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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.
| 101. |
(4) inversely proportional to R but directly proportional to vIf S' is stress and Y' is Young's modulus of material of a wire, the energy stored in the wire per unitvolume isAIEEE-2005, 31225,-112Y(4)YS2 |
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Answer» Energy stored per unit volume, E = 1/2 * stress * strain We know that, E = 1/2* stress * stress/Y = 1/2 .S2/Y |
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| 102. |
b) What do you mean by inflamÄąmable substances?15. Explain the types of pollution. |
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Answer» 1. Thesubstanceswhich have very low ignition temperature andcaneasily catch fire with a flamearecalled inflammable substances. Examples ofinflammable substances arepetrol, alcohol, Liquified Petroleum Gas (LPG), etc Air Pollution. Water Pollution. Soil pollution. Noise pollution. Radioactive pollution. Light pollution. |
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| 103. |
15. Explain Potential Energy and derive its value of electric dipole. |
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Answer» Electric potential energy, orelectrostatic potential energy, is apotential energy(measured in joules) that results from conservative Coulomb forces and is associated with the configuration of a particular set of point charges within adefinedsystem |
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| 104. |
Que14 Why do Star twinkle?Que.15 Explain why the planets do not twinkle. |
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| 105. |
Two points A and B in space have the co-ordi(2,-1,3) and (4,2,5) respectivelyFind vector AB.and (4,2,5) respectively |
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Answer» Vector AB= (4-2)î+(2-(-1))ĵ+(5-3)k̂ => Vector AB=2î+3ĵ+2k̂ |
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| 106. |
1. By increasing the temperature, the specific resistance of a conductor and a semiconductor(a) increases for both(b) decreases for both(c) increases, decreases respectively(d) decreases, increases respectively |
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Answer» Thus resistivity ofconductor increaseswith increaseintemperature. |
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| 107. |
If the displacement is given 6t³ + 4t² - 2t + 5 m find average acceleration in 3rd sec. |
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| 108. |
Average angular acceleration: |
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Answer» Theaverage angular accelerationis the change in theangularvelocity, divided by the change in time. Theangular accelerationis a vector that points in a direction along the rotation axis. The magnitude of theangular accelerationis given by the formula below. The unit ofangular accelerationis radians/s2. |
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| 109. |
19. A car travels the first one-thirddistance at a speed of 10 km/h,the next one-third distance at20 km/h and the last one-thirddistance at 60 km/h. Then, theaverage speed of the car is(a) 30 km/h(c) 18 km/h(b) 24 km/h(d) of these |
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Answer» Let the total distance be 3x km. Now, average speed is given by average speed = (total distance travelled) / (total time taken) So, let times taken for the first, second and third halves of the journeys be t1, t2, and t3 respectively. Then, t1 = x/10 t2 = x/20 t3 = x/60 So, average speed = x / (t1 + t2 + t3) = x / (x/10 + x/20 + x/60) = 3/ ((6 + 3 + 1)/60) = (3×60)/10 = 18 km/hr Like my answer if you find it useful! |
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| 110. |
Kinetic Energy) A body freely falls from a certain height ontothe ground in a time 't'. During the firstone third of the interval it gains a kineticenergy AK, and during the last one third ofthe interval, it gains a kinetic energy ΔΚ2.The ratio ΔΚ. AK, is1:5uin the mass of a boy and has |
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Answer» speed gained from t = 0 , to t = t/3 => V = gt/3 so, K.E = 1/2*(m)*(gt/3)² = mg²t²/18 speed when t =2t/3 v = 0+g*2t/3 = 2gt/3 now speed gained from t=2/3 to t= t V = gt/3 +g2t/3 = gt so, K.E increased in last third = 1/2*m(g²t²) -1/2*m*(2gt/3)²= 1/2*m*g²t² -4/18m*g²*t² = 5/18mg²t². so, ∆k1:∆k2 = 1/18:5/18 = 1:5 option 4 |
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| 111. |
A ball is thrown vertically upwards with a given velocity 'u' such that it rises for T seconds (T> 1), Whatthe distance traversed by the ball during the last one second of ascent (in meters)? (Acceleration duegravity is g m/s'.) |
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Answer» the bollis throw inth veritically or hai answers velocity at the toppest point (time) us zeroes there fore v=u -gt 0=u-gtu=gtt=u/g |
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| 112. |
35. A ball is dropped on the floor from a height of( 50 m10 m. It rebounds to a height of 2.5 m. If the ballis in contact with floor for 0.01 s, then theaverage acceleration during contact is nearly |
