite point on the bank of a stream iswimmingwith a speed of 0.5m/s at an angleof 120° with the direction of flow of waterThe speed of water in the stream is |
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Answer» Vᵣ = velocity of river or stream = ? Vs = velocity of swimmer = 0.5 m/s V= resultant velocity given angle is total angle ie 90+30= 120 From figure in triangle Δ sin θ = p/ h sin 30⁰ = Vᵣ / Vs 1/2 = Vr/ 0.5 so Vr= 1/4 ie Vr = 0.25 m/s so velocity of river is 0.25 m/ s hit like if you find it useful |
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| 4054. |
Unsolved Numerical Problems for Practice1.Calculate the amount of current which flows through a conductor if 10 electaint in 10 s |
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Answer» Current= charge/timeso, I=Q/there number of electrons is 10and time, t= 10so charge, Q= 10* 1.6*10^-19 C therefore, I=10* 1.6*10^-19 C/10I = 1.6*10^-19 A. (answer) |
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| 4055. |
Attempt the following questions : (Any five)(1) Write about artificial food colours, the substances used in them and theirharmful effects.(2) Aniket from Std. X uses spectacles. The power of the lenses in his spectaclesis-0.5 D |
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Answer» .Artificial food colurs makes the food colourful and more desirable. .They are responsible for the bright colors of candy, sports drinks and baked goods..beta carotene and beet extract natural food colurants.erythrosine, allura red, tartrazine are chemical food colurant.Harmful effects.linked with cancer.allergies.hyperactivity.irritability.hearing impairment.aggressiveness |
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| 4056. |
6. Write the equation of continuity ior the flow of incompressible fluid (water)?7. What is stream line flow? |
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Answer» A1V1=A2V2flowof a fluid in which its velocity at any point is constant or varies in a regular manner. It can be represented bystreamlines |
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| 4057. |
(13)Whenthe radiation coming out from certainradioactive substance is passed through anelectric field, marks are found at three placeson the photographic plate placed in its path.(iv) |
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Answer» Because the radiation coming from the nuclear substance consists of alpha rays beta rays and gamma rays alpha rays are positively charged therefore they deflect towards the negatively charged Cathode..Beta rays are negatively charged and therefore moves towards the positively charged anode.And Gamma rays are electrically neutral therefore they are not deflected by the electric field..Therefore a single ray coming from the nuclear substance gets divided into three rays and strike the photographic plate at three places.. |
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| 4058. |
A radioactive nucleus X converts into stable nucleus Y. Halfradioactive sample when the ratio of X and Y is 1:15.life of X is 50 years. Calculate the age of2 |
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Answer» Initial concentration of X is 1.Final concentration of X is 1/(1+15) = 1/16.Given: Half life of X is 50yrs.Therefore, X -> X/2 ( 50 years ) ,X/2 -> X/4 ( + 50 yrs ),X/4 -> X/8 ( + 50 yrs ),X/8-> X/16 ( + 50 yrs ) Therefore, age of radioactive sample is 50+50+50+50 = 200 years |
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| 4059. |
Recognize the type of filter given below?503+1(A) LPF(B) HPF(C) BPF(D) BSF |
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| 4060. |
Recognize the type of filter given below?LA(A) LPF(B) HPF(C) BPF(D) BSF |
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| 4061. |
Recognize the type of filter given below?von LA(A) LPF(B) HPF(C) BPF(D) BSF |
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| 4062. |
24. Three identical metal plates with large surface areakept parallel to each other as shown in figure (30-E8). Theleftmost plate is given a charge Q, the rightmost a charge-20 and the middle one remains neutral. Find the chargeappearing on the outer surface of the rightmost plate.--20Figure 30-E8 |
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| 4063. |
changedwhen:(i)slitwidthisb. How is the angular width of central bright maximumdecreased, (ii) the distance between the slit and screen is increased, (ii ight of smallerwavelength is used? Justify your answer in each case. |
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| 4064. |
Recognize the type of filter given below?(A) LPF(B) HPF(C) BPF(D) BSF |
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| 4065. |
98. In the folowing circuit, the resistance of the voltmeteris 10,000 Ω and that of the ammeter is 20 Ω . If thereading of the ammeter is 0.1 A and that of the 10voltmeter is 12 V, then the value of R is :-(1) 122 Ω(2) 100 Ω(3) 118Ω(4) 116210 |
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Answer» reading in the Voltmeter will give the potential difference across combination of R and A so, V = 12v , and since current is 0.1A so, 12 = 0.1(R+20) => 120 = R+20=> R = 120-20 = 100Ω option 2 |
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| 4066. |
७3५०० 3 o..\ \‘ RV A w o |
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Answer» A metal rod or wire fixed to an exposed part of a building or other tall structure to divert lighting harmlessly into the ground. |
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| 4067. |
Curet entes at A and ve at D14%rv |
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Answer» all the resistance are in series in the both side of AD so 1+1+1= 3ohmand same for lower Branch = 1+1+1 = 3ohmnow two resistance of 9ohm are connected in parallelso result will be 3*3/3+39/63/2= 1.5 ohm |
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| 4068. |
