40048 + Interview Questions in Physics in Class 11

1.	. . . . . isgiven by the slope of velocity -time graph at any instant
Answer» SPEED acceleration retardation velocity Answer :B

Discussion

2.	A rod with rectangular cross section oscillates about a horizontal axis passing through one of its ends and it behaves like seconds pendulum, its length will be
Answer» 1.5m 1 m 3 m 2 m ANSWER :A

Discussion

3.	A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm . What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ?
Answer» 0.25 rad `s^(-2)` 25 rad `s^(-2)` 5 m `s^(-2)` `25 m s^(-2)` Solution :`m = 3` kg, r= 40 CM = `40 xx 10^(-2) m , F = 30` N Moment of inertia of hollow cylinder about its axis = `mr^(2) = 3 kg xx (0.4)^(2) m^(2) = 0.48 kg m^(2)` The torque is given BYT = `I alpha` where I = moment of inertia , `alpha`= angular acceleration In the given case , `tau = r F` , as the force is acting perpendicularly to the radial vector . `therefore alpha = (tau)/(I) =(Fr)/(mr^(2)) = (F)/(mr) = (30)/(3 xx 40 xx 10^(-2)) = (30 xx 100)/(3 xx 40)` `alpha = 25 rad s^(-2)`

Discussion

4.	The magnitude of average velocity is equal to average speed when the particle moving with ........
Answer» VARIABLE SPEED CONSTANT velocity variable velocity constant acceleration SOLUTION :constant velocity

Discussion

5.	A body while being acted upon by a force vec(F )(x)=(3x^(2)-2x+7)hat(i) N gets displaced from x = 0 to x = 10 m in the direction of X -axis .Find the work done [int x^(n) dx = (x^(n+1))/(n+1)]
Answer» SOLUTION :`W = int_(0)^(10) Fdx` ` :. W = int_(0)^(10) (3x^(2)-2X+7)dx` ` :. W = [(3x^(3))/3]_(0)^(10) -[(2x^(2))/2]_(0)^(10) +[7X]_(0)^(10)` ` :. W = [x^(3)]_(0)^(10) -[-x^(2)]_(0)^(10)+[7x]_(0)^(10)` ` = 1000 - 100 + 70 ` ` :. W = 97 J `

Discussion

6.	Two points masses of mass 4m and m respectively separated by distance d are revolving under mutual force of attraction. Ratio of their kinetic energies will be
Answer» ` 1 : 4` `1 : 5` `1 : 1` `1 : 2` ANSWER :A

Discussion

7.	The equation for the displacement oif a particle executive SHM is y={5 sin 2 pi t} cm. Find the (i) the velocity at 3 cm from the mean positon, (ii) acceleration after 0.5 s after leaving the mean position
Answer» `8picms^(-1)`, ZERO `6picm^(-1)`, zero `4picms^(-1)`, zero `2picms^(-1)`, zero Answer :A

Discussion

8.	Underline the correct alternative : In an inelastic collision of two bodies, the quantities which do not change after the collision are the total energy/total linear momentum/total energy of the system of two bodies.
Answer» Solution :Total LINEAR momentum, and also total energy (if the SYSTEM of TWO bodies is isolated).

Discussion

9.	An air bubble released at the bottom of a lake, rises and on reaching the top, its radius found to be doubled. If the atmospheric pressure is equivalent to H metre of water column, find the depth of the lake (Assume that the temperature of water in the lake is uniform)
Answer» <P> SOLUTION :Volume of the air bubble at the bottom of the lake `(V_(1) ) = (4)/(3) pi r^(3)` Volume of the air bubble at the surface of the lake `(V_(2)) = (4)/(3) pi (2r)^(3)` Pressure at the surface of the lake `(P_(2))`=H METRE of water COLUMN. If .h. is the depth of the lake.the pressure at the bottom of the lake `(P_(1)) `= (H+h) metre of water column. Since the temperature of the lake is uniform, According to Boyle.s LAW, `P_(1) V_(1) = P_(2) V_(2)` `(H + h) ((4)/(3) pi r^(3)) = H [ (4)/(3) pi (2r)^(3) ] ` ( H + h ) = 8Hh = 7H

