11 Physics Sample Papers Solved 05

8/13/2019 11 Physics Sample Papers Solved 05

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Solved Paper 5

Class 11, Physics

Time: 3 hours Max. Marks 70

General Instructions1. All questions are compulsory. Symbols have their usual meaning.2. Use of calculator is not permitted. However you may use log table, if required.3. Draw neat labeled diagram wherever necessary to explain your answer.4. Q.No. 1 to 7 are of very short answer type questions, carrying 1 mark each.5. Q.No.8 to 19 are of short answer type questions, carrying 2 marks each.6. Q. No. 20 to 27 carry 3 marks each. Q. No. 28 to 30 carry 5 marks each.

1. Arrange four fundamental forces in decreasing order of strength.

2. State the number of significant figures in the following:

(a) 0.007 m 2 (b) 2.64 10 24 kg

3. Draw Velocitytime graph for motions with positive constant acceleration inpositive Direction.

4. For what value of a does the vectors

2 A ai j k = +

and

2 4 B ai aj k = +

areperpendicular to each other.

5. The casing of a rocket in flight burns up due to friction. At whose expense is theheat energy required for burning obtained? The rocket or the atmosphere?

6. Give the location of the centre of mass of a cube of uniform mass density

7. There were two fixed points in the original Celsius scale as mentioned above whichwere assigned the number 0 C and 100 C respectively. On the absolute scale, oneof the fixed points is the triple-point of water, which on the Kelvin absolute scale isassigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale ?

8. We measure the period of oscillation of a simple pendulum. In successivemeasurements, the readings turn out to be 2.63 s, 2.56 s, 2.42 s, 2.71s and 2.80 s.Calculate the absolute errors, relative error or percentage error.


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9. An aircraft is flying at a height of 3400 m above the ground. If the angle subtended

at a ground observation point by the aircraft positions 10.0 s apart is 30, what is thespeed of the aircraft?

10. A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of theshell is 80 ms 1, what is the recoil speed of the gun?

11. The speed-time graph of a particle moving along a fixed direction is shown in Fig.Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2 s to6 s.

What is the average speed of the particle over the intervals in (a) and (b)?

12. A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s 1.The radius of the cylinder is 0.25 m. What is the kinetic energy associated with therotation of the cylinder? What is the magnitude of angular momentum of thecylinder about its axis?

13. A rain drop of radius 2 mm falls from a height of 500 m above the ground. It fallswith decreasing acceleration (due to viscous resistance of the air) until at half itsoriginal height, it attains its maximum (terminal) speed, and moves with uniformspeed thereafter. What is the work done by the gravitational force on the drop in thefirst and second half of its journey? What is the work done by the resistive force inthe entire journey if its speed on reaching the ground is 10 m s 1?

14. Two wires of diameter 0.25 cm, one made of steel and the other made of brass areloaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that ofbrass wire is 1.0 m. Compute the elongations of the steel and the brass wires.


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15. A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 C and

then placed on a large ice block. What is the maximum amount of ice that can melt?(Specific heat of copper = 0.39 J g 1 K1; heat of fusion of water = 335 J g 1).

16. In changing the state of a gas adiabatically from an equilibrium state A to anotherequilibrium state B, an amount of work equal to 22.3 J is done on the system. If thegas is taken from state A to B via a process in which the net heat absorbed by thesystem is 9.35 cal, how much is the net work done by the system in the latter case?(Take 1 cal = 4.19 J)

17. What amount of heat must be supplied to 2.0 10 2 kg of nitrogen (at roomtemperature) to raise its temperature by 45 C at constant pressure? (Molecular massof N 2 = 28; R = 8.3 J mol

1 K1.)

18. Explain why

(a) To keep a piece of paper horizontal, you should blow over, not under, it(b) When we try to close a water tap with our fingers, fast jets of water gush

through the openings between our fingers

19. Let us assume that our galaxy consists of 2.5 10 11 stars each of one solar mass.How long will a star at a distance of 50,000 ly from the galactic centre take to

complete one revolution? Take the diameter of the Milky Way to be 105

ly.