The reading of voltmeter is200Ί100V300V ) 60001002(1) 50V(2) 60 V(3) 40V(4) 80 VThe meter-hridne io D |
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| 4069. |
Consider the circuit shown in below figure.0.5 mHThe circuit shown is a,(a) Low pass filter(b) High pass filter(c) Wide Bandpass filter(d) Narrow bandpass filter |
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Answer» The circuit bis a high pass filter as the output voltage is taken across a resistor,Option b is correct. thqso much |
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| 4070. |
3)Consider the circuit shown in below figure.0.5 mH10ΊThe circuit shown is a(a) Low pass filter(b) High pass filter(c) Wide Bandpass filter(d) Narrow bandpass filter |
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Answer» The circuit is a high pass filter as the output voltage is taken across a resistor,Option b is correct. |
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| 4071. |
22. For a circuit shown in figure. calculateValue of effective resistance of circuit.(ii) Total current in the circuit(ii) Current in each resistance12 V |
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| 4072. |
For the circuit shown find I: |
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Answer» 6A will be the value of currentincoming current = outgoing current at each node we did it with the help of Kirchoff current law |
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| 4073. |
133. Find R. for the circuit shown in figure.rv50 20 |
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| 4074. |
18 . If the velocity of a body is increased 4% then what is the approximate% change in its K. E. ?Ans. 8%. |
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| 4075. |
eons the hills in terms of90 slopeb aspect |
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Answer» (i) elevation Mountains have more steeper slope than Hills and mountains have higher elevations than Hills. |
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| 4076. |
un willl a stick, dust com |
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Answer» इसके बारे में सबसे महत्वपूर्ण हिस्सा न केवल यह है कि वस्तुएं एक-दूसरे पर खींचती हैं, बल्कि यह है कि दो वस्तुएं एक-दूसरे को एक बल के साथ आकर्षित करती हैं जो उनके द्रव्यमान के उत्पाद के आनुपातिक हैं और उनके बीच की दूरी के वर्ग के व्युत्क्रमानुपाती हैं। इसे न्यूटन के सार्वभौमिक गुरुत्वाकर्षण के नियम के रूप में जाना जाता है। |
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| 4077. |
16. The rate of radiation from a black body at 0°C isQ J s^-1. Find the rate of radiation by the same blackbody at 273°C.Ans. 16 Q J s^-1 |
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| 4078. |
I am very weak in physics please any intelligent on give me ur WhatsApp number please |
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Answer» Contact the founders of Scholr I am unable to understand what u are saying you give me ur number If u have any query in the steps in the answer provided by us..u are free to ask.. why u are not giving ur phone number |
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| 4079. |
A black body, at temperature T K emits radiationat the rate of 81 W/m^2. f the temperature falls toT/3 K, then the new rate of thermal radiation will be |
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| 4080. |
Electric charges of +10pC,+54AC,-34C and +8C areplaced at the comers of a square of side 2 m. the potentialat the centre of the square isKCET (Engg Med.) 1999: MP PET 2006 |
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Answer» Let P be the center of square ABCDThen diagonal of square of length sqrt(2) m = 2m PA = PB = PC = PD = 1/2 diagonal = 1 m Potential at P due to all the charges= 9 * 10^9 (10/1 + 5/1 - 3/1 + 8/1)= 9*10^9 (23 - 3)= 180 * 10^9 |
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| 4081. |
9) A parallel plate air condenser has a capacity of20μF. What will be the new capacity ifthe distance between two plates is doubled?and(a)(b) a marble slab of dielectric constant 8 isintroduced between the two plates ?(0 10 uF, (b) 160 LF) |
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Answer» a) we know that C = AE°/d so, when distance d is doubled , the value of C' will be AE°/2d = C/2 b) where there is a di electric having K = 8, the value of C is kAE°/d = 8AE°/d = 8C. |
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| 4082. |
20. Complete the following table:Driver XReaction time 0'20 sDriver YReaction time 0 30Car ModelA (deceleration Speed 64 km/h Speed 72 km/hon hard braking Braking distance Braking distance- 60 m/s)Total stoppingdistanceTotal stoppingdistanceB (deceleration Speed - 54 km/on hard braking Braking distance7.5 m/s)Speed72km/hBraking distanceTotal stoppingdistanceTotal stoppingdistance |
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| 4083. |
(a) Give any two important sources of background radiation. |
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Answer» Cosmic rays – radiation that reaches the Earth from space.Rocks and soil – some rocks are radioactive and give off radioactive radon gas.Living things – plants absorb radioactive materials from the soil and these pass up the food chain. Explanation:One common source of background radiation is cosmic rays. This type of radiation comes from space and makes it through our atmosphere in small amounts. Certain rocks and parts of the Earth's surface give off background radiation, though the amount varies by location. |
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| 4084. |
How does a microwave oven heatup food even though it emits nothermal radiation? |
