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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.
| 4601. |
Recalling IdeasI. Select the correct option:1. Which of the following cannot be charged easily by friction?(a) A ball of wood(c) An inflated balloon(- ) (b) A glass rod⥠(d) An ebonite rod |
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Answer» A ball of wood cannot be easily charged by friction because it doesn't have free electrons. |
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| 4602. |
Which of the following cannot be charged easily by friction?(a) A plastic scale(b) A copper rod(c) An inflated balloond) A woollen cloth. |
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Answer» I think copper rod .l hope it helps for you |
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| 4603. |
b A large steel wheel is to be fitted on to a shaft of the same material. At 27theouter diameter of the shaft is 8.70 cm and the diameter of the central hole in thewheel is 8.69 cm. The shaft is cooled using 'dry ice. At what temperature of theshaft does the wheel slip on the shaft? Assume coefficient of linear expansion ofthe steel to be constant over the required temperature range:asteel=1.20 × 10-s K-1. |
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Answer» thanks bro can ask this?? |
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| 4604. |
Detailed line sketch of 4-stroke (Petrol &Diesel) engine |
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| 4605. |
A motor-cyclist moving with a velocity of 72 kmh1 on a flat road takes a turnat a point where the radius of curvature of the road is 20 m. In order to avoidsliding, he must not bend with respect to the vertical by an angle e greaterthan (g 10 ms-2):(a) tan-16(c) tam 25.92(b) tan 12(d) tam14 |
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Answer» We know ,tanةµ = v^2/(r*g) < where v= velocity of motorcycle, r= radius of the curvature & g= acceleration due to gravity>tanةµ= 20^2/(20*10)tanةµ=2ةµ=tan−1 (2) |
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| 4606. |
alsory.AllquestioncarryequalmarksQ.I. (a) Find the value of constant 'a' such thatAa +4 (sinz-3p/-( +4cos.y)k is solenoidal. |
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Answer» When vector is solenoidal the divergence is 0.So, take divergence of vector we get a+ (-3) -0 = 0 so a = 3 will be the answer |
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| 4607. |
TuuS TUUthen L is000000210 uF 502100 v, 500 Hz1) 0.1 H3) 0.0.2 H2) 0.2 H4) 0.01 H |
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Answer» Xl= XCf= 1/2π√l/csoafter putting the valuesl=0.01H the answer could be option 4 |
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| 4608. |
C, -10FCalculate:C-10uFC, -10uFal Equivalent Capacitanceb) The charge on each capacitorCy-100FSoov |
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Answer» Upper 3 capacitor are in sériesnow equivalentof those 3 will be = 1/C = 1/10+1/10+1/101/C = 3/10C = 10/3now this 10/3 is parallel with 10 so overallanswer = 10+10/3= 40/3 c1and c3 are parallel thenresultant c5=10+10=20 now c2,c4 and c5 are in series then resultant1/c=1/10+1/10+1/201/c=(2+2+1)/201/c=5/201/c=1/4c=4uF. ans. |
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| 4609. |
A car moving with velocity of 25 ms-1. It takesturn in 5 s. The average acceleratoin in this time..ms2a) 2.5(c) 7,5(b) 5.0(d) 10 |
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| 4610. |
Q.32 A cannon ball is fired with a velocity 200m/secat an angle of 60° with the horizontal. At thehighest point of its flight. It explodes into 3equal fragments, one going vertically upwardswith a velocity 100m/sec, the second onefalling vertically downwards with a velocity100 m/sec. The third fragment will be movingwith a velocity(A) 100 m/sec in the horizontal direction(B) 300m/sec in the horizontal direotion(C) 300 m/sec in a direction making an angleof 60° with the horizontal(D) 200 m/sec in a direction making an angleof 60° with the horizontal |