Discussion

10.	An unmanned satellite A and a spacecraft B are orbiting around the earth in the same circular orbit as shown. The spacecraft is ahead of the satellite by some time. Let us consider that some technical problem has arisen in the satellite and the astronaut from B has made it correct. For this to be done docking of two (A and B) is required (in layman terms connecting A and B). To achieve this, the rockets of A have been fired in forward direction and docking takes place as shown in the figure below: Take mass of the earth =5.98xx10^(24)kg Radius of the earth =6400 km Orbital radius =9600km Mass of satellite A=320 kg Mass of spacecraft =3200 kg Assume that initially spacecraft B leads satellite A by 100s, i.e., A arrives at any particle position after 100 s of B’s arrival. Based on the above information answer the following questions The initial total energy and time period of satellite are, respectively,
Answer» `-6.65xx10^(10)J, 9358s` `-6.65xx10^(9)J, 9358s` `-6.65xx10^(10)J, 9140s` `-6.65xx10^(9)J, 9140s` Solution :Time period of `A` and `B` in a given circular orbit is `T=(2pir^(3/2))//(sqrt(GM))` where `r=9600km, G=6.67xx10^(-11)Nm^(2)kg^(-2), M=5.98xx10^(24)kg,` which gives `T=9357.79s=9358s` INITIALLY, TOTAL energy of` A` is `E=-(GMM)/(2R)=-6.65xx10^(9)`

Discussion

11.	Find the (i) instantaneous position and (ii) velocity of the centre of mass of a system of particles of masses 1 kg and 3 kg which are at (2hati + 5hatj + 13hatk)m and (-6hati + 4hatj - 2hatk)possessing velocities (10hati - 7hatj - 3hatk)ms^(-1)and (7hati - 9hatj + 6hatk)ms^(-1)respectively.
Answer» ANSWER :(i) `1/4(-16hati + 17hatj + 7hatk) m; (II) 1/4(3HATI - 34hatj + 15hatk) MS^(-1)`

Discussion

12.	An unmanned satellite A and a spacecraft B are orbiting around the earth in the same circular orbit as shown. The spacecraft is ahead of the satellite by some time. Let us consider that some technical problem has arisen in the satellite and the astronaut from B has made it correct. For this to be done docking of two (A and B) is required (in layman terms connecting A and B). To achieve this, the rockets of A have been fired in forward direction and docking takes place as shown in the figure below: Take mass of the earth =5.98xx10^(24)kg Radius of the earth =6400 km Orbital radius =9600km Mass of satellite A=320 kg Mass of spacecraft =3200 kg Assume that initially spacecraft B leads satellite A by 100s, i.e., A arrives at any particle position after 100 s of B’s arrival. Based on the above information answer the following questions. To dock A and B in the above-described situation, one can use the rocket system of either one, i.e., either of A or of B. To accomplish docking in the minimum possible time which is the best way?
Answer» To USE ROCKET system of `A`. To use rocket system of `B`. Either (a) or (b). Information insufficient Solution :As it is not known that by how MUCH the velocity has changed, which will cause the change in orbital radius and hence time speed, so we can use either rocket system to carry out the docking in the minimum possible time, though the extent of the firing rockets may DIFFER.

Discussion

13.	A boy is standing at the centre of a boat which is free to move on water. If the masses of the boy and the boat are m1 and m2 respectively and the boy moves a distance of 1 m forward then the movement of the boat is ...... Metres
Answer» `m_(1)/(m_(1)+m_(2))` `m_(2)/(m_(1)+m_(2))` `m_(1)/m_(2)` `m_(2)/m_(2)` ANSWER :A

Discussion

14.	An unmanned satellite A and a spacecraft B are orbiting around the earth in the same circular orbit as shown. The spacecraft is ahead of the satellite by some time. Let us consider that some technical problem has arisen in the satellite and the astronaut from B has made it correct. For this to be done docking of two (A and B) is required (in layman terms connecting A and B). To achieve this, the rockets of A have been fired in forward direction and docking takes place as shown in the figure below: Take mass of the earth =5.98xx10^(24)kg Radius of the earth =6400 km Orbital radius =9600km Mass of satellite A=320 kg Mass of spacecraft =3200 kg Assume that initially spacecraft B leads satellite A by 100s, i.e., A arrives at any particle position after 100 s of B’s arrival. Based on the above information answer the following questions. After once returning to the original point, i.e., the place from where the rockets have been fired, in which direction and with what extent the rockets have to be fired from the satellite to again come back in the original orbit?
Answer» Forward DIRECTION with the same extent. Backward direction with the same extent. Forward direction with the higher extent. Backward direction with the higher extent. SOLUTION :To BRING back the satellite in the same orbit, we have to make the SPEED of the satellite as initial one for which we have to increase `v_(F)` by such amont so that it becomes equal to `v_(i)` for which we have to fire the rocket in backward direction with the same extent.