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Answer» A microwave oven does emit thermal radiation to heat up food. Microwave radiation is thermal radiation. For some reason, pre-college teachers and books have a mistaken notion that thermal radiation = infrared radiation. All frequencies of the electromagnetic spectrum carry energy, from radio waves, microwaves, infrared waves, visible light, ultraviolet, and X-rays to gamma rays. All frequencies of radiation heat up an object that they strike and therefore can be thermal radiation. When physicists use the term "thermal radiation", they either mean radiation that has the ability to heat up an object it strikes. Or they mean a broad spectrum of frequencies with a certain shape that depends on the emitter's temperature. |
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| 4085. |
citors of capacitances3HFo edo a potential of 12Veach They arThey areith the positive6reected to each other vith theined to the negtive plate of theplate of eachotential difference across eachwillKCET 20021V 2)Zero )6V(4)4v(36v RCw |
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Answer» how charge has come |
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| 4086. |
Problem 72. What happens if the galvanometer acell are interchanged at the balance point of the bridgWould the galvanometer show any current? |
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Answer» no change because they have balamcing ratio (p/q=r/s) here p,q,r,s are the resistance value of wheatstone bridge |
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| 4087. |
2) 2046 J 344 J (4 ZeroWork done in time t on a body of mass m whichis accelerated from rest to a speed v in time t, asa function of time t is given by 2as1 V +2V 12 S(1) 2mtV 12(4) 2mlf mV, 12unduith |
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Answer» W = F × sandF = m × asoW = m × a × s"The body is accelerated from 0 to v in time t1"sov = 0 + a × t1a = v / t1The distance traveled in a certain time t is W = m × (v / t1) × ( (v × t²) / (2 × t1) )Rearranging the equation,W = (m × v² × t²) / (2 × t1²)Or, W = ( m × v² / 2 ) × ( t / t1 )² |
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| 4088. |
Resistance in two gaps of a meter bridge are 10 2 and 30 2 respectively. If the resistanare interchanged the balance point shifts by:a) 33.3 cmb) 66.67 cmc) 25 cmd) 50 cm |
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| 4089. |
ORb) Apply KVL to find the voltage between points A and B in the given circuit.632V12 V6V +413 12V3423109 |
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| 4090. |
calculate value of gravitation mass of earth= 6×10^24radius = 6.4×10^6gravitation constant=6.67×10^_11 |
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Answer» whats your namw |
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| 4091. |
CALCULATE THE VALUE OF g |
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| 4092. |
20. Even though current flows through this potentiometer circuit why does null point won't beobtained? |
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Answer» Because voltage diffrence is there and current will keep flowingtill the violtage is not samw |
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| 4093. |
LUNGIME32. (a) With the help of a circuit diagram, deduce the equivalent resistance of two resistances connected inseries.(b) Two resistances are connected in series as shown in the diagram:K10VF6V-|50V volt(1) What is the current through the 5 ohm resistance ?!(ii) What is the current through R?(ii) What is the value of R?(in) What is the value of V?the formula for the resultant resistance of three resistors connected in |
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| 4094. |
220} 52212V6VFind II I2 I3 |
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Answer» Let the ratio factor be x.then A = 5x and B = 7xnow,5x = 185x = 185/5 = 37 B = 7x = 7*35 = 245B = 245 |
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| 4095. |
522in 522ETICOLESOSCON11V512592i1)0.6 A, 0.2A3) 0.8 A, 0.4A2) 0.4A, 0.6A4) 0.2A, 0.4A |
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Answer» 4) 0.2A, 0.4A is the correct answer |
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| 4096. |
what is microwave |
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Answer» Ans :- Microwaves are a form of electromagnetic radiation with wavelengths ranging from about one meter to one millimeter; with frequencies between 300 MHz and 300 GHz. Different sources define different frequency ranges as microwaves; the above broad definition includes both UHF and EHF bands. |
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| 4097. |
\vec { P } = \hat { i } + 2 \hat { j } - \hat { k } \quad \vec { Q } = 2 \hat { i } - \hat { j } - 3 \hat { k } \text { find } i ) P + Q \quad \text { ii } ) P - |
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| 4098. |
7Ĺ.A person A does 500 J work in 10 minutes and another person B does 600 J of work in 20 minutes. Let thapower delivered w A and B be P, and P2 respectively, then1] P2(2) P, > P(3) P1 < P2(4) P, znd P2 are undefined |
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Answer» p1>p2 p=work/time P1 is one minute work= 500/10=50j P2 is one minute work=600/20=30j P1>P2 please post my questions soon |
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| 4099. |
\begin{array} { l } { \text { If } \vec { P } = 2 \hat { i } + 3 \hat { j } - \hat { k } \text { and } \vec { Q } = 2 \hat { i } - 5 \hat { j } + 2 \hat { k } } \\ { \text { find } ( i ) \vec { P + Q } \text { (ii) } \vec { 3 P } - 2 \vec { Q } \text { . } } \end{array} |
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Answer» thank u |
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| 4100. |
Two particles P and Q are moving with velocities of (i+ j) and (-i+2) respectively. At time t-0,P is at origin and Q is at a point with position vector (21 + j). Then the shortest distance between P & Q is :-4/552V53V5 |
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