40048 + Interview Questions in Physics in Class 11

1.	Find the moment of inertia and radius of gyration of uniform cross-section rod of mass M and length l about an axis through a point at distance (l)/(3) from one end and perpendicular to the length of rod.
Answer» SOLUTION :`I=(ML^(2))/(9),k=(L)/(3)`

Discussion

2.	Molar specific heat at constant volume is C for a monoatomic gas is
Answer» `3/2R` `5/2R` 3R 2R Answer :A

Discussion

3.	S is a piston of mass M which can slide inside a cylinder without friction. The walls of the cylinder is adiabatic and the piston is diathermic and area of cross section of cylinder is A. Length of the cylinder is 2l_0. Initial pressure of each chamber is P_0. Volume of the chambers are equal. Initially spring is unstretched and has spring constant k. The spring is fixed with piston and wall of the cylinder as shown in the figure. Find out its time period when piston is given small displacement.
Answer» SOLUTION :`T=2pisqrt((MI_0)/(2P_0 A+Kl_0))`

Discussion

4.	A body at temperature 40^(@)C is kept in a surrounding of constant temperature 20^(@)C. It is observed that its temperature falls to 35^(@)C in 10 minutes. Find how much more time will it taken for the body to attain a temperature of 30^(@)C.
Answer» Solution :from equation `(14.3)` `Delta theta_(1) = Delta theta_(i)e^(-kt)` for the interval in which temperature falls from `40 to 35^(@)C` `(30 -20) = (40 -20)e^(-k.10)` `rArr e^(-10k) = (3)/(4)` `rArr k =("ln"(4)/(3))/(10)` for the next interval `(30 - 20) =(35 - 20)e^(-kt)` `rArr e^(-kt) = (2)/(3) rArr kt = "ln"(3)/(2)` `rArr (("ln"(4)/(3))t)/(10) ="ln"(3)/(2)` `rArr t = 10 (("ln"(4)/(3)))/(("ln"(4)/(3)))` minute `- 14.096 MIN` ALITER: (by approximate method) for the interval in whcih temperature falls form `40 to 35^(@)C` `lt theta gt = (40 +35)/(2) = 37.5^(@)C` from equation `(14.4) ltlt(d theta)/(dt)gtgt =-k(lt thetagt -theta_(0))` `rArr ((35^(@)C-40^(@)C))/(10(min)) =- K(37.5^(@)C-20^(@)C)` `rArr K = (1)/(35) (min^(-1))` for the interval in which temperature falls from `35^(@) to 30^(@)C` `lt theta gt = (35 +30)/(2) = 32.5^(@)C` from equation `(14.4)` `((30^(@)C-35^(@)C))/(t) =- (32.5^(@)C-20^(@)C)K` `rArr` required TIME, `t = (5)/(12.5) xx 35 min = 14 min`

Discussion

5.	There is a limit beyong whichthe polishibng of a surface increases frictional resistancerather than decreasing it why
Answer» Solution :Friction is a contact force. Friction is directly proportional to area of contact. In the normal SURFACES there are bumps to interlock the surfaces in contact. But the surfaces are POLISHED beyond CERTAIN limit. The area of contact will be increased and the MOLECULES come closer to each other. It increases electrostatic force between the molecules. As a result it increases friction.

Discussion

6.	State principle of superposition.
Answer» Solution :When NUMBER of waves travel through a medim SIMULTANEOUSLY, the resultant displacement of any PARTICLE of the medium at any given TIME is equal to the algebraic sum of the displacement due to indvidual waves.

Discussion

7.	Light year is :
Answer» LIGHT EMITTED by the SUN in one year. The time taken by light to travel from sun to earth. The distance travelled by light in free space in one year. The time taken by earth to GO once around the sun. Solution :Light year is the distance travelled by light in free space in one year.

Discussion

8.	Two skaters approach each other as shown in figure and lock hans. Given each has an initial speed of 2.50m//s relative to ice. Each has a mass of 70.0kg and their centres of mass are 0.800m from their locked hands. You may approximate their momerits of inertia to be that of point masses at this radius:
Answer» FINAL angular velocity is `3.12rad//s` Skaters begin to ROTATE about CM of system There is a no LOSS of energy If skaters pull each other and reduce their separation by half thei energy is increased Answer :A::B::D

Discussion

9.	Rain is falling vertically with a speed of 35 ms^(-1). Winds starts blowing after sometime with a speed of 12ms^(-1) in east to west direction. In which direction should a boy waiting at a bus stop hold his umbrella ?
Answer» Solution :The VELOCITY of the rain and the wind are represented by the vectors `v_(r)` and `v_(w)` in the figure and are in the direction specified by the problem. USING the rule of vector addition. We see that the resultant of `v_(r)` and `v_(w)` is R as shown in the figure. The magnitude of R is The direction `theta` that R MAKES with the VERTICAL is given by `tan theta = (v_(w))/(v_(r)) = (12)/(35) = 0.343` or, `theta = tan^(-1)(0.343) = 19^(@)` Therefore, the boy should hold his umbrella in the vertical plane at the angle of about `19^(@)` with the vertical towards the east.