Discussion

15.	A rod of length l(lt2R0 is kept inside a smooth sperical shell as shwon in fifure .mass of the rod is m The normal reaction when l=R is
Answer» `(MG)/(2)` `(mg)/(4)` `(mg)/(2sqrt(3))` `(mg)/(SQRT(3))` ANSWER :d

Discussion

16.	An unmanned satellite A and a spacecraft B are orbiting around the earth in the same circular orbit as shown. The spacecraft is ahead of the satellite by some time. Let us consider that some technical problem has arisen in the satellite and the astronaut from B has made it correct. For this to be done docking of two (A and B) is required (in layman terms connecting A and B). To achieve this, the rockets of A have been fired in forward direction and docking takes place as shown in the figure below: Take mass of the earth =5.98xx10^(24)kg Radius of the earth =6400 km Orbital radius =9600km Mass of satellite A=320 kg Mass of spacecraft =3200 kg Assume that initially spacecraft B leads satellite A by 100s, i.e., A arrives at any particle position after 100 s of B’s arrival. Based on the above information answer the following questions. By doing the above operation, now by how much time is the satellite leading the spacecraft?
Answer» `38s` `138s` Lags by `38 s` Lags by `138 s` SOLUTION :Initially the time periods of the satellite and the spacedraft were same equal to `9358 s` and after firing rockets of `A`, it acquires, slower and faster orbit and `A` will come to initial POSITION in time PERIOD of `9220 s` while `B` will take `9358 s` to come to its initial position or `(9358-100)s` to come to the ORIGINAL position of `A` as `100 s` is the initial time gap between the two. It means now `A` LEADS `B` by `(9358-100-9220)=38s`.

Discussion

17.	Find the force of attraction between parallel glass plates separated by a distance d=1/10 mm, after a water drop of mass m=80 mg is introduced between them. The wetting is assumed to be complete. Surface tension of water =73mN/m.
Answer» Answer :`F=(2mT)/(rho d^(2))=1.17xx10^(6)` N

Discussion

18.	An unmanned satellite A and a spacecraft B are orbiting around the earth in the same circular orbit as shown. The spacecraft is ahead of the satellite by some time. Let us consider that some technical problem has arisen in the satellite and the astronaut from B has made it correct. For this to be done docking of two (A and B) is required (in layman terms connecting A and B). To achieve this, the rockets of A have been fired in forward direction and docking takes place as shown in the figure below: Take mass of the earth =5.98xx10^(24)kg Radius of the earth =6400 km Orbital radius =9600km Mass of satellite A=320 kg Mass of spacecraft =3200 kg Assume that initially spacecraft B leads satellite A by 100s, i.e., A arrives at any particle position after 100 s of B’s arrival. Based on the above information answer the following questions. If from the base station, the rocket system of A has been operated, so that its rocket has been fired in forward direction thus reducing the speed of A by 0.5%, then which of the following will happen?
Answer» Its orbit becomes elliptical with SEMI-major axis `9505.3 km`. its total energy becomes `-6.714xx10^(9)J` Its new time period becomes `9219.67s`. All of the above. Solution :Speed of a before firing the rocket, `v_(i)=sqrt(GM//r)=644.582ms^(-1)` After firing the rocket speed of `A` is `V_(f)=0.995, v_(i)=6413.6ms^(-1)` Energy of the SATELLITE after firing, `E_(f)=2E+(mv_(r)^(2))/w=-6.714xx10^(9)J` as due to firing potential energy doens not change. From `E=-(GMM)/(2a),` the new semi major axis is given by `a=-(GMm)/(2E_(f))impliesa=9505.3km` New time period `T'=(2pia^(32//3))/(sqrt(GM))=9219.67s`

Discussion

19.	A spring is made to oscillate horizontally and vertically. Will there be any change in the period ?
Answer» SOLUTION :No CHANGE in PERIOD, as T is INDEPENDENT of G.

