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13551.

A point charge 8.5 u C is kept at origin. Magnitude of strength of field at (4m, 3m) isa) 3.06 x 103 N/c b) 2.1 x 10' N/c c) 4.52 x 103 N/c d) 2.59 x 10 N/c

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The correct answer for this following question is option (C).

Option c is the correct answer

Option C is the correct answer

the correct answer is option c

option c is the correct answer

option c is answer that question

the correct answer is (C)

the correct answer is C

13552.

16. Find the angle of banking for a cyclist moving with velocity V to negotiate a safe turn of radius r. neglect theforce of friction of safe turn.

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13553.

12. Why are passengers thrown forward from their seats when a speechstops suddenly?

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The passengerwill appear tosuddenlylurchforwardsuntil theystopdue to friction withtheir seat, or because theystopthemselves by holding on, for instance.Bus comes in to rest ,butpassengersdue to Newton's 1st law tries to stay intheirmotion. So they face aforwardjerk while stoppingbus.

13554.

5. Ball rolling on the floor stops due to.a) gravitational force b) magnetic forcec) frictional forced) muscular force

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option c is correct answer according to your question

c is correct option to question

ball rolling on the floor stop due to answer is muscular force

c. frictional force is correct answer

13555.

2. Gravitational force acts on all objects in proportion to theirmasses. Why then, a heavy object does not fall faster than alight object?

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13556.

(2) zero(3) K/4(4) K/24.An athlete in the olympic gamesestimated to be in the range(1)2 x 103 J-3 x 105 Jcovers a distance of 100 m in 10 s. His kinetic energy can beAIEEE 2008; 3/105,-11(3) 2,000 J-5,000 J(2) 20,000 J-50,000 J(4) 200 J-500 J

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13557.

What is the magnitude of the gravitational force between theearth and a 1 kg object on its surface? (Mass of the earth is6 × 1024 kg and radius of the earth is 6.4 × 106 m.)

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13558.

2. Ablack hole is an object whose gravitational fieldis so strong that even light cannot escape from it.To what approximate radius would earth(mass 5.98 x 1024 kg) have to be compresed tobe a black hole?AIPMT-2014](1) 10-9 m(3) 10-2 m(2) 10-6 m(4) 100 m

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13559.

II3. I1.2 km SMass of the moon is 7.34 x 102 kg and mean radius174 x 10 m. If G 6.67 x 10N m2 kg 2, then1.74 xalculate the escape velocity on the moon's surface.

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13560.

uacts the earth? Why?If the moon attracts the earth, why does the earth not movetowards the moon?

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13561.

The velocity of a car changes from 40 km h-1 to 60km h l. The mass of the car is 1000 kg. Calculate thechange in kinetic energy.

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KE = m*v*v/2 initial KE = 1000*40*40/2 = 800000J updated KE = 1000*60*60/2 = 1800000 J so difference is 1800000-800000 = 1000000 J

13562.

26. What is the force of gravity on a body of mass 150 kg lying on the surface of the earth?(Mass of earth = 6 x 1024 kg; Radius of earth = 6.4 × 106 m; G = 6.7 x 10-11 Nm?/kr)

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13563.

illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. why dies the bib eventually come to rest? what happens to its energy eventually? is it a violation of the law of conservation of energy?

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In physics, the law of conservation of energy states that the total energy of an isolated system remains constant and that energy can neither be created nor destroyed; rather, it can transform from one form to another.

When a pendulum moves from its main position to either of its extremes,it rises through a height above the mean level.At this point,the K.E of the bob changes into P.E..As it moves back to the main position,its P.E decreases progressively and K.E. increases.As the bob reaches the main position,it has only K.E.This process is repeated as long as the bob oscillates.

The bob eventually comes to rest because the air resistance resists its motion and the pendulum loses its K.E. to overcome this friction and stops after sometime.

The law of conservation of energy isn't violated as the energy lost by the pendulum is gained by its surroundings

13564.