20. A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s 1 to3.5 m s 1 in 25 s. The direction of the motion of the body remains unchanged. Whatis the magnitude and direction of the force?

21. The position of a particle is given by

Where t is in seconds and the coefficients have the proper units for r to be in metres.(a) Find the v and a of the particle?(b) What is the magnitude and direction of velocity of the particle at t = 2.0 s?

22. The oxygen molecule has a mass of 5.30 10 26 kg and a moment of inertiaof 1.9410 46 kg m 2about an axis through its centre perpendicular to the lines

joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500m/s and that its kinetic energy of rotation is two thirds of its kinetic energy oftranslation. Find the average angular velocity of the molecule.


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23. Two inclined frictionless tracks, one gradual and

the other steep meet at A from where two stonesare allowed to slide down from rest, one on eachtrack. Will the stones reach the bottom at the

same time? Will they reach there with the samespeed? Explain. Given 1 = 30, 2 = 60, and h =10 m, what are the speeds and times taken by thetwo stones?

24. A brass boiler has a base area of 0.15 m 2 and thickness 1.0 cm. It boils water at therate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the partof the flame in contact with the boiler. Thermal conductivity of brass = 109 J s 1 m1 K1; Heat of vaporisation of water = 2256 10 3 J kg 1 .

25. Answer the following questions:

(a) Time period of a particle in SHM depends on the force constant k andmass m of the particle:

. A simple pendulum executes SHM approximately. Why then is thetime period of a pendulum independent of the mass of the pendulum?

(b) The motion of a simple pendulum is approximately simple harmonic for small

angle oscillations. For larger angles of oscillation, a more involved analysis

shows that T is greater than . Think of a qualitative argument toappreciate this result.

(c) A man with a wristwatch on his hand falls from the top of a tower. Does thewatch give correct time during the free fall?

(d) What is the frequency of oscillation of a simple pendulum mounted in acabinthat is freely falling under gravity?

26. Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in itslevels in the two limbs of the tube? Surface tension of water at the temperature ofthe experiment is 7.3 10 2 N m 1. Take the angle of contact to be zero and densityof water to be 1.0 10 3 kg m 3 (g = 9.8 m s 2).

27. A spaceship is stationed on Mars. How much energy must be expended on thespaceship to launch it out of the solar system? Mass of the space ship = 1000 kg;mass of the Sun = 2 10 30 kg; mass of mars = 6.4 10 23 kg; radius of mars = 3395km; radius of the orbit of mars = 2.28 10 8kg; G= 6.67 10 11 m2kg 2.


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28. A thin circular loop of radius R rotates about its vertical diameter with an angular

frequency . Show that a small bead on the wire loop remains at its lowermost point

for .What is the angle made by the radius vector joining the centre to the

bead with the vertical downward direction for ?Neglect friction.

29. (i) A metre-long tube open at one end, with a movable piston at the other end,shows resonance with a fixed frequency source (a tuning fork of frequency 340Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound inair at the temperature of the experiment. The edge effects may be neglected.

(ii) You have learnt that a travelling wave in one dimension is represented by afunction y = f ( x, t )where x and t must appear in the combination x v t or x + v

t , i.e. y = f ( x v t ). Is the converse true? Examine if the following functionsfor y can possibly represent a travelling wave:

(a) ( x vt )2

(b)

(c)

30. A gas in equilibrium has uniform density and pressure throughout its volume. Thisis strictly true only if there are no external influences. A gas column under gravity,for example, does not have uniform density (and pressure). As you might expect, itsdensity decreases with height. The precise dependence is given by the so-called lawof atmospheres

n2 = n1 exp [- mg (h2 h1)/ k BT ]Where n2, n1 refer to number density at heights h2 and h1 respectively. Use thisrelation to derive the equation for sedimentation equilibrium of a suspension in aliquid column:n2 = n1 exp [- mg N A( - P ) (h2 h1)/ ( RT )]Where is the density of the suspended particle, and that of surrounding medium.[ N A is Avogadros number, and R the universal gas constant.] [Hint: UseArchimedes principle to find the apparent weight of the suspended particle.]