Discussion

20.	Which one of the following statements is true for the speed v and the acceleration a of a particle executing simple harmonic motion?
Answer» When v is MAXIMUM, a is maximum Value of a is zero, whatever MAY be the value of v When v is zero, a is zero When v is maximum, a is zero Solution :At the mean POSITION, `UPSILON` is maximum and a is zero.

Discussion

21.	Briefly discuss the concept of frictional forces ?
Answer» SOLUTION :1 (a) 9

Discussion

22.	A 500 kg satellite is in a circular orbit at an altitude of 500 km above the earth.s surface. Because of air friction, the satellite eventually falls to the Earth.s surface, where is hit the ground with a speed of 2km/s The work done by air friction is [M_(E ) = 6 xx 10^(24) kg, R_(E) = 6400 km]
Answer» `1.67 xx 10^(11)` J `6 xx 10^(9)` J `-6 xx 10^(9)` J `-1.57 xx 10^(10)` J ANSWER :D

Discussion

23.	A straight rod of length L extends from x= a to x= L+a. The linear mass density of the rod varies with x- coordinate is lamda=A +bx^(2). The gravitational force experienced by a point mass m at x= 0 is
Answer» `GM[(A)/(a) + BL]` `Gm[BL + (A)/(a+L)]` `Gm[A[(1)/(a)-(1)/(a+L)+BL]` `Gm[(BL)/(a) + (A)/(a^(2))]` ANSWER :C

Discussion

24.	The velocity of a boat in still water is 10 m/s. If water flows in the river with a velocity of 6 m/s what is the difference in times taken to cross the river in the shortest path and the shortest time. The width of the river is 80 m.
Answer» 1s 10s `(SQRT(3))/(2)` 2s Answer :D

Discussion

25.	Hooke's law states that
Answer» Stress is INVERSELY proportional to STRAIN Stress is directly proportional to strain Stress is INDEPENDENT of strain Stress is proportional to elastic modulus Answer :B

Discussion

26.	Given that y = A sin((2pi)/(lamda) (ct - x)) . Where y and x are measured in metres. Which of the following statements is true?
Answer» The unit of `lamda` is same as that of x and A The unit of `lamda` is same as that of x but not of A The unit of C is same as that of `2pi//lamda` The unit of (CT - x) is same as that `2pi//lamda` Solution :The unit of `lamda` is same as that of x and A

Discussion

27.	The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16 seconds, (i) What is its angular acceleration (assume the acceleration to be uniform) (ii) How many revolutions does the wheel make during this time?
Answer» SOLUTION :`alpha=4pi RAD s^(-2)` `n=576`

Discussion

28.	A 10 KW drilling machine is used to drill a bore in a small aluminium block of mass 8 kg. How much is the rise of temperature in 2.5 minutes assuming 50% of the power is used up in heating the machine itself and remaining is lost to the surroundings. (specific heat of aluminium is 0.9Jgm^(-1)K^(-1))
Answer» `92^@`C `103^@`C `77^@`C `122^@`C Answer :B

Discussion

29.	When a disc rotates with uniform angular velocity , which of the following is not true ?
Answer» The sense of rotation remains same The ORIENTATION of the AXES of rotation remains the same The speed of rotation is non-zero and remains same The ANGULAR ACCELERATION is non-zero and remains same Answer :D

Discussion

30.	A rod of length 'l' lying horizontally on a smooth horizontal surface is moving with centre of mass velocity 'v' and angular velocity omega=(2u)/(l)/ The distance civered by one of the end points of the rod during the time the rod completes one full rotation is nl. Find the value of 'n'
Answer» 2 3 4 1 Answer :C

Discussion

31.	Velocity and acceleration vectors of charged particle moving perpendicular to the direction of magnetic field at a given instant of time are vec(v)=2hat(i)+c hat(j) and bar(a)=3hat(i)+4hat(j) respectively. Then the value of .c. is
Answer» 3 `1.5` `-1.5` `-3` ANSWER :C

Discussion

32.	The mass and diameter of a planet are twice that of the earth. What will be the time period of oscillation of a pendulum on this planet, if it is a seconds pendulum on earth ?
Answer» `sqrt(2)` SEC 2 sec `1/(SQRT2)`sec `2sqrt(2)` sec Answer :D

Discussion

33.	We beat a blanket with stick to remove dust particles.Why ?
Answer» Solution :DUE to inertia of rest, the dust particles REMAIN at their position and the blanket comes into motion on BEATING with a STICK. HENCE the dust particles fall down.

