MOCKUP Solved Paper−1 Class 11, Physics I wish you all the best, remember if no one else believes in you. I believe in you.

Prof Nanda : 931793SE1711

Time: 3 hours Max. Marks 70 General Instructions 1. 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 19are 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 increasing order of strength.

2. Express light year in parsec

3. Can an object be at rest as well as in motion at the same time?

4. Find the value of λ if vectors ˆˆ ˆ2A i j kλ= + +�

and ˆˆ ˆ4 2 2B i j k= − −�

are perpendicular to

each other.

5. Comets move around the sun in highly elliptical orbits. The gravitational force on thecomet due to the sun is not normal to the comet’s velocity in general. Yet the workdone by the gravitational force over every complete orbit of the comet is zero. Why ?

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

7. The absolute temperature (Kelvin scale) T is related to the temperature tc on theCelsius scale by tc = T – 273.15. Why do we have 273.15 in this relation, and not273.16 ?

8. Find the dimension of a b× in equation2b l

Pat

−= , where P is Power, l is length & t is

time.

9. A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. Byadjusting its angle of projection, can one hope to hit a target 5.0 km away? Assumethe muzzle speed to the fixed, and neglect air resistance.

10. A batsman deflects a ball by an angle of 45° without changing its initial speed whichis equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15kg.)

yCBSE

.com

11. Two towns A and B are connected by a regular bus service with a bus leaving ineither direction every T minutes. A man cycling with a speed of 20 km h–1 in thedirection A to B notices that a bus goes past him every 18 min in the direction of hismotion, and every 6 min in the opposite direction. What is the period T of the busservice and with what speed (assumed constant) do the buses ply on the road?

12. Show that the area of the triangle contained between the vectors a and b is one half ofthe magnitude of a × b.

13. The blades of a windmill sweep out a circle of area A. (a) If the wind flows at avelocity vperpendicular to the circle, what is the mass of the air passing through it intime t?(b) What is the kinetic energy of the air? (c) Assume that the windmill converts25% of the wind’s energy into electrical energy, and that A = 30 m2, v = 36 km/h andthe density of air is 1.2 kg m–3. What is the electrical power produced?

14. Figure shows the strain-stress curve for a given material. What are (a) Young’smodulus and (b) approximate yield strength for this material?

15. Two absolute scales A and B have triple points of water defined to be 200 A and 350B. What is the relation between TA and TB?

16. Explain why

(a) Surface tension of a liquid is independent of the area of the surface(b) Water with detergent dissolved in it should have small angles of contact.

17. Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on ahorizontal table. What is the gravitational force and potential at the mid point of theline joining the centers of the spheres? Is an object placed at that point in equilibrium?If so, is the equilibrium stable or unstable?

18. You have learnt that a travelling wave in one dimension is represented by afunction y = f (x, t)wherex and t must appear in the combination x – v t or x + v t, i.e. y

yCBSE

.com

= f (x ± v t). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave:

(a) (x – vt)2

(b)

(c)

19. Estimate the mean free path and collision frequency of a nitrogen molecule in acylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of anitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time themolecule moves freely between two successive collisions (Molecular mass of N2 =28.0 u).

20. A constant retarding force of 50 N is applied to a body of mass 20 kg moving initiallywith a speed of 15 ms–1. How long does the body take to stop?

21. A particle starts from the origin at t = 0 s with a velocity of and moves in

the x-y plane with a constant acceleration of .

(a) At what time is the x-coordinate of the particle 16 m? What is the y-coordinate of the particle at that time?

(b) What is the speed of the particle at the time?

22. From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre ofthe hole is atR/2 from the centre of the original disc. Locate the centre of gravity ofthe resulting flat body.

23. A family uses 8 kW of power. (a) Direct solar energy is incident on the horizontalsurface at an average rate of 200 W per square meter. If 20% of this energy can beconverted to useful electrical energy, how large an area is needed to supply 8 kW? (b)Compare this area to that of the roof of a typical house.

24. Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube ofradius 1.00 mm made of this glass is dipped in a trough containing mercury. By whatamount does the mercury dip down in the tube relative to the liquid surface outside?Surface tension of mercury at the temperature of the experiment is 0.465 N m–1.Density of mercury = 13.6 × 103 kg m–3.

yCBSE

.com

25. Two stars each of one solar mass (= 2× 1030 kg) are approaching each other for a headon collision. When they are a distance 109 km, their speeds are negligible. What is thespeed with which they collide? The radius of each star is 104 km. Assume the stars toremain undistorted until they collide. (Use the known value of G).