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Answer» For V before hitting the ground, v^2=u^2+2gh=0^2+2×9 8×10(it us preferable here to use g=9.8) Therefore v=14ms Now after bounce, V^2=U^2+2gH(here g is -ve) Hence V=-7m's Now a=(v-V)/t a=2100ms When it is dropped from 10m,Initial height = 10minitial velocity = 0velocity just before hotting ground =√2gh = √2*9.8*10 = 14.07 m/s (downward) after rebound,maximum height reached = 2.5mfinal velocity at top = 0initial velocity(just after rebound) =√2gh =√2*9.8*2.5 =√49 = 7 m/s (upward) assuming downward as positive directionSo velocity just before hitting ground = +14.07 m/svelocity just after hitting ground = -7 m/schange in velocity = +14.07 - (-7) = 21.07 m/stime = 0.01sacceleration = change in velocity/time = 21.07/0.01 = 2107 m/s² ~ 2100m/s² |
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| 113. |
The functions y - sin x and y - sin(x +a) + b, forconstants a and b, are graphed in the standard (x,y)coordinate plane below. The functions have the samemaximum value. One of the following statementsabout the values of a and b is true. Which statement isit?A. a < 0 and b-0B. a < 0 and b>0C. a -0 and b> 0D. a> 0 and b< 0E. a> 0 and b> 0 |
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Answer» The only difference between our function graphs is a horizontal shift, which means that our b value (which would determine the vertical shift of a sine graph) must be 0. Just by using this information, we can eliminate every answer choice but A, as that is the only answer with b=0. For expediency's sake, we can stop here. Our final answer is A, a<0 and b=0 |
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| 114. |
11. A charged particle moves in a gravity free spacewithout change in velocity. Which of the followingis not possible in that space?(1) E 0, B 0(3) E 0, B# 0(2) E* 0, B 0(4) E 0, B 0 |
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| 115. |
Ex. 11.4.4 Two tuning forks A and B whensounded together produces 4 beats per second.The frequency of A is 256. When B is loadedwith wax and sounded again with A, produces8 beats. Calculate the frequency of B |
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Answer» Frequency of B = 256Hz 8/4=2 Frequency=No of beats/Time Frequency is directly proportional to no of beats Frequency of B = 256Hz×2 = 512Hz |
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| 116. |
a tuning fork produces 4 beats per second when sounded with a fock of frequency 512 hz. The same tuning fock when sounded with another fock of same frequency produces 6 beats per sec. Find the frequency of tuning fock. |
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Answer» tell me the answer please hurry uptomorrow is my exam please be fast. The language of the question is unclear. Is the tuning fork loaded before sounding it again? Or filed? Because if two tuning forks have the same frequency, they cannot produce beats, please provide a snapshot of the question if possible. sir mujha bhi asa hi lag rha hai sir mana tabhi oh pucha questioni think question is incomplete Tuning forks are usually loaded because it's a reversible process conserving the original tuning fork, so I'll solve the question assuming the tuning fork is loaded with wax. |
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| 117. |
3) Two tuning forks A and B produce notes offrequencies 250 Hz and 260 Hz respectively.Whenan unknown note is sounded with A it produces xbeats/sec. But when the same note is soundedwith B, the number of beats produced per secondbecomes 4x. The unknown frequency could be |
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| 118. |
24. The frequencies of two tuning forks A and B are 250 Hz and 255 Hz respectively. Botlsounded together. How many beats will be heard in 5 second |
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Answer» Beat frequency = f1-f2= 255-250= 5 Hz Hence, 1 beat will be heard in 5s |
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| 119. |
Threecoherentwaveshavingamplitudes12mm,6mmand4mmarriveatagivenpointwithsuccessive phase difference of π/2 . Then, the amplitude of the resultant wave is(a) 7 mm(b) 10 mm(c) 5 mm(d) 4.8 mm |
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| 120. |
Solved Problems1. The wavelength of two notes in air are 83 m and 83 m. Each of these notes172170produces 4 beats per second with a third note of a fixed frequency. Find the speed ofsound in air. |
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Answer» Let the third note be having L as wavelength Hence the frequency = c/L Given c / (85/172) - c/L = 4-------------(1) And c/L - c/(85/170) = 4 Adding c (2/85) = 8 So c = 340 m/s Hence the third note would have a frequency 340/L From (1) 340*172 / 85 - 340/L = 4 So 340/L = 684 Hz Frequency of third note is 684 Hz and velocity of sound is 340 m/s.. |
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| 121. |
10. A boy projects a stone vertically perpendicular to the trolley car with a speed u. If the trolly car moves with a constantvelocity u, the time of flight of the stone is:B)C)24D) none of these |
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Answer» t = 2vsin theta/g = 2v/g as theta is 90 degree. |
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| 122. |
Find the ratio of the potential differences that mustbe applied across the parailel and the senescombnation of twa capactorsand with theirapactances in the rato1 so that the enengysturedn the awe cases, becomes the sume |
