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12801.

A man weighing 50 kg f supports a body of 25 kg fon his head. What is the work done when he movesa distance of 20 m up an incline of 1 in 10 ?

Answer»
12802.

A man weighing 50 kg f supports a body of 25 kg fon his head. What is the work done when he movesa distance of 20 m up an incline of 1 in 10 ? TakeFig.

Answer»

Incline of one in 10 means that in moving 20 m he reached a height of H= 2 m.

Total weight of man and block = 75kg

Thus work done is: mgH

75*9.8*2 = 1470J

12803.

A car travelling at a speed of30 kmph is brought torest in 8m by applying brakes. If the same car istravelling at 60 kmph, in what distance can it bebrought to halt with the same braking force?

Answer»

S = u^2 / (2a)

Same car will cause same retardation (a) in both cases.

∴ S ∝ u^2

S1 / S2 = (u1 / u2)^2S2 = S1 * (u2 / u1)^2S2 = 8 * (60 / 30)^2S2 = 32 m

It will come to rest after 32 m

12804.

A man weighing 60kg runs along the rails with a velocity of 18kmph and jumps into a car ofmass 1 quintal standing on the rails. Calculate the velocity with which the car will starttravelling along the rails.

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Mass of man(m1) = 60kg Initial velocity of man (u1)= 18 km/h=18x5/18=5m/sFinal velocity of man (v1)= (v2) = vMass of car (m2)= 100kgInitial velocity of car (u2) = 0 (as it is in rest)Final velocity of car (v2) = vAccording to law of conservation of momentum,m1u1 +m2u2 = m1v1 + m2v260x5 + 100x0 = 60xv + 100xv300+0 = 160vv = 300/160v = 1.875 m/s

12805.

A man weighing 60 kg runs along the rails with avelocity of 18 km/h and jumps into a car of mass 1quintal standing on the rails. Calculate the velocitywith which the car will start travelling along the rails.

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12806.

पक Detdne oleckte व Totenilly ¢ Blate % onl* endoe.om en hoo no g es

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The force experienced by aunitpositive charge placed at that point is termed as theelectric field intensity. The SIunitofelectric field intensity= NC–1.

Electric fieldat any equatorial point of a dipole is given by, At the mid-point of the dipole, r = 0.

12807.

Cive Reasonshoo suciesive tenth rails Explained A small gap is Left behoeen

Answer»

The rails Bend due to thermal expansion and thetraingoes off therailway trackto avoid. ... and Ted eachrailis connected to the other. in such a way that asmall gapis left inbetweenthem. thesesmall gapsprovidespacefor easy mention of therailwhich happens during summer as well.

Thanks so much for

12808.

1 +3 Work done for a certain spring when stretched through 1 m is 10 joule. The amountbe done on the spring to stretch it further by 1 mm is(a) 30 J(b) 40J() 10 J(d) 20J11.11.17

Answer»

c) is correct answer

a) 30 j is the right answer

D) 20 j is the correct answer

(c) is right answer like my answer

12809.

Q.9. A hockey ball of mass 200 g travelling at 10 ms is struck by a wood piece of mass 5kg andmbine with it. After that both move in straight line. Find the total momentum before collision andafter collision. Calculate the change of momentum occurred in the motion of hockey ball by the forceapplied by the hockey stick.

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12810.

Calculate the Force required to move a stationary object of mass 3kg from rest to velocityat 8 m/s in 2 seconds.

Answer»

given data:v = 8m = 3kgt = 2 secondsu = 0

by using 1st equation of motion, we will find acceleration firstlyv = u+at8 = 0+2aa = 4 m/s^2

force = mass×accelerationforce = 4×3 = 12N

12811.

calculate the force required to move an object of having mass 5kg from rest to velocity of 8m/s in 2 minutes.

Answer»

F=ma=5*8/120=40/120=0.33N

Given:mass=5kg,u=0,v=8m/s,t=2min=120sAs F=maBy first equation of motion v=u+at8=0+a.120(. means multi)8/120=a As F=maF=2.8/120

12812.

The torrespending digplacenvent tine graphul be 1 1oe body starts ono yestbody starts fromOptions 4t(s)H(s)bgte )

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since initially acceleration is +ve and constant so, speed will increase continuously...so displacement will also be increasing but with increasing slope ( slope = velocity)

and when acceleration will be -ve.. the speed will reduce... so, displacement will increase but slowly.. and then when speed is completely 0. the object stops.

so, option A. is correct.

12813.

