40048 + Interview Questions in Physics in Class 11

1.	A body of mass 5kg starts from origin with a velocity baru = (30hati +40hatj)ms^(-1) . If a constant force F=-(hati+5hatj)Nacts on the body, then pick up the correct statements
Answer» y - component of the velocity becomes zero at t = 40 second X- component of the velocity becomes zero at t = 180 second body will NEVER attain permanent rest the direction of motion CHANGES only once along x and y- directions Answer :A::B::C::D

Discussion

2.	Temperature of substances A and B are 727^(@)C and 327^(@)C respectively, then ratio of rate of their radiant energy H_(A):H_(B)=. . . . .
Answer» `727:327` `5:3` `25:9` `625:81` Solution :`T_(A)=273+727=1000K` `T_(B)=273+327=600K` Rate of RADIANT energy `HpropT^(4)` `:.(H_(A))/(H_(B))=(T_(A)^(4))/(T_(B)^(4))` `=((1000)^(4))/((600)^(4))` `=((10)/(6))^(4)` `=((5)/(3))^(2)` `=(625)/(81)` `:.H_(A):H_(B)=625:81`

Discussion

3.	Consider twoo concentric sphere of radius R_(1) at temperature T_(1) and radius R_(2) at temperature T_(2). A material of thermal capacity conductivity K is filled in the space between shells. What is the distance from the centre of the system of shells, where the temperature is mean of T_(1) and T_(2). (T_(1)gtT_(2)) q is the heat transferred per second
Answer» `(R_(2)Q)/(q+2pi R_(1)K(T_(2)-T_(1)))` `((R_(2)-R_(1))q)/(q+2piR_(2)K(T_(2)-T_(1)))` `(R_(1)q)/(q+4piR_(1)K(T_(2)-T_(1)))` `(R_(1)q)/(q+2piR_(1)K(T_(2)-T_(1)))` Answer :D

Discussion

4.	A mass attached to a spring is free to oscillate, with angular velocity omega in a horizontal plane without friction or damping. It is pulled to a distance x_0 and pushed towards the centre with a velocity v_0 at time t=0. Determine the amplitude of the resulting oscillations in terms of the parameters omega, x_0 and v_0
Answer» Solution :x= A `cos (OMEGA x + theta) `, velocity , `(DX)/(dt) =- A omega sin(OMEGAT + theta)` When t=0 , `x=x_0 and (dx)/(dt) = -v_0` `THEREFORE x_0 = A cos theta ""……..(i)` `-v_0 = - A omega theta ` (or) ` A sin theta = v_0//omega ""......(ii)` Squaring and adding (i) and (ii) , we get , `A^2 (sin^2 theta + cos^2 theta) = (v_0^2 //omega^2) + x_0^2 , A= [v_0^2 .//omega^2 + x_0^2J^(1//2)]`

Discussion

5.	Moment of inertia of a uniform rod of length L and mass M , about an axis pasing through L/4 from one end and perpendicular to its length is
Answer» `(7)/(36) ML^2` `(7)/(48) ML^2` `(11)/(48) ML^2` `(ML^2)/(12)` Solution :APPLYING theorem of PARALLEL AXES , `I = I_(0) + M((L)/(4))^(2) = (ML^2)/(12) + (ML^2)/(16) = (7)/(48) ML^2`

Discussion

6.	In a resonance tube, the first resonance is obtained at 40 cm length, using a tuning fork of frequency 450 Hz. Ignoring end correction, the velocity of sound in air is
Answer» 620 m/s 720 m/s 820 m/s 1020 m/s SOLUTION :`(lambda)/(4) = 40` CM = 0.4 `implies therefore lambda = 4 xx 0.4 = 1.6 m , f = 450` Hz Velocity of sound `V = f lambda = 450 xx 1.6 = 720` m/s

Discussion

7.	In the question number 56, the maximum horizontal distance covered by the ball will be
Answer» `160sqrt(3)`m `140sqrt(3)` m `120sqrt(3)` m `100sqrt(3)` m Solution :Horizontal range, `R = (U^(2)sin2theta)/(G)` `R = ((56)^(2) SIN60^(@))/(9.8) = (56 xx 56 xxsqrt(3))/(9.8 xx 2) = 160sqrt(3)m`