Discussion

10.	The system is released from rest with the spring initially stretched 75 mm. Calculate the velocity of the block after it has dropped 12 mm. The spring has a stiffness of 1050 Nm^(-1). Neglect the mass of the small pulley.
Answer» `0.371 MS^(-1)` `0.45 ms^(-1)` `5 ms^(-1)` `2.2 m s^(-1)` ANSWER :A

Discussion

11.	A 20 g particle movesin SHM with a frequency of 3 oscillations per second and amplitude of 5 cm. Through what total distance does the particle move during one oscillation? What is its average speed?
Answer» 20CM, 20 CM/sec 40 cm, 40 cm/sec 40 cm, 60cm/sec 20 cm, 60cm/sec Answer :D

Discussion

12.	Answer the followings, (a)You can shield a change from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means? (b) An astronaut inside a small spaceship orbiting around the earth cannot delect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity ? (c ) If you compare the gavitational force on the earth due to sun to that due to the moon, you would find that the sun's pull is greater than the moon's pull. However, the effect of the moon's pull is greater than tidal effect of sun. Why?
Answer» Solution :(a) We cannot shield a body from the gravitational influence of nearby matter, because the gravitational force on a body due to nearby matter is independent of the intervening medium or the presence of other matter in their way, whereas it is not so in case of electrical forces. It MEANS the graviational screens are not POSSIBLE. (b) Yes, if the zero of the spaceship orbiting around the earth is large enough, an astronaut inside the sapceship can detect the variation in `g`. (c ) TIDAL effect depends inversely on the cube of the distance, unlike force which depends inversely on the square of the distance. Since the distance of Moon from the ocean water is very as compared to the distance of SUN from the ocean water on Earth. THEREFORE, the ideal effect of Moon's pull is greater than the ideal effect of the sun.

Discussion

13.	State the law of conservation of mechanical energy. Show that total mechanical energy of a body falling freely under gravity is conserved.
Answer» Solution :Energy can NEITHER be created nor destroyed but can be transformed into other forms of energy. Let a particle be at 'C'. Initially the particle will be zero velocity. Let 'h' be the HEIGHT of the particle. `P.E_(C)=mgh,""K.E_(C)=0`, `T.E_(C)` = mgh At B, `K.E_(B)=1//2mv_(B)` `=1//2m(2gx)` = mgx `P.E_(B)=mg(h-x)` Hence `T.E_(B)=(mgh-mgx)+(mgx)` = mgh At A, `v_(A)^(2)=2gh` So that `K.E_(A)=(1)/(cancel2)mcancel2gh` = mgh `P.E_(A)=0` Hence at all points. A, B and C, the total energy is CONSERVED.

Discussion

14.	(A) : A piece of ice floats in water. The level of water remains unchanged when the ice melts completely. (R ) : According to Archimede's principle, the loss in weight of the body in the liquid is equal to the weight of the liquid displaced by the immersed part of the body.
Answer» Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A) Both (A) and (R ) are true and (R ) is not the correct explanation of (A) (A) is true but (R ) is FALSE Both (A) and (R ) are false Answer :A

Discussion

15.	Define a vector. Give examples.
Answer» Solution :VECTOR is a quantity which is described by the both magnitude and direction. Geometrically a vector is directed line segment. Example FORCE, VELOCITY and displacement.

Discussion

16.	A small block of mass M moves with velocity 5 m/s towards another block of same mass M placed at a distance of 2m on a rough horizontal surface. Coefficient of friction between the blocks and ground is 0.25. Collision between the two blocks is elastic, the separation the blocks, When both of them come to rest, is (g=10 m//s^(2))
Answer» Solution :Retardation DUE to friction `= mu g = (0.25)(10)=2.5 m//s^(2)` Collision is elastic, i..e., after collision first block comes to rest and the second block acquires the velocity of first block. Or we can understand it in this manner that second block is PERMANENTLY at rest while only the first block moves. Distance travelled by it will be `s=(v^(2))/(2A)=((5)^(2))/((2)(2.5))=5m` `THEREFORE` FINAL separation will be `(5-2)=3m`

Discussion

17.	A simple pendulum is making oscillations with its bob immersed in a liquid of density n times less than the density of the bob. What is its period?
Answer» `2pi sqrt((l)/(mg))` `2pi sqrt((l)/((1 + 1/N)g))` `2pi sqrt((nl)/(g))` `2pi sqrt((nl)/((n - 1)g))` ANSWER :B

Discussion

18.	All harmonics are overtones but all overtones are not harmonics. How?
Answer» SOLUTION :The overtones with FREQUENCIES that are integral multiples of the fundamental are called HARMONICS. Hence all harmonics are overtones. But overtones which are non-integral multiples of the fundamental are not harmonics.