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Answer» i think A option is correct answer In this problem it is given that the flight explodes into 3 equal fragments so the mass of the fragments must also be same. BY USING THE METHOD OF CENTER OF MASS WE GET V1+V2+V3/3=VELOCITY OF CENTER OF MASS WHERE V1= VELOCITY FRAGMENT MOVING VERTICALLY UPWARDS V2=VELOCITY OF FRAGMENT MOVING VERTICALLY DOWNWARDS V3 = VELOCITY OF UNKNOWN FRAGMENT. HERE AS THE TWO FRAGMENTS ARE MOVING IN VERTICALLY COMPONENT THEIR HORIZONTAL COMPONENT IS ZERO.AND VERTICALLY COMPONENT OF THE FLIGHT WHEN IT EXPLODES IS ZERO AS IT IS AT THE MAX HEIGHT (I THINK YOU KNOW THE BASICS OF PROJECTILE).... 1).SO VELOCITIES ALONG VERTICALLY V1y+V2y+V3y/3=velocity of center of mass Let us take take the upward coordinate axis as positive and downward as negative.... So 100-100+V3y/3=0,this is equal to zero because vertical component of the flight at max height is zero. Therefore this gives V3y=0 2).velocity along horizontal V1x+V2x+V3x/3=VELOCITY OF horizontal component of center of mass (here center of mass is of the flight ) V1x=V2x=0 as they are moving vertically and not horizontally or at some angle . And velocity of center of mass = vcos (theta) ( BECAUSE AT MAX HEIGHT HORIZONTAL COMPONENT IS VCOS (THETA) , WHERE V=VELOCITY OF PROJECTION,AND THETA=ANGLE OF PROJECTION) THEREFORE VCOS (THETA)=200 (COS60)=200×1/2=100m/s Therefore V1x+V2x+V3x/3=100 0+0+V3x=100 Therefore V3x=100 Therefore net velocity of the third fragment is resultant of the vertices and horizontal component V3=root (Vx2+Vy2) V3=rroot (10000+0) V3=100m/s |
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| 4611. |
A man weighs 70 kg. He stands on a weighing machine ina lft, which is moving(a) Upwards with a uniform speed of 10 m/s.(b) Downwards with a uniform acceleration of 5 m/s2(c) Upwards with a uniform acceleration of 5 m/s2What would be readings on the scale in each case? |
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Answer» ok, thanks |
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| 4612. |
(0) becT0. A balloon is moving along with constant upwardacceleration of I m/s, A stone isthrown from the balloon downwardswith speed 10 m/s with respect to 20 m/sthe balloon. At the time ofprojection balloon is at height 120m from the ground and is movingwith speed 20 m/s. Find the timerequired to fall on the ground by the stone after theprojection.(a) 6 sec(c) 3 sec1 m(b) 4 sec(d) 4.5 sec |
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| 4613. |
d. 4i+ 2j3. A stone is thrown upwards at a speed of 20m/s, what is it's speed (in m/s) and direction after 3secondsa. 10, upwardsb. 20, downwardsC. 10, downwardsd. 0 |
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Answer» Velocity at maximum height (v) = 0Acceleration = -g(-ve sign because it’s in the direction opposite to motion of stone) v² - u² = 2aS S = v² - u² / (2a)= [0 - (20 m/s)²] / [2 × (-9.8 m/s²)]= 20.40 m It will reach a height of 20.40 m before it begins to fall. answer is b part. |
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| 4614. |
7. A car takes 25 s to stop after the application of brakes.What is the distance traveled by the car if the brakesproduce a uniform retardation of 0.4 m s2 |
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Answer» A=-0.4m/s^2t=25secv=0m/su?s?v=u+atu=v-atu=0-(-0.4×25)u=10m/s s=v^2-u^2/2as= 0-100/-0.8s=125m |
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| 4615. |
2) Figure (8-E14) shows a light rod of length I rigidlyattached to a small heavy block at one end and a hookat the other end. The system is released from rest withthe rod in a horizontal position. There is a fixed smoothring at a depth h below the initial position of the hoolkand the hook gets into the ring as it reaches there, Whatshould be the minimum value of h so that the blockmoves in a complete circle about the ring?Figure 8-E14 |
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| 4616. |
Match the Following (no negative marking) Q.10(8 marks 6 min,)(08, 06). A force which varies with time t as F (3ti +5jN acts on a body due to which its position varies withtime tas š (2t21-5j) where t is in seconds. Work done by this force in initial 2s is:(A) 23 J1,(B) 32 J(C) zero(D) can't be obtained |
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Answer» but the options are A) 23JB) 32JC) 0D) Can't be determined and 48i^ - 25j^ isn't any of these |
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| 4617. |
Describe the 'Green House Effect' in your own words. |