Discussion

8.	A satellite is orbiting at a certain height in a circular orbit. If the mass of the planet is reduced to half the initial value, then the satellite would
Answer» FALL on the planet go to the ORBIT of SMALLER radius go to the orbit of LARGER radius escapes from the planet Answer :D

Discussion

9.	A metallic sphere is heated from 225 K to 325 K. On heating, the volume increases by 2% of original value. Find the coefficients of linear, superficial and cubical expansion of metal.
Answer» SOLUTION :`2.09 XX 10^(-4) K^(-1) , 4.18 xx 10^(-4) K^(-1) , 6.27 xx 10^(-4) K^(-1)`

Discussion

10.	A person measured the time period of a simple pendulum inside a stationary lift and finds it to be .T.. If the lift starts accelerating down wards with an acceleration g/3, the time period of the pendulum will be
Answer» `SQRT(3)T` `sqrt((3)/(2))T` `(T)/(sqrt(3))` `T"/"3` ANSWER :B

Discussion

11.	A drop of liquid having radius 2mm has a terminal velocity 20"cms"^(-1), the terminal velocity of a drop 1 mm radius will be
Answer» `40 "cms"^(-1)` `20 "cms"^(-1)` `10 "cms"^(-1)` `5 "cms"^(-1)` ANSWER :D

Discussion

12.	Same force acts on two bodies of masses m, and m_(2)(m_(2) gt m_(1)) After travelling the same distance
Answer» Kinetic ENERGY of `m_(2)` is GREATER than kinetic energy of `m_(1)` Kinetic energy of `m_(1)` is greater than kinetic energy of `m_(2)` RATIO of kinetic ENERGIES of those two is equal to ratio of their momentum Kinetic energy of `m_(1)` is equal to kinetic energy of `m_(2)` Answer :D

Discussion

13.	Tube A has both ends open, while B has on end closed otherwise the two tubes are identical. What is the ratio of fundamental frequency of the tubes A and B ?
Answer» SOLUTION :Thefundamental frequency for tube A with both ends open is `f_A = V/(2L)` The fundamental frequency for tube B with ONE end closed is `f_B = v/(4L)` `f_A/f_B = (v//2L)/(v//4L) =2 RARR f_A : f_B = 2 : 1`

Discussion

14.	Briefly explain centrifugal force with suitable examples
Answer» Solution :Centrifugalforce : It is a pseudoforce . Ituniformcircularmotiondue toacceleratedframeof referencewhichis ALONGTHE radiusand directedaway from THECENTRE . EG1 :Whena car takesa turnin a curvedroadpersoninsidethe car feelsan outwardforcewhichpushes the personaway. EG2 :A personstandingon arotatingplatformfeelsan outwardcentifugalforceand islikelyto bepushedawayfrom the PLATFORM. to avoidthisusuallythe outeredgeof theplatformislittleinclined upwardswhich exertsa normalforceon thepersonthatpreventstheperson from falling.

Discussion

15.	During blood transfusion the needleis inserted in a vein where the gauge pressure is 2000 Pa . At what height must the blood container be placed so that blood may just enter the vein ? [Use the density of whole blood from Table 10.1]
Answer» Solution :Here gauge PRESSURE `P=2000Pa` Density of BLOOD `rho=1.06xx10^(3)kgm^(-3)` `g=9.8ms^(-2)` The HEIGHT of vein of blood container =h Gauge pressure `P=grhog` `h=(P)/(rhog)` `=(2000)/(1.06xx10^(3)xx9.8)` `=192.5xx10^(-3)` `=0.192m` `thereforeh=2.0m`

Discussion

16.	The M.I. of a thin rod about a normal axis through its centre is I. It is bent at the centre such that, the two parts are perpen-dicular to each other and perpendicular to the axis. The M.I. of the system about the same axis will be
Answer» 2I I I/2 4I Answer :B

Discussion

17.	The Young's modulus of steal is 210 GPa. Its poissions ratio is 0.3. Then its rigidity modulus is
Answer» 8.8GPa 80.8GPa 19.6GPa 6.6GPa Answer :2

