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.
| 1401. |
1 kg-lf 3 equal masses of 1 kg as shown in the figure are 11 kg-accelerated with an acceleration a = 2 m/s2, then the trtemtension T2; is:(A) 2N(B) 6 N(C) 4 N(D) none of these |
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Answer» given the net acceleration is 2m/s². so, force equation for left is => T2 = m*a = T2 = 1.2 = 2N |
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| 1402. |
A body of mass m is accelerated uniformly from rest to a speed v in a time T. The instantaneous powerdelivered to the body as a function of time, is given by :[AIEEE 2005, 4/300]mu?1 my 21 mv2(3).12212 |
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Answer» option 3 is the correct answer option 3 is correct answer option 3 is the right answer 3 part is right answer option 3 is the correct answer of the given question |
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| 1403. |
What is the difference between non uniformcircular motion and uniform circular motion?Give examples. |
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| 1404. |
(19) What is the difference between non uniformcircular motion and uniform circular motion?Give examples |
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| 1405. |
3. How would you ususe an electroscope to determine the nature of the charge (positive or negative)on a body? |
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| 1406. |
Q. 2. What is vertical circular motion? Show that the motion of an object revolving in vertical circle is non uniformcircular motion.tion and hottom nosition far an object |
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Answer» The motion in the vertical circle involves the concepts of centripetal force ,conservation of energy and the kinetic energy. its all about the constraint that is in simple words a system of parameters that the system must obey. eg. a stone sliding down through mountain. so motion in a vertical circle means that constraint of motion in a vertical circle must be equal to the centripetal of remained circle and when the mass moved down it must be satisfied with the demands of conservation of energy as gravitational potential energy when converted to kinetic energy . Velocity and tension in string when body moves in Vertical Circle. Energy at Point P = Energy at Point L 1/2 m v sq. + m g h =1/2 m u sq. Hence v sq. + 2gh = u sq. Hence v sq. = u sq. - 2gh Hence v = underroot u sq. - 2gh ...............................( 1 ) Hey Mate.. Consider the centripetal force at Point T T - mg Cos0 = mv sq. / r ......... ( 2 ) Hence mg Cos0 + mv sq. / r Cos0 = ( r - h ) / r Subsituting in eq ( 2 ) T = mg ( r - h ) / r + mv sq. / r Hence T = m / r ( g ( r - h ) + v sq. ) But v sq. = u sq. - 2 gh T = m / r ( gr - gh + u sq. - 2 gh ) T = m / r ( u sq. - 3 gh + gr ) |
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| 1407. |
transferred?) What is an electric current? What makes an electric current flow in a wire ? |
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| 1408. |
A. Very Short Answer Questions1. Define electric current. Give its unit.2. What is the conventional direction of electric current?3. Is electric current a scalar or a vector? |
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Answer» 1)The SIunitfor measuring anelectric currentis the ampere, which is the flow ofelectriccharge across a surface at the rate of one coulomb per second.2)from positive terminal to negative terminal3)current is a scalar quantity |
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| 1409. |
Electric resistanceElectric currentElectric potential differenceF Electric power |
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Answer» 1.The electrical resistance of an object is a measure of its opposition to the flow of electric current2.An electric current is a flow of electric charge. In electric circuits this charge is often carried by moving electrons in a wire.3.Voltage, electric potential difference, electric pressure or electric tension is the difference in electric potential between two points4.Electric power is the rate, per unit time, at which electrical energy is transferred by an electric circuit |
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| 1410. |
5.When a plastic rod rubbed with wool is broughtnear the knob of a negatively charged gold leafelectroscope, the gold leaves(1) Contract(2) Dilate(3) Start oscillating (4) Collapse completely |
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Answer» The answer is dilate. Reason : The rod repels negative charge from the top of the electroscope. This creates an excess of negative charge at the leaves. The leaves having the same charge, are repelled, and spread. the answer is dilate because the rod repels negative charge from the top of the electroscope |
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| 1411. |
What is an electroscope? |
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| 1412. |
what is the use of electroscope??? |
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Answer» djdjulS slaps elysians ska all cd LDL isliye is all of aap electroscope is an early invention of instrument used to detect the flow of electrical charge. The electroscope was the first electrical measuring instrument |