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| 123. |
21. A ball is thrown at different angles with the same speed u and from the same points and it has same rangein both the cases. If y, and y2 be the heights attained in the two cases, then yitvIt(B) 242g |
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| 124. |
93. An ideal inductor is in turn put across 220 V. 50 Hz and 220 V, 100 Hz supplyCompare the currents flowing in two cases. |
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| 125. |
8.Whataretheadvantagesanddisadvantages of using a Solar Cooker ? Two points each ofadvantage and disadvantage.12) |
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| 126. |
Escompile... A partile starts with an initial recesity Bom/8alon festere a direction and it acceluatia miferingat the rate 0.5 m/s Let Ind the distance Leveledday it in the first test second . (6) How much IOdees it take a beach the vent, tam/s ? |
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| 127. |
6. The frequency of radiations emitted whenelectron falls fromn 4 to n1 in H-atomwould be (Given E, for H 2.18 x 10-18 Jatomand h 6.625 10-34 Js.)(a) 1.54 x 1015s (b) 1.03 x 1015 s-1(c) 3.08 x 1015 s (d) 2.0 x 1015 s-112004] |
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| 128. |
What is the momentum of a photon having frequency 1.5 x 1013Hz-(a) 3.3 x 10 29 kg m/s(c) 6.6 x 1018 kg m/s40.(b) 3.3x 1034 kg m/s(d) 6.6 x 103kgm/s |
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Answer» p =hv/c => p = 6*64×10^-34*1.5×10¹³/3×10^8=> p = 3.3 ×10^-29 |
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| 129. |
18. A small metal plate (work function = 1.17 eV) isplaced at a distance of 2 m from a monochromaticlight source of wavelength 4.8 x 107 m and power20 watt. The light falls normally on the plate. Thenumber of photons striking the metal plate persecond per sq metre area will be: (h 6-6 x 10-34 Js)(a) 9.64 x 1018(c) 9.64 x 1016(b) 9.64 x 1017(d) 9.64 x 1015mum kinetic enerov |
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Answer» Suppose the source emits n photons of frequency f per second. Therefore , n hf = nh(c/Lambda)=20J. ( Because a 20W source emits 20J energy per second). Therefore, n=[20x 4.8x10^-7]/[(6.62x10^-34)(3.0x10^8)]=4.8338x10^19 This number of photons are incident on the spherical surface of radius of 2 m per second . Hence, the number of photons arriving on unit area per second will be 4.8338x10^19/[4 pi 2^2]=4.8338x10^19/(16x3.14)=0.096x10^19=9.6x10^17 photons per meter square per second hence, correct option is (b) |
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| 130. |
12. Two neutrons of mass 1.67 x 1027 kg are atdistance of 10-15 m from each other. Thegravitational force of attraction betweenthem is:(a) 4.86 x 10 34 N (b) 5.86 x 10 34 N(c) 1.86 x 1034 N (d) 2.86 x 10-34 N |
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| 131. |
驸苣:(a) 12 qu(c) 48 ueThe time of revolution around the earth ofCommunication Satellite INSAT-11B is:(b)(d)2430 |
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Answer» Geostationary orbit means it take 1 day to orbit the earth. Option -B option B is the best👍💯 one.. 24 hours |
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| 132. |
30. The figure given below shows an electric circuit in which current flows from a 6 V battery through tworesistors.6 Vi2017 1130(a) Are the resistors connected in series with each other or in parallel ?(b) For each resistor, state the p.d. across it.(c) The current flowing from the battery is shared between the two resistors. Which resistor will have biggershare of the current ?(d) Calculate the effective resistance of the two resistors.(e) Calculate the current that flows from the battery. |
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| 133. |
Three charges, each + q, are placed at thecorners of an isosceles triangle ABC of equalsides BC and AC 2 a, Fig. D and E are the midpoints of BC and CA. The work done in taking acharge Q from D to E isC(q)2a 2a3 q Q(3) zero (4) 4 TCo a A)B(q) |
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| 134. |
Two resistors of 12 Ω and 3 Ω arefonnected in parallel in a circuit having 2n andra resistors inseries, I the current flowing in the circuit is 1.95 A when a potential difference of 9 V is applied, Calculatethe value of X. |
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| 135. |
he thangle?[Ans 8 23. A parallel combination of three resistors takes a currentof 7.5 ampere from a 30 V battery. If two of the resistorshave a resistance of 10 ohm and 12 ohm, find theAns 15 ohmresistance of the third resistor. |
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Answer» The resistance of the group of 3 resistors is equal to R = 30 V / 7.5 A =4 ohm. Let R3 be the resistance of the third resistor, then 1/R =1/10 + 1/12 + 1/R3 so, 1/4 = 1/10 + 1/12 + 1/R3 1/R3 = 1/4- 1/10 - 1/12= 15/60 - 6/60 - 5/60= 4/60 hence R3 = 60/4 = 15 ohm thank |
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| 136. |
: PHYSELECTRICITYALtively29. A 5 V battery is connected to two 20 2 resistors which are joined together in series.(a) Draw a circuit diagram to represent this. Add an arrow to indicate the direction of conventional currentflow in the circuit.(b) What is the effective resistance of the two resistors ?(c) Calculate the current that flows from the battery.(d) What is the p.d. across each resistor ? |