Discussion

18.	The rotational analogue of force is
Answer» energy work inertia torque Answer :D

Discussion

19.	Two droplets each radius r of mercury mergeand form a bigger drop . If surface tension of mercury is T.Then what will be the surface energy ?
Answer» `2^((5)/(3))pir^(2)T` `4pir^(2)T` `2pir^(2)T` `2^((8)/(3)pir^(2))T` Solution :Radius of small droplet is R and volume is `V_(1)` when TWO droplets are merged and form a bigger drop having radius R and volume is `V_(2)` `thereforeV_(2)=V_(1)xx2` `(4)/(3)piR^(3)=2XX(4)/(3)pir^(3)` `R^(3)=2r^(3)` `R=2^((1)/(3))xxr`...(1) Now , surface enerfy of bigger drop, `E=4piR^(2)T=4pi(2^((2)/(3))*r^(2))T=pixx2^(2)xx2^((2)/(3))xxr^(2)T` `=pixx2^(2+(2)/(3))xxr^(2)T` `thereforeE=2^((8)/(3))pir^(2)T`

Discussion

20.	The amount of work done in pumping water out of a cubical vessel of height 1 m is nearly
Answer» 5000 J 10000 J 50 J 10 J Solution :`W = 1 m xx 1 m xx 1m xx 1000 kg m^(-3) xx 10 ms^(-2) xx 1/2 m` `=5000 J`

Discussion

21.	Which unit can be used for measurement of very small masses ?
Answer» 1 atomic MASSS unit (a.m.u) `= 1.66xx10^(-27)KG.` 1 atomic masss unit (a.m.u) `= 1.66xx10^(27)kg.` 1 atomic mass unit (a.m.u) `= 1.66xx10^(28)kg.` 1 atomic mass unit (a.m.u) `= 1.7xx10^(26)kg.` Answer :B

Discussion

22.	A person walking in a straight line, covers half of the distance to be travelled with a speed of v_(0). For half of the time, required to complete the remaining distance, he walks with a speed of v_(1)and for the remaining half time his speed is v_(2)What is the person's average speed during this complete walk ?
Answer» Solution :SUPPOSE, the total walking DISTANCE is d, time taken to travel half of the distance `((d)/(2))` is `t _(1)` and time taken to travel remaining half distance is 2t. Now, distance TRAVELLED = average speed `xx` time `therefore d/2 = v _(0) t _(1)` and HENCE `t _(1) = (d)/(2v _(0)) ""...(1)` and `d/2 = v _(1) t + v _(2)t` (for remaining half distance) `= (v_(1) +v_(2))t ` `therefore 2t = (d)/(v _(1) + v _(2))` `(because t = 2t` for that half distance) ...(ii) Thus, tota time taken to walk total distance d is `t +2t.` Now the person.s average speed `lt v GT = (d)/(t _(1) + 2t)` `= (d)/(((d)/(2 v _(0))) +((d)/( v _(1) + v _(2))))` (From eq. (i) & (ii)) `therefore lt v gt = (2 v _(0) (v _(1) + v _(2)))/( v _(1) + v _(2) + 2v _(0))`

Discussion

23.	A real gas behaves as an ideal gas at
Answer» HIGH TEMPERATURE and high PRESSURE high temperature and LOW pressure low temperature and low pressure low temperature and high pressure Answer :A::D

Discussion

24.	A disc of mass 'm' and radius R has a concentration hole of radius 'r'. Its M.I. about an axis through centre and normal to plane is
Answer» `m/2(R-r)^(2)` `m/2(R^(2)-r^(2))` `m/2(R+r)^(2)` `m/2(R^(2)+r^(2))` ANSWER :D

Discussion

25.	A thread is tied slightly loose to a wire frame as in figure and the frame is dropped into a soap solution and taken out. The frame is completely covered with the film. When the portion A is punctured with a pin, the thread
Answer» BECOMES CONCAVE towards A Becomes convex towards A Either (a) and (b) DEPENDING on the size of A with respect to B Remains in the INITIAL position Answer :A

Discussion

26.	A force of 98 N is just required to move a mass of 45 kg on a rough horizontal surface. Find the coefficient of friction and angle of friction?
Answer» SOLUTION : `F= 48 N, R = 45 xx 9.8 = 441 N` ` mu = (F.)/(R ) = 0.22` ANGLE of friction ` theta = TAN^(-1) mu= tan^(-1) 0.22 =12^@ 24.`

Discussion

27.	A girl is swinging in the sitting position.How will the period of the swing change if she stands up?
Answer» Solution :The girl and the swing together constitute a PENDULUM of TIME period `T=2pisqrti/g`. As the girl stands up,her CENTRE of gravity is raised.The DISTANCE between the point of suspension and the centre of gravity decreases.i.e. lenght I decresese.Hence the time period T decreases.