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| 4618. |
Q.6-Section-The escape velocity v of a body depends upon the accelaration due to gravity of aplanet and (ii) Radious of the planet R. Derive formula diamensionallyT |
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Answer» The answer of this Question is 2 |
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| 4619. |
11. Explain why the planets do not twinkle.12. Why does the Sun appear reddish early in the morning?13. Why does the sky appear dark instead of blue to an astronaut? |
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Answer» 11. The planets are much closer to the earth, and are thus seen as extended sources. If we consider a planet as a collection of a large number of point-sized sources of light, the total variation in the amount of light entering our eye from all the individual point-sized sources will average out to zero, thereby nullifying the twinkling effect. Hence, the planets do not twinkle.12. It is due to scattering of light. Light near the horizon passes thicker layers of air and larger distance of atmosphere hence, most of the blue light is scattered away and longer wavelengths reach our eyes giving rise to reddish colour of sun.13.Because there is no atmosphere in the space. |
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| 4620. |
By A simple pendulum of dength I has time period toand is Susbended Muam "O' such that an obstadetouches the string herticalles below at a distanceequilibrium position. Pendulum isy and itsdeflected slightly to the might and Meleased itcomes back to initial position aften timeFind the value ofy(A)Ice BECB152CD) 34 |
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Answer» D. 3L/4 is a correct answer can you please give the solution process |
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| 4621. |
39. In the circuit shown in figure heat developed across 222,4 Ω and 3Ω resistances are in the ratio ofτ (a) 2:4:3(b) 8 : 4: 12(c) 4:8:274Ω Ι(d) 8: 4:272 ΩΛΜ3 Ω |
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Answer» the answer is a no a (d) 8: 4: 27is a correct answer |
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| 4622. |
24. The current flowing through a resistor connected in an electrical circuit and the potential (2difference developed across its ends are shown in the given ammeter and voltmeter. Findthe leastgiven resistor?count of the voltmeter and ammeter. What is the voltage and the current across the |
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Answer» LC of ammeter = 0.2/10a = A = 0.02 A LC of Voltmeter = 1/10 = 0.1 V Current = 15 x 0.02 A = 0.3 A Potential difference = 21 x 0.1 V = 2.1 V |
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| 4623. |
Answer the following :An electric iron of resistance 20heat developed in 30 seconds.takes a current of 5A. CalculateExplain the process of digestion of food in small intestine of man. |
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Answer» Given that Resistance(R) = 20 Ω, Electric current (I)= 5 A and time(t)= 30 s Now we know that H= I2Rt = 5×5×20×30 = 15000J=1.5×104J ∴The heat developed in 30s is1.5×104J Digestionbegins in the mouth with the secretion of saliva and itsdigestiveenzymes. The partiallydigested foodenters the duodenum as a thick semi-liquid chyme. In thesmall intestine, the larger part ofdigestiontakes place and this is helped by the secretions of bile, pancreatic juice andintestinaljuice. |
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| 4624. |
The current flowing through a resistor connected in a circuit and thepotential difference developed across its ends are as shown in thediagram by milliammeter and voltmeter readings respectively :(a) What are the least counts of these meters ?(b) What is the resistance of the resistor ?200100 This++mA |
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Answer» (a) Least count of milliammeter = 100/10 = 10 mALeast count of voltmeter = 1/10 = 0.1 V(b) Voltmeter reading = 2+(4*0.1) = 2.4 VMilliammeter reading = 200+(5*10) = 250 mAResistance = V/I = 2.4V / 250mA= 96 ohm Hope it helps.Mark as best if I am correct. 96 is the answer ok. 96 is the answer ok......... |
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| 4625. |
