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.
| 401. |
oja sold a T.V forofit %?15,000 and earned profit of2500, Fin |
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| 402. |
Example 12.2 A person clapped his handsnear a cliff and heard the echo after5 s. What is the distance of the linefrom the person if the speed of thesound, v is taken as 346 m si? |
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Answer» Hi jogendra , aapke question ka answer yaha hai........ The sound traveled from the place the person stood, to the cliff and then rebounded back to the person , so the person heard the resound/echo. So the sound traveled twice the distance between him/her and the cliff. Distance between cliff and person = 346 m/sec * 5 seconds / 2 = 865 meters |
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| 403. |
Q.2 Consider the circuit shown below4225 2W100 mof the ideal source supplies aIf the ideal source supplies 1000w, half of which is delivered to the 100 2 load, then the valueof bis(A)1.5(B).89(C).56(D).67 |
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| 404. |
Q.2 Consider the circuit shown below4 8225 2w1b0 000000100 V1002of the ideal source suIf the ideal source supplies 1000w, half of which is delivered to the 100 load, then the valueof bis(A)1.5(C).56(D).67(B).89 |
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| 405. |
2 Consider the circuit shown below42252w100 Vis0000001002the ideal source supplies 1000w, half of which is delivered to the 100 a load, then the valueFb isa)1.5(B).89(C).56(D).67 |
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| 406. |
five 100 w bulbs are used 10 hours daily for 30 days.if price of of per unit is rs 4.what is the total bill? |
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Answer» Total bill will be1000 watt = 1 unitso for 150000 unit150*4600 rupees |
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| 407. |
I. Calculate the current through the 10. k Ω resistor shown in Fig. T.221. Assume thatthe Zener diodes have ideal voltage-source characteristics.lAns. 1 mA]60 Ω10 knsvずv-Fig. T.2.2.1 |
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Answer» since zener diode here is in reverse biased form , so maximum potential around 10kΩ is the sum of potential of both the zener diode = 5+5 = 10v now current through 10kΩ = v/R = 10/10kΩ = 1mA. |
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| 408. |
WIN MT Dombay)Passage l:An ideal source of emf V and a perfectly conducting wire are provided with the circuit.RaReŽR3If the conducting wire is not used and the source is connected between the terminals A & B, the heat power liberated is40 W. If the conducting wire Connects C&D and the source is connected between the terminals A & B, the heat powerliberated is 60 W. If the conducting wire is not used and the source is connected between the terminals C&D, the heatpower liberated is 30 W. If the conducting wire Connects A & B and the source is connected between the terminals C&D, the heat power liberated is P1. Considering the two cases, when the conducting wire is not used, R(B) 3R, -4R2(C) 4R2 - 3R(D) 4R, -3R(A) 3R2 - 4R2. Considering the two cases, when conducting wires is used, P: 60 =(C) (R2 + R3) : (R+R)(A) (R2-? + R3-1): (R? + R3-4)(D) (R+R): (R2 + R3)(B) (R-1 + R3-1): (R2-1 + R3-1)(C) 45 W(D) 80 W(A) 20 W(B) 40 W3. PE |
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| 409. |
Two electric bulbs, rated at (25 W, 220 V)and (100 W, 220 V), are connected in seriesacross a 220 V voltage source. If the 25 Wand 100 W bulbs draw powers P, and P2respectively, then: |
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| 410. |
a) 100 W(b) 75 W(c) 50 W(d) 25 Wconductingwires of the same material and of equal lengths and equal diametersTwoare first connected in series and thendifference. The ratio of heat produced in series and parallel combinations would beparallel in a circuit across the same potentialhl 2:1 |
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Answer» Heat produced in the circuit is inversely proportional to the resistanceR. LetRSbe the equivalent resistances of the wires if connected in series andRPbe the equivalent resistances of the wires if connected in parallel ThenRS=R+R=2R And 1/RP= 1/R +1/R 1/RP= 2/R OrRP=R/2 Now the ratio of heat produced is given by Hs/Hp=V^2/Rs*t/V^2/Rp*thenceHs/Hp=1/4 |
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| 411. |
\begin{array} { l } { \text { An electric iron of } 1100 \mathrm { W } \text { is } } \\ { \text { operated for } 2 \text { hrs daily. What will } } \\ { \text { be the electrical consumption } } \\ { \text { expenses for that in the month of } } \\ { \text { April? (The electric company } } \\ { \text { charges Rs } 5 \text { per unit of energy). } } \\ { \text { (Ans: Rs 30) } } \end{array} |
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Answer» thanks for the answer. thanks for the answer |
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| 412. |