Discussion

19.	A man of mass 60 kg is standing on a horizontal conveyor belt. When the belt is given an acceleration of 1 ms^(-2), the man remains stationary with respect to the moving belt. If g=10 ms^(-2), The net force acting on the man is
Answer» zero 120 N 60 N 600 N Solution :( c) `F+("NET")="MASS"XX"ACCELERATION"` `=60xx1=60N`

Discussion

20.	(A) : A unitless quantity cannot have a dimensional formula. (R) : A dimensionless quantity cannot have a unit.
Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is false Both (A) and (R) are false Answer :C

Discussion

21.	A closed bottle containig water 30°C is carried to the moon in sa spaceship.If it is placed on the surface of the moon,What will happen to the water as soon as the lid is opened?
Answer» WATER will boil Water will freeze nothing will HAPPEN to it it will DECOMPOSE into `h_2O_2` ANSWER :A

Discussion

22.	Three closed vessels A, B and C are at the same temperature T and contain gases which obey Maxwellian distribution of velocities. Vessel A contain O_2. B only N_2 and C mixture of equal quantities of O_2 and N_2. If the average speed of the O_2 molecules in the vessel A is V_1, that of N_2 molecules in the vessel B is B_2, the average speed of the O_2 molecules in vessel C is
Answer» `((v_1 + v_2))/( 2)` `v_1` `SQRT( 4v_1)` `sqrt(4"kT/m")` ANSWER :B

Discussion

23.	Find the height of a satellite from the surface of the earth if its orbital velocity is 4 km/s.
Answer» SOLUTION :`1.8612 XX 10^4 KM`

Discussion

24.	The density of a material in c.g.s system is 2g/cc. In a system of units in which units of length is 2cm, and unit of mass is 4g, what is the numerical value of density of the material is
Answer» 4 2 1 0.5 Answer :A

Discussion

25.	There are two pendulums of length l_(1) and l_(2) start vibrating. At some instant, the two are in mean position in the same phase. Calculate after how many vibrations of shorter pendulum, the two wll be in phase in the mean position? [(l_(1)gtl_(2)),l_(1)=121cm, l_(2)=100cm]
Answer» 11 10 9 8 Solution :`T_(1)=2pisqrt(((l_(1))/(g))),T_(2)=2pisqrt(((l_(2))/(g)))(l_(1)gtl_(2))` Let SHORTER PENDULUM makes n vibrations, then the longer will make one less than n to COME in phase again. `nT_(1)=(n-1)T_(1)` `2pisqrt(((l_(2))/(g))n)=(n-1)2pisqrt(((l_(1))/(g)))` `THEREFORE"n=11"`

Discussion

26.	A string 1 m long can just support a weight of 16 kg. A mass of 1 kg is attached to one of its ends. The body is revolved in a horizontal circle about the other fixed end of the stringy Find the greatest number of revolutions made by the mass per seconds ? (Take g=10 ms^(-2)) The maximum tension the string can withstand = weight of 16 kg.
Answer» SOLUTION :`MG = 16 xx 10 N = 160 N` Centripetal force `=F = mromega^(2) = MR (2pin)^(2)` `F = 160 N, m = 1kr, r = 1m, n=?` `n^(2) = F/(mr 4pi^(2)) = 160/(1 xx 1 xx 4pi^(2)) = 4`

Discussion

27.	A spring of force constant K is cut into two parts whose lengths in the ratio 2 :3. The part having larger force constant is
Answer» SHORTER spring Longer spring Both have EQUAL FORCE constants Nothing can be said Answer :A

Discussion

28.	Two particles are projected from the same point with the same speed at different angles theta_(1) and theta_(2) to the horizontal. If their respectively times of flights are T_(1) and T_(2) and horizontal ranges are same then (a) theta_(1) + theta_(2) = 90^(@)(b) T_(1) = T_(2)tan theta_(1) (c) T_(1) = T_(2) tan theta_(2) (d) T_(1) sin theta_(2) = T_(2) sin theta_(1)
Answer» a, B, d are correct a, C, d are correct b, c, d are correct a, b, c are correct Answer :A