The current flowing through a resistor connected in a circuit and thepotential difference developed across its ends are as shown in thediagram by milliammeter and voltmeter readings respectively(a) What are the least counts of these meters?What is the resistance of the resistor?200100300(b)mA |
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Answer» Least count = 200-100/10= 100/10= 10 for ameeterand for Voltmeterits2-1/10= 0.1nowb) resistance= v/i= 2.4/250*10^-3= 9.6 ohm |
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| 4626. |
uiigHow does a DBMS provide Data Security? |
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| 4627. |
NMMS(E)99. Which of the following system gives 7N as resultant force14N7NISNISN3N(a23N7N7N7N100. An ohiect mose |
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Answer» 1) 14 from both direction so resultant 02) 23 from one and 30 from other direction so resultant 73) 7 from both direction so resultant 04) 7 from both direction so resultant 0 |
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| 4628. |
The potential dillererfigure isotential difference between A and B in the followingIKCET 20084V12V 992(a) 24 V(c) 321(b) 14 V480 |
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Answer» potential difference is 24 |
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| 4629. |
Which of the following sets of concurent forcesbe in equilibrium?(1) F, 3N, F2 -5N, F, - 1NIKCET 2000(4) F.- 3N, F.5N, F,-15N |
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| 4630. |
an electric bulb of 60w is used for 6h perday.Calculate the unit of energy consumed in one day by the bulb. |
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Answer» Electrical energy consumed by the bulb per day = P x t = 60 W x 6h = 360 Wh1 unit = 1 kW h = 1000 W h1 W h = (1/1000) unit360 W h = 0.360 unitElectrical energy consumed by the bulb per day = 0.36 unit |
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| 4631. |
W N T = e बिन्दु होतास्पष्ट दृष्टि की न्यूनतम दूरी है- i(2007 HW, 10 FD)MM AT |
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Answer» स्पष्ट दृष्टि की न्यूनतम दूरी 25 सेंटीमीटर होती है |
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| 4632. |
(o2- Ratio between maximum range and square of time of flight10- Ain projectile motion is(b) 49:10(d) 10:98are thrown form the top of anda) 0:49c) 98:10bree marticles A B and |
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| 4633. |
darpan kishe kheate hain |
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Answer» mirror ko / sisa ko kehete hain |
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| 4634. |
With the switch K open, the ammeter reads 0.6 A. What will be the ammeter reclosed ?en tading whMultiple Choice Questions (MCQs)36. The figure given below shows three resistors6Ω6Ω2ΩTheir combined resistance is(a) 110(b) 14 Ω(c) 62Ω(d) 7 Ω |
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Answer» C is correct one. 6Ω and 2Ω are in parallel so their Req = 6*2/(6+2) = 12/8 = 3/2 Ω now 6Ω is again in series with Req so , Req final = 6+3/2 = 15/2Ω = 7(1/2)Ω c is not correct .. D is correct yes d is correct thanx |
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| 4635. |
who wrote the Aryabhatiya? |
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Answer» Aryabhata I |
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| 4636. |
B) increasesdecreasesDi becomes zero5)When a mercury drop of radius R, breaks into n droplets of equal size, the radius (t) of each droplet isA)r=-D)71n1/31l6) The surfane tensinof a tiquict in Tr |
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| 4637. |
IA liquid drop of a diameter' D ‘ is split into 27 droplets . IfT is surface tension of theliquid the change in energy is1) 2n D48.4) zero |
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| 4638. |
73. A water drop is divided into 8 equal droplets. Thepressure difference between inner and outer sidesof the big drop(1) will be the same as for smaller droplet(2) will be half of that for smaller droplet(3) will be one-forth of that for smaller droplet(4) will be twice of that for smaller droplet. |
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| 4639. |
| Nl&B (u)wojun जपा o) “aAnum5/ U1 $nq 241 Jo voTEIEE5.बा |