Relative Motion)oat covers certain distance between two spots5in a river taking t, hrs going downstream and t2hrsq upstream. What time will be taken by boat togoincover same distance in still water?(2) 2(t2-t,)22t1t21.22

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12814.

s double slit experiment usingmonochromatic light, the fringe pattern shifts bycertain distance on the screen when mica sheetof refractive index 1.6 and thickness 1microns is introduced in the path of one of theinterfering waves. The mica sheet is thenremoved and the distance between the slits andscreen is doubled. It is found that the distancebetween successive maxima now is same as theobserved fringe shift upon the introduction ofthe mica sheet. Calculate the wavelength of themonochromatic light used in the experiment.

Answer»

explain plz.

12815.

2 A force 'F' stops a body of mass 'm' movingwith a velocity 'u' in a distance 's'. The forcerequired to stop a body of double the massmoving with double the velocity in the samedistance is-(A) 2F (B) 4F (C) 6F (D) 8F

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thanku

12816.

find the focal length of a convex mirror is 50 cm. what is its radius of curvature.

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f= 50 cm R = f/2= 50/ 225 ans

12817.

1) Derive an expression for the capacity of a parallel-plate capacitor filled with a dielectrics.

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12818.

The breaking force for a wire is F. The breakingforces for a single wire of double thickness is(2) 2 F(4) 0.5F(3) 4 F

Answer»

Stress = F/A

=> when A→2A

F should be twice → 2F

option 2

answer is opt. 3

breaking force of wire is F . breaking force of single wire i.e. F of double thickness i.e.2 will be breaking force =2F

but i think option 2 is fair enough.. check your answer again.

umm...ansr is 4F.... i have checked it again

12819.

The force between two charges qand q, is F, when they are sepa-rated by a certain distance in freespace. Calculate the force if (a)distance between them isdoubled and (b) distance ishalved.1) F/4, 4F3) F/8, 8F2) F/2, 2F4) F/6, 6Fcld strenoth re-

Answer»

F 1= (k x q1 x q2)/r^2

F2= (k x q1 x q2)/ (2r)^2

F2= (k x q1 x q2)/ 4r^2

Therefore

F1 / F2= 1/1/4

Therefore: F2= F1/4

option 3

12820.

The outer surface of a hollow sphere of aluminium of radius 50 cm is to beused as a mirror. What will be the focal length of this mirror? Which type ofspherical mirror will it provide?

Answer»
12821.

I1 : A slab of material of dielectric constant K has same area as the plates of a parallel plateactor but has thickness 34(d), where d is the separation of plates. How is the capacitancechanged, when the slab is inserted between the plates?

Answer»

C =eo *A/(d-t + t/k)

C =eo *A/ (d- 3d/4 + 3d/4k)

C = (4k/k+3) (eo *A /d)

C = (4k/k+3) * Co

12822.

A uniform disc of radius R is put over another uniform disc of radius 2R made of same material havingsame thickness. The peripheries of the two discs touches each other. Locate the centre of mass of thesystem taking center of large disc at origin.5.

Answer»

post a clear image

12823.

A uniform disc of radius R is put over another unifom disc of radius 2R made of same material havingsame thickness. The peripheries of the two discs touches each other. Locate the centre of mass of thesystem taking center of large disc at onigin.

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unable to understand

12824.

A wire is stretched by 0.01m by a certain force F another wire of same material whose diameterand lengths are double to original wire is stretched by the same force then its elongation will be

Answer»
12825.

A circular plate of diameter 'a' is kept in contactwith a square plate of side a as shown. Thedensity of the material and the thickness are sameeverywhere. The centre of mass of compositesystem will be(1) Inside the circular plate(2) Inside the square plate(3) At the point of contact(4) Outside the system

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12826.

.What is magnetic permeability ?

Answer»

The magnetic permeability is defined as the property of the material to allow the magnetic line of force to pass through it.

12827.

The load versus elongation graph for four wires of the same material and same length is shown in the figure. Thethinnesf wire is represented by the line(1) OA(3) ОС(2) OB(4)ODelongation

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12828.

Aconductor has resistance of 15 Ω at 10°C and 18 Ω at 400°C Find the temperature cof resistance of the material.

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12829.

The magnetic field B and the magnetic intensityHin a material are found to be 1.6 T and 1000 A/mrespectively. Calculate the relative permeabilityur, and the susceptibility 'χ, of the material.

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12830.

(7)The magnetic field B and the magnetic intensityH in a material are found to be 1.6 T and1000 A/m respectively. Calculate the relativepermeability ur' and the susceptibility 'x' of thematerial ?(Ans. : 1.273 x 103, 1272)

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12831.

1. A plot of magnetic flux (0) versus current () is shown in the figure for twoinductors A and B Which of the two has larger value of self inductance?

