40048 + Interview Questions in Physics in Class 11

1.	Electric displacement is given by D=epsilonE Here epsilon= electric permittivity E= electric field strength, Find the dimensions of electric displacement.
Answer» `[AML^(-2)T]` `[AL^(-2)T^(-1)]` `[AL^(-2)T]` None of these Solution :(c) `D=epsilonE` UNIT of `D=(C^(2))/(NM^(2)) N/C :. [D]=C/(m^(2))=(At)/(L^(2))[AL^(-2)T]`

Discussion

2.	We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to the 2.63 sec, 2.56 sec, 2.42 se, 2.71 sec, 2.80 sec
Answer» The absolute errors in the measurements are 0.01 SEC, 0.06 sec, 0.20 sec, 0.20 sec, 0.09 sec, 0.18 sec The MEAN absolute error is 0.11 sec Then mean time period can be reported as `(2.62+0.11)` sec. The time period can actually reported as `(2.6+0.1)` sec, with a PERCENTATGE error of 4% Answer :A::B::C::D

Discussion

3.	An internal force cannot change the state of motion of centre of mass of a body. How does the internal force of the brakes bring a vehicle to rest?
Answer» Solution :In this case the force which BRING the VEHICLE to REST is friction, and it is an EXTERNAL force.

Discussion

4.	A wire 4 m long and cross section 3xx10^(-6)m^(2)is stretched by Imm. If its Young.s modulus is 2xx 10^(11) N//m^(2), find the energy stored in the wire.
Answer» ANSWER :`0.075J`

Discussion

5.	Time interval between two successive heart beat is in the order of ……………
Answer» `10^(@)` s 10 s `10^(2)` s `10^(-3)` s Solution :`10^(@)` s

Discussion

6.	The spectra of all stars show a continuous spectrum like the solar spectrum, in which dark absorption lines are superimposed. The photosphere (inner layer of the star) emits radiations of all wavelengths, producing a continuous spectrum. On passing through the outer (comparatively cooler) layer of the star, radiations of certain wavelengths are selectively absorbed by this layer. This is the cause behind the dark lines in the star's spectrum. These dark lines are characteristics of the substances present in the outer layer of the star. For measuring the surface temperature of the star, we need to measure the wavelength at which the intensity of emitted radiation is maximum and then using Wien's displacement law, given below lamda_m = T = b where b is the Wien's constant and its experimentally determined value is 2.89 xx 10^(-3) mKWhich of the following options is true for spectrum of light received from a star?
Answer» CONTINUOUS EMISSION spectrum ABSORPTION line spectrum emission line spectrum emission band spectrum Solution :absorption line spectrum

Discussion

7.	The spectra of all stars show a continuous spectrum like the solar spectrum, in which dark absorption lines are superimposed. The photosphere (inner layer of the star) emits radiations of all wavelengths, producing a continuous spectrum. On passing through the outer (comparatively cooler) layer of the star, radiations of certain wavelengths are selectively absorbed by this layer. This is the cause behind the dark lines in the star's spectrum. These dark lines are characteristics of the substances present in the outer layer of the star. For measuring the surface temperature of the star, we need to measure the wavelength at which the intensity of emitted radiation is maximum and then using Wien's displacement law, given below lamda_m = T = b where b is the Wien's constant and its experimentally determined value is 2.89 xx 10^(-3) mKThe study of dark lines in the spectrum of a star shows that the atmosphere of stars contain
Answer» nitrogen HELIUM hydrogen uranium Solution :helium

Discussion

8.	The spectra of all stars show a continuous spectrum like the solar spectrum, in which dark absorption lines are superimposed. The photosphere (inner layer of the star) emits radiations of all wavelengths, producing a continuous spectrum. On passing through the outer (comparatively cooler) layer of the star, radiations of certain wavelengths are selectively absorbed by this layer. This is the cause behind the dark lines in the star's spectrum. These dark lines are characteristics of the substances present in the outer layer of the star. For measuring the surface temperature of the star, we need to measure the wavelength at which the intensity of emitted radiation is maximum and then using Wien's displacement law, given below lamda_m = T = b where b is the Wien's constant and its experimentally determined value is 2.89 xx 10^(-3) mKThe factor on which the colour of a star depends is
Answer» mass SURFACE TEMPERATURE gaseous content SIZE SOLUTION :surface temperature