Discussion

28.	An object moving along a straight line covers equal distances in equal intervals of time, it is said to be in uniform motion along a straight line.(c) Which of the folloming statements is/ are true?
Answer» An object-with CONSTANT VELOCITY has always INSTANT speed. An object with constant speed has always constant velocity. An object with zero velocity has always zero ACCELERATION An object with zero acceleration has always zero velocity. Answer :A::B::C::D

Discussion

29.	The value of coefficient of volume expansion of glycerin is 5 xx 10^(-4) K^(-1). The fractional change in the density of glycerin for a rise of 40^@ C in its temperature is _____
Answer» `0.010` `0.015` `0.020` `0.025` Solution :SUPPOSE final DENSITY is `d_f` and initial density is `d_i`. `therefore d_f=(di)/(1+gammaDeltaT)rArr d_i/d_f =(1-gammaDeltaT)`…(1) Fractional change in density =`(d_i-d_f)/d_i` `=1-d_f/d_i` `=1-(1-gamma DeltaT)^(-1)` USING binomial theorem for EXPANSION from equation (1), `=1-(1-gammaDeltaT)` `=gamma DeltaT` `=5xx10^(-4)xx40` `=2XX10^(-2)` =0.020

Discussion

30.	An object moving along a straight line covers equal distances in equal intervals of time, it is said to be in uniform motion along a straight line. The position-time graph of an object in uniform motion is…….
Answer» a STRAIGHT line PARALLEL to the time AXIS . straight line parallel to the position axis a straight line inclined to the time axis a parabola Answer :A::C::D

Discussion

31.	A bullet fired at an angle of 60^(@) with the vertical hits the ground at a distance of 2 km. Calculate the distance at which the bullet will hit the ground when fired at an angle of 45^(@), assuming the speed to be the same.
Answer» ANSWER :`(2.31 KM)`

Discussion

32.	Three samples of the same gas x, y and z, for which the ratio of specific heats is r=3/2, have initially the same volume. The volumes of each sample is doubled , by adiabatic process in the case x, by iosbaric process in the case of y and by isothermal process in the case of z. If the initial pressures of the samples x,y and z are in the ratio 2 sqrt2 :1 :2, then the ratio of their final pressure is
Answer» `2:1` `1:1:1` `1:2:1` `1:1:2` ANSWER :B

Discussion

33.	When a source of light and an observer from each other, the apparent change in the wavelenght of light is called ________ shift . [Fill in the blank]
Answer» SOLUTION :N/A

Discussion

34.	A girl is swinging on a swing in the sitting position. How will the period of swing be affected if she stands up?
Answer» Solution : This can be explained using the concept of a simple pendulum. We know that the time period of a simple pendulum is given by ` T = 2PI sqrt(l/g) i.e., T PROP sqrtl` When the girl stands up, the distance between the POINT of suspension and the centre of mass of the swinging BODY decreases i.e., I decreases, so I will also decrease.

Discussion

35.	Obtain the formula of mechanical energy of damped oscillation for b lt lt sqrt(km).
Answer» Solution :Mechanical energy of damped oscillatior, in `THEREFORE E(t)= (1)/(2) KA^(2) e^(-(bt)/(m)), b lt lt sqrt(km)` then `(b)/(sqrt(km)) lt lt 1`, by TAKING b=0 `therefore E(t) = (1)/(2) kA^(2)`.

Discussion

36.	If S is the stress and Y is young's modulus of the material of a wire , then the energy stored in the wire per unit volume is
Answer» `1/2S^2/Y` `2Y/S^2` `S/2Y` `2S^2Y` ANSWER :A

Discussion

37.	How does the energy of a simple pendulum vary as it moves from one extreme to the other during its oscillations?
Answer» Solution :The total energy of a body in S.H.M. is constant. But the CONTENT of PE and KE changes at various positions. At extreme POSITION PE is maximum and KE is zero. At MEAN position KE is maximum and PE is zero.