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Answer» Convert km/hr into m/sec . That's 90*5/18 which comes to 25m/s Now using kinematics equation 25 =. 0+a*30 It comes to 25/30 = 0.834 m/sec^2 Like my answer if you find it useful! |
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| 4640. |
How does the force of gravitation between twoobjects change when distance between them isreduced to half? |
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Answer» oo is my first post in a bit of a joke that |
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| 4641. |
(10)- How does the force of gravitation between twoobjects change when distance between them isreduced to half? |
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| 4642. |
How does the force of gravitation between two objects changewhen the distance between them is reduced to half?1. |
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| 4643. |
1.How does the force of gravitation between two objects changewhen the distance between them is reduced to half? |
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| 4644. |
Two sphere each of mass M and radius - are2connected with a massless rod of length 2R asshown in the figure. What will be the moment ofinertia of the system about an axis passingthrough the centre of one of the spheres andperpendicular to the rod:R/2R/2 |
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Answer» Moment of inertia of sphere A about the axis PQ is, IA= (2/5)M(R/2)2= (1/10)MR2 Moment of inertia of sphere B about its own axis is, IB= (2/5)M(R/2)2= (1/10)MR2 Using parallel axes theorem the moment of inertia of sphere B about the axis PQ is, IB/= M(2R)2+ (1/10)MR2= (4 + 1/10)MR2= (41/10)MR2 So, moment of inertia of the system is, I = IA+ IB/= (1/10)MR2+ (41/10)MR2 => I = (42/10)MR2= (21/5)MR2 |
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| 4645. |
L. The momentum of a body of mass 5 ig is 500kgmsFind its KEAns 2.5x 1 |
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Answer» momentum =500mv=500m=5kghence V=500/5=100m/shence ke=1/2mv^2=1/2*5*100*100=2.5*10^4J |
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| 4646. |
(a) 4NA heavy block of mass M hangs in equilibrium at the end of a rope of mass m and length 2 connected toa ceiling. Determine correct variation of tension T as a function of xb) 6IN34.g.T.MgT.(d) (M+ m)gig(c) (M+ m)g |
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Answer» Option b). because tension will vary linearly according to length of string. |
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| 4647. |
35.(C) a parabola (D) an ellipseThe electric field at (30, 30) cm due to acharge of -8 nC at the origin in NC is(A) -400(i+1) (B) 400 (i+j)(C)200,2 (i+] (D) 200/2 (1+1)Techarges 4 x 10°C and -16 x 10°C |
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Answer» correct Option C Distance of the charge from the origin = d= The electric field strength is in the direction of Ф where Tan Ф = 30 cm/ 30 cm = 1 => 45 deg. with X- axis. Cos Ф = 1/√2 and Sin Ф = 1/√2Since E is negative, the Ф is in the 3th quadrant. ie., Ф = 180+45 deg. = 225 deg The vector form of electric field = |
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| 4648. |
An infinite line charge produces a field of 9 x 104 NC-1 at a distance of 4 cmCalculate the linear charge density |
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| 4649. |
6. Fll in the blanks in the following(e) Aperson i m in front of a plane mÄąror seens to beA person 1 m in front of a plane mirror seems to beaway from his imageInear with right hand in front of a planein the mirror that your right ear is touched with(b) if you touch yourmirror it will be seen(c) The size of the pupil becomes(d) Night birds havewhen you see in dim light.cones than rods in their eyes. |
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Answer» thanks |
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| 4650. |
Aperson 1 m in front of a plane mirror seems toaway from his image.be(a)ear with right hand in front of a planemirror it will be seen in the mirror that your right ear is touched with(b)If you touch your(c)The size of the pupil becomesyou see in dim light.(d)Night birds havecones than rods in their eyes. |
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