26. A metre-long tube open at one end, with a movable piston at the other end, showsresonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when thetube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at thetemperature of the experiment. The edge effects may be neglected.

27. Figure shows plot of PV/T versus Pfor 1.00×10–3 kg of oxygen gas at two differenttemperatures.

(a) What does the dotted plot signify? (b) Which is true: T1 > T2 or T1 < T2? (c) What is the value of PV/T where the curves meet on the y-axis? (d) If we obtained similar plots for 1.00 ×10–3 kg of hydrogen, would we get the

same value of PV/Tat the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low pressure high temperature region of the plot)? (Molecular mass of H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J mo1–1 K–1.)

28. (i) The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.

(ii) A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s–2. Calculate the initial thrust (force) of the blast.

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

(ii) A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36° C, calculate the coefficient of performance.

yCBSE

.com

30. (i) Answer the following questions:

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

. A simple pendulum executes SHM approximately. Why then is the time 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 to appreciate this result.

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

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

(ii) An air chamber of volume V has a neck area of cross section a into which a ball of mass m just fits and can move up and down without any friction (Fig.). Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal [see Fig.].

yCBSE

.com

Solved Paper−1 Class 11, Physics

Solutions

1: Fg:Fw:Fe:Fs::1:1025:1036:1038

2: light year/parsec = 9.46×1015/3.08×1016 =0.31

3: yes in reference to one body an object can be in motion but in reference to another body it may be at rest.

4: _given A B⊥� �

( ) ( )ˆ ˆˆ ˆ ˆ ˆ2 4 2 2 0

8 2 2 0

3

i j k i j kλ

λλ

⇒ + + − − =

⇒ − − =⇒ =

i

5: Gravitational force is a conservative force. Since the work done by a conservative force over a closed path is zero, the work done by the gravitational force over every complete orbit of a comet is zero.

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

7: The temperature 273.16 K is the triple point of water. It is not the melting point of ice. The temperature 0°C on Celsius scale is the melting point of ice. Its corresponding value on Kelvin scale is 273.15 K.

Hence, absolute temperature (Kelvin scale) T, is related to temperature tc, on Celsius scale as: tc = T – 273.15

8: 2_ _ _ _dimension of b dimension of l=

yCBSE

.com

( )( )

( )

2

22 2

1 2

1 2 2

_ _

_ __ _

_ _ _ _

_ _

_ _

[ ]

dimension of b L

dimension of bdimension of P

dimension of a dimension of t

LML T

dimension of a T

dimension of a M T

a b M L T

−

−

−

=

=

=

=× =

9: No

Range, R = 3 km Angle of projection, θ = 30° Acceleration due to gravity, g = 9.8 m/s2 Horizontal range for the projection velocity u0, is given by the relation:

The maximum range (Rmax) is achieved by the bullet when it is fired at an angle of 45° with the horizontal, that is,

On comparing equations (i) and (ii) , we get:

Hence, the bullet will not hit a target 5 km away.

10: The given situation can be represented as shown in the following figure. yCBSE

.com

Where, AO = Incident path of the ball OB = Path followed by the ball after deflection ∠AOB = Angle between the incident and deflected paths of the ball = 45° ∠AOP = ∠BOP = 22.5° = θ Initial and final velocities of the ball = v Horizontal component of the initial velocity = vcos θ along RO Vertical component of the initial velocity = vsin θ along PO Horizontal component of the final velocity = vcos θ along OS Vertical component of the final velocity = vsin θ along OP The horizontal components of velocities suffer no change. The vertical components of velocities are in the opposite directions. ∴Impulse imparted to the ball = Change in the linear momentum of the ball

Mass of the ball, m = 0.15 kg Velocity of the ball, v = 54 km/h = 15 m/s ∴Impulse = 2 × 0.15 × 15 cos 22.5° = 4.16 kg m/s

11: Let V be the speed of the bus running between towns A and B.

Speed of the cyclist, v = 20 km/h Relative speed of the bus moving in the direction of the cyclist = V – v = (V – 20) km/h

The bus went past the cyclist every 18 min i.e., (when he moves in the direction of the bus).