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| 1413. |
Describe the working of an electroscope |
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| 1414. |
2. Describe the method of charging an electroscope by the method of conduction |
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| 1415. |
27. An elelctric bulb of 60 W is used for 6 h per dayCalculate the uits of energy connsumed in oneday by the bulb. |
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| 1416. |
34. An electron makes 3 x 10 revolutions per second in acircle of radius 05 angstrom. Find the magnetic field IBat the centre of the circle |
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| 1417. |
2. Water flows through a horizontal tube of variable cross section as Ashown in figure. The area of cross section at A and B are 4 mm2 and2 mm2 respectively. If 1 cc of water enters per second through A, find (a)the speed of water at A, (b) the speed of water at B and (c) the pressureFig. 164difference pA PB- |
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| 1418. |
A wheel is at rest. Its angular velocity increasesuniformly and becomes 80 radian per second after 5second. The total angular displacement is :(A) 800 rad(C) 200 rad0400 rad(D) 100 rad |
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| 1419. |
(B) 12A wheel is at rest. Its angular velocity increases uniformly and becomes 80 radian per second after 5 secondThe total angular displacement is(A) 800 rad(B) 400 racd(C) 200 rad(D) 100 rad |
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Answer» thanks yrr |
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| 1420. |
12 Fa he t13. A giri weighing 25 kg stands on the floor.She exerts a downward force of 250 N on the floor. What hadoes the floor exert on her ? |
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| 1421. |
A boat1oaatitis depressed by 3 cm. The mass of theing length 2m and width 1m isna lake. When a man stands on the 10boat haboat, it isman issokg2) 55kg3) 60 kgf wood supporting 200g mass iust4) 70 kg |
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Answer» Volume of water displace=2×1×3/100=6/100=•06m^3mass of man =volume of water displace ×density of waterM.=•06×1000=60kgmass of man=60kg |
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| 1422. |
Two plane mirrors are inclined to each other such that aray of light incident on the first mirror and parallel to thesecond is reflected from the second mirror parallel to thefirst mirror. The angle between the two mirrors is(1)30°(2) 45°(3) 60°(4) 75° |
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Answer» thanks |
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| 1423. |
io.For what angle between the two plane mirrors, a light ray, after reflection once fromeach of the mirrors becomes parallel to the incident ray direction?D) 90 |
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| 1424. |
6. If a body' displaces through (3i +4m on application of a force F(10i-3j) N, then the work done will be-a) 10 w(b) 12W(c) 16W(d) 18W |
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Answer» Use F. S =w............ 18joule. |
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| 1425. |
Initially the capacitor was uncharged. Current in the capacitor just after switching on wll be ?switch2R |
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| 1426. |
value.WhalnaingeWllUto half of its formerit?Why are coils of electric toasters and electric irons made of an alloyrather than a pure metal? |
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| 1427. |
A force F=10i- 6j+ 8 k Newton produces an acceleration of 1 m/s2 in body of mass m calculate m |
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| 1428. |
Distance and DisplacementQ.A runner completes one round of a circular pathof radius r in 40 seconds. His displacement after2 minutes 20 seconds will be(A) Zero(B) 2Ttr(D) 7Tur |
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Answer» in 40 second he completes one circular path..so in 2 minutes 20 seconds he will complete 5 complete revolution and half revolutionso displacement=2r thanks |
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| 1429. |
A particle is moving along a circular path of radius6 m with uniform speed of 8 ms-1. The averageacceleration when the particle completes one halfof the revolution is(A) m/s2(B) m/s2(C)m/s2(D) of these |
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Answer» acceleration=change in velocity/time =8-(-8)/πr/8 =16*8/6π 64/3π |
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| 1430. |
Example 2/A body moves in acircular path of radius 20 cm. If itcompletes two and half revolution alonthe circular path, then find distance anddisplacement of the body |
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| 1431. |
A particle moves on a circular path of radius 'Y, It completes one revolution in 40 s. Calculate distance anddisplacement in 2 min 20 s.lumina nn a cimular in around of radius 200 m start froin a point P on the edge of the ground and reachhot in the marmitude of the disc |
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Answer» what is your question |
