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.
| 12601. |
A block slides down a slope of angle θ withconstant velocity. It is then projected up witha velocity of 10ms-1, g=10ms-2 & θ= 30°. Themaximum distance it can go up the planebefore coming to stop is |
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| 12602. |
(a) ¥A-2ada (0) यंत्र आधारित(c) sitGor जाकारित (d) N1 ua 2l |
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Answer» ans is (A) Yantra swatantra |
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| 12603. |
Two fixed frictionless inclined planes makinangle 30° and 60° with the horizontal as showrthe figure. Two blocks A and B are placed on thtwo planes. What is the relative verticaacceleration of A with respect to B30°(1) 4.9 m/s2 in vertical direction(2) 4.9 m/s2 in horizontal direction(3) 9.8 m/s2 in vertical direction(4) Zero |
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Answer» Hence, option (a) is correct. thanks bro |
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| 12604. |
Match the pairsA' Group1. Free electrons a. V/R2. CurrentB' Groupb. Increases theresistance in thecircuitstivity4. Resistances inc. Weakly attachedd. VALIseries |
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| 12605. |
17.When no current is passed through a conductor,(a)(b)the free electrons do not movethe average speed of a free electron over a large periodof time is not zerothe average velocity ofa free electron over a large periodof time is zerothe average of the velocities of all the free electan instant is non zero(c)(d) |
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| 12606. |
3. The elevator shown in figure 15-E5) isdescending wian acceleration of 2 ms The masa of the block A05 kg. What foree is exerted by the block A on theblock BT2 m/s2Figure 5-ES |
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| 12607. |
Figure -8413. The elevator shown in figure (5-E5) is descending witban acceleration of 2 m/'s. The mass of the block A is05 kg. What force is exerted by the block A os theblock B?2 m/s261Figure 5-E5 |
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Answer» Answer:If the elevator was not moving the force applied by the block A on the block B would have been equal to its weight = mg = 0.5 kg x 9.8 m/s²= 4.9 N. But in this case the elevator is descending with an acceleration of a=2 m/s². So in order to apply the Newton's Laws of motion with respect to the elevator we apply a pseudo force equal to ma on the block A in the direction opposite to the acceleration ie upwards. The net force on the block is = mg-ma= 4.9 - 0.5x2 = 4.9-1 = 3.9 N ≈ 4.0 N |
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| 12608. |
5. A current i flows in a wire of circular cross-sectionwith free electrons travelling with a drift velocity v.What will be the drift velocity of electrons when acurrent 2i flows in another wire of twice the radius[Ans. v/2]and of the same material |
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| 12609. |
2. Find out the number of electrons passing through awire per second if the wire carries a current of 5 A.Given : charge on electron1.59 x 10^-19 C.[Ans. 3.125 x 10^19] |
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Answer» number of electrons=current/charge of one electron=5/(1.59×10^(-19))=3.125×10^19 |
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| 12610. |
Fluid Mechani24. A wooden block of mass 0-5 kg and density 800 kg m 32is fastened to the free end of a vertical spring of springconstant 50 N m1 fixèd at the bottom. If the entiresystem is completely immersed in water, find (a) theelongation (or compression) of the spring in equilibriumand (b) the time-period of vertical oscillations of theblock when it is slightly depressed and released. |
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| 12611. |
13. The elevator shown in figure (5-E5) is descending withan acceleration of 2 m/s ". The mass of the block A is0-5 kg. What force is exerted by the block A on theblock B?2 m/s2 |
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| 12612. |
PHYSICSChoose the correct answerA 1000 kg rocket is fired from the surface of earthso that its exhaust speed is 1000 m/s relative torocket. If initial acceleration of the rocket is5 m/s2 then mass of fuel burning per second is(Take g 10 m/s2)(1) 15 kg/s(3) 10 kg/s49-5kg/s (-(4) 20 kgls6.2The magnitude of tension in the horizontal string |
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Answer» dm/dt*V = m*a => dm/dt = 1000*5/1000 = 5kg/s |
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| 12613. |
2. A current of 2-0 A exists in a wire of cross sectional area10 mm If each cubic metre of the wire contains60 x 10 free electrons, find the drift speed. |
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Answer» i=2A he apne 1A likha he or answer bhi galat he |
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| 12614. |
7. A rocket set for vertical launching, has a mass of50 kg and contains 450 kg of fuel. It can havemaximum exhaust speed of 2 km/s. If g 10 m/s?the minimum rate of fuel consumption just to lift itoff the launching pad will be(1) 2.5 kg/s(3) 5 kg/s(2) 10 kg/s(4) 20 kg/s |