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Solved Paper 5

Class 11, Physics

Solutions

1: F g:Fw:Fe:Fs::1:1025:10 36:10 38

2: (a) Answer: 1The given quantity is 0.007 m 2.If the number is less than one, then all zeros on the right of the decimal point(but left to the first non-zero) are insignificant. This means that here, two zerosafter the decimal are not significant. Hence, only 7 is a significant figure in thisquantity.

(b) Answer: 3The given quantity is 2.64 10 24 kg.Here, the power of 10 is irrelevant for the determination of significant figures.Hence, all digits i.e., 2, 6 and 4 are significant figures.

3:

4: _given A B

( ) ( )2

2

2 2 4 0

2 2 4 0

2 0

2, 1

ai j k ai aj k

a a

a a

a

+ + =

=

=

=

i

5: The burning of the casing of a rocket in flight (due to friction) results in thereduction of the mass of the rocket.

According to the conservation of energy:


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The reduction in the rockets mass causes a drop in the total energy. Therefore, theheat energy required for the burning is obtained from the rocket.

6: The centre of mass (C.M.) is a point where the mass of a body is supposed to beconcentrated. For the given geometric shapes having a uniform mass density, theC.M. lies at their respective geometric centres.

7: The absolute zero or 0 K is the other fixed point on the Kelvin absolute scale.

8: 2.63 2.56 2.42 2.71 2.805

t + + + + +

=

1

2

3

4

5

2.624 2.62

_ _ _

2.63 2.62 0.01

2.56 2.62 0.06

2.42 2.62 0.20

2.71 2.62 0.09

2.80 2.62 0.18

_ _

0.01 0.06 0.20 0.09 0.185

0.11 0.1

_

m

m

t s

error in the measurment

t

t

t

t

t

mean absolute error

t

t s s

relative err

=

= =

= =

= =

= =

= =

+ + + + =

=

0.10.04

2.62% _ 100 4%

m

m

t or

t t error

t

= = =

= =

9: The positions of the observer and the aircraft are shown in the given figure.


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Height of the aircraft from ground, OR = 3400 mAngle subtended between the positions, POQ = 30Time = 10 sIn PRO:

PRO is similar to RQO. PR = RQPQ = PR + RQ= 2PR = 2 3400 tan 15= 6800 0.268 = 1822.4 m

Speed of the aircraft

10: Mass of the gun, M = 100 kg

Mass of the shell, m = 0.020 kgMuzzle speed of the shell, v = 80 m/sRecoil speed of the gun = V Both the gun and the shell are at rest initially.Initial momentum of the system = 0Final momentum of the system = mv MV

Here, the negative sign appears because the directions of the shell and the gun areopposite to each other.According to the law of conservation of momentum:Final momentum = Initial momentummv MV = 0


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11: (a) Distance travelled by the particle = Area under the given graph

Average speed =(b) Let s1 and s2 be the distances covered by the particle between timet = 2 s to 5 s and t = 5 s to 6 s respectively.Total distance ( s) covered by the particle in time t = 2 s to 6 ss = s1 + s2 (i)For distance s 1: Let u be the velocity of the particle after 2 s and a be the acceleration of theparticle in t = 0 to t = 5 s.Since the particle undergoes uniform acceleration in the interval t = 0 to t = 5 s,

from first equation of motion, acceleration can be obtained as:v = u + at Where,v = Final velocity of the particle12 = 0 + a 5

Again, from first equation of motion, we havev = u + at = 0 + 2.4 2 = 4.8 m/sDistance travelled by the particle between time 2 s and 5 s i.e., in 3 s

For distance s 2: Let a be the acceleration of the particle between time t = 5 s and t = 10 s.From first equation of motion,v = u + at (where v = 0 as the particle finally comes to rest)0 = 12 + a 5

Distance travelled by the particle in 1s (i.e., between t = 5 s and t = 6 s)


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From equations (i), (ii), and (iii), we gets = 25.2 + 10.8 = 36 m

12: Mass of the cylinder, m = 20 kg

Angular speed, = 100 rad s 1 Radius of the cylinder, r = 0.25 mThe moment of inertia of the solid cylinder:

Kinetic energy

Angular momentum, L = I = 6.25 100= 62.5 Js

13: Radius of the rain drop, r = 2 mm = 2 10 3 m

Volume of the rain drop,

Density of water, = 10 3 kg m 3

Mass of the rain drop, m = V

=Gravitational force, F = mg

=The work done by the gravitational force on the drop in the first half of its journey:W I = Fs

= 250= 0.082 J


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This amount of work is equal to the work done by the gravitational force on the dropin the second half of its journey, i.e., W II, = 0.082 JAs per the law of conservation of energy, if no resistive force is present, then thetotal energy of the rain drop will remain the same.

Total energy at the top: E T = mgh + 0

= 500 10 5 = 0.164 JDue to the presence of a resistive force, the drop hits the ground with a velocity of10 m/s. Total energy at the ground:

Resistive force = E G E T = 0.162 J

14: Elongation of the steel wire = 1.49 10 4 m

Elongation of the brass wire = 1.3 10 4 mDiameter of the wires, d = 0.25 m

Hence, the radius of the wires, = 0.125 cmLength of the steel wire, L1 = 1.5 mLength of the brass wire, L2 = 1.0 mTotal force exerted on the steel wire:F 1 = (4 + 6) g = 10 9.8 = 98 N

Youngs modulus for steel:

Where, L1 = Change in the length of the steel wire

A1 = Area of cross-section of the steel wireYoungs modulus of steel, Y 1 = 2.0 10

11 Pa


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Total force on the brass wire: F 2 = 6 9.8 = 58.8 NYoungs modulus for brass:

Elongation of the steel wire = 1.49 10 4 mElongation of the brass wire = 1.3 10 4 m

15: Mass of the copper block, m = 2.5 kg = 2500 g

Rise in the temperature of the copper block, = 500C

Specific heat of copper, C = 0.39 J g1

C1

Heat of fusion of water, L = 335 J g 1 The maximum heat the copper block can lose, Q = mC = 2500 0.39 500= 487500 JLet m1 g be the amount of ice that melts when the copper block is placed on the iceblock.The heat gained by the melted ice, Q = m1 L

Hence, the maximum amount of ice that can melt is 1.45 kg.


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16: The work done ( W ) on the system while the gas changes from state A to state B is

22.3 J.

This is an adiabatic process. Hence, change in heat is zero. Q = 0W = 22.3 J (Since the work is done on the system)From the first law of thermodynamics, we have:Q = U + W Where,U = Change in the internal energy of the gas U = Q W = ( 22.3 J)U = + 22.3 JWhen the gas goes from state A to state B via a process, the net heat absorbed by thesystem is:Q = 9.35 cal = 9.35 4.19 = 39.1765 JHeat absorbed, Q = U + Q W = Q U = 39.1765 22.3= 16.8765 JTherefore, 16.88 J of work is done by the system.

17: Mass of nitrogen, m = 2.0 10 2 kg = 20 g

Rise in temperature, T = 45CMolecular mass of N 2, M = 28Universal gas constant, R = 8.3 J mol 1 K1

Number of moles,

Molar specific heat at constant pressure for nitrogen,

The total amount of heat to be supplied is given by the relation:Q = nC P T = 0.714 29.05 45= 933.38 JTherefore, the amount of heat to be supplied is 933.38 J.


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18: (a) When air is blown under a paper, the velocity of air is greater under the paper

than it is above it. As per Bernoullis principle, atmospheric pressure reducesunder the paper. This makes the paper fall. To keep a piece of paper horizontal,one should blow over it. This increases the velocity of air above the paper. As

per Bernoullis principle, atmospheric pressure reduces above the paper andthe paper remains horizontal.

(b) According to the equation of continuity:Area Velocity = Constant

For a smaller opening, the velocity of flow of a fluid is greater than it is when theopening is bigger. When we try to close a tap of water with our fingers, fast jets ofwater gush through the openings between our fingers. This is because very smallopenings are left for the water to flow out of the pipe. Hence, area and velocity areinversely proportional to each other.