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| 1432. |
9. A runner completes a circular path of 1000m inn 5 s. What is its average speed and averagevelocity?2mi |
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Answer» average speed = 1000/125 = 8 m/s |
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| 1433. |
6. A particle starts from the origin, goes along X-axis to thepoint (20 m, 0) and then returns along the same line to thepoint (-20 m, 0) The distance and displacement of theparticle during the trip are(a) 40 m, 0(c) 40 m,-20 m(b) 40 m, 20 m(d)-60 m,-20 m |
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| 1434. |
Example 9.7 A girl of mass 40 kg jumpswith a horizontal velocity of 5 ms ontoa stationary cart with frictionlesswheels. The mass of the cart is 3 kgWhat is her velocity as the cart startsmoving? Assume that there is noexternal unbalanced force working inthe horizontal direction. |
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Answer» . we will use conservation of linear momentum initial momentum of the system is =mass of girl × velocity = 40× 5 = 200kg-m/s and when she will jump the net mass of the system will be 40+3 = 43 kg and both will have same final Velocity V so, final momentum will be 43*(V) now mi = mf => 200 = 43v => v = 200/43 = 4.65 m/s |
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| 1435. |
Example 9.7 A girl of mass 40 kg jumpswith a horizontal velocity of 5 m s1 ontoa stationary cart with frictionlesswheels. The mass of the cart is 3 kg.What is her velocity as the cart startsmoving? Assume that there is noexternal unbalanced force working inthe horizontal direction. |
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Answer» No external force => momentum is constant. momentum of girl = 40 * 5 = 200 kg-m/sec momentum of girl and cart = total momentum of the girl + momentum of cart = 200 + 0 = 200 kg m/sec speed = 200 / (40 + 3) = 4.65 m/sec |
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| 1436. |
Example 9.7 A girl of mass 40 kg jumpswith a horizontal velocity of 5 m s1 ontoa stationary cart with frictionlesswheels. The mass of the cart is 3 kgWhat is her velocity as the cart startsmoving? Assume that there is noexternal unbalanced force working inthe horizontal direction. |
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Answer» no external force => momentum is constant. momentum of girl = 40 * 5 = 200 kg-m/sec momentum of girl and cart = total momentum of the girl + momentum of cart = 200 + 0 = 200 kg m/sec speed = 200 / (40 + 3) = 4.65 m/sec tq |
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| 1437. |
An object experiences a net zero external unbalanced force. Isit possible for the object to be travelling with a non-zero velocity?If yes, state the conditions that must be placed on themagnitude and direction of the velocity. If no, provide a reason. |
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| 1438. |
1. An object experiences a net zero external unbalanced force. Isit possible for the object to be travelling with a non-zero velocity?f yes, state the conditions that must be placed on themagnitude and direction of the velocity. If no, provide a reasorn. |
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Answer» When a net zero external unbalanced force is applied on the body, it is possible for the object to be travelling with a non-zero velocity. In fact, once an object comes into motion and there is a condition in which its motion is unopposed by any external force; the object will continue to remain in motion. It is necessary that the object moves at a constant velocity and in a particular direction. |
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| 1439. |
Under a force of 101-3 j +6 k newton, a body ofmass 5 kg is displaced from the position61+ 5-3K to the position 10i-2j + 7kAns. 121 J)Calculate the work done. |
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| 1440. |
e Balance and unbalanced forceDerive the Newton's second law of motion in terms of linear momeWhat would be the force required to produce an acceleration of 2m/st on a body of2 |
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Answer» Newton's Second law of motion :- The rate of change of momentum is directly proportional to the force applied on the system.Force applied is directly proportional to the product of mass and acceleration .Let be the initial and final momentums respectively.According to newton's second law :-pf - pi / t ∝ FWe know that,Momentum ( P) = mv .Let v be the final and u be the initial velocity .Now,mv - mu / t ∝ FF ∝ m ( v-u) /tF ∝ ma.F = kma.Here, K is the proportionality constant. It's value is 1 .Units of Force are given by the units of mass and acceleration. Units of force is Kgm/s² .In accordance to honour the contributions of Newton, 1 kgm/s² is termed as 1 Newton. |
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| 1441. |
b)2 3When 99.8 cm3 water at 4°C changes its temperature to 0°C, its volumebecomes 100 cm'. Calculate average value of coefficient of volumeexpansion of water in the range 0°C to 4°C.3+2 |
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| 1442. |