Discussion

34.	Speed of moving object can never be negative. Why ?
Answer» SOLUTION :The distance covered by moving OBJECT can never be NEGATIVE. Thus, SPEED can never be negative

Discussion

35.	Water is flowing through a horizontal tube of radius 5 cm and length 5 cm at the rate of 30 lt/s. Find the pressure required to maintain the flow of water if coefficient of viscosity of water is 0.002 Pa s.
Answer» ANSWER :`1.22 XX 10^(5)"PASCAL"`

Discussion

36.	Is the work done by friction on a body sliding down an inclined plane positive or negative ?
Answer» Solution :Negative, SINCE FRICTION ACTS in a direction opposite to the direction of MOTION.

Discussion

37.	A person walks up a stationary escalator in time t1. If he reamains stationary on the escalator, then it can take him up in time t_(2). How much time would it take him to walk up the moving escalator?
Answer» Solution :LET L be the length of escalator. Speed of MAN w.r.t. escalator is `v_(me)=(L)/(t_(1))`. Speed of escalator `v_(e)= (L)/(t_(2))`. Speed of man with RESPECT to ground would be `v_(m)=v_(me)+v_(e)=L((1)/(t_(1))+(1)/(t_(2)))` `:. "The desired time is" t = (L)/(v_(m))=(t_(1)t_(2))/(t_(1)+t_(2))`

Discussion

38.	Statement I: It is easier to spray water in which some soap is dissolved. Statement II: Soap is easier to spread.
Answer» STATEMENT I is true, statement II is true , statement II is a correct EXPLANATION for statement I. Statement I is true, statement II is true , statement II is not a correct explanation for statement I. Statement I is true, statement II is FALSE. Statement I is false, statement II is true. Answer :C

Discussion

39.	SI unit of Thermal resistance is
Answer» `W K^(-1)` `W^(-1) k` `W^(-1) k^(-1)` `W m^(-1) k^(-1)` ANSWER :B

Discussion

40.	The molar specific heat of a gas at constantvolume is 20 Joule/gm mol/K. The value of gammafor it will be
Answer» `11//10` `7//5` `5//3` `3//2` ANSWER :B

Discussion

41.	What is the power needed to maintain uniform circuler motion?
Answer» Solution :No POWER is NEEDED to maintain uniform circuler motion.Power=`overlineF.overlinev=fvcos(pi/2)=0`

Discussion

42.	The maximum focal length of the eye lens of a person is greater than its distance from the retina The eyeis
Answer» ALWAYS strained in LOOKING at an object strained for OBJECTS at LARGE distances only strained for objects or SHORT distances only unstrained for all distances Answer :A

Discussion

43.	"Proper inflation of tyres can save fuel". Explain how ?
Answer» Solution :Area of contact between the TYRE and the ground is reduced. This reduces ROLLING friction. In that case the VEHICLE will cover GREATER distance for the same AMOUNT of fuel consumed.

Discussion

44.	The amplitude and time period of a particle of mass 0.1 kg executing simple harmonic motion are 1m and 6.28s, respectively. Find its (i) angular frequency, (ii) acceleration and (iii) velocity at a displacement of 0.5 m.
Answer» Solution :`sqrt(3)/2 MS^(-1)`

Discussion

45.	If Z=A^(4) then relatie error in Z is .....
Answer» `(DELTAA)^(4)` `4((DeltaA)/(A))` `((DeltaA)^(4))/(A)` `((DeltaA)/(A))^((1)/(4))` SOLUTION :`Z=A^(4)` `:.(DeltaZ)/(Z)=4(DeltaA)/(A)`

Discussion

46.	If a simple harmonic oscillator has got a displacement of 0.02m and acceleration eaual to 2.0 m"/"s^(2) at any time, the angular frequency of the oscillator is equal to…….
Answer» `10 rad"/"s` `0.1 rad"/"s` `100 rad"/"s` `1 rad"/"s` Solution :`OMEGA =SQRT(("ACCELERATION")/("DISPLACEMENT"))` `omega = sqrt((2.0)/(0.02))= 10 rad"/"s`.