19: Mass of our galaxy Milky Way, M = 2.5 10 11 solar mass

Solar mass = Mass of Sun = 2.0 10 36 kgMass of our galaxy, M = 2.5 10 11 2 10 36 = 5 10 41 kgDiameter of Milky Way, d = 10 5 lyRadius of Milky Way, r = 5 10 4 ly1 ly = 9.46 10 15 m r = 5 10 4 9.46 10 15 = 4.73 10 20 mSince a star revolves around the galactic centre of the Milky Way, its time period isgiven by the relation:


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20: 0.18 N; in the direction of motion of the body

Mass of the body, m = 3 kgInitial speed of the body, u = 2 m/sFinal speed of the body, v = 3.5 m/sTime, t = 25 sUsing the first equation of motion, the acceleration ( a ) produced in the body can becalculated as:v = u + at

As per Newtons second law of motion, force is given as:F = ma = 3 0.06 = 0.18 NSince the application of force does not change the direction of the body, the netforce acting on the body is in the direction of its motion.

21: (a)

The position of the particle is given by:

Velocity , of the particle is given as:

Acceleration , of the particle is given as:

(b) 8.54 m/s, 69.45 below the x-axis

The magnitude of velocity is given by:


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The negative sign indicates that the direction of velocity is below the x-axis.

22: Mass of an oxygen molecule, m = 5.30 10 26 kg

Moment of inertia, I = 1.94 10 46 kg m 2 Velocity of the oxygen molecule, v = 500 m/sThe separation between the two atoms of the oxygen molecule = 2 r

Mass of each oxygen atom =Hence, moment of inertia I , is calculated as:

It is given that:


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23: No; the stone moving down the steep plane will reach the bottom first

Yes; the stones will reach the bottom with the same speedvB = vC = 14 m/st 1 = 2.86 s; t 2 = 1.65 sThe given situation can be shown as in the following figure:

Here, the initial height (AD) for both the stones is the same ( h). Hence, both willhave the same potential energy at point A.As per the law of conservation of energy, the kinetic energy of the stones at points Band C will also be the same, i. e.,

v1 = v2 = v, sayWhere,m = Mass of each stone

v = Speed of each stone at points B and CHence, both stones will reach the bottom with the same speed, v.For stone I:Net force acting on this stone is given by:

For stone II:

Using the first equation of motion, the time of slide can be obtained as:

For stone I:

For stone II:


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Hence, the stone moving down the steep plane will reach the bottom first.The speed ( v) of each stone at points B and C is given by the relation obtained fromthe law of conservation of energy.

The times are given as:

24: Base area of the boiler, A = 0.15 m 2

Thickness of the boiler, l = 1.0 cm = 0.01 mBoiling rate of water, R = 6.0 kg/minMass, m = 6 kgTime, t = 1 min = 60 sThermal conductivity of brass, K = 109 J s 1 m1 K1 Heat of vaporisation, L = 2256 10 3 J kg 1 The amount of heat flowing into water through the brass base of the boiler is givenby:

Where,T 1 = Temperature of the flame in contact with the boilerT 2 = Boiling point of water = 100CHeat required for boiling the water: = mL ( ii)Equating equations ( i) and ( ii), we get:


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Therefore, the temperature of the part of the flame in contact with the boiler is237.98C.

25: (a) The time period of a simple pendulum,

For a simple pendulum, k is expressed in terms of mass m, as:k m

= ConstantHence, the time period T , of a simple pendulum is independent of the mass ofthe bob.

(b) In the case of a simple pendulum, the restoring force acting on the bob of thependulum is given as:F = mg sin Where,F = Restoring forcem = Mass of the bobg = Acceleration due to gravity = Angle of displacementFor small , sin For large , sin is greater than .This decreases the effective value of g.Hence, the time period increases as:

Where, l is the length of the simple pendulum

(c) The time shown by the wristwatch of a man falling from the top of a tower isnot affected by the fall. Since a wristwatch does not work on the principle of a


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simple pendulum, it is not affected by the acceleration due to gravity duringfree fall. Its working depends on spring action.