(BSEB, 2013)(a)Cm-2(e) Am(b)N-C(d) Vm-12, feR翹弱弱制(BSEB, 2010)fa)wet (Conservative)(b)研 Non-conservative)4drdV5dVdr |
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Answer» the unit of electric flux density is C m^-2 static electric fields are conservative in nature. |
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| 413. |
4) 10 s15 s68. A weight lifter of power 1960awatt lits e load of mass M from the ground to a height of 2m in 3 sacondsc 1M is(1) 100 kg2) 200 kg300 kg(4) 400 kg |
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Answer» thanks for this question Counting in twos write from 400 to 410 |
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| 414. |
weight lifter jerks 220 kg vertically thrometers and holds still at that heighto minutes. The works done by him in lifd in holding it still are respectively2201, 330J2334 J, 10 J52) 3234 J, 04) 0, 3234 J |
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Answer» Dear Student ,Here in this case the work done by lifting the weight 220 kg through 1.5 m is equal to the change in potential energy .So the work done = change n potential energy = mgh = 220 x 10 x 1.5 = 3300 JBut the work done in holding the weight at the same height for two minutes is zero . thanks |
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| 415. |
2)0.29 and 3.5 m (3) 0.29 and 6.5 mB-3.8.person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 timesAssume that the potential energy lost each time he lowers the mass is dissipated. How much fat will heuse up considering the work done only when the weight is lifted up ? Fat supplies 3.8 x 10' J of energper kg which is converted to mechanical energy with a 20% efficiency rate. Take g 9.8 msSectiJEE (Main) 2016; 4/120,-11C-2(1) 6.45 x 10 kg (2)9.89 x 10* kg(3) 12.89 x 10 kg (4)2.45 x 10 kg |
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| 416. |
A person trying to lose weight (dieter) lifts a 10 kg mass through 0.5 m, 1000 times. Assume that thepotential energy lost each time she lowers the mass is dissipated. (a) How much work does she do againstthe gravitational force? (b) Fat supplies 4 x 107 J of energy per kilogram which is converted to mechanicalenergy with a 20% efficiency rate. How much fat will the dieter use up? (use g 10 m/s)100 Ni m-1 as |
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| 417. |
A porter lifts a luggage of 15kg from the ground and puts it on hishead 1.5 m above the ground. Calculatethe work done by him on the luggage. |
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Answer» Work done by the porter on the luggage = force×displacement =mgh= 15 kg x (9.8 m/s2) x 1.5 m= 220.5 J |
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| 418. |
â LongAnswer Type QuestionsDescribe the three equations for uniform accelerated |
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Answer» 1)v=u+at2)v^2 =u^2 +2as3)s=ut+(1/2)at^2 where v:final velocityu:initial velocitya:uniform acceleration t:time takens:distance travelled |
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| 419. |
Long Answer Type Questions3. Desmibe the three equations for uniform accelerated |
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| 420. |
When the momentum of body is increasedby three times, its K.E. becomes :(a) Twice(b) Half(c) Four times (d) Nine times |
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Answer» Relationship between kinetic energy and momentum is like:- k.E= P²/ m p= (momentum) m= (mass) by keeping mass constant we have:- k.E= p² k.E= (3)² K.E= 9 times☺ Relationship between kinetic energy and momentum is like:- k.E= P²/ m p= (momentum) m= (mass) by keeping mass constant we have:- k.E= p² k.E= (3)² K.E= 9 timesAnswerOption- D |
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| 421. |
If the linear momentum of a sphere is doubled, its kinetic energy will becomea) twiceb) four timesc) eight timesd) sixteen times |
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Answer» p = (2*kinetic energy*m)^0.5=> (p2/p1)^2 = k2/k1=> (2/1)^2 = k2/k1=> k2/k1 = 4/1Here, k represents Kinetic energy and p represents momentum.Hence, with doubling of momentum, the kinetic energy becomes four times. Please hit the like button if this helped you. |
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| 422. |
(4) Te8. A car is travelling with linear velocity v on a circular road of radius r. If the speed is increasing at tof 'a' metre/sec2, then the resultant acceleration will be -v2+ a2VL |
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Answer» IThe instantaneous acceleration pointing towards the center of the circle is called centripetal acceleration. It is denoted by 'a₍r₎'It's magnitude is given asa₍r₎ = v²/rwhere 'r' is the radius of curvature.The tangential component cause a change in its speed and it's magnitude is given asa₍t₎ = dv/dt = a m/s²Hence, tangential acceleration is,a₍t₎ = a m/s²Now,Resultant acceleration = a² = a₍r₎² + a₍t₎²Putting the values , we get:Resultant acceleration = a² = (v²/r)² + a²Resultant acceleration = a² = v⁴/r² + a²Resultant acceleration = a = √ (v⁴/r² + a²)which is the required answer. |