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| 12615. |
Conceptsuniform field of 2 0 NC exists in space in x-direction.Taking the potential at the origin to be zero, writeexpression for the potential at a general pointy, 2). (b) At which points, the potential is 25 V? (c) Ifpotential at the origin is taken to be 100 V, whatbe the expression for the potential at a generalt? (d) What will be the potential at the origin if thential at infinity is taken to be zero? Is it practicaloose the potential at infinity to be zero?much work has to he done in assembling thre |
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| 12616. |
14.atare me rend DEVE potential direrence, work done and charge moved(b) Calculate the work done in moving a charge of 4 coulombs from a point at 220 volts to anothe |
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| 12617. |
9. Acubical vessel of height 1 m is full of water. The minimum work done in taking water-out fromcubical vessel of height 1 m is full of water. The(1) 5000 J(3) 5J(2) 10000 J(4) 10 J |
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| 12618. |
AVERAGE LIFE OR MEAN LIFE OF RADIOACTIVE ELEMENT |
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Answer» If in a givenradioactive element, half of itselementshave decayed after one halflife, some well-definedaverage lifeexpectancy can be assumed which is themean lifeof the atoms. |
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| 12619. |
2.23 The voltage gain A of the circuit shown below13.7 Volts312 ΚΩ100 ΚΩΣβ = 1001Ο ΚΩ |
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| 12620. |
2. A fixed pulley is driven by a 100 kg mass falling ata rate of 8.0 m in 4-0 s. It lifts a load of 75.0 kgf.Calculate :(a) the power input to the pulley taking the force ofgravity on 1 kg as 10 N.(b) the efficiency of the pulley, and(c) the height to which the load is raised in 40 s. |
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| 12621. |
120and upper fixedater is 90 mm.ower fixed point,1) 130 - 2020/The distance between the lower and upperpoints of a Fahrenheit thermometer is 90 mmercury rises 30 mm above the lower fixedfind the corresponding temperature.1) 65°F 2) 92°F 3) 62°F 4) 60021 The upper and lower fixed points of a thermomet |
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| 12622. |
//2.23 The voltage gain A of the circuit shown below13.7 VoltsEw312k22100 k22100 kawwwHB = 10010 k2 |
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| 12623. |
8. gy The wok done against gravity in taking 10 ka mass at 1 m height il b a(1)49 J(2) 98 J(3) 196 J(4) of these |
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Answer» work done = mgh = 10*9.8*1 =98J |
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| 12624. |
0. Study the velocity time graph and calcualte25 -201510-1 2 3 4(a) The acceleration from A to B.(b) The acceleration from B to C.(c) The distance covered in the region ABE.(d) The average velocity from C to D.(e) The distance covered in the region BCFE |
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| 12625. |
A stone of mass 1 kg is tied to the end of a string1 m long. It is whirled in a vertical circle. If the veloc-ity of stone at the top be 4 m/s. What is the tensionin the string at the lowest point? Take g-10 m/s2(1) 6 N(3) 16N(2) 66 N(4) 76 N |
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Answer» We know that, Tension= mv²/r-mg Tension= 1×(16/1)-1×(1×10) Tension= 6N The answer is option (a) Sorry, but the answer is 66 N |
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| 12626. |
eedg 10 m/s23. Find the accelerations a, a,. aof the three blocksshown inapplied on (a) 2 kg block, (b) 3 kg block, (eTake g 10 m/sfigure (6-ES) if a horizontal force of 10 N is(c) 7 kg block7 kgFigure 6-ES |
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| 12627. |
18. A particle is thrown vertically upwards. If its velocity at half of the maximum height is 10 m/s, thenmaximum height attained by it is (Take g 10 m/s')(A) 8 m(B) 10 m(C) 12 m(D) 16 m |
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| 12628. |
en drops a 10 ka rock from the top of a 5 m ladder. What is its speed just before it hits the ground? What[Ans. V = 10 m/s. K.E. = 500 J]is its K.E. when it reaches the ground? (take g = 10 m/s) |
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Answer» thanks allot sir |
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| 12629. |
6. The speed of a car as a function of time is shown infigure (3-El). Find the distance travelled by the car in8 seconds and its acceleration2 20100 2 4 610Time in seconaFigure 3-E1 |
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| 12630. |
Two forces F1 and F2 acting on a body have a resultant R, if F2 is doubled the new resultantis Rforms are right angle with F. Prove that R, and F2 have the same magnitude. |
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Answer» Let θ be the angle betweenf1andf2 (Quantities in bold represents vectors) Given, f1+f2=r1----- 1 f1+ 2f2=r2 ----2 Taking dot product on itself of eqn. 1 r1. r1=f1.f1+f1.f2+f2.f1+f2.f2 => r12= f12+ f22+ 2f1f2cosθ ---3. Taking dot product off1with eqn. 2 f1.(f1+ 2f2)=f1.r2 Since,f1andr2are perpendicular to each otherf1.r2= 0 => f12+ 2f1f2cosθ = 0 ---4. By eqn. 3 and 4. r12= (f12+ 2f1f2cosθ) + f22 => r12= 0 + f22 => r1= f2 QED. |