Discussion

29.	The potential energy of a particle in a force filled is U=(A)/(r^(2))-(B)/(r ) where A and B are positive constants and .r. is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is
Answer» A/B B/A `(B)/(2A)` 2A/B Answer :D

Discussion

30.	Obtain an expression for the power delivered by torque.
Answer» `P = VEC(TAU).vec(THETA)` `P = vec(tau) xxvec(theta)` `P = tau theta SIN theta` P = 0 (zero always) Answer :a

Discussion

31.	Name the basic forces in nature. Given some examples of gravitational, electromagnetic and nucler forces from daily life experiences.
Answer» SOLUTION :1 (a) 9

Discussion

32.	(A) : Space rockets are usually launched in the equatorial line from west to east. (R) : The acceleration due to gravity is minimum at the equator.
Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is false Both (A) and (R) are false Answer :B

Discussion

33.	A fighter plane flying horizontally at an altitude of 2 km with speed of 540 kmph passes directly over head an antiaircraft gun. If the gun can fire a bullet at the muzzle speed of 500ms^(-1) , at angle with the vertical the gun should fire the bullet so that the bullet hits the plane ?
Answer» `COS^(-1)(3//10)` `Sin^(-1)(3//10)` `TAN^(-1)(10//3)` `45^(@)` ANSWER :B

Discussion

34.	Pick out the correct statements about the strong nuclear force from the following. S1: It is charge independent. S2: It is the strongest force in nature. S3: Its range is very large. S4: It is responsible for the stability of nuclei.
Answer» S1 and S3 S1, S2 and S3 S1, S2 and S4 S2 and S3 Solution :The range of strong NUCLEAR FORCE is very small, og about nuclear dimensions `(10^(-15)m)`.

Discussion

35.	A car starting from rest, moves some distance with constant acceleration 3 ms^(-2). Then it moves with constant deceleration 2 ms ^(-2) and finally becomes stationary. If the total time taken for the complete journey is 100 s, what was maximum velocity of the car e iu during the journey ?
Answer» 100 80 140 120 Solution :`t _(1) + t _(2) = 100 s` For first case,` v . =v _(0) + a _(1) t _(1)` `v.=0 +3t _(1) therefore t _(1) = (v.)/(3) ""…(1)` For second case, `v=v. + a _(2) t _(2) ` `0=v. therefore t _(2) = (v.)/(2) ""…(2)` `therefore t _(1) + t _(2) = (v.)/(3) + (v.)/(2)` `therefore 100 = (5v .)/(6)` `therefore v . = 120 ms ^(-1)`

Discussion

36.	A constant volume gas thermometer works on
Answer» BOYLE's law Gay LUSSAC's law Pascal's law Stefan's law Answer :B

Discussion

37.	Assertion : The Angular velocity of a planet in its orbit around the sun increases when it is nearer to the sin, as the moment of Inertia of the planet about the sun decreases. This is an example for conservation of Angular momentum. Reason : The nearest position of the planet to the sun is called apogee and the farthest position is perigee.
Answer» ASSERTION and Reason are correct and Reason is correct EXPLANATION of Assertion Assertion and Reason are true but Reason is the FALSE explanation of the Assertion Assertion is true but Reason is false Assertion is false but Reason is true Answer :c

Discussion

38.	A circular metal disc of mass 200 g and radius 20 cm makes 10 revolutions per second about an axis passing through the centre and perpendicular to its plane. Find a) Moment of inertia about the axis, b) Torque required to change its angular velocity from 10 rps to 20 rps in 5 s.
Answer» Solution :Moment of inertia`I=(Mr^(2))/(2)=(200xx10^(-3)(20xx10^(-2)))/(2)=4xx10^(-3) kgm^(-2)` b)Torque `taualpha=(I(omega_(2)-omega_(1)))/(t) omega_(1)=2pin_(1)=2pixx10=20pi` `omega_(2)=2pin_(2)2pixx20=40pi,t=5 sectherefore TAU=(4xx10^(-3)(40 pi-20pi))/(5)` `=(4xx10^(-3)xx20pi)/(5)=16xxpixx10^(-3),tau=0.05 Nm`

Discussion

39.	What stops the collapse of a white dwarf and a neutron star ?
Answer» SOLUTION :Electron degeneracy PRESSURE STOPS the collapse of white dwarf whereas neutron degeneracy pressure stops the collapse of neutron STARS.