Discussion

9.	A stone is thrown vertically up at 20 m/s from a tower 80 m height. The speed with which it hits the ground is [g=10m//s^(2)].
Answer» `14.4m//s` `28.14m//s` `80m//s` `SQRT(5)20m//s` ANSWER :D

Discussion

10.	A particle is moving on a straight line. It is initially at rest v= instantaneous velocity P = instantaneous powerS = displacement F= forcet = time Match the possible expression of the quantities in column-1 with the situation in column-2 {:(,"Column-1",,"Column-2"),("(A)",v^(3) prop S,"(P)","P = constant"),("(B)",v^(2) prop t,"(Q)",P prop v),("(C)",v^(2) prop S,"(R)","F = constant"),("(D)",v prop t,"(S)",F prop (1)/(v)),(,,"(T)",P prop t):}
Answer» ANSWER :A-P, S; B-P, S; C-R, T; D-Q, R, T

Discussion

11.	When a body is projected from the surface of a planet with a velocity less than the orbital velocity, what pathdoes it follow when it returns to the earth ?
Answer» SOLUTION :PARABOLIC PATH.

Discussion

12.	Suppose universal gravitational constant starts to decrease, then a) length of the year will increase b) earth will follow a spiral path of decreasing order c) kinetic energy of earth will decrease
Answer» only a and B are true only a and C are true only b and c are true all are true Answer :B

Discussion

13.	A heavy, flexible, uniform claim of length pi r and mass lambda pi r lies in a smooth semicircular tube AB of radius r. Assuming a slight disturbance to start the claim in motion, find the velocity w with which it will emerge from the end B of the tube.
Answer» Solution :Since friction is absent, we can apply the law of conservation of energy. Centre of gravity of a SEMICIRCULAR are is at a distance `((2r)/(pi))` from centre. Initial potential energy `= (lambda pi r) g((2r)/(pi))`, Final potential energy `= (lambda pi r) g((-pir)/(2))` when the CHAIN COMPLETELY slips off the TUBE, all the links of the chain have the same velocity v. Kinetic energy of chain `= (1)/(2) (lambda pi r) v^(2)` , From COE, `lambda pi r((2r)/(pi))g = (lambda pi r)g((-pi r)/(2))+ (1)/(2) (lambda pi r)v^(2)` From which we find `v = sqrt(2gr ((2)/(pi) + (pi)/(2)))`

Discussion

14.	A Carnot engine working between 300 K and 600 K has work output of 800 J per cycle. What isamount of heat energy supplied to the engine from source per cycle
Answer» 1800 J/cycle 1000 J/cycle 2000 J/cycle 1600 J/cycle Answer :D

Discussion

15.	100 g of water and 100 g of iron are heated up to the same temperature and are separately added to 50 g of water at a lower temperature, kept in two identical vessels. Compare the temperature changes in the two vessels.
Answer» Solution :TEMPERATURES of both the vessels will increase. Specific heat of iron is much less than that of WATER `(s_("water")=1 and s_("iron")=0.117)`. The initial temperature of the given specimens of 100 g of water and 100 g of iron are equal. But, due to its higher specific heat, water will lose more heat than iron will. HENCE, the final temperature of the vessel in which 100 g of HOT water is added will be higher.

Discussion

16.	A bird of mass 0.1 kg rising vertically with an acceleration 0.2 ms^(-2). The muscular force exerted by it is (g=9.8 ms^(-2))(a) 1 Newton(b) 9.8 kg. wt(c ) 1//9.8 kg. wt(d) 980 newtons.
Answer» a is CORRECT a and B are correct a and C are correct b and d are correct ANSWER :C

Discussion

17.	Calculate the length of the arc of a circle of radius 31.0 cm which subtends an angle of (pi)/(6) at the centre.
Answer» SOLUTION :Plane ANGLE `theta=(l)/(R)` radian `:.(pi)/(6)=(l)/(3l)( :. theta=(pi)/(6) and r=31 cm)` `:.l=31xx(pi)/(6) cm=(31xx3.14)/(6) cm` `16.22 cm` By ROUNDING off to THREE significant figures, it would be `16.2 cm`

Discussion

18.	The velocity of projectile at the point of projection is 2hat(i) + 3hat(j)m//s. The velocity at a point where the projectilereaches is
Answer» `-2HAT(i) + 3HAT(J)m//s` `2hat(i) - 3hat(j) m//s` `2hat(i) + 3hat(j) m//s` `-(2hat(i) + 3hat(j))m//s` ANSWER :B