Distance covered by the bus = … (i)Since one bus leaves after every T minutes, the distance travelled by the bus will be

equal toBoth equations (i) and (ii) are equal.

Relative speed of the bus moving in the opposite direction of the cyclist = (V + 20) km/h

Time taken by the bus to go past the cyclist

yCBSE

.com

From equations (iii) and (iv), we get

Substituting the value of V in equation (iv), we get

12: Consider two vectors and , inclined at an angle θ, as shown in the following figure.

In ∆OMN, we can write the relation:

= 2 × Area of ∆OMK

∴Area of ∆OMK

13: Area of the circle swept by the windmill = A

Velocity of the wind = v Density of air = ρ (a) Volume of the wind flowing through the windmill per sec = Av

Mass of the wind flowing through the windmill per sec = ρAv Mass m, of the wind flowing through the windmill in time t = ρAvt

yCBSE

.com

(b) Kinetic energy of air

(c) Area of the circle swept by the windmill = A = 30 m2 Velocity of the wind = v = 36 km/h Density of air, ρ = 1.2 kg m–3 Electric energy produced = 25% of the wind energy

14: (a) It is clear from the given graph that for stress 150 × 106 N/m2, strain is 0.002.

∴Young’s modulus, Y

Hence, Young’s modulus for the given material is 7.5 ×1010 N/m2. (b) The yield strength of a material is the maximum stress that the material can

sustain without crossing the elastic limit. It is clear from the given graph that the approximate yield strength of this material is 300 × 106Nm/2 or 3 × 108 N/m2.

15: Triple point of water on absolute scaleA, T1 = 200 A

Triple point of water on absolute scale B, T2 = 350 B Triple point of water on Kelvin scale, TK = 273.15 K

yCBSE

.com

The temperature 273.15 K on Kelvin scale is equivalent to 200 A on absolute scale A. T1 = TK 200 A = 273.15 K

The temperature 273.15 K on Kelvin scale is equivalent to 350 B on absolute scale B. T2 = TK 350 B = 273.15

TA is triple point of water on scale A. TB is triple point of water on scale B.

Therefore, the ratio TA : TB is given as 4 : 7.

16: (a) Surface tension is the force acting per unit length at the interface between the plane of a liquid and any other surface. This force is independent of the area of the liquid surface. Hence, surface tension is also independent of the area of the liquid surface.

(b) Water with detergent dissolved in it has small angles of contact (θ). This is because for a small θ, there is a fast capillary rise of the detergent in the cloth. The capillary rise of a liquid is directly proportional to the cosine of the angle of contact (θ). If θ is small, then cosθ will be large and the rise of the detergent water in the cloth will be fast.

17: 0;

–2.7 × 10–8 J /kg;Yes; Unstable Explanation: The situation is represented in the given figure:

Mass of each sphere, M = 100 kg Separation between the spheres, r = 1m

yCBSE

.com

X is the mid point between the spheres. Gravitational force at point X will be zero. This is because gravitational force exerted by each sphere will act in opposite directions. Gravitational potential at point X:

Any object placed at point X will be in equilibrium state, but the equilibrium is unstable. This is because any change in the position of the object will change the effective force in that direction.

18: No;

(a) Does not represent a wave (b) Represents a wave (c) Does not represent a wave The converse of the given statement is not true. The essential requirement for a function to represent a travelling wave is that it should remain finite for all values of 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.

19: Mean free path = 1.11 × 10–7 m

Collision frequency = 4.58 × 109 s–1 Successive collision time ≈ 500 × (Collision time) Pressure inside the cylinder containing nitrogen, P = 2.0 atm = 2.026 × 105 Pa Temperature inside the cylinder, T = 17°C =290 K

yCBSE

.com

Radius of a nitrogen molecule, r = 1.0 Å = 1 × 1010 m Diameter, d = 2 × 1 × 1010 = 2 × 1010 m Molecular mass of nitrogen, M = 28.0 g = 28 × 10–3 kg The root mean square speed of nitrogen is given by the relation:

Where, R is the universal gas constant = 8.314 J mole–1 K–1

= 508.26 m/s The mean free path (l) is given by the relation:

Where, k is the Boltzmann constant = 1.38 × 10–23 kg m2 s–2K–1

= 1.11 × 10–7 m

Collision frequency

= 4.58 × 109 s–1 Collision time is given as:

= 3.93 × 10–13 s Time taken between successive collisions:

= 2.18 × 10–10 s

∴

Hence, the time taken between successive collisions is 500 times the time taken for a collision.

20: Retarding force, F = –50 N

Mass of the body, m = 20 kg

yCBSE

.com

Initial velocity of the body, u = 15 m/s Final velocity of the body, v = 0 Using Newton’s second law of motion, the acceleration (a) produced in the body can be calculated as: F = ma –50 = 20 × a

Using the first equation of motion, the time (t) taken by the body to come to rest can be calculated as: v = u + at

= 6 s

21: Velocity of the particle,

Acceleration of the particle Also,

But,

Integrating both sides:

Where,

= Velocity vector of the particle at t = 0

= Velocity vector of the particle at time t

Integrating the equations with the conditions: at t = 0; r = 0 andat t = t; r = r

Since the motion of the particle is confined to the x-y plane, on equating the

coefficients of , we get:

yCBSE

.com

(a) When x = 16 m:

∴y = 10 × 2 + (2)2 = 24 m

(b) Velocity of the particle is given by:

22: R/6; from the original centre of the body and opposite to the centre of the cut portion.

Mass per unit area of the original disc = σ Radius of the original disc = R Mass of the original disc, M = πR2

σ The disc with the cut portion is shown in the following figure:

Radius of the smaller disc =

Mass of the smaller disc, M’ =

yCBSE

.com

Let O and O′ be the respective centres of the original disc and the disc cut off from the original. As per the definition of the centre of mass, the centre of mass of the original disc is supposed to be concentrated at O, while that of the smaller disc is supposed to be concentrated at O′. It is given that:

OO′= After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses. The two masses are: M (concentrated at O), and

–M′ concentrated at O′ (The negative sign indicates that this portion has been removed from the original disc.) Let x be the distance through which the centre of mass of the remaining portion shifts from point O. The relation between the centres of masses of two masses is given as:

For the given system, we can write:

(The negative sign indicates that the centre of mass gets shifted toward the left of point O.)

23: (a) 200 m2

(a) Power used by the family, P = 8 kW = 8 × 103 W Solar energy received per square metre = 200 W Efficiency of conversion from solar to electricity energy = 20 % Area required to generate the desired electricity = A As per the information given in the question, we have:

(A × 200)

yCBSE

.com

(b) The area of a solar plate required to generate 8 kW of electricity is almost equivalent to the area of the roof of a building having dimensions 14 m × 14 m.

24: Angle of contact between mercury and soda lime glass, θ = 140°

Radius of the narrow tube, r = 1 mm = 1 × 10–3 m Surface tension of mercury at the given temperature, s = 0.465 N m–1 Density of mercury, ρ =13.6 × 103 kg/m3 Dip in the height of mercury = h Acceleration due to gravity, g = 9.8 m/s2 Surface tension is related with the angle of contact and the dip in the height as:

Here, the negative sign shows the decreasing level of mercury. Hence, the mercury level dips by 5.34 mm.

25: Mass of each star, M = 2 × 1030 kg

Radius of each star, R = 104 km = 107 m Distance between the stars, r = 109 km = 1012m For negligible speeds, v = 0 total energy of two stars separated at distance r

Now, consider the case when the stars are about to collide: Velocity of the stars = v Distance between the centers of the stars = 2R

Total kinetic energy of both stars

Total potential energy of both stars

yCBSE

.com

Total energy of the two stars = Using the law of conservation of energy, we can write:

26: Frequency of the turning fork, ν = 340 Hz

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

Such a system produces odd harmonics. The fundamental note in a closed pipe is 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

27: (a) The dotted plot in the graph signifies the ideal behaviour of the gas, i.e., the

ratio is equal. µR (µ is the number of moles and R is the universal gas constant) is a constant quality. It is not dependent on the pressure of the gas.

yCBSE

.com

(b) The dotted plot in the given graph represents an ideal gas. The curve of the gas at temperature T1 is closer to the dotted plot than the curve of the gas at temperature T2. A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore, T1 > T2is true for the given plot.

(c) The value of the ratio PV/T, where the two curves meet, is µR. This is because the ideal gas equation is given as: PV = µRT

Where, P is the pressure T is the temperature V is the volume µ is the number of moles R is the universal constant Molecular mass of oxygen = 32.0 g Mass of oxygen = 1 × 10–3 kg = 1 g R = 8.314 J mole–1 K–1

∴

= 0.26 J K–1 Therefore, the value of the ratio PV/T, where the curves meet on the y-axis, is 0.26 J K–1.

(d) If we obtain similar plots for 1.00 × 10–3 kg of hydrogen, then we will not get the same value of PV/T at the point where the curves meet the y-axis. This is because the molecular mass of hydrogen (2.02 u) is different from that of oxygen (32.0 u). We have:

R = 8.314 J mole–1 K–1 Molecular mass (M) of H2 = 2.02 u

m = Mass of H2

∴

yCBSE

.com

= 6.3 × 10–2 g = 6.3 × 10–5 kg Hence, 6.3 × 10–5 kg of H2 will yield the same value of PV/T.

28: (i) Initial speed of the three-wheeler, u = 36 km/h = 10 m/s

Final speed of the three-wheeler, v = 0 m/s Time, t = 4 s Mass of the three-wheeler, m = 400 kg Mass of the driver, m' = 65 kg Total mass of the system, M = 400 + 65 = 465 kg Using the first law of motion, the acceleration (a) of the three-wheeler can be calculated as: v = u + at

The negative sign indicates that the velocity of the three-wheeler is decreasing with time. Using Newton’s second law of motion, the net force acting on the three-wheeler can be calculated as: F = Ma = 465 × (–2.5) = –1162.5 N The negative sign indicates that the force is acting against the direction of motion of the three-wheeler.

(ii) Mass of the rocket, m = 20,000 kg

Initial acceleration, a = 5 m/s2 Acceleration due to gravity, g = 10 m/s2 Using Newton’s second law of motion, the net force (thrust) acting on the rocket is given by the relation: F – mg = ma F = m (g + a) = 20000 × (10 + 5) = 20000 × 15 = 3 × 105 N

29: (i) 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 = 0 ∆W = –22.3 J (Since the work is done on the system) From the first law of thermodynamics, we have: ∆Q = ∆U + ∆W

yCBSE

.com

Where, ∆U = Change in the internal energy of the gas ∴ ∆U = ∆Q – ∆W = – (– 22.3 J) ∆U = + 22.3 J When the gas goes from state A to state B via a process, the net heat absorbed by the system is: ∆Q = 9.35 cal = 9.35 × 4.19 = 39.1765 J Heat absorbed, ∆Q = ∆U + ∆Q ∴∆W = ∆Q – ∆U = 39.1765 – 22.3 = 16.8765 J Therefore, 16.88 J of work is done by the system.

(ii) Temperature inside the refrigerator, T1 = 9°C = 282 K

Room temperature, T2 = 36°C = 309 K

Coefficient of performance =

Therefore, the coefficient of performance of the given refrigerator is 10.44.

30: (i) (a) The time period of a simple pendulum,

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

= Constant Hence, the time period T, of a simple pendulum is independent of the mass of the bob.

(b) In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as: F = –mg sinθ Where, F = Restoring force m = Mass of the bob g = Acceleration due to gravity θ = Angle of displacement For small θ, sinθ

yCBSE

.com

For large θ, sinθ is greater than θ. 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 is not affected by the fall. Since a wristwatch does not work on the principle of a simple pendulum, it is not affected by the acceleration due to gravity during free fall. Its working depends on spring action.

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

(ii) Volume of the air chamber = V

Area of cross-section of the neck = a Mass of the ball = m The pressure inside the chamber is equal to the atmospheric pressure. Let the ball be depressed by x units. As a result of this depression, there would be a decrease in the volume and an increase in the pressure inside the chamber. Decrease in the volume of the air chamber, ∆V = ax

Volumetric strain

Bulk Modulus of air,In this case, stress is the increase in pressure. The negative sign indicates that pressure increases with a decrease in volume.

The restoring force acting on the ball, F = p × a

In simple harmonic motion, the equation for restoring force is: F = –kx … (ii)

yCBSE

.com

Where, k is the spring constant Comparing equations (i) and (ii), we get:

Time period,

yCBSE

.com