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We know self inductance L= Magnetic flux / I .. given plot is of magnetic flux versus I . Therefore the slope will be equal to L (self inductance) . Since slope of A is more than B => A has higher value of self inductance.

12832.

2.A 2000 kg car has to go over a furn whose radius is750 m and the angle of slope is'5°. The coefficient offriction between the wheels and the road is 0.5.What should be the maximum speed of the car sothat it may go over the turn without slipping?(Ans. 67.2 ms 1)

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12833.

magnetic susceptibility of diamagnetic substance is:(a) small and positive(c) smal and negative(b) large and positive(d) large and negagive

Answer»

Option d ) is correct.

please explain it

12834.

72cm34. A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. its centre of gravity is 1.05 m behindthe front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.(3675 N; 5145 N

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12835.

eu oyue ChargerQ2.Draw a schematic diagram of a circuit consisting of a battery of six cell of 1.5V each, three resistoreach of 3ohm in series and a plug key.

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12836.

Ouf State laus of oflettion of sand Eyplaio thyplication hbased drn

Answer»

What is reflection?

Reflection is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated. Common examples include the reflection of light, sound and water waves.

Do you know how sound propagates?

Sound propagates through air as a longitudinal wave. The speed of sound is determined by the properties of the air, and not by the frequency or amplitude of the sound. If a sound is not absorbed or transmitted when it strikes a surface, it is reflected. The law for reflection is the same as that of light, ie., the angle ofincidenceof a sound waveequals the angle of reflection, just as if it were produced by a 'mirror image' of the stimulus on the opposite side of the surface.

How do we describe the reflection of sound?

When sound travels in a given medium, it strikes the surface of another medium andbounces back in some other direction, this phenomenon is called the reflection of sound. The waves are called the incident and reflected sound waves.

What are incident and reflected sound waves?

The sound waves that travel towards the reflecting surface are called the incident sound waves. The sound waves bouncing back from the reflecting surface are called reflected sound waves. For all practical purposes, the point of incidence and the point of reflection are the same point on the reflecting surface.

A perpendicular drawn on the point of incidence is called the normal. The angle which the incident sound waves makes with the normal is called the angle of incidence, "i". The angle which the reflected sound waves makes with the normal is called the angle of reflection, "r".

Let’s look at the two laws of reflection

The following two laws of reflection of light are applicable to sound waves as well:

The incident wave, the normal to the reflecting surface and the reflected wave at the point of incidence lie in the same plane.

The angle of incidence ∠i is equal to the angle of reflection∠r.

Learning Outcomes

Student will understand the First and Second Laws of Reflection.

First Law of Reflection:The incident wave, the reflected wave, and the normal at the point of incidence lie on the same plane.Law of Reflection:The angle of incidence is equal to the angle of reflection.

12837.

6. Explain the principle and working of a transistor as an oscillator. Draw a schematic diagramshowing its various parts

Answer»

Transistor as Oscillator

When we use atransistorin a circuit, it continuously produces undamped oscillations at the output terminals of the circuit. Here we can explain you how touse a transistor as an oscillator with the help of circuit diagram.

Oscillator circuit:

The diagram ofTransistor oscillator circuitis shown below. This circuit is divided into three sections:

(i) Tank Circuit:This circuit produces oscillations which are amplified by the transistor and produces amplified output in the collector side.

(ii) Amplifier circuit:This circuit basically amplifying the small sinusoidal oscillations present in the base-emitter circuit and generates output in the amplified form.

(iii) Feedback circuit:This is the important section of the circuit because to amplify the oscillations for the amplifier we need some energy at the tank circuit. For this purpose, we fed back the energy of the collector circuit to the base circuit through the phenomenon ofMutual induction. We fed back the energy from output to input with the help of this circuit.

12838.

Temb Electrical Engineering • VolumeWhich of th(a) 1, 2 ar(c) 2 and1.15 Assertion (A): The energy band diagram of areal semiconductor can not be represented byflat conduction and valence bands.Reason (R): The presence of energy states inthe forbidden energy gap results in carriertrapping, giving rise to finite curvature in thebands.(a) Both A and R are true and R is the correctexplanation of A(b) Both A and R are true but R is NOT the correctexplanation of A(c) A is true but R is false(d) A is false but R is true1.19 The concextrinsic se(a) directly(b) directly(c) inversConcei(d) inversconcet

Answer»

Here assertionand reason both are correctas energy states the gap result finite curvature

12839.

Explain intrinsic motivation.Explain the term “Realistic" in goal setting principles.alactivities onhances self-est

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Realistic means aim of the person should be achievable. He should set a realistic goal and accept the challenges that lie ahead with enthusiasm. It should be within the capacity of the individual, neither too high nor too low.