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| 12631. |
6. A body of mass 1 kg is allowed to fall freely undergravity. Find the momentum and kinetic energy ofthe body 5 seconds after it starts falling. Take(Ans. 50 kg ms1, 1250 J)g 10 ms 2 |
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| 12632. |
A body is dropped from certain height H. If the ratio of the distances travelled by it in (n3) seconds to (n -3)nd second is 4 :3, find H. (Take g 10 m s2) |
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| 12633. |
(4)2g20. A uniform rope of mass M and length L is fixed aits upper end vertically from a rigid support. Thenthe tension in the rope at the distance I from therigid support isMg (L-I)(1) Mg L+I(2) L(3) Mg(4) Mg |
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Answer» If the chain is held vertically downward two forces will be acting along the surface of chainone of them is the mass of the chain that is mg acting vertically downward and other is the tension of chain acting vertically upward and canceling the effect of mass pulling it downwardsince acceleration is zero soF(net)=0so mass of chain is m=(L-x)T=mgm=(L-x)/L*mT=(L-x)/LM*g |
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| 12634. |
In the figure, the blocks A, B and C cachof massm have accelerations a,, a, and a, respectively.F, and F2 are external forces of magnituand mg respectively. Thernde 2 mg2mF1 2mgF2-mg |
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| 12635. |
26. Give reasons:) Colour of the clear sky is blueil) Danger Signals are red |
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Answer» The colour of the clear sky is blue because molecules in the air scatter bluelightfrom the sun more than they scatterred light. When we look towards the sun at sunset, we seeredandorangecolours because the blue lighthas been scattered out and away from the line of sight. The primary reason why the colorredis used fordanger signalsis thatredlight is scattered the least by air molecules. The effect of scattering is inversely related to the fourth power of the wavelength of a color. Soredlight is able to travel the longest distance through fog, rain, and the alike. |
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| 12636. |
178. In a short transmission line, voltage regulation is zero what thepower factor angle of the load at the receiving end side is equal to |
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Answer» It is equal to feraanti voltage asat receiving end voltage is higher due to feraanti effect. |
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| 12637. |
The volume of a solid of mass 500 mg is 350cm2.(i) what will be the density of the solid?(ii) What will be the relative density of the solid?07.(iii) Will it float or sink in water? |
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| 12638. |
Integrate by using the substitution suggested in bracket2π4. |
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| 12639. |
air quality standard is measured by (a) TLV (b)TVL (c) TVC (d) none of these |
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Answer» TLV measure the quality |
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| 12640. |
Explain the process of receiving the sound signals to brain |
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Answer» Sound funnels into the ear canal and causes the eardrum to move. Sound vibrations move through the ossicles to the cochlea. Sound vibrations cause the fluid in the cochlea to move. ... The auditory nerve sends signals to the brain where they are interpreted as sounds.1]Sound funnels into the ear canal and causes the eardrum to move.2]The eardrum vibrates with sound.3]Sound vibrations move through the ossicles to the cochlea.4]Sound vibrations cause the fluid in the cochlea to move.5]Fluid movement causes the hair cells to bend. Hair cells create neural signals which are picked up by the auditory nerve. Hair cells at one end of the cochlea send low pitch sound information and hair cells at the other end send high pitch sound information.6]The auditory nerve sends signals to the brain where they are interpreted as sounds. |
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| 12641. |
laced two blocks in contactthe 10 kg blochs State br' S oN two10 kg block. (ii) If the abov e this blocke 20 kgsmooth table are placed5. On a() Ahorizontal force of5.0 N is applied on the 20 kgNthe 10 kg block.the rher sState by what force this block presseswforce is applied onthe other sideof the hen by 5.020ね10 tagwhat force the20 kg blockwill press the 10 kg block?Ans. (i) (10/6) N towards right,(ii) (206) N towards right. |
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Answer» F=ma => That is your magnitude my dear friend. But acceleration is a vector, (x,y,z) direction. If you have a space between the blocks and then contact is made, then you're dealing with momentum. p=mv. So, that is then p=m((v-u)/t). |
