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Samacheer Kalvi Class 11 Physics Practice Question Papers

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Brain Grain · braingrain.in
Physics — Practice Paper · Set 1
Class: 11Samacheer KalviMax Marks: 93
Name: ____________________Reg No: ____________
Part I — Multiple Choice Questions 15 × 1 = 15

Choose the correct answer. (Answer all questions.)

1.The angle between (\(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)) and (\(\overrightarrow{\mathrm{A}}\) – \(\overrightarrow{\mathrm{B}}\)) can be – (a) only 0° (b) only 90° (c) between 0° and 90° (d) between 0° and 180°[1]
2.The dimensional formula of planck’s constant h is _____________. [AMU, Main, JEE, NEET](a) [ML² T -1 ](b) [ML²T -3 ](c) [MLT -1 ](d) [ML 3 T -3 ][1]
3.\(\frac{1}{12}\) of the mass of carbon 12 atom is ….. (a) 1 TMC (b) mass of neutron (c) 1 amu (d) mass of hydrogen[1]
4.In hot summer after a bath, the body’s:(a) internal energy decreases(b) internal energy increases(c) heat decreases(d) no change in internal energy and heat[1]
5.The study of forces acting on bodies whether at rest or in motion is …..(a) classical mechanics(b) quantum mechanics(c) thermodynamics(d) condensed matter physics[1]
6.A particle undergoes uniform circular motion. The angular momentum of the particle remain conserved about, _______ [IIT 2003] a) the center point of the circle. b) the point on the circumference of the circle. c) any point inside the circle. d) any point outside the circle.[1]
7.The dimensional formula for (ε 0 ) permittivity of free space _______.(a) M -1 L 3 T 4 A 2(b) M -1 L 3 A 2(c) M -1 L -3 T 4 A 2(d) M 1 L 3 T -4 A -2[1]
8.A Screw gauge gives the following reading when used to measure the diameter of the wire. Main scale reading = 0. Circular scale reading = 52 divisions Given that 1 mm on the main scale corresponds to 100 division on a circular scale. The diameter of the wire is _______.(a) 0.52 cm(b) 0.052 cm(c) 0.0026 cm(d) 0.005 cm[1]
9.The dimensional formula for the coefficient of viscosity is _______.(a) M 0 L -1 T -1(b) M 1 L 1 T 1(c) M 1 L -1 T 1(d) M 2 L 2 T 0[1]
10.A man of weight 80 kg, stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 5 ms -2. What would be the reading on the scale? [g = 10 ms -2 ]. a) 400 N b) 1200 N c) 800 N d) zero[1]
11.If a ball is thrown vertically upwards with a speed u the distance covered during the last ‘t’ seconds of its ascent is _______. a) 1/2 gt² b) ut – 1/2gt² c) (u – gt)t d) ut[1]
12.The force acting on a body moving along x-axis varies with the position of the particle as in fig _______. The body in stable equilibrium at a) x = x 1 b) x = x 2 c) both x 1, and x 2 d) neither x 1 nor x 2[1]
13.If \(\overline{A}\) = 2\(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\), \(\overline{B}\) = \(\hat{i}\) + 2\(\hat{j}\) + 3\(\hat{k}\) and \(\overline{C}\) = 6\(\hat{i}\) – 2\(\hat{j}\) – 6\(\hat{k}\) then angle between \(\overline{A}\) + \(\overline{B}\) and \(\overline{C}\) will be _______. a) 30° b) 45° c) 60° d) 90°[1]
14.A force of 3 N and 4 N are acting perpendicular to an object, the resultant force is-(a) 9 N(b) 16 N(c) 5 N(d) 7 N[1]
15.Which of the following combinations have the dimensions of time? L, C, R represent inductance, capacitance, and resistance respectively. (a) RC (b) \(\sqrt{LC}\) (c) L/R (d) C/L[1]
Part II — Short Answer Questions 14 × 2 = 28

Answer briefly. (Answer all questions.)

16.What is the relation between torque and angular momentum?[2]
17.A piece of wood of mass m is floating erect in a liquid whose density is ρ. If it is slightly pressed down and released, then executes simple harmonic motion. Show that its time period of oscillation is T = 2π\(\sqrt{\frac{m}{\mathrm{Ag} \rho}}\)[2]
18.Convert the vector \(\overline{r}\) = 3\(\hat{i}\) + 2\(\hat{j}\) into a unit vector.[2]
19.Define molar specific heat capacity.[2]
20.In an adiabatic expansion of the air, the volume is increased by 4%, what is percentage change in pressure? (For air γ = 1.4)[2]
21.The time period for small vertical oscillations of block of mass m when the masses of the pulleys are negligible and spring constant k 1 and k 2 is: (a) T = 4π\(\sqrt{m\left(\frac{1}{k_{1}}+\frac{1}{k_{2}}\right)}\) (b) T = 2π\(\sqrt{m\left(\frac{1}{k_{1}}+\frac{1}{k_{2}}\right)}\) (c) T = 4π\(\sqrt{m\left(k_{1}+k_{2}\right)}\) (d) T = 2π\(\sqrt{m\left(k_{1}+k_{2}\right)}\)[2]
22.Define terminal velocity.[2]
23.A ball of mass 1 kg and another of mass 2 kg are dropped from a tall building whose height is 80 m. After, a fall of 40 m each towards Earth, their respective kinetic energies will be in the ratio of _______. (AIPMT model 2004) a) \(\sqrt{2}\): 1 b) 1: \(\sqrt{2}\) c) 2: 1 d) 1: 2[2]
24.Three identical solid spheres move down through three inclined planes A, B, and C all same dimensions. A is without friction, B is undergoing pure rolling and C is rolling with slipping compare the kinetic energies E A, E B and E c at the bottom.[2]
25.Write down the expression for Stoke’s force and explain the symbols involved in it.[2]
26.Suppose we go 200 km above and below the surface of the Earth, what are the g values at these two points? In which case, is the value of g small?[2]
27.What is the effect of temperature on elasticity?[2]
28.Can two non-zero vectors give zero resultant when they multiply with each other?[2]
29.A planet moving along an elliptical orbit is closest to the Sun at distance r 1 and farthest away at a distance of r 2. If v 1 and v 2 are linear speeds at these points respectively. Then the ratio \(\frac{v_{1}}{v_{2}}\) is: (NEET 2016) (a) \(\frac{r_{2}}{r_{1}}\) (b) (\(\frac{r_{2}}{r_{1}}\))² (c) \(\frac{r_{1}}{r_{2}}\) (d) (\(\frac{r_{1}}{r_{2}}\))²[2]
Part III — Long Answer Questions 10 × 5 = 50

Answer in detail. (Answer all questions.)

30.Can we measure the temperature of the object by touching it?[5]
31.In an empty room, why is it that a tone sounds louder than in the room having things like furniture, etc.[5]
32.Which one of these is more elastic, steel or rubber? Why?[5]
33.There is a limit beyond which the polishing of a surface increases frictional resistance rather than decreasing it. Why?[5]
34.Explain Wien’s law and why our eyes are sensitive only to visible rays?[5]
35.What are stationary waves? Explain the formation of stationary waves and also write down the characteristics of stationary waves.[5]
36.How will you differentiate motion in one dimension, two dimensions, and in three dimensions?[5]
37.A spring balance shows wrong readings after using for a long time. Why?[5]
38.Why is the energy of a satellite (or any other planet) negative?[5]
39.What is a sonometer? Give its construction and working. Explain how to determine the frequency of tuning fork using a sonometer.[5]
🔑 Show Answer Key — Set 1
  1. 1. (d) between 0° and 180°
  2. 2. (a) [ML² T -1 ]
  3. 3. (d) mass of hydrogen
  4. 4. (a) internal energy decreases
  5. 5. (a) classical mechanics
  6. 6. a) the center point of the circle.
  7. 7. (c) M -1 L -3 T 4 A 2
  8. 8. (b) 0.052 cm
  9. 9. (c) M 1 L -1 T 1
  10. 10. b) 1200 N
  11. 11. a) 1/2 gt²
  12. 12. b) x = x 2
  13. 13. d) 90°
  14. 14. (c) 5 N
  15. 15. (b) \(\sqrt{LC}\)
  16. 16. Rate of change in angular momentum is equal to torque. τ = \(\frac { d(L) }{ dt }\)
  17. 17. When a wood is pressed and released,
  18. 18. A vector divided by its magnitude is a unit vector
  19. 19. Heat energy required to increase the temperature of one mole of substance by IK or 1°C.
  20. 20. Percentage of increased volume = 4% \(\frac { ∆V }{ V }\) x 100 = 4% γ = 1.4 In adiabatic process PV γ = Constant
  21. 21. (a) T = 4π\(\sqrt{m\left(\frac{1}{k_{1}}+\frac{1}{k_{2}}\right)}\) Hint: T = 2π\(\frac { m }{ k }\) The given arrangement is similar to the combination of springs in series.
  22. 22. The maximum constant velocity acquired by a body while falling freely through a viscous medium is called the terminal velocity V T.
  23. 23. d) 1: 2 This online Velocity Calculator is used to find the velocity of water in a pipe with the flow rate and diameter of the pipe.
  24. 24. A possesses translational K.E B possesses sum of rotational K.E + Translational K.E C possesses more rotation than translational KE
  25. 25. F= 6πηrv * Radius (r ) of the sphere * Velocity (v) of the sphere and * Coefficient of viscosity η of the liquid.
  26. 26. Given: depth (d) = height(A) = 200 km We know that R – 6400 km Formula:
  27. 27. As the temperature of the substance increases, its elasticity decreases.
  28. 28. If yes condition for the same. Yes. for example, the cross product of two non-zero vectors will be zero when θ = 0 or θ = 180°.
  29. 29. (a) \(\frac{r_{2}}{r_{1}}\) Hint: v = rw ∴ v ∝ r \(\frac{v_{1}}{v_{2}}\) = \(\frac{r_{1}}{r_{2}}\)
  30. 30. No, we can’t measure the temperature of the object touching it. Because the temperature is the degree of hotness or coolness of a body. Only we can sense the hotness or coolness of the object.
  31. 31. When a room has furniture reverberation time can be suitably decreased since furniture has a large absorption coefficient of sound. So a tone sounds with lesser amplitude and intensity. Whereas in an empty room, reverberation time will be more than a room having things like furniture.
  32. 32. Steel is more elastic than rubber. If equal stress is applied to both steel and rubber, the steel produces less strain. So Young’s modulus is higher for steel than rubber. Hence steel is more elastic than rubber.
  33. 33. When surfaces are highly polished the area of contact between them increases. As result of this a large number of atoms and molecules lying on both surfaces start exerting strong attractive forces on each other. Therefore the frictional force increases.
  34. 34. Wien’s law states that, ‘the wavelength of maximum intensity of emission of a black body radiation is inversely proportional to the absolute temperature of the black body’. λ m ∝ \(\frac { 1 }{ T }\) (or) λ m = \(\frac { b }{ T }\) … (1) It is implied that if temperature of the body increases, maximal intensity wavelength (λ m ) shifts towards lower wavelength (higher frequency) of electromagnetic spectrum. Wien’s law and Vision: Why our eye is sensitive to only visible wavelength (in the range 400 nm to 700nm)? The Sun is approximately considered as a black body. Since any object above 0 K will emit radiation, Sun also emits radiation. Its surface temperature is about 5700 K. By substituting this value in the equation (1), It is the wavelength at which maximum intensity is 508nm. Since the Sun’s temperature is around 5700K, the spectrum of radiations emitted by Sun lie between 400 nm to 700 nm which is the visible part of the spectrum. The humans evolved under the Sun by receiving its radiations. The human eye is sensitive only in the visible.
  35. 35. When the wave hits the rigid boundary it bounces back to the original medium and can interfere with the original waves. A pattern is formed, that is known as standing waves or waves stationary. Let us consider two harmonic progressive waves (formed by strings) that have the same amplitude and same velocity but move in opposite directions. Then the displacement of the first wave (incident wave) is y 1 = A sin(kx – ωt) … (1) (waves move toward right) The displacement of the second wave (reflected wave) is y 2 = A sin(kx + ωt) … (2) (waves move toward left) both will interfere with each other by the principle of superposition, the net displacement is y = y 1 + y 2 … (3) By substituting equation (1) and equation (2) in equation (3), we get y = \(\left\{\begin{array}{l} \mathrm{A} \sin (k x-\omega t) \\ +\mathrm{A} \sin (k x+\omega t) \end{array}\right.\) … (4) Using trigonometric identity, we rewrite equation (4) as y(x, t) = 2A cos(ωt) sin(kx) … (5) This represents a stationary wave or standing wave, It is meant that this wave does not move either forward or backward, whereas progressive or travelling waves will move forward or backward. In addition, the displacement of the particle in equation (5) can be written in more compact form, y(x, t) = A’cos(cot) where, A’ = 2A sin(Ax). It is implied implying that the particular element of the string executes simple harmonic motion with amplitude equals to A’. The maximum of this amplitude occurs at positions for which sin(kx) = 1 ⇒ kx = \(\frac { π }{ 2 }\), \(\frac { 3π }{ 2 }\), \(\frac { 5π }{ 2 }\), … = nπ where m takes half-integer or half-integral values. The position of maximum amplitude is known as an antinode. Expressing wave number in terms of wavelength, let us represent the anti-nodal positions as x m = (\(\frac { 2m+1 }{ 2 }\)) \(\frac { λ }{ 2 }\) … (6) where, m = 0, 1, 2 …. For m = 0 we have maximum at x 0 = \(\frac { λ }{ 2 }\) For m = 1 we have maximum at x 0 = \(\frac { 3λ }{ 4 }\) For m = 2 we have maximum at x 2 = \(\frac { 5λ }{ 4 }\) and so on. The distance between two successive antinodes can be computed by, Similarly, the minimum of the amplitude A’ also occurs at some points in the space, and these points can be determined by setting sin(kx) = 0 ⇒ kx = 0, π, 2π, 3π, … = nπ where n takes integer or integral values. It is noted that the elements at these points do not vibrate (not move), and the points are called nodes. The n th nodal positions is given by, x n = n\(\frac { λ }{ 2 }\) … (7) where, n = 0, 1, 2, … For n = 0 we have minimum at x o = 0 For n = 1 we have minimum at x 1 = \(\frac { λ }{ 2 }\) For n = 2 we have maximum at x 2 = λ and so on. The distance between any two successive nodes can be calculated as x n – x n-1 – n\(\frac { λ }{ 2 }\) – (n-1)\(\frac { λ }{ 2 }\) = \(\frac { λ }{ 2 }\) Characteristics of stationary waves: * Stationary waves are characterized by the confinement of a wave disturbance between two rigid boundaries. It is meant that the wave does not move forward or backward in a medium (does not advance), it remains steady at its place. Hence, they are called “stationary waves or standing waves”. * Certain points in the region in which the wave exists have maximum amplitude, called anti-nodes. At certain points, the amplitude is minimum or zero, called nodes. * The distance between two consecutive nodes (or) anti-nodes is \(\frac { λ }{ 2 }\). * The distance between a node and its neighboring anti-node is \(\frac { λ }{ 4 }\). * The transfer of energy along the standing wave is zero.
  36. 36. Motion in one dimension: One-dimensional motion is the motion of a particle moving along a straight line. Example: An object falling freely under gravity close to the earth. Motion in two dimensions: If a particle is moving along a curved path in-plane, then it is said to be in two-dimensional motion. Example: Motion of a coin in a carom board. Motion in three dimensions: A particle moving in usual three-dimensional space has three-dimensional motion. Example: A bird flying in the sky.
  37. 37. When a spring balance has been used for a long time, it develops elastic fatigue, the spring of such a balance takes a longer time to recover its original configuration and therefore it does not give correct measurement.
  38. 38. The negative sign of the total energy implies that satellite is bound to the Earth’s gravitational force. So it cannot escape from the Earth. At large distances satellite is not bound to Earth. It is completely free from gravitational force of the Earth.
  39. 39. Sonometer is used for sound-related measurements. Using this device, The following quantities can be determined. (i) The frequency of the tuning fork or frequency of the alternating current. (ii) The tension in the string. (iii) The unknown hanging mass. Construction: The sonometer is made up of a hollow box that is one meter long with a uniform metallic thin string attached to it. One end of the string is connected to a hook and the other end is connected to a weight hanger through a pulley as shown in Figure. Since only one string is used, it is also known as monochord. The weights are added to the free end of the wire to increase the tension of the wire. Two adjustable wooden knives are put over the board, and their positions are adjusted to change the vibrating length of the stretched wire. Procedure: A transverse stationary or standing wave is produced. So, at the knife edges P and Q, nodes are formed. In between the knife edges, anti-nodes are formed. If the length of the vibrating element is l then l = \(\frac { λ }{ 2 }\) ⇒ λ = 2l Let f be the frequency of the vibrating element, T the tension in the string, and p the mass per unit length of the string. Then using the the equation v = \(\sqrt{\frac{\mathrm{T}}{\mu}}\) measured in ms -1, we get, f = \(\frac { v }{ λ }\) = \(\frac { 1 }{ 2l }\)\(\sqrt{\frac{\mathrm{T}}{\mu}}\) in Hertz. Let ρ be the density of the material of the string and d be the diameter of the string. Then the mass per unit length μ,
Brain Grain · braingrain.in
Physics — Practice Paper · Set 2
Class: 11Samacheer KalviMax Marks: 93
Name: ____________________Reg No: ____________
Part I — Multiple Choice Questions 15 × 1 = 15

Choose the correct answer. (Answer all questions.)

1.If an object is dropped from the top of a building and it reaches the ground at t = 4s, then the height of the building is (ignoring air resistance) (g = 9.8ms -2 ) a) 77.3m b) 78.4m c) 80.5 d) 79.2m[1]
2.A couple produces, _______. [AIPMT 1997] a) pure rotation b) pure translation c) rotation and translation d) no motion[1]
3.The youngs modulus of a material of the wire is 12.6 x 10 11 dyne/cm 2. Its value is MKS system is _______.(a) 12.6 x 10 12 N/M 2(b) 12.6 x 10 10 N/M 2(c) 12.6 x 10 6 N/M 2(d) 12.6 x 10 8 N/M 2[1]
4.According to Kepler, planet move in(a) Circular orbits around the Sun(b) Elliptical orbits around the Sim with Sun at the exact centre(c) Straight lines with constant velocity(d) Elliptical orbits around the Sun with Sun at one of its foci.[1]
5.The work done by the conservative force for a closed path is _______. a) always negative b) zero c) always positive d) not defined[1]
6.A block is kept on a frictionless inclined surface with the angle of inclination α. The incline is given an acceleration ‘a’ to keep the block stationary. Then a is equal to _______. a) g b) g tan α c) g / tan α d) g cosec α[1]
7.Six vectors \(\vec{a}\) through \(\vec{f}\) have magnitudes and directions as indicated in figure. Which of the following statement is true? a) \(\overline{b}\) + \(\overline{e}\) = \(\overline{f}\) b) 3\(\hat{i}\) – 2\(\hat{j}\) + \(\hat{k}\) \(\hat{b}\) + \(\hat{c}\) = \(\hat{f}\) c) \(\hat{d}\) + \(\hat{c}\) = \(\hat{f}\) d) \(\hat{d}\) + \(\hat{e}\) = \(\hat{f}\)[1]
8.The coefficient of restitution e for a perfectly elastic collision is _______. a) 1 b) 0 c) ∞ d) -1[1]
9.An object of mass 4kg falls from rest through a vertical distance of 20m and reaches with velocity 10ms -1 on ground. The work done by air friction is. a) 800J b) – 800J c) 600J d) – 600J[1]
10.A ball of mass 200g is attached to a string 50 mm and a force F is applied as shown. The W.D by this force if the string makes an angle 60° with vertical is? [at initial and final positions speed of the ball is zero.] a) 1J b) 0.5J c) 0.05J d) 0.25J[1]
11.If one object is dropped vertically downward and another object is thrown horizontally from the same height, then the ratio of vertical distance covered by both objects at any instant t is _______. a) 1 b) 2 c) 4 d) 0.5[1]
12.The centrifugal force appears to exist _______. a) only in inertial frames b) only in rotating frames c) in an accelerated frame d) both in inertial and non-inertial frames[1]
13.If the velocity is \(\overline{V}\) = \(2 \hat{i}+t^{2} \hat{j}-9 \hat{k}\) then the magnitude of acceleration at t = 0.5s is _______. a) 1ms -2 b) 2 ms -2 c) zero d) -1ms -2[1]
14.Dimensional formula for work done is(a) MLT -1(b) ML 2 T 2(c) M -1 L -1 T 2(d) ML 2 T -2[1]
15.The work done by the Sun’s gravitational force on the Earth is:(a) always zero(b) always positive(c) can be positive or negative(d) always negative[1]
Part II — Short Answer Questions 14 × 2 = 28

Answer briefly. (Answer all questions.)

16.Define heat engine.[2]
17.What are the rotational equivalents for the physical quantities? * mass * force[2]
18.Write down the difference between simple harmonic motion angular simple harmonic motion.[2]
19.What is meant by the force constant of a spring?[2]
20.State Hooke’s law of elasticity.[2]
21.Consider two organ pipes of the same length in which one organ pipe is closed and another organ pipe is open. If the fundamental frequency of closed pipe is 250 Hz. Calculate the fundamental frequency of the open pipe.[2]
22.An object at an angle such that the horizontal range is 4 times the maximum height. What is the angle of projection of the object?[2]
23.Two bodies of masses m and Am are placed at a distance r. Calculate the gravitational potential at a point on the line joining them