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.
| 3451. |
India and Guatemala has recently agreed to support each other for Non-permanent membership of UNSC. Guatemala will support India’s candidature for which year? |
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Answer» 2021-22] Notes: India and Guatemala have signed a pact for strengthening diplomatic cooperation through Respective Foreign Service Institutes in the educational field. The agreement was recently inked after a meeting between Vice President M Venkaiah Naidu and his Guatemala counterpart, Jafeth Cabrera Franco in Guatemala City. Beside this, both the countries agreed to support each other’s candidature for Non-Permanent Membership in the United Nations Security Council (UNSC). Guatemala will support India’s candidature for UNSC membership for 2021-22 while New Delhi will do so for 2031-32 for Guatemala. During the talks, Indian side also agreed to the Guatemala’s request for supplying solar panels to its airports. After discussions and in the presence of both the Vice Presidents, both the sides signed a Memorandum of Understanding (MoU) for training of diplomats and a Letter of Intent for training Guatemalan English Teachers in India. These agreements will be valid for three years and will be extended thereafter, if required. |
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| 3452. |
Who given the formula of intensity of magnetic field? |
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| 3453. |
The friction coefficient between the horizontal surfaceand each of the blocks shown in figure (9-E20) is 0 20.The collision between the blocks is perfectly elastic. Findthe separation between the two blocks when they cometo rest. Take g 10 m/s 21.0 m/s4 kg2 kg16 cm |
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| 3454. |
5.For the arrangement shown in the figure, the reading of spring balance is(a) 50 N(b) 100 N(c) 150(d) of the above5 Kg10 Kg |
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Answer» The reading in the spring , should be sum of both the forces acting on the block = (5+10)g = 15g = 150N |
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| 3455. |
11, यदि किसी चालक से 5 सेकेण्ड तक 3 mA की धारा प्रवाहित ।होती है तो चालक पर आवेश होगी -| (a) 3 C (b) 5C (c) 15 C (d) 15 mC ।Bu |
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Answer» we know thatQ=iTSO 5*3mA15mC |
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| 3456. |
Derive an expression for magnetic field intensity at a point on hemagnetic dipole.axis of a |
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| 3457. |
9. Figure (3-E4) shows the graph of the x-coordinate of aparticle going along the X-axis as a function of time.Find (a) the average velocity during 0 to 10 s,(b) instantaneous velocity at 2, 5, 8 and 12sx (m)10050Time (second)2.5 5.0 7.5 10.0 12.5 15.0Figure 3-E4 |
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| 3458. |
in9. Figure (3-E4) shows the graph of the x-coordinate of aparticle going along the X-axis as a function of time.Find (a) the average velocity during 0 to 10 s,(b) instantaneous velocity at 2, 5, 8 and 12s.x (m)10050Time (second)2.5 5.0 7.5 10.0 12.5 15.0Figure 3-E4 |
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| 3459. |
4.A particle of mass 2 kg moves on a ciraular path with constant speed 10 m/s. Find change in speed andmagnitude of change in velocity. When particle completes half revolution. |
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Answer» what is that relation between same magnitued and opposite direction so that u given the answer 20m/s |
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| 3460. |
A force of 50 N is inclined to the vertical at an angleof 300. Find the acceleration it produces in a body ofmass 2 kg which moves in the horizontal direction. |
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Answer» in horizontal direcn.50sin30 = 2*aa = 50/4 = 25/2 = 12.5 m/s^2hence the accln. of 2 kg mass is 12.5 m/s^2. in horizontal direcn.50sin30 = 2*aa = 50/4 = 25/2 = 12.5 m/s^2hence the accln. of 2 kg mass is 12.5 m/s^2. |
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| 3461. |
The forces acting on an object of mass 2 kg are shownin the fig. If the body moves horizontally then findacceleration if force P is 400 N.300 N2 kgP(A) 50 m/s2(C) 350 m/s2(B) 10 m/s?(D) 60 m/s? |
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| 3462. |
73. Three charges are arranged on the vertices ofequilateral triangle as shown in figure (29-E6). Find thedipole moment of the combination.-q2q-qFigure 29-E6 |
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| 3463. |
Q.29. Explain the origin of magnetism on the basis of circulating charges.Define magnetic axis and pole strength. |
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Answer» The product of the distance (2 l) between the twopolesand thepole strengthof eitherpoleis calledmagneticdipole moment. Its SI unit is 'joule/tesla' or 'ampere-metre^2'. Its direction is from southpoletowards northpole. Where θ is angle between the dipoleaxisandmagneticfield. |
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| 3464. |
Thepotential difference appliedacross a given resistor is altered, sothat the heat produced per secondincreases by a factor of 9. By whatfactor does the applied potentialerence change?In the figure shown, an ammeter Aand a resistor of 4 Ί are connected tothe terminals of the source. The emfof the source is 12 V having aninternal resistance of 2 Ί. Calculatethe voltmeter and ammeter readingsAll India 2017 |
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Answer» For question b , kindly post the image. |
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| 3465. |
On centigrade scale the temperature of a BUuyincreases by 30 degrees. The increase intemperature on Fahrenheit scale is |