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| 423. |
Find the external workdone by the system inK cal, when 12.5 k cal of heat is suppliedto the system and the correspondingincreasing in internal energy is 10500 J(J 4200 J/kcal) |
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| 424. |
When a piece of aluminium wire of finite length is drawn through a series of dies to reduceits diameter to half its original value, its resistance will becomea) Two times b) Four times c Eight times d) Sixteen timesAIIMS 1997] |
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| 425. |
e unlm sperdcen dare |
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| 426. |
A parallel plate capacitor of area 1sqm and dielectric constant 7 is charged toa pot. Of 300V. Il the distance b/w the plates is 0.01cm. Find the cnergystored in the capacitor |
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| 427. |
101. The velocity of a particle is v = 61 + 2-2k. Thecomponent of the velocity parallel to vectora = i + j + k in vector form is(a) 6i+ 2j+2k(b) 2i+2j +2k(d) 6+ 2j-2k^ |
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| 428. |
The (x, y, z) coordinates of two points A andB are given respectively as (0, 4, -2) and (2, 8, -4). The displacement vector from A toB is-(25)(a) -2i+4j-2k(b) 2i-4j+2k(c) 2i+4j-2k (d) -2i-4j-2k |
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Answer» the displacement vector will be B - A =(-2-0)i +(8-4)j+(-4+2)k = -2i+4j-2k operation A |
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| 429. |
The angular velocity of a particle rotating in a cir-cular orbit 100 times per minute is10.47 rad/s C1.66 rad/s(3) 10.47 deg/s(4) 60 deg/s |
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Answer» N = Number of rotation per minute N = 100 per minute so the frequency F = N/60 per second =100/60 per second. Thus the angular velocity, W = 2πF so W = 2 x 3.14 x (100/60) = 10.466 radian per second |
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| 430. |
10A particle of mass m is tied to a light string of length / and rotatedin horizontal circle with constant speed v /29 Tension in thestring is |
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Answer» T = mV²/r here r = l , and v = √2g so, T = m*2g/l = 2mg/l. check my other questions plus Bro plsss |
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| 431. |
A particle is moving with constant speed v along x-axis in positive direction. Find the angularvelocity of the particle about the point (0, b), when position of the particle is (a, 0). |
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Answer» namanpreet bhai plz bta dena ye shi h ya galat plz W=v(perpendicular)/r ( where r is the distance between the position of the particle and the point about which the angular velocity is to be calculated) thus,W=vcos/r {m using '' for theta} From the triangle ABC,sin=b/root of (a^2+b^2) where = 90-, Thus sin(90-)= cos, Therefore, cos=b/root of (a^2+b^2). So,W={v*b/root of (a^2+b^2)}/root of(a^2+b^2). which is equal to vb/a^2+b^2. Thus angular velocity = vb/a^2+b^2. shi h bhai shukriya bhai apka |
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| 432. |
A particle is moving with constant speed v in xyplane as shown in figure. The magnitude of itsangular velocity about point O is(0,b)(a,0)a2 + b2vb(a2 + b2 ) |
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Answer» aur iska answer v/(√a^2+b^2) matlab 1st option hai kya answer bhai? 3rd h iska 4 th h? 3rd The component of V perpendicular to the line joining O and V , will give the angular velocity and angle can be found out by cos∅ = b/(√a²+b²)and R = √a²+b² so, angular velocity is Vcos∅/r = vb/(√a²+b²)²=vb/a²+b². option 3 |
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| 433. |
24. A particle is moving with constant speed v along x-axis in positive direction. Find the anguilarvelocity of the particle about the point (0, b), when position of the particle is (a, 0) |
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Answer» copy paste?? |
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| 434. |
A body revolves with constant speed v in a circularpath of radius r. The magnitude of its averageacceleration during motion between two points indiametrically opposite direction is41.2(1) Zero222r |
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Answer» Average acceleration is change in velocity by either change in its magnitude or change in its direction. But here for a uniform circular motion magnitude being constant (speed) direction changes which causes change in acceleration and by definition from NCERT acceleration is change in velocity during the course of overall interval. Therefore following ans should be the quick answer for the given question : Here v1 and v2 are two velocities say at point P and Q diametrically opposite points. |
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| 435. |
17. A particle moves on a given straight line with a constantspeed v. At a certain time it is at a point P on its straightline path. O is a fixed point. Show that OPxusindependent of the position |