Discussion

47.	An impulse is applied to a moving object with the force at an angle of 20^@with respect to velocity vector. The angle between the impulse vector and the change in momentum vector is
Answer» Solution :Impulse and CHANGE in MOMENTUM are along the same DIRECTION. Therefore angle between these TWO vectors is ZERO.

Discussion

48.	A uniform chain of mass.m. and length .L. is kept on a horizontal table with half of its length hanging from the edge of the table. Work done in pulling the chain on to the table so that only 1/5 th of its length now hangs from the edge is
Answer» `(MGL)/8` `(mgl)/50` `(mgl)/18` `(21mgl)/100` ANSWER :D

Discussion

49.	Derivean expressionfor theworkdoneduringthe isothermalexpansionof an ideal gas.
Answer» Solution :Considern molesof an idealgascontainedin acylinderhavingconductingwallsandprovidedwithfrictionless andmovablcpiston. P - pressureofthe gas , A - areaof thepiston , dV-smallincreasesin thevolumeof thegas Workdoneby the gaswhenthepiston movesup THROUGHA smalldistance DW = PA DX. = Pd V Supposethe gasexpandsisothermallyfrom `P_(1)V_(1)` to `P_(2) V_(2)` The totalamountworkdone `W_("iso" ) =UNDERSET( V_(1))overset( V_(2)) PdV` FornmolesPV= nRT(or )`P = (n Rt)/( V)` `W _("iso") = underset(V_(1))overset(V_(2))(int)(n RT)/(V ).dV= n Rtunderset(V_(1))overset(V_(2))(int).(1)/(V).dV` `n Rt[ logV ] _(V_(1))^(V_(2))= nRt [ logV_(2)- log_(e ) V_(1)^(t) ]` `W _("iso") = nRt["log".(V_(2))/(V_(1))]= 2.303nRt"log" .((V_(2))/(V_(1)))` `= 2.303n Rtlog .((P_(1))/(P_(2)))`

Discussion

50.	In two separate collisions, the coefficients of retitutions, 'e_(1)' are in the ratio 3:1. In the first collision, the relative velocity of approach is twice the relative velocity of separation. Then the ratio between the relative velocity of approach velocity of separation in the second collision is
Answer» `1:6` `2:3` `3:2` `6:1` ANSWER :D

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

. . . . . isgiven by the slope of velocity -time graph at any instant

A rod with rectangular cross section oscillates about a horizontal axis passing through one of its ends and it behaves like seconds pendulum, its length will be

A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm . What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ?

The magnitude of average velocity is equal to average speed when the particle moving with ........

A body while being acted upon by a force vec(F )(x)=(3x^(2)-2x+7)hat(i) N gets displaced from x = 0 to x = 10 m in the direction of X -axis .Find the work done [int x^(n) dx = (x^(n+1))/(n+1)]

Two points masses of mass 4m and m respectively separated by distance d are revolving under mutual force of attraction. Ratio of their kinetic energies will be

The equation for the displacement oif a particle executive SHM is y={5 sin 2 pi t} cm. Find the (i) the velocity at 3 cm from the mean positon, (ii) acceleration after 0.5 s after leaving the mean position

Underline the correct alternative : In an inelastic collision of two bodies, the quantities which do not change after the collision are the total energy/total linear momentum/total energy of the system of two bodies.

An air bubble released at the bottom of a lake, rises and on reaching the top, its radius found to be doubled. If the atmospheric pressure is equivalent to H metre of water column, find the depth of the lake (Assume that the temperature of water in the lake is uniform)

Find the (i) instantaneous position and (ii) velocity of the centre of mass of a system of particles of masses 1 kg and 3 kg which are at (2hati + 5hatj + 13hatk)m and (-6hati + 4hatj - 2hatk)possessing velocities (10hati - 7hatj - 3hatk)ms^(-1)and (7hati - 9hatj + 6hatk)ms^(-1)respectively.