When 99.8 cm water at 4°C changes its temperature to 0°C, its volumebecomes 100 cm. Calculate average value of coefficient of volumeexpansion of water in the range 0°C to 4°C.3+2 |
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| 1443. |
Why nitrogen gas is used to fill electric bulbs? |
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Answer» A bulb is filled with an inert gas such as argon ornitrogen slows down the evaporation of the tungsten filament compared to operating it in a vacuum. This allows for greater temperature and therefore results in greater efficiency of filament life. Inert gases reduce the rate of evaporation rate of tungsten metal used in Filament of bulbs. Thus it increases the efficiency of bulbs. N2 gas is an inert gas. That's why the bulbs are filled with nitrogen. |
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| 1444. |
What are the benefits of CFLs over electric bulbs? |
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Answer» CFLs are up to four times more efficient than incandescent bulbs. You can replace a 100-watt incandescent bulb with a 22-watt CFL and get the same amount of light. While initially theycostmore, CFLs are less expensive in the long run because they last much longer than incandescent bulbs. |
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| 1445. |
1. 1 g of water at 373 k is converted into steam at the same temperature at normal pressare. Thevolume of 1 em of water becomes 1671 em on boiling. Calculate the change in the internalnerer of the system if the heat of vaporization is 540 cal gT. Ans 2098 Jsrie of heat and |
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Answer» here we shall use the first law of thermodynamics ΔQ = ΔU + ΔW so, change in internal energy will be ΔU = ΔQ -ΔW here, ΔQ = mL =1g x 540Cal/g = 540 Cal ΔU = change in internal energy (to be calculated) ΔW = change in work done so, ΔW = pΔV where ΔV = 1671cm3- 1cm3= 1670cm3= 1670 x 10-6m3 p = hρg = 0.76 x 13600 x 9.8 ~ 101293 N/m2 so, ΔW = 101293 x 1670x10-6= 169.16J or ΔW = 169.16 / 4.2 = 40.3 Cal thus, we have ΔU = 540 Cal - 40.3 Cal so, ΔU = 499.7 Cal = 2098.74 J |
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| 1446. |
what is absolute pressure |
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Answer» Absolute pressure is the pressure measured with absolute zero taken as a reference. Hence, a perfect vacuum, called absolute zero, is the datum. Gauge pressure is the pressure measured with atmospheric pressure taken as its datum. Absolute pressureis zero-referenced against a perfect vacuum, so it is equal to gaugepressureplus atmosphericpressure. Gaugepressureis zero-referenced against ambient airpressure, so it is equal toabsolute pressureminus atmosphericpressure. Negative signs are usually omitted. |
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| 1447. |
100. Even Carnot engine CANNOT give 100% efficiency, because we CANNOT(A) ellminate friction(B) prevent radiation(C) reach absolute zero temperature (D) find ideal sources |
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Answer» find ideal sources D find ideal sources Efficiency of a carnot engine is given by η = 1 - Q2/Q1 In order to achieve 100% efficiency (η=1), Q2 must be equal to 0 which means that all the heat form the source is converted to work. The temperature of sink means a negative temperature on the absolute scale at which the temperature is greater than unity. This would be violation of second law of thermodynamics. Hence negative temperature of absolute scale is impossible and we cannot reach absolute 0 temperature. The correct option is C. |
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| 1448. |
Explain absolute Pressure, atmospheric pressure and gauge pressure. |
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Answer» Ans :- Absolute pressureis zero-referenced against a perfect vacuum, so it is equal to gaugepressureplus atmospheric pressure. Gaugepressureis zero-referenced against ambient airpressure, so it is equal toabsolute pressureminus atmosphericpressure. Atmospheric pressure is the force that the weight of air exerts on a surface. It is usually estimated through hydrostatic pressure, which is produced by the total weight of air above the spot to be measured. This type of pressure can be applied in many industrial processes, including corrosion protection. Gauge pressure is zero-referenced against ambient air pressure, so it is equal to absolute pressure minus atmospheric pressure. ... Gauge pressure is measured with the standard atmospheric pressure as the baseline. |
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| 1449. |
9:45 an10:0010.110.102. A farmer moves along theboundary of a square field of side10 m in 40 s. What wll be themagnitude of displacement of thefarmer at the end of 2 minutes 20seconds from his initial position? |
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| 1450. |
A farmer moves along the boundary of a square field of side 10 min 40 s. What will be the magnitude ofdisplacement of the farmer at the end of 2 minutes 20 seconds from his initial position ? |
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Answer» displacement after 2m 20s is 20meter |
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