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| 137. |
Example 2. Force on a Charge at Centroid ofTriangle Consider three charges Q,.,,Q, eachequal to Q at the vertices of an equilateral triangleof side a. What is the force on a charge q (with thesame sign as q) placed at the centroid of thetriangle?NCER |
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| 138. |
3) How much force is required in Newton to produce an aeceleration of 2ms+ in abody of mass 50 gm? |
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Answer» 1 newton = 1 m/s^2 * 1kg 2m/s^2 * 50/1000 kg = 0.1 newtons hit like if you find it useful |
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| 139. |
Q.3. Stationary waves are produced in 10 m long stretched string fiked at two ends.If the string vibrates in 5segments and wave velocity is 20 m/s, the frequencyof waves isd) 2 Hz |
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Answer» 10hz is the answer but not fully sure |
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| 140. |
riod of oscillation if a body vibrates 150 times in a25 Find the time period of oscillation if abody vibratesminute? |
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Answer» frequency = 1/T( in second)in 60 seconds= 150 oscillationin 1 second= 150/602.5 oscillation per second will be the frequency |
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| 141. |
In an experiment, the string vibrates in 4 loopswhen 50 gm wt is placed in pan of weight 15gm. To make the string vibrate in 6 loops theweight that has to be removed from the pan isapproximately |
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| 142. |
2. If an insect produces sound when it vibrates 200vibrations per second. What is the time periodof the vibration? |
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Answer» time period =1/frequency of oscillation =1/200 =0.005 why did you divided with 1 that's the formula The time required to complete one oscillation is known as time period. It is given by the inverse of the frequency |
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| 143. |
(ii) The frequency of a stretched and vibrating string ndepends uponthe length /of the string, tension T of the string and mass m of theper unit length of the string. Establish the equation for frequency nby dimensional analysis method |
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| 144. |
Short Answer Type QuestionsQ1. Three resistors are ceA is flowing.are connected as shown in the figure. Through the resistor 5 92 a constant current of(3 Marks)1022R522R 31512 R2(1) What is the current through the other two resistors?on What is the potential difference across AB and across AC?(iii) What is the total resistance? |
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Answer» sorry jsj jjs jsis jjsks |
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| 145. |
/4/d 6 Ω be connectedHow can three resistors of resistances 2 Ω, 3 Ω, anto give a total resistance of (a) 4 Ω, (b) I Ω?anto give a total resistanceof |
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| 146. |
01212,352 ahu o sa respectively so as to obtain a resuntan resistance of2.0 12 Draw the diagram to show the arrangement.48. How will you connect three resistors of resiand (b) 1 2?you connect three resistors of resistances 20.30 and 6 2 to obtain a total resistance of 31249. What is (a) highest, and (b) lowest resistance which can be obtained by comofollowing resistances ?resistance which can be obtained by combining four resistors having the |
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Answer» Highest resistance when they are in seriesb) lowest when they are in parallel To obtain 1 ohm6 and 3 in paraller soequivalent= 3*6/3+6= 2now 2 parallel with 2= 2*2/2+2= 4/4= 1 |
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| 147. |
OLVO arrangement.48. How will you connect three resistors of resistances 22, 3 Q and 6 22 to obtain a total resistance of: (a) 4and (b) 1 2 ? |
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Answer» yes it is correct answer |
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| 148. |
2) Polerud LUCICIILE CU ULICUIU16. Three resistors are connected as shown in the diagram.10521 amp.1522Through the resistor 5 ohm, a current of 1 ampere is flowing.(i) What is the current through the other two resistors ?(ii) What is the p.d. across AB and across AC?(iii) What is the total resistance ?17. For the circuit shown in the diagram below:612 12 |
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| 149. |
3. How long would it take for a radiowave of frequency 6 x 103 per sec to travel from mars to earth, distanceof 8 x 107Km.(Ans = 266s] |
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Answer» Radio waves travel at the speed of light. So, in this case we have to use the formula of distance = speed x time Time=8×10^7Km/ 3×108ms−1=8×10^10m/ 3×108ms−1=266.66sec |
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| 150. |
The work function of a metal is 4.2eV, its threshold wavelength will be(a) 4000 A(b) 3500 A(c) 2955 A(d) 2500/4 |
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