Discussion

19.	A force 2hati+hatj+hatk newtons acts on aa body which is initially at rest. At the end of 20 seconds the velocity of the body is 4hati+2hatj-2hatk m//s. What is the mass of the body?
Answer» SOLUTION :`m=10kg`

Discussion

20.	Find out the coefficient of linear expansion of the material of an open pipeso that the frequency of any tone emitted from itdoes not very with temperature .
Answer» Solution :The fundamental FREQUENCIES at ` 0^(@)` C and` t^(@)` C are respectively , `n_(1) = (V_(0))/(2l_(0)) and n_(2) (V_(t))/(2l_(t))` Here,`V_(0)`= velocity of SOUND at ` 0^(@) C l_(0)`= length of the pipeat ` 0^(@) C , V_(t)` = velocity of sound at ` t^(@) C , l_(t)` = length of the pipe at ` t^(@) C ` . According to the question, ` n_(1) = n_(2)` so , ` (V_(0))/(2l_(0)) = (V_(t))/(2l_(t))` or , `V_(0) l_(t) = V_(t) l_(0) or V_(0) l_(0) (1 + alpha t) = V_(0) ( 1 + 0 .00183 t) l_(0)` [`alpha = ` coefficient of LINEAR expansion] or, ` alpha = 0 . 00183^(@) C^(-1)` .

Discussion

21.	One end of a string of length L and cross sectional area A is fixed to a support and the other end is fixed to a bob of mas m. The bob is revolved in a horizontal circle of radius r with an angular velocity omega such that the string makes an angle theta with the vertical. Young's modulus is Y. The stress in the string is
Answer» `(MG)/A` `(mg)/A(1-r/l)` `(mg)/A(1+r/l)` `(mg)/A(r/l)` ANSWER :A

Discussion

22.	A wire 10 m long has a cross-sectional area 1.25 xx 10^(-4) m^(2). It is subjected to a load of 5 kg. If Young's modulus of the material is 4 xx 10^(10) N m^(-2), calculate the elongation produced in the wire. Take g = 10 ms^(-2)
Answer» Solution :We know that`(F)/(A)=YXX(DeltaL)/(L)` `DeltaL=(F/A)(L/Y)=((50)/(1.25xx10^(-4)))((10)/(4xx10^(10)))=10^(-4)m`

Discussion

23.	If the temperature of sun were to increase from T to 2T and its radius from R to 2 R, then ratio of radiant energy received on earth to what it was previously will be
Answer» 4 16 32 64 Answer :D

Discussion

24.	The amount of work done in lifting a body of mass 'm' from the surface of the earth to a height equal to twice the radius of the earth is,
Answer» `(2GMm)/(3R)` `(3GMm)/(2R)` `(5GMm)/(3R)` `(3GMm)/(5R)` ANSWER :A

Discussion

25.	A solid sphere is rolling down an inclined plane. Thenthe ratio of its translational kinetic energy to its rotational kinetic energy is
Answer» 2.5 1.5 1 0.4 Solution :Let M and R be the mass and RADIUS of the solid sphere . As it is rolling down an inclinedplane , `therefore` Its TRANSLATIONAL kinetic energy is `K_(t) = (1)/(2) M v_(cm)^(2)` (where `v_(cm)` is the velocity of centre of mass) and its rotational kinetic energy is `K_(r) = (1)/(2) I omega^(2)` But `I = (2)/(5) MR^(2)` (For solid sphere) and `omega = (v_(cm)) /(R) therefore K_(r) = (1)/(2) ((2)/(5) MR^(2)) ((v_(cm))/(R))^(2) = (1)/(5) M v_(cm)^(2)` Their corresponding ratio is `(K_(i))/(K_(r)) = ((1)/(2) M v_(cm)^(2))/((1)/(5) M v_(cm)^(2)) = (5)/(2) = 2.5`

Discussion

26.	Two boys are standing at the ends A and B of a ground where AB = a. They boy at B starts runing in a direction perpendicular to AB with velocity v_(1). The boy at A starts runing simultaneously with velocity v and catches the other in a time t, where t is
Answer» `(a)/(SQRT(V^(2) + v_(1)^(2)))` `(a)/(v+v_(1))` `(a)/(v-v_(1))` `sqrt((a^(2))/(v^(2) - v_(1)^(2)))` ANSWER :D