A boy is standing at the centre of a boat which is free to move on water. If the masses of the boy and the boat are m1 and m2 respectively and the boy moves a distance of 1 m forward then the movement of the boat is ...... Metres

A rod of length l(lt2R0 is kept inside a smooth sperical shell as shwon in fifure .mass of the rod is m The normal reaction when l=R is

Find the force of attraction between parallel glass plates separated by a distance d=1/10 mm, after a water drop of mass m=80 mg is introduced between them. The wetting is assumed to be complete. Surface tension of water =73mN/m.

A spring is made to oscillate horizontally and vertically. Will there be any change in the period ?

Which one of the following statements is true for the speed v and the acceleration a of a particle executing simple harmonic motion?

Briefly discuss the concept of frictional forces ?

A 500 kg satellite is in a circular orbit at an altitude of 500 km above the earth.s surface. Because of air friction, the satellite eventually falls to the Earth.s surface, where is hit the ground with a speed of 2km/s The work done by air friction is [M_(E ) = 6 xx 10^(24) kg, R_(E) = 6400 km]

A straight rod of length L extends from x= a to x= L+a. The linear mass density of the rod varies with x- coordinate is lamda=A +bx^(2). The gravitational force experienced by a point mass m at x= 0 is

The velocity of a boat in still water is 10 m/s. If water flows in the river with a velocity of 6 m/s what is the difference in times taken to cross the river in the shortest path and the shortest time. The width of the river is 80 m.

Hooke's law states that

Given that y = A sin((2pi)/(lamda) (ct - x)) . Where y and x are measured in metres. Which of the following statements is true?

The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16 seconds, (i) What is its angular acceleration (assume the acceleration to be uniform) (ii) How many revolutions does the wheel make during this time?

When a disc rotates with uniform angular velocity , which of the following is not true ?

A rod of length 'l' lying horizontally on a smooth horizontal surface is moving with centre of mass velocity 'v' and angular velocity omega=(2u)/(l)/ The distance civered by one of the end points of the rod during the time the rod completes one full rotation is nl. Find the value of 'n'

Velocity and acceleration vectors of charged particle moving perpendicular to the direction of magnetic field at a given instant of time are vec(v)=2hat(i)+c hat(j) and bar(a)=3hat(i)+4hat(j) respectively. Then the value of .c. is

The mass and diameter of a planet are twice that of the earth. What will be the time period of oscillation of a pendulum on this planet, if it is a seconds pendulum on earth ?

We beat a blanket with stick to remove dust particles.Why ?

Speed of moving object can never be negative. Why ?

Water is flowing through a horizontal tube of radius 5 cm and length 5 cm at the rate of 30 lt/s. Find the pressure required to maintain the flow of water if coefficient of viscosity of water is 0.002 Pa s.

Is the work done by friction on a body sliding down an inclined plane positive or negative ?

A person walks up a stationary escalator in time t1. If he reamains stationary on the escalator, then it can take him up in time t_(2). How much time would it take him to walk up the moving escalator?

Statement I: It is easier to spray water in which some soap is dissolved. Statement II: Soap is easier to spread.

SI unit of Thermal resistance is

The molar specific heat of a gas at constantvolume is 20 Joule/gm mol/K. The value of gammafor it will be

What is the power needed to maintain uniform circuler motion?

The maximum focal length of the eye lens of a person is greater than its distance from the retina The eyeis

"Proper inflation of tyres can save fuel". Explain how ?

The amplitude and time period of a particle of mass 0.1 kg executing simple harmonic motion are 1m and 6.28s, respectively. Find its (i) angular frequency, (ii) acceleration and (iii) velocity at a displacement of 0.5 m.

If Z=A^(4) then relatie error in Z is .....

If a simple harmonic oscillator has got a displacement of 0.02m and acceleration eaual to 2.0 m"/"s^(2) at any time, the angular frequency of the oscillator is equal to…….

An impulse is applied to a moving object with the force at an angle of 20^@with respect to velocity vector. The angle between the impulse vector and the change in momentum vector is

A uniform chain of mass.m. and length .L. is kept on a horizontal table with half of its length hanging from the edge of the table. Work done in pulling the chain on to the table so that only 1/5 th of its length now hangs from the edge is

Derivean expressionfor theworkdoneduringthe isothermalexpansionof an ideal gas.