U 11 Unu Imf. The masses of the earth and moonare 6 x 1024 kg and 7.4x1022 kg,respectively. The distance betweenthem is 3.84 x 105 km. Calculate thegravitational force of attractionbetween the two?Use G = 6.7 x 10-11 N m² kg-2Ans: 2 x 1020 N

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ans: 2*10^20N is the right answer

2*10*20N is right answer

Use Newton law of gravity F=Gm1m2/r^2m1=6×10^24kgm2=7.4×10^22kgr=3.84×10^8mG=6.7×10^-11Nm^2kg^-2

13565.

36.Whatistheforceofgravity on a body of mass 150 kg lying on the surface of the earth ?(Mass of earth = 6 x 1024 kg Radius of earth, 6.4 x 106 m: G-67-x 10-11 Nm3/kch

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13566.

Example 10.1 The mass of the earth is6 x 1024 kg and that of the moon is7.4x 10 kg. If the distance betweenthe earth and the moon is 3.84105 km,calculate the force exerted by the earthon the moon. G6.7x10-11 N m2 kg

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13567.

When two moles of a gas is heated from O to10°C at constant volume, its internal energychanges by 420J. The molar specific heat ofthe gas at constant volume

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13568.

man ?Mass of earth is 6 x 1024 kg and mass of a man is 60 kg. Earth pulled a man by force 600 NHow much force applied by man on earth and how will be acceleration of earth towardsb acceleation due to

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Force of Man on Earth - is his weight i.e 60×9.81=588.6 N and acceleration of Earth towards Man will be very small i.e nearly equal to zero due to it's bulky mass.

13569.

15. Illustrate the law of conservation of energy by discussing theenergy changes which occur when we draw a pendulum bob toone side and allow it to oscillate. Why does the bob eventuallycome to rest? What happens to its energy eventually? Is it aviolation of the law of conservation of energy?

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13570.

36 A body of mass m collides against a wall withthe velocity u and rebounds with the samespeed. Its change of momentum is-(A) 2 mu(C) mu(B) mu(D) 0

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let the velocity of body moving towards the wall be 'v'after collision the velocity would be (-v), i.e. in the opposite direction.so change in momentum, p= mv-(-mv)= 2mvso answer is (A)

thanku

13571.

3. What is the magnitude of the gravitational force between theearth and a 1 kg object on its surface? (Mass of the earth is6 x 1024 kg and radius of the earth is 6.4 x 10° m.)

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13572.

INILAT UUESTTUNS S.LThe mass of the earth is 5.97 x 1024 kg and its mean radius is bCalculate the value of g at the surface of the earth.100 kg and its mean radius is 6.371 x 106 m.

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The MassOf TheEarth Is 5.97 X 10 24Kg, AndThe MassOf The Moon Is 7.35X 1022 Kg. The Distance Of Separation, Measured Between Their Centers, Is 3.84X108 M. Locate The Center OfMassOf TheEarth–Moon System As Measured From The Center Of TheEarth.

value of g at the surface on the surface of earth = 9.8 m/s2

surface of earth=9.8m/s2

surface of earth =9.8m/s2

13573.

. What is the magnitude of the gravitational force between theearth and a 1 kg object on its surface? (Mass of the earth is6 x 1024 kg and radius of the earth is 6.4 x 106 m.)

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13574.

1030 kg and the mass of earth is 6 x 104 kg. If the average distance between the sunThe mass of sun is 2 × andand the earth be 1.5 x 108 km, calculate the force of gravitation between them

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13575.

Q4. Write the condition for a particle to show simple harmonic motion

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Necessary conditions for particle to show SHM are:

1) In system there should be an elastic restoring force(A force which rises towards to an equilibrium.If the system is disturbed away from the equilibrium, the restoring force will tend to bring the system back towards equilibrium) which is acted on the system.

2) The acceleration of the system should be directly proportional to the displacement and is always directed towards mean position.

13576.

2: Which of the following functions of timerepresent (a) simple harmonic (b) periodic butnot simple harmonic and (c) non periodicmotion? Give period for each case of periodicmotion (oo is any positive constant)

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13577.