Discussion

38.	Two sound waves moves in the same direction in the same medium. The pressure amplutides of the waves are equal but the wavelength of the first wave is double the second. Let the average power transmitted across a cross section by the first wave be P_1 and that by the second wave be P_2 then.
Answer» `P_1=P_2` `P_1=4P_2` `P_2=2P_1` `P_2=4P_1` ANSWER :A

Discussion

39.	A particle is projected vertically upwards. Prove that it will be at3//4 of its greatest height at times which are in the ratio 1:3
Answer» Solution :If U is the initial velocity of a PARTICLE while going vertically upwards then maximum height attained , `h=u^(2)/(2g)` If t is the time when particle reaches at a height `(3//4)`h, then using the RELATION `s=ut+1/2at^(2) ` we have `3/4h=1/2ut+(-g)t^(2)` (or)`3/4(u^(2)/(2g))=ut-1/2"gt"^(2)"(or)" t^(2)-(2u)/"gt"+3/4(u^(2))/(g^(2))=0` Solving it for t , we have `t(2u)/g+-sqrt((4U^(2)/g^(2)-4xx1xx3/4u^(2)/g^(2))/2)=u/g+-u/(2g)` Taking negative sign , `t_(1)=u/g-u/(2g)=(3u)/(2g)` Taking positive sign `=t_(2)=u/g+u/(2g)=(3u)/(2g)` `:.t_(1)/t_(2)=((u/2g)/(3u/2g))=1/3`

Discussion

40.	Distinguish between Geocentric model and Heliocentric model.
Answer» SOLUTION :

Discussion

41.	A projectile moving vertically upwards with a velocity of 200 m/s breaks into two equal parts at the height of 490 m.one part starts moving vertically upwards with a velocity of 400 m/s.How much time after the break up wil the other part hit the ground?
Answer» `SQRT10 s ` `2sqrt10 s ` 5s 10s ANSWER :D

Discussion

42.	A body is projected from height of 60 m with a velocity 10 ms^(-1) at angle 30^(@) to horizontal. The time of flight of the body is [g = 10ms^(-2)].
Answer» 1s 2s 3s 4s Answer :D

Discussion

43.	Consider the quantites: pressure, power, energy, impulse, gravitational potential, electrical charge, temperature, area. Out of these, the only vector quantities are
Answer» IMPULSE, PRESSURE and AREA Impulse and area Area and gravittational potantial IMPLUSE and pressure Answer :B

Discussion

44.	Water is flowing through a horizontal pipeof variable cross-sectional. The rate of volume flow is 0.2 lit/sec. The pressure is 30,000 Pa. at the point where the area of cross - section is 5xx10^(-5)m^(2). The pressure at the point where the area of cross-section is 10^(-4) is
Answer» 6000 Pa 36000 Pa 24000 Pa 30000 Pa Answer :B

Discussion

45.	The span area of the wings of an aeroplane is 50 m^(2). When it is running on the runway the pressure of air at the bottom and top of the wings are found to be 0.75x10^(5) pascal and 0.70x10^(5) pascal respectively. Find the dynamic lift on the wings.
Answer» SOLUTION :`2.5xx10^(5)N`

Discussion

46.	The correct order in which the dimensions of time decreases in the following physical quantities is 1. Stefan's constant 2. Coefficient of volume expansion 3. Work done 4. Velocity gradient
Answer» 2,4,3,1 1, 2, 3, 4 4,3,2,1 1, 2, 4, 3 Solution :Stefan.s coefficient `=("Energy")/("Time" xx "Area" xx ("Temperature)^(4))` `= ([ML^(2)T^(-2)])/([T][L^(2)][K]^(4))= [ML^(0)T^(-3)K^(-4)]` 2. Coefficient of volume expansion= `("Change in volume")/("Original Volume" xx "Temperature")= ([L^(2)])/([L^(2)][K])= [M^(0)L^(0)T^(0)K^(-1)]` 3. `"Work done" = "force" xx "Distance" = [MLT^(-2)][L]= [ML^(2)T^(-2)]` 4. VELOCITY gradient= `("Velocity")/("Distance")` `= ([LT^(-1)])/([L])= [T^(-1)]= [M^(0)L^(0)T^(-1)]` The correct order is (a): 2,4, 3, 1