Discussion

40.	A point P lies on the axis of a ring of mass M and radius a, at a distance a from its centre C. A small particle starts from P and reaches C under gravitational attractiononly. Its speed at C will be
Answer» `SQRT((2GM)/a)` `sqrt((2GM)/a(1-1/sqrt2))` `sqrt((2GM)/a (sqrt2-1))` 0 Answer :B

Discussion

41.	How does a gas do work when its expands adiabatically ?
Answer» SOLUTION :when a gas expands adiabatically it does work against the surrondingat the COST of its INTERNAL ENERGY .

Discussion

42.	An open glass tube is immersed in mercury in such a way that the length 8 cm extends above the mercury level. Now the open ,end of the tube is closed by a finger and raisedfurlher by 44 cm. What will be the length of air column above mercury in the tube. Take atmospheric pressure to he 76cm of mercury. Neglect capillary effect.
Answer» <P> Solution :Figure-(a) shows the tube in initial state then it is closed at top and raised up so that its length above MERCURY becomes 8 + 44 = 52 cm as shown in figure (b) . As initially tube is open to atmosphere, pressure of air inside is 76CM of Hg and its volume is 8A, if A is the cross-sectional area of the tube. When its upper end is closed and raised up, situtation is shown in figure (b ). LET the air column be of length x then its final volume will be xA. Pressure at the mercury level in the container is equal to atJnospheric pressure 76 cm of Hg and the air column is separated by another mercury column of height 52- x above this level which will oppose the atmospheric pressure inside the column will be 76-(52-x) = (24+x)cm of mercury. As during the process temperature of system remains constant thus we can use Boyle.s law for the initial and final states as `P_(1) V_(1) = P_(2) V_(2) , 76 xx 8A = (24 + x ) xx x A `or 608 = 24 x + `x^(2) " or " x^(2) + 24x - 608 = 0 ` On solving we get x = 15.4 cm or x = -39.4 cm Thus the acceptable VALUE of final air column is 15.4cm.

Discussion

43.	The mechanical equivalent of heat J is :
Answer» a constant a PHYSICAL quantity a CONVERSION FACTOR none of the above Solution :a conversion factor

Discussion

44.	An object of mass 2 kg is taken to a height 5 m from the ground (g = 10ms^(-2)). (a) Calculate the potential energy stored in the object (b) Where does this potential energy come from? (c) What external force must act to bring the mass to that height? (d) What is the net force that acts on the object while the object is taken to the height h?
Answer» Solution :(a) `"The POTENTIAL energy U"=mgh =2xx10xx5=100J`Here the positive sign implies that the energy is stored on the MASS. (b) This potential energy is TRANSFERRED from external agency which applies the force on the mass (c) The external applied force `vecF_(a)`, which takes the object to the height 5 m is, `vecF_(a)=-vecF_(g)` `vecF_(a)=-(-mghatj)mghatj` where, `HATJ` represents unit vector along vertical upward direction (d) From the definition of potential energy, the object must be moved at constante So the net force acting on the object is zero `vecF_(g)+vecF_(a)=0`

Discussion

45.	A liquid drop of diameter 'D' is split into 27 droplets. If T is surface tension of the liquid the change in energy is
Answer» `2 PI D ` `2 pi D^2` `2 pi D^2 T` Zero Answer :C

Discussion

46.	For a man who wants to cross the river, with the shortest path AB, find the anglet theta (see figure)
Answer» `sin^(-1)(3/4)` `sin^(-1)(4/3)` `30^(@)` Given situation is not possible Solution :(d) For the man to CROSS the RIVER frm the shortest path, `V_(m) sin THETA=v_(r)` `sin theta=(v_(r))/(v_(m))` `sin theta=4/3` `theta =sin^(-1)(/3)` But it is not possible, because `sin theta` could not be GREATER than `t`. Hence the man could not cross the river from the shortest path `AB`.

Discussion

47.	In a football match a ball is kicked by a player with a force of 25 N for 0.2 sec and then by another player with a force of 70 N for 0.1 sec in the same direction. If the foot ball gains a velocity of 24 m/s after the two kicks, the mass of the foot ball is nearly
Answer» 5 KG 10 kg 2.5 kg 0.5 kg ANSWER :D

Discussion

48.	A light body and a heavy body have equal kinetic energies. Which one has greater momentum?
Answer» SOLUTION :`P= sqrt(2mK)` So HEAVY BODY has GREATER momentum.

Discussion

49.	Air is heated by
Answer» conduction convection eradiation none of these Answer :B

Discussion

50.	The resultant of two equal forces is 141.4N when they are mutually perpendicular. When they are inclined at an angle 120^(@), then the resultant force will be
Answer» 100 N 141.4 N 196 N Zero Answer :A

Discussion

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