Discussion

27.	A man walks east with certain velocity . A car is travelling along a road which is 30^(@) West of north . While a bus is travelling another road which is 60^(@) South of West . Find the angle between velocity vector of (a)man and car (b)car and bus (c)bus and man.
Answer» Solution : Form the diagram the ANGLE between VELOCITY VECTOR of man and car is `90^(@)+30^(@)=120^(@)` The angle between velocity vector of car and bus is `60^(@)+60^(@)=120^(@)` The angle between velodcity vector of bus and man is `30^(@)+90^(@)=120^(@)`

Discussion

28.	The distances of two planets from the sun are 10^(12)m and 10^(10)m respectively. The ratio of the time periods of these plantes is
Answer» `1000:1` `100 :1` `10:1` `1:1` ANSWER :A

Discussion

29.	A solid sphere of mass m rolls without slipping on an inclined plane of inclination theta. The linear acceleration of the sphere is
Answer» `(7)/(5) G SIN theta` `(2)/(7) g sin theta` `(3)/(7) g sin theta` `(5)/(7) g sin theta` Solution :The lnear ACCELERATION of the SPHERE, `a=(g sin theta)/(1+(I)/(mR^(2)))=(g sin theta)/(1+(2)/(5))=(5)/(7) g sin theta`.

Discussion

30.	A horizontal force of 150 N produces an acceleration of 2m//s^(2) in a body placed on a horizontal surface. A horizontal force force of 200 N produces an acceleration of 3m//s^(2). The mass of the body and coefficient of kinetic friction are (g=10 ms^(-2))
Answer» 50 KG, 0.1 25 kg, 0.1 50 kg, 0.5 50 kg, 0.2 ANSWER :A

Discussion

31.	A body of mass 'm' dropped from a height 'h' reaches the ground with a velocity sqrt((gh)/(2)). The work done by air resistance is
Answer» `(MGH)/(2)` `(mgh)/(4)` `(2mgh)/(4)` `(3mgh)/(4)` ANSWER :D

Discussion

32.	The work done in forming a soap film of size 10cmx10cm will be, if the surface tension of soap solution is 3 xx 10^(-2) N/m
Answer» `3 xx 10^(-4) J` `3 xx 10^(-2)J` `6 xx 10^(-4) J` `6 xx 10^(-2) J` ANSWER :C

Discussion

33.	Reel of thread in the form of solid cylinder is allowed to unroll by holding the loose end of the thread in hand. The acceleration with which the reel falls down is (g=9.9 ms^(-2))
Answer» `9.9ms^(-2)` `3.3 MS^(-2)` `6.6 ms^(-2)` `1.1 ms^(-2)` ANSWER :C

Discussion

34.	A drop of liquid pressed between two glass plates spreads to a circle of diameter 10cm. Thickness of liquid film is 0.5mm and coefficient of surface tension is 70 xx 10^(-3) "Nm"^(-1). The force required to pull them a part is
Answer» `4.4 N` `1.1 N` `2.2 N` `2.6 N` ANSWER :C

Discussion

35.	A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific hears gamma . It is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by ,
Answer» `((gamma -1))/(2(gamma+1)R) mv^2 R` `((gamma -1))/(2gamma R) mv^2 K` `(gamma Mv^2)/(2R)` K `((gamma -1 ))/(2R) Mv^2 K` ANSWER :D

Discussion

36.	The length of a rubber cord is l_1 metres when the tension in it is 4N and l_2metres when the tension is 5N. Then the length in meters when the tension is 9 N is
Answer» Solution :Let L be the original length and `l_3`be the length of the wire when the TENSION is 9N. ` rArr (l_1 - l) prop 4......(1), (l_2-l) prop 5........(2)` `(l_3 - l) prop 9...........(3)` From (1) and (2) `(l_1- l)/(l_2 - l) = 4/5 rArr 5l_1- 5l= 4l_2 - 4l` `l = 5l_1 - 4l_2`...(4) From (1) and (3) `(l_1 - l)/(l_3 - l) = 4/9 rArr (l_1 - (5l_2 - 4l_2))/(l_3 - (5l_1 - 4l_2)) = 4/9` `rArr (l_2 - 5l_1 + 4l_2)/(l_3 - 5l_1 + 4l_2) = 4/9 rArr (4l_2 - 4l_1)/(l_3 - 5l_1 + 4l_2) = 4/9` ` rArr 36l_2 - 36l_1 = 4l_3 = 20l_2 + 16l_2` ` 20l_2 - 16l_1= 4l_3 ` `l_3 = 5l_2 - 4l_1`