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| 436. |
17. A particle moves on a given straight line with a constantspeed v. At a certain time it is at a point P on its straightline path. O is, a fixed point. Show that OPxu isindependent of the position P. |
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| 437. |
5. At what angle must the 2 forces (xey) and (-y)act so that the resultant may be x2 + y2a) cos -202-2b) cos-12x2-y2)x2+y22(x2-y2)2(x2+y2)x2+y2c) cos 2(x2-y2)x2+y2 |
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Answer» thank you |
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| 438. |
8.5.3 EQUATION FOR POSITION-VELOCITYRELATION |
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| 439. |
2 IX-CLASS MATHEMATICsxample-y. If x-3, y 2 is a solutionrite the resultant equation.ion of the equation 5x-7y- k, find the value of k andequationinlution of the equation |
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| 440. |
32.Choose the wrong statement(s)"(a) A dimensionally correct equation may be correct(c) A dimensionally incorrect equation may be incorrect(b) A dimensionally correct equation may be incorrect(a) A dimensionally incorrect equation may be incorrect |
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Answer» a) A dimensionally correct equation may be correct.(b) A dimensionally correct equation may be incorrect.(d) A dimensionally incorrect equation may be incorrect. It is not possible that a dimensionally incorrect equation is correct. All the other situations are possible. (a) A dimensionally correct equation may be correct.(b) A dimensionally correct equation may be incorrect.(d) A dimensionally incorrect equation may be incorrect. It is not possible that a dimensionally incorrect equation is correct. All the other situations are possible. (a) A dimensionally correct equation may be correct.(b) A dimensionally correct equation may be incorrect.(d) A dimensionally incorrect equation may be incorrect. It is not possible that a dimensionally incorrect equation is correct. All the other situations are possible. |
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| 441. |
The equation areax velocity -conslant is called(a) Bemoulli's theorem(c) Equation of continuity(b) Stokes equationd) Reynold's number |
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Answer» Equation Area× velocity = constant is called as Bernoulli's equation. |
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| 442. |
24. Find the area bounded under the curve y 3x +6x +7and the X-axis with the ordinates at x 5 and x 10. |
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| 443. |
What is perfume? How is it made? |
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Answer» Perfume is a mixture of fragrantessential oilsoraroma compounds,fixatives andsolvents, used to give the human body, animals, food, objects, and living-spaces an agreeablescent. Natural ingredients—flowers, grasses, spices, fruit, wood, roots, resins, balsams, leaves, gums, and animal secretions—as well as resources like alcohol, petrochemicals, coal, and coal tars are used in the manufacture of perfumes. |
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| 444. |
Name three liquids, whichwhen tested in the mannershown in Fig.14.9, maycause the magnetic needleto deflect.714 |
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| 445. |
3 Name three liquids, whichwhen tested in the mannershown in Fig.14.9, maycause the magnetic needleto deflect |
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Answer» thank you bro |
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| 446. |
3. Name three liquids, whichwhen tested in the mannershown in Fig. 14.9, maycause the magnetic needleto deflect.Fig. |
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| 447. |
14. परिपथ (circuit) में, वोल्टमापी का पठन12/2002120 2LWWW |
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Answer» Voltmeter = 12*(100/300)= 4 volt 4 volt का पठन करेगा |
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| 448. |
2. When the free ends of a tester are dipped into a solution, the magaeticneedle shows deflection. Can you explain the reason? |
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Answer» what is the meaning of the figix please give me the answer |
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| 449. |
VEWLON Flete6. Aof focal length 20 cm. Theconvex lens of focal length 10 cm is being placed in contact with a concave lensfocal length of the combination is.A) -20 cnmC) 10 cmB) 20 cmD) -10 cm |
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| 450. |
36. A plano convex lens has a focal length f = 20 cm. Ifits plane surface is silvered, then the new focallength will be(a) 10 cm (b) 15 cm (c) 20 cm (d) 25 cm |
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Answer» Its focal length will be 10cm and it will act as a concave lens |
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