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| 12642. |
1.Whichof the following is not aIScharacteristic of sound?a. QualityC. PitchWhich part of the ear is locateexternally and visible?b. Frequed. Loudn2. |
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Answer» Quality is not a characteristic of sound. Frequency , pitch and loudness is the characteristics of sound. |
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| 12643. |
12. The normal duration of I.Se. Physics practical period inIndian colleges is 100 minutes. Express this period inmicrocenmany microcentuturies. I microcentury-10 " × 100 years. Howries did you sleep yesterday ? |
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| 12644. |
Q.1A wire of length 3.in the wire.0m has a percentage strain of 0.015% under a tensile force. Determine the extension |
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Answer» Here, original length, L = 3.0 mStrain =∆L/L = 0.015% = 0.015/100 ∆L = Strain x L or ∆L = extension = 0.015 / 100 x L= 0.015 x 3.0/100 = 0.00045m thanks |
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| 12645. |
When I bump one end of a longmetal bar, the other endinstantaneously moves. Can I usethis to send messages faster thanlight? |
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Answer» When you bump one end of a long metal bar, the other end does not instantaneously move. The movement would be instantaneous if the bar were perfectly rigid, but perfectly rigid materials are fundamentally impossible in the real world. Although the movement of the bar may seem uniform and instantaneous to our human eyes, it is really not. Our human eyes are simply too slow to notice the quick, but non-instantaneous, sequence of events that happens. Public Domain Image, source: Christopher S. Baird.When you bump one end of a bar, you only locally deform that end and the rest of the bar is unaffected at first. But by inwardly deforming the near end of the bar, you have created a region of high pressure in the bar surrounded by regions of lower pressure. Said another way, you have forced the atoms in the near end of the bar to get closer to each other than the equilibrium positions of their chemical bonds. Said more simply, you have knocked the first few layers of atoms against the next few layers of atoms. What happens next? |
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| 12646. |
7) Derive the ideal gas eq" PV = nRT |
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Answer» There are several ways to derive theIdeal Gas Law, but the simplest way is to use the three simplegas laws. AVOGADRO'S LAWstates the volume of a gas is directly proportional to the number of moles.V ∝ n BOYLE'S LAWstates that the volume of a gas is inversely proportional to its pressure.V ∝ 1/P CHARLES'S LAW states that the volume of a gas is directly proportional to its Kelvin temperature.V ∝ T If we combine these laws, we getV ∝ nT/P We covert the proportionality to an equalityV = knT/P We replace k with the universal gas constant R and getV = nRT/P This can be rearranged to give the IDEAL GAS LAW PV = nRT |
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| 12647. |
(2M)I(A)(I) Define ideal gas and write ideal gas equation in molecular form. |
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Answer» Ans :- An ideal gas is a theoretical gas composed of many randomly moving point particles whose only interactions are perfectly elastic collisions. The ideal gas model tends to fail at lower temperatures or higher pressures, when intermolecular forces and molecular size becomes important. PV=nRTn= m/Mm: mass ,M: Molar mass |
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| 12648. |
9.A constant volume gas thermometer works on(A) Archimedes Principle(B) Pascal's law(C) Boyle's law(D) Charles's law |
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Answer» A constant volume gas thermometerususally consists of a bulb filled with a fixed amount of a dilutegaswhich is attached to a mercury manometer. The manometer is used to measure variation in pressure. Thisthermometer workson the principle of Law of Charles lawoption d Yes d shi hai confirm hai na |
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| 12649. |
& GAS LAWS AND IDEAL GAS EQUATIONFind the approximate number of moleculescontained in a vessel of volume 7 litres at 0°C at1.3 x 105 pascals |
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Answer» 1.3 × 10^5 pascal in pressure in atm is = 1.283 atm , 0° C = 273 K now using PV = nRT => 1.283*(7) = n*(0.082057 L atm mol-1K-1)× 273 => n = 0.401 moles no..of molecules = 0.4Na |
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| 12650. |
Two parallel rail tracks run north-south. Train Amoves north with a speed of 54 kmh-1and trainB moves south with a speed of 90 km h-1. Whatis the (a) relative velocity of B with respect to A?(b) relative velocity of ground with respect to B ?and(c) velocity of a monkey running on the roof of thetrain A against the motion (with a velocity of18 km h1with respect to the train A as observedby a man standing on the ground? |
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