Discussion

37.	At the top of a mountain a thermometer reads 7^(@)C and a barometer reads 70 cm of Hg. At the bottom of the mountain they read 27^(@)C and 76 cm of Hg respectively. Compare the density of air at the top with that at the bottom.
Answer» <P> Solution :At the top of the MOUNTAIN , Pressure `P_(1)=70cm` of Hg , Temperature `T_(1)=7+273=280K` Density of the air `=d_(1)`, At the bottom of the mountain , Pressure `P_(2)=76` CM of Hg Temperature `T_(2)=27+273=300K` , Density of the air = `d_(2)` From the GAS equation, `P_(1)/(d_(1)T_(1))=P_(2)/(d_(2)T_(2))" or "d_(1)/d_(2)=P_(1)/T_(1) times T_(2)/P_(2)=70/280 times 300/76=0.9868`.

Discussion

38.	Acceleration due to gravity is independent of mass of the earth/mass of the body.
Answer» SOLUTION : MASS of the BODY

Discussion

39.	A copper block of mass 1 kg slides down on a rough inclined plane of inclination 37^@at a constant speed. Find the increase in the temperature of the block as it slides down through 60cm assuming that the loss in mechanical energy goes into the copper block as thermal energy. (Specific heat of copper = 420 J kg^(-1)K^(-1), g = 10ms^(-2) )
Answer» `6.6 xx10^(-3)""^@C` `7.6 xx10^(-3)""^@C` `8.6 xx10^(-3)""^@C` `9.6 xx10^(-3)""^@C` ANSWER :C

Discussion

40.	Which of the following have no units and dimensions ?
Answer» Emissive POWER Absorptive power Emissivity Radiancy ANSWER :B::C

Discussion

41.	If the excess of pressure inside a spherical soap bubble of radius 1 cm is balanced by that due to a column of oil of specific gravity 0.9gmcm^(-3) height. If g=10mS^(-2) find the surface tension of the soap water.
Answer» ANSWER :`3.06xx10^(-2)N//m`

Discussion

42.	A particle is projected from the ground with velocity u making an angle thetawith the horizontal. At half of its maximum heights,
Answer» its HORIZONTAL velocity is `U cos THETA ` its vertical velocity is `(u SIN theta)/(SQRT2)` its velocity is `u((1+ cos^2 theta)/(2))^(1/2)` all the above are true Answer :D

Discussion

43.	A body starts from rest from a point distance R_(0) from the centre of the earth. The velocity acquired by the body when it reaches the surface of the earth will be (Rto Radius of the earth) (R_(0) gt R)
Answer» `2GM((1)/(R )-(1)/(R_(0)))` `SQRT(2GM((1)/(R)-(1)/(R_(0))))` `GM((1)/(R)-(1)/(R_(0)))` `2GMsqrt(((1)/(R)-(1)/(R_(0))))` Answer :B

Discussion

44.	Two particle A and B are moving as shown in the figure Their total angular momentum about the point O is
Answer» `9.8 kg m^(2) s^(-1)` ZERO `52.7 kg m^(2) s^(-1)` `37.9 kg m^(2) s^(-1)` Solution :Total angular momentum about O is GIVEN as, `L=L_(1)+L_(2)=m_(1)v_(1)r_(1)+m_(2)v_(2)r_(2)` `=-6.5 xx2.2xx1.5+3.1xx3.6 xx2.8` `=-21.45 + 31.248 = 9.8 "KGM"^(2) s^(-1)`.