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Answer» We know that C = 5/9*(F-32) => F = 9C/5 +32 ....(1) so, when c is increased to C+30 => F' = 9(c+30)/5 +32 =9C/5 +9*30/5+32 = 9C/5 +32+ 9*6 => F' = F+54 ....... ( since 9c/5 +32 = F) so, in Fahrenheit ,the increase is = 54°F it's wrong check it again.. just missed a 9 to be multiplied with 6, while expanding the formula.. |
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| 3466. |
On centigrade scale the temperature of a bodyincreases by 30 degrees. The increase intemperature on Fahrenheit scale is |
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| 3467. |
Tuo point objects of masses 15g and 25g respedtively are atradistance of 16 cm apart, the centre of gravity is at a distarcexfom the objerd of mass1.5gwhene x isb) 6cmcrn(d) 3 cm |
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Answer» thank |
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| 3468. |
40. The masses and radii of the earth and moon are M, Rj and M2, R2 respectively. Their centre are at adistance r apart. Find the minimum speed with which the particle of mass m should be projected from apoint midway between the two centres so as to escape to infinity |
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| 3469. |
On centigrade scale the temperature of a body increases by 30 degrees. The increase intemperature on Fahrenheit scale is1) 502) 40°3) 30°4) 54 |
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Answer» (30°C× 9/5) + 32 = 86°FAs farehneit scale starts from 32°Hence Reading will be 86-32=54°F |
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| 3470. |
Lengthof a minute hand of a dock is 4.5 cm. Find the average velocity of the tip of minute'shandbetween6 A.M. to 6.30 A.M. & 6 A.M. to 6.30 P.M. |
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| 3471. |
Q.8 Two spheres of masses 2 Kg and 3 Kg having their centre 50 cm apart find the centre ofmass of system |
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Answer» Center of mass= 3*50+2*50= 150+100250/5= 50 |
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| 3472. |
(c) 3 mgs M is hanging from a rigidope at a distance x from theThree blocks of masses 2 kgto each other with lightfrictionless surface as shown in thapulled by a force F IONses 2 kg, 3 kg and 5 kg are17.as shown in the bhe10N, then tension Tstring and are then placedand are thenThe soT2 |
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Answer» i think t1 is 8N thanks 😀😀 is it correct? ya...... |
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| 3473. |
7. Four point masses are placed at the cot masses are placed at the corners of a square of side 2 m as shown in the figure.Find the centre of mass of the system w.rt. the centre of square.4 kg1 kg3 kg2 m2 kg |
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| 3474. |
Two bodies of masses 2 kg and 4 kg are connectedto the ends of a string of negligible mass and thenpassed around a frictionless pulley. Calculate (i) theacceleration of the system and (ii) the tension inthe string, (Take g 10m/s) |
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| 3475. |
[Ans: 1.6 kg] Et 4.54. Two masses 8 kg and 12 kg are connected at the twoends of a light inextensible string that goes over africtionless pulley. Find the acceleration of themasses and the tension in the string when themasses are released.ite[Ans: 2 ms2 and T-96 NJngle of 45° |
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| 3476. |
A 10 V cell of negligible internal resistance is connected in parallel acrossa battery of emf 200 V and internal resistance 38 2 as shown in thefigure. Find the value of current in the circuit.ŃĐľ10 V38 92200 V |
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| 3477. |
7. A 10 V cell of negligible internal resistance is connected in parallel across abattery of emf 200 V and internal resistance 38 2 as shown in the figure.Find the value of current in the circuit.10 V200 V38 Ί |
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| 3478. |
CBSEA 10V cell of negligible internal resistance is connected in parallel acrossa in thcicitinternal resistance 3832 as shown ina battery of emf 200 V and12]the figure. Find the value of current in the circui10V200V 382OR |
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Answer» Let i be the current flowing in the circuit By Kirchhoff’s law 38i = 200 – 10 => i = 190/38 => i = 5A |
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| 3479. |
34. An ideal gas goes from the state i to the state fas shown in figure. The work done by the gas duringthe processis positive(3) is zero(2) is negative(4) cannot be obtained from this information |
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| 3480. |
Calculate the Wattage of soldering iron with a hot resistance 4K Ω and operativoltage of 200V.I.(Ans: 10W) |
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| 3481. |
30,A 10 μ fcapacitor is connected across a200V, 50 HzAC supply. The peak currentthrough the circuit is:(A) 0.6 /2 A(B) 0.6A(C) 0.3 t AD) 0.6 v24A |
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Answer» We know, Capacitive reactance(Xc) =1÷(2*3.14*f*C) =1÷(2*3.14*50*10*10^-6) {~1 uF=10^-6 F~} =318 ohms(approx.) Again voltage across capacitor =200V So, Current across capacitor =200÷318=0.7 A *The calculation was done assuming peak voltage =200 V. If 200V is the R.M.S value then peak voltage would be 200*root over of 2=283 V So current =283÷318= 0.89A = 0.6√2 A it is not in option and solution not understandable |
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| 3482. |
If A 21+3jand8-51+7j, then the vectorhaving the magnitude of A and direction of Bwould be1127413(21 + 3j)(2)74 (5i +7j)) 1/2(3) (137(51+77) (4) 5 13 i +7 13 |
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Answer» thank you ... your answer is correct.... |
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| 3483. |