Discussion

47.	The change in internal energy of the gas andworkdone by the gas in an adiabatic process are
Answer» UNEQUAL EQUAL Only NUMERICALLY equal No relation. Answer :C

Discussion

48.	If unit of mass is taken as 1 kg, of time a 1 minute and that acceleration due to gravity is taken as 9.81 ms^(-2) . What is the unit of energy?
Answer» Solution :New unit of ENERGY `=E_1` New unit of mass `M_(1) = 1 kg` New unit of time `T_(1) = 1 "MINUTE" = 60 "sec"` New unit of LENGTH `= L_(1)` `E_(1) = [M_(1) L_(1) T_(1)^(-2) ]` New unit of acceleration due to gravity, `g_(1) = 9.81 ms^(-2)` `g_(1) = L_(1)T_(1) ^(-2)L_(1) = g_(1) T_(1)^(-2) = 9.81 xx (60)^(2) "meter" ` `E_(1) = [ 1 kg xx (9.81 xx 60^(2) )^(2) m^(2) (60)^(-2) S^(-2) ]` `E_(1) = (1 xx 9.81 xx 60 xx 60 xx 9.81 xx 60 xx 60)/( 60 xx 60) "kgm"^(2) s^(2)` `E_(1) = 3.464 xx 10^(5) "kg" m^(2) sec^(-2)` `THEREFORE` The new unit of energy `E_(1) = 3.464 xx 10^(5)` joule.

Discussion

49.	The velocity-time graph of a particle moving along a srtaight line is as shown in fig. The rate of acceleration and deceleration is constant and it is equal to5ms^(-2).If the avrage velocity during the motion is 20ms^(-1), then The maximum velocity of the particle is
Answer» 5 s 10 s 20s `5sqrt(2)s` ANSWER :A

Discussion

50.	The velocity-time graph of a particle moving along a srtaight line is as shown in fig. The rate of acceleration and deceleration is constant and it is equal to5ms^(-2).If the avrage velocity during the motion is 20ms^(-1), then The distance travelled with uniform velocity is
Answer» 375m 125 m 300 m 450 m Answer :A

Discussion

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A bullet fired at an angle of 60^(@) with the vertical hits the ground at a distance of 2 km. Calculate the distance at which the bullet will hit the ground when fired at an angle of 45^(@), assuming the speed to be the same.

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A particle is projected vertically upwards. Prove that it will be at3//4 of its greatest height at times which are in the ratio 1:3

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A body is projected from height of 60 m with a velocity 10 ms^(-1) at angle 30^(@) to horizontal. The time of flight of the body is [g = 10ms^(-2)].

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Water is flowing through a horizontal pipeof variable cross-sectional. The rate of volume flow is 0.2 lit/sec. The pressure is 30,000 Pa. at the point where the area of cross - section is 5xx10^(-5)m^(2). The pressure at the point where the area of cross-section is 10^(-4) is

The span area of the wings of an aeroplane is 50 m^(2). When it is running on the runway the pressure of air at the bottom and top of the wings are found to be 0.75x10^(5) pascal and 0.70x10^(5) pascal respectively. Find the dynamic lift on the wings.

The correct order in which the dimensions of time decreases in the following physical quantities is 1. Stefan's constant 2. Coefficient of volume expansion 3. Work done 4. Velocity gradient

The change in internal energy of the gas andworkdone by the gas in an adiabatic process are

If unit of mass is taken as 1 kg, of time a 1 minute and that acceleration due to gravity is taken as 9.81 ms^(-2) . What is the unit of energy?

The velocity-time graph of a particle moving along a srtaight line is as shown in fig. The rate of acceleration and deceleration is constant and it is equal to5ms^(-2).If the avrage velocity during the motion is 20ms^(-1), then The maximum velocity of the particle is

The velocity-time graph of a particle moving along a srtaight line is as shown in fig. The rate of acceleration and deceleration is constant and it is equal to5ms^(-2).If the avrage velocity during the motion is 20ms^(-1), then The distance travelled with uniform velocity is