Discussion

45.	A cartravelsa distancex kmwitha speed36km/ h and another x km with 45 km / h . Theaveragespeedofthe caris
Answer» `40.5km //h` `40KM //h` `42km//h` DEPENDENT on x Solution :IF a= angleofprojectionand u-intialvelocity, horizontalcomponentofvelocity`=u COS alpha` . Thehorizontalcomponentof velocity of aprojectilleremainsunchangedthroughoutitspathbutat thetopmostpointthe verticalcomponentof itsvelocitybecomeszero. Given`, u COSALPHA = ( u )/(2) ` or `cos alpha=1/2= cos60 ^@ " or ", a= 60 ^(@)`

Discussion

46.	A 4kg mass is resting on a horizontal surface. For this surface mu_(s)=0.6 & mu_(k)=0.2. Force required to move the body with 5ms^(-2) acceleration is (g=10 ms^(-2))
Answer» 20 N 24 N 28 N 32 N Answer :C

Discussion

47.	Figure shows a cylinder containing helium gas at atmospheric pressure with a tightly fitting piston of weight, one fourth the force acting due to atmospheric pressure. The cylinder is jammed at its bottom. A block B is hung by means of a massless string whose other end is attached to the piston. It is observed that the piston just starts sliding. next, the block B is directly placed above the piston. Assuming the piston to move slowly, find the fractional volume of the gas compressed
Answer» `1/2` `1/3` `1/4` `1/8` ANSWER :B

Discussion

48.	A bullet fired at an angle of 30^(@) with the horizontal hits the ground 3.0km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away ? Assume the muzzle speed to be fixed, and neglect air resistance.
Answer» Solution :We are GIVEN that angle of projection with the horizontal, `theta=30^(@)`, horizontal range `R = 3km`. As `R=(v_(0)^(2)SIN2THETA)/(g),3=(v_(0)^(2)sin60^(0))/(g)=(v_(0)^(2))/(g)xx(sqrt(3))/(2)` or `(v_(0)^(2))/(g)=2sqrt(3)km` SINCE the muzzle speed `(v_(0))` is fixed, `R_(max)=(v_(0)^(2))/(g)=2sqrt(3)=2xx(1.732=3.464km` Obviously, it is not possible to hit the target 5km away.

Discussion

49.	A mass M is broken into two parts m and (M-m) What should be the ratio betweenand M. So that the force of gravitation between them is maximum
Answer» `2:1` `1:2` `1:1` `1:3` ANSWER :B

Discussion

50.	The magnitude of gravitational intensity at a point is 0.7 N/kg. What would be the magnitude of the gravitational force on a body of 5 kg mass at this point ?
Answer» Solution :`IMPLIES`F = GRAVITATIONAL intensity (I) `xx` Mass of body (m) `:.F = 0.7 xx 5 ` ` :.F = 3.5 N`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Electric displacement is given by D=epsilonE Here epsilon= electric permittivity E= electric field strength, Find the dimensions of electric displacement.

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A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific hears gamma . It is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by ,

The length of a rubber cord is l_1 metres when the tension in it is 4N and l_2metres when the tension is 5N. Then the length in meters when the tension is 9 N is

At the top of a mountain a thermometer reads 7^(@)C and a barometer reads 70 cm of Hg. At the bottom of the mountain they read 27^(@)C and 76 cm of Hg respectively. Compare the density of air at the top with that at the bottom.

Acceleration due to gravity is independent of mass of the earth/mass of the body.

Which of the following have no units and dimensions ?

If the excess of pressure inside a spherical soap bubble of radius 1 cm is balanced by that due to a column of oil of specific gravity 0.9gmcm^(-3) height. If g=10mS^(-2) find the surface tension of the soap water.

A particle is projected from the ground with velocity u making an angle thetawith the horizontal. At half of its maximum heights,

A body starts from rest from a point distance R_(0) from the centre of the earth. The velocity acquired by the body when it reaches the surface of the earth will be (Rto Radius of the earth) (R_(0) gt R)

Two particle A and B are moving as shown in the figure Their total angular momentum about the point O is

A cartravelsa distancex kmwitha speed36km/ h and another x km with 45 km / h . Theaveragespeedofthe caris

A 4kg mass is resting on a horizontal surface. For this surface mu_(s)=0.6 & mu_(k)=0.2. Force required to move the body with 5ms^(-2) acceleration is (g=10 ms^(-2))

A bullet fired at an angle of 30^(@) with the horizontal hits the ground 3.0km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away ? Assume the muzzle speed to be fixed, and neglect air resistance.

A mass M is broken into two parts m and (M-m) What should be the ratio betweenand M. So that the force of gravitation between them is maximum

The magnitude of gravitational intensity at a point is 0.7 N/kg. What would be the magnitude of the gravitational force on a body of 5 kg mass at this point ?