4. An electric bulb is rated at 220 V, 100 W. What is its resistance?5. What io thn1 thofTd nelectric power? |
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| 3484. |
7. An electric bulb is rated 250w-230v.What information does this convey? |
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Answer» It is the power rating for the bulb.(A power rating is a measurement of the maximum amount of power that can be used with a specific tool or device.) Power = 250 WVoltage = 230 V It indicates two factors:(i) Maximum power that the bulb can be used with and (ii) Amount of energy that the bulb consumes. Multiply the number of hours the bulb is used on an average day by the wattage. This will give you the watt-hours or energy consumed per day. it means that the amount of electricity bulb can transform into light over a given amount of time. here the |
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| 3485. |
68. If one mole of a mono-atomic gas (g 5/3) is mixedwith one mole of a diatomic gas (g 7/5), the value ofg for the mixture is(3) 1.53(4)3.07 |
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Answer» Mean gamma = Mean Cp/Mean Cv = 3/2 = 1.5 |
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| 3486. |
In an NPN transistor the collector cu24 mA. If 80% of electrons reach collbase current in mA is:(1) 36(3) 16(2) 264) 6 |
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| 3487. |
6.7×10-11×6×1024×7.4×1022(3 84 x 108)2 |
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| 3488. |
roltihe man uomooalso calalota a dug sQSO |
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Answer» This answer is from quora I am class 9 |
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| 3489. |
Q.12.How many electrons would have to be removed from a coin to leave it with a charge of 10-7?[e= 1.6x10-19] |
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Answer» This number of electrons has to be removed. |
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| 3490. |
Q.12 How many electrons would have to be removed from a coin to leave it with a charge of 10-7?[e= 1.6x10-19] |
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Answer» This number of electrons has to be removed. |
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| 3491. |
4 ball is projected from ground at an angle 45° withhorizontal from distance d, from the foot of a poleand just after touching the top of pole it the falls onground at distance d2 from pole on other side, theheight of pole is(2 102d,+d24(1) 2 d,d2d, d22d, d2(4) d,+02 |
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| 3492. |
2P=abt, where p4. The dimension ofin the equationP =, where Pis pressure, x is distance and t is time, is |
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Answer» a]=[t2]=T2 [P]=T2/[bx] [b]=T2/[P][x]=T2/ML-1T-2L=M-1T4 a/b=T2/M-1T4=MT-2 |
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| 3493. |
If the difference in the molar specific heats of a gas is equal to R what will be the difference in the molar specific heats if there are n mole? |
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Answer» R is the difference . |
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| 3494. |
The volume (V) of a monatomic gas varies with itstemperature (T), as shown in the graph. The ratioof work done by the gas, to the heat absorbed byit, when it undergoes a change from state A to[NEET-2018]state B, is→T2 |
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| 3495. |
Define the molar specific heats of a gas and deduce the relation between them. |
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| 3496. |
3 One mole of an ideal diatomic gas undergoes atransition from A to B along a path AB as shownin the figureP (in kPa) t5246V (in m)The change in internal energy of the gas during thetransition is[AIPMT-2015] |
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| 3497. |
(m + 2M)Four point masses, each of value m, are placed at the corners of a square ABCD of side (moment of inertia about an axis passing through A and parallel to BD is [AIEEE-2006, 4/180](1) m2(m+ 2M)(m+M). The(2) 2m2(3) V3 m(2(4) 3mCorporate Office: CG Tower, A-46 & 52. IPİA Near City Mall, Jhalawar Road, Kota Raj324005rsnance ag.in l E-mail: contactresonance.ac.in |
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| 3498. |
. The current gain in a common-emitter transistor is 90.the change in base current is 200 μΑ, the change in collectorcurrent is |
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| 3499. |
k)ăŽh-20seconds,uwhat is theahan |
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Answer» Case 1:Initial velocity = 36km/h=10 m/sFinal velocity = 72km/h= 20m/sTime taken = 5 sAcceleration =( 20 -10)/5= 10/5= 2m/s^2Case 2:Initial velocity = 72km/h= 20m/sFinal velocity = 0m/sAcceleration = (0-20)/20= -1m/s^2So, retardation = 1m/s^2 |
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| 3500. |
cooton + 0 800, ० कित तल! «०2 220.UWhat 1४ नर कपल (कीकत 9 5010 28094 2८-20.2 1:00131 2 हलवा) 9 |
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Answer» An electric motor converts electrical energy into mechanical energy.Principle;It works on the principle of the magnetic effect of current. A current-carrying coil rotates in a magnetic field. The following figure shows a simple electric motor. Working: When a current is allowed to flow through the coil MNST by closing the switch, the coil starts rotating anti-clockwise. This happens because a downward force acts on length MN and at the same time, an upward force acts on length ST. As a result, the coil rotates anti-clockwise. Function of split ring:The split ring in the electric motor also known as a commutator reverses the direction of current flowing through the coil after every half rotation of the coil. Due to this the coil continues to rotatein the same direction |
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