(d) When a simple pendulum mounted in a cabin falls freely under gravity, itsacceleration is zero. Hence the frequency of oscillation of this simple

pendulum is zero.

26: Diameter of the first bore, d 1 = 3.0 mm = 3 103 m

Diameter of the second bore, = 6.0 mm

Surface tension of water, s = 7.3 10 2 N m 1 Angle of contact between the bore surface and water, = 0Density of water, =1.0 10 3 kg/m 3 Acceleration due to gravity, g = 9.8 m/s 2 Let h1 and h2 be the heights to which water rises in the first and second tubesrespectively. These heights are given by the relations:

The difference between the levels of water in the two limbs of the tube can becalculated as:

Hence, the difference between levels of water in the two bores is 4.97 mm.

27: Mass of the spaceship, ms = 1000 kg


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Mass of the Sun, M = 2 10 30 kgMass of Mars, mm = 6.4 10

23 kgOrbital radius of Mars, R = 2.28 10 8 kg =2.28 10 11mRadius of Mars, r = 3395 km = 3.395 10 6 m

Universal gravitational constant, G = 6.67 1011

m2

kg2

Potential energy of the spaceship due to the gravitational attraction of the

SunPotential energy of the spaceship due to the gravitational attraction of

MarsSince the spaceship is stationed on Mars, its velocity and hence, its kinetic energywill be zero.

Total energy of the spaceship

The negative sign indicates that the system is in bound state.Energy required for launching the spaceship out of the solar system= (Total energy of the spaceship)

28: Let the radius vector joining the bead with the centre make an angle , with thevertical downward direction.


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OP = R = Radius of the circle

N = Normal reactionThe respective vertical and horizontal equations of forces can be written as:mg = N cos ... ( i)

ml2

= N sin ( ii)In OPQ, we have:

l = Rsin ( iii)Substituting equation ( iii ) in equation ( ii), we get:m( Rsin ) 2 = N sin mR 2 = N ... ( iv)Substituting equation ( iv) in equation ( i), we get:mg = mR 2 cos

... ( v)

Since cos 1, the bead will remain at its lowermost point for , i.e.,

for

For orOn equating equations ( v) and ( vi), we get:

29: (i) Frequency of the turning fork, = 340 Hz

Since the given pipe is attached with a piston at one end, it will behave as apipe with one end closed and the other end open, as shown in the given figure.


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Such a system produces odd harmonics. The fundamental note in a closed pipeis given by the relation:

Where,

Length of the pipe,

The speed of sound is given by the relation:= 340 1.02 = 346.8 m/s

(ii) No;

(a) Does not represent a wave

(b) Represents a wave(c) Does not represent a waveThe converse of the given statement is not true. The essential requirement for afunction to represent a travelling wave is that it should remain finite for all valuesof x and t .Explanation:

(a) For x = 0 and t = 0, the function ( x vt )2 becomes 0.Hence, for x = 0 and t = 0, the function represents a point and not a wave.

(b) For x = 0 and t = 0, the function

Since the function does not converge to a finite value for x = 0 and t = 0,it represents a travelling wave.

(c) For x = 0 and t = 0, the function

Since the function does not converge to a finite value for x = 0 and t = 0,it does not represent a travelling wave.

30: According to the law of atmospheres, we have:

n2 = n1 exp [- mg (h2 h1)/ kBT ] ( i)Where,n1 is thenumber density at height h1, and n2 is the number density at height h2 mg is the weight of the particle suspended in the gas columnDensity of the medium = '

Density of the suspended particle = Mass of one suspended particle = m'


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Mass of the medium displaced = m Volume of a suspended particle = V According to Archimedes principle for a particle suspended in a liquid column, theeffective weight of the suspended particle is given as:

Weight of the medium displaced Weight of the suspended particle= mg m'g

Gas constant, R = k B N

( iii)Substituting equation ( ii) in place of mg in equation ( i) and then using equation ( iii),we get:n2 = n1 exp [- mg (h2 h1)/ k BT ]

= n1 exp [- ( h2 h1) ]

= n1 exp [- ( h2 h1) ]

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11 Physics Sample Papers Solved 05