12840.

Q3. Write the difference between direct band gap semiconductor and indirect band gap semiconductor

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Indirectbandgap semiconductor the electron “rising” from valence band to the conduction band will only change it’s potential(energy).

Inindirectbandgap semiconductor the electron “rising” from valence band to the conduction band will change it’s potential (energy) andmomentum

12841.

What is the difference between an n-typeand a p-type extrinsic semiconductor?Delhi 2012C

Answer»

Ans :- n-type Semi-Conductor

It is an extrinsic semi-conductor which is obtained by doping the impurity atoms of Vth group of the periodic table to the pure Ge and Si semi-conductor.

The impurity atoms added, provide extra electrons in the structure and are called donor atoms.

The electrons are majority carriers and holes are minority carriers.

The electrons density (ne) is much greater than the hole density (nh) i.e.>> nh.

The donor energy level is close to the conduction band and for away from the valence band.

The Fermi-energy level lies in between the donor energy level and conduction band.

P-type Semi-Conductor

It is an extrinsic semi-conductor which is obtained by doping the impurity atoms of III group of the periodic table to the pure Ge and Si semi-conductor.

The impurity atoms added, create vaccines of electrons (i.e. holes) in the structure and are called acceptor atoms.

The holes are majority carriers and electrons are minority carriers.

The holes density (nh) is much greater than the electrons density (ne) i.e.>> ne.

The acceptor energy level is close to the Valence band and for away from the Conduction band.

The Fermi-energy level lies in between the acceptor energy level and valence band.

12842.

gaseous mixture of H2 and CO2 gases contains 66 mass % of CO2. The vapour density of theixture is:a) 6.1(b) 5.4(c) 2.7(d) 10.8

Answer»

Let total moles of mixture be 100gMoles of CO2 = 66gThen hydrogen will be = 34gNumber of moles of CO2=66/44=1.5Numbsr of moles if H2=34/2=17Total moles =1.5+17=18.5Formula to be used:Mavg= Total mass/ total moles= 100/18.5=5.40VAPOUR DENSITY= MAvg/2=5.40/2=2.70So, VAPOUR density of mixture is 2.7

12843.

(1) Define intrinsic semiconductor and extrinsisemiconductor.

Answer»

Anintrinsic semiconductoris an undopedsemiconductor. Thismeansthat holes in the valence band are vacancies created by electrons that have been thermally excited to the conduction band, as opposed to doped semiconductorswhere holes or electrons are supplied by a “foreign” atom acting as an impurity.

Anextrinsic semiconductoris asemiconductordoped by a specific impurity which is able to deeply modify its electrical properties, making it suitable for electronic applications (diodes, transistors, etc.) or optoelectronic applications (light emitters and detectors).

12844.

A wheel of radius 40 cm rests against a step ofheight 20 cm as shown fig. If the mass of thewheel is 2 kg, then the minimum horizontalforce, applied perpendicular to the axle, requiredto make the wheel climb the step is40 cm20 cmTITTITNIITITITTITTITITIII

Answer»
12845.

differentiate bt. heat capacity and specific heat capcity?(plz answer in simple language)

Answer»

Heat capacityia heat required to raise​ the temperature of a substance by 1 degree Celsius.

Specific heatis heat required to raise the temperature of a unit mass of a substance by 1 degree Celsius.

On the other hand heat capacitydepends on mass of the substance. If I have 5kg of that substance then heat capacity will be 5KJ/Kg, if I have 10kg then heat capacity becomes 10KJ/kg . Because more the mass, more it will take to raise the temperature by 1 degree Celsius.

thank you

12846.

At a certain temperature and a total pressureof 105 Pa, iodine vapour contains 40% by volume of I atoms.I2 (g)21(g)Calculate Kp for the equilibrium

Answer»

thanks too much

12847.

thenas isume.38. Two mole of a diatomic gas is contained in a 20 liter vessel at2x10 N/m' pressure. The temperature is-(1) 250K(2) 240K(3) 230K(4) 220K

Answer»

3) is the correct answer

(3)is the write answer

(3)230k is right answer babes

12848.

34.For a diatomic gas, change in internal energy for unitchange in temperature for costant pressure and vol-ume is U, and U,respectively. U, :U, is:(2)7:5(4)5:7

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12849.

6.1Which one of the following semiconductor devices does not need any reverse bias forits operation ?(A) Zener diode(B) LED(C) Photodiode(D) Solar cell

Answer»

Solar Cell does not need any reverse bias for its operations.

12850.

6. Which of the electromagnetic waves are used to detect flaws in metal sheets?7

Answer»

Ultraviolet rays and x rays can be used.