DWhat is meant by simple harmonicmotion ? Find the velocity of a particleexecuting S.H.M. at a point on its path.

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When an object moves to and fro along a line, the motion is called simple harmonic motion.

13578.

(114 cm12020.13.The focal length of a concave mirror is 12 cmWhere should an object of length 4 cm be placedso that a real image of 1 cm length is formed ?SANO OPTICAL INSTRUMENTOZ-EXERCISE FOS(1) 48 cm(2) 3 cm(B) 60 cm(4) 15 cm 3

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60cm is the correct answer

60 cm is the right answer.

3 is the correct answer 12×4 =60

13579.

-1112. Two conducting spheres of radii R, and R, are kept widelyseparated from each other. What are their individualcapacitances ? If the spheres are connected by a metal wire,what will be the capacitance of the combination? Think interms of series-parallel connections13 Fech of the canacitors shown in figure (31-E6) has a

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13580.

2. Two conducting spheres of radii R, and R, are kept widelyseparated from each other. What are their individualcapacitances? If the spheres are connected by a metal wire,what will be the capacitance of the combination ? Think interms of series-parallel connections.

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13581.

8 Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s。rlikle and mbound with the auttne speed. What is the inipuise ingarted to each ball dueto the other ?

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13582.

a bus of mass 10000kg is moving with a velocity of 60 km/h,calculate the work done to stop this work

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work done= change in K.E.=1/2mv^2=1/2(60×10000)=300000 j=300 kj

13583.

30.25. Consider the situation shown in figure (12-E101. Showthat if the blocks are displaced slightly in oppositedirections and released, they will execute simpleharmonic motion. Calculate the time period.Figure 12-E10

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13584.

An object 4 cm in size, is placed at 25cm in front of a concavemirror of focal length 15cm. At what distance from the mirrorshould a screen be placed in order to obtain a sharp image? Findthe nature and size of the image.11/2

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focal length give 15 cm

13585.

Two cars moving in opposite directions cover same distance in one hourIf the cars were moving in north-south direction. what will be thedisplacement in one hour?

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there will be no displacement they are moving to opposite direction there for the displacement will be zero

13586.

Example 8. Two trains 320 m and 430 mare moving in the opposite directions at a speedof 50 km/h and 40 km/h, respectively. Howlong will it take the trains to pass each other?

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Ans :- Given:

lengths of two trains = 320m and 430m.

and their respective speed = 50km/hr and 40km/hr

Trains are moving in opposite direction. so to crossing each othertotal distance = 320 + 430 = 750m

and total speed = 50km/hr + 40km/hr = 90km/hr

13587.

МЕ 523Score20 Go through the circuit given below240 V500 W240 V40 W240 V Fuse(a) Which lamp will brighter(b) What will be the amperage of the fuse in the circuit?be brighter when the circuit is switched on ? Justify your answer

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The one with more power that is240V and 500W will light up brighter.

13588.

have taken I noTess T0I hohours. The tap of larger diameter takes 10hours less than the smallerone to fill the tank separately. Find the time in which each tar9. Two water taps together can fill a tank in 9can separately fill the tank.nassenger train to travel 132 km between

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13589.

32. Two cyclists cycle along circular tracks of radii R, and R2 at uniform rates. If both of them take same timeto complete one revolution, then their angular speeds are in the ratio(2) R2 R1(4) R,R2: 1

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13590.

4. If Ry and R2 are the resitance of tatements of a 400 W and 200 W lamp designedto operate at same voltage then:(a) R2R(C R4R,(b) R2 2R,(d) R,-R2

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13591.

54 x 10, xl . 5x1048лж3N along CA,en two resistors are connected in series with a cell of emf 2 V and negligible internalresistance, a current of 2 A lows in the circuit. When the resistors are connected inWh34.5parallel the main current is A Calculate the resistances.35

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In seriesV=IRI=V/RI=2/0.4I=5AIn parallelI=2/1.666I=1.199A

13592.

35. A popular game in Indian villages is goli which is playedwith small glass balls called golis. The goli of one playeris situated at a distance of 20 m from the goli of thesecond player. This second player has to project his goliby keeping the thumb of the left hand at the place ofhis goli, holding the goli between his two middle fingersand making the throw. If the projected goli hits the goliof the first player, the second player wins. If the heightfrom which the goli is projected is 196 em from theground and the goli is to be projected horizontally, withwhat speed should it be projected so that it directly hitsthe stationary goli without falling on the ground earlier?speed oftime tohalf thepeed ofe timeake in

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13593.

o \\%\\ [ N S VL S ८. 5 15 omiodaN D) Qonen\axe R« MoL 60" ४ 220 ८. Ve vecaoeeds Bata ik vt N SR

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Here, Initial temperature ( Ti) = 80°C Final temperature ( Tf) = 50°C Temperature of the surrounding ( To) = 20°C t = 5 min A/C to Newton's law of cooling

Rate of cooling ( dT/dt) = K[ (Ti+Tf)/2 - To] ( Tf - Ti)/t = K[ ( 80 + 50)/2 - 20]( 80-50)/5 = K[ 65 - 20]6 = K× 45 K = 6/45 = 2/15

in second condition, initial temperature ( Ti) = 60°C Final temperature ( Tf) = 30°C Time taken for cooling is t A/C Newton's law of cooling ( 60 - 30)/t = 2/15 [ (60+30)/2 -20]30/t = 2/15 × 25 30/t = 50/15 = 10/3 t = 9 min

13594.

.The length of a seconds pendulum is increasedby 1%. How many seconds it will lose per day?(a) 432s(d) 3737 s (e) 3927 s(b) 864 s(c) 3427 s

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Pendulum period in secondsT ≈ 2π√(L/g)L is length of pendulum in metersg is gravitational acceleration = 9.8 m/s²

there are 86400 seconds in a day.

dT/dL = (1/2)2π√(1/g)√1/L) = π√(1/Lg)dT = π/√(gL) dL

dL/L = 0.01 (ie, 1%)dL = 0.01LdT = π/√(gL) (0.01)LdT = (0.01)(π/√g) L/√LdT = (0.01)(π/√g)√LdT = (0.01)(π√(L/g))dT = (0.01)(1/2)(2π√(L/g))dT = (0.005)TdT = (0.005)(86400) = 432 seconds

13595.

). The position vector of a body of mass 2 gm is(2 1+31) meter and that of mass 5gm is(4 1-5)meter. Find the position vector of the centre of mass.

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position vector will.be m1r1+m2r2/M1+m2=2(2i+3j)+5(4i-5j)/2+5=4i+6j+20i-25j/7=24i-19j/7

13596.

2. (a) Calculate the heat capacity of a copper vessel ofmass 150 g if the specific heat capacity ofcopper is 410 J kg 1 K-(b) How much heat energy will be required toincrease the temperature of the vessel inpart (a) from 25°C to 35°C?Ans. (a) 61-5 J K-1, (b) 615 J

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Heat capacity = Mass × Specifc heat capacity= 150 × 10^-3 kg × 410 J / (kg K)= 61.5 J / K

Heat capacity of copper is 61.5 J/K

13597.

101.The velocity of a particle is v6i 2j -2k. Thecomponent of the velocity parallel to vectora-i +j + k in vector form is(a) 6i +2+ 2k(c)+j+k(b) 21 +2j+2k(d) 6i+ 23 -2k^

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Option (d) is correct.

13598.

la nuclear fusion One gram hydrogen isconverted into 0.993m. If the efficiencyof the generator be 5%.then the energyobtained in KWH is

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the energu is also increase by 5℅

It will also increase by 5%

13599.

Hydrogen bomb is based on theprinciple ofA) nuclear fissionB) nuclear fusionC) natural radioactivityD) artificial radioactivity

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B) nuclear fusion is the correct answer.

hydrogen bomb is based on the principle of nuclear fusion

13600.

The frequency a of an oscillating liquid ddrop ma2.The frequency n of an oscillating liquidepend upon the radius r of the drop, desurface tension S of the liquid. Obtain a formthe frequency by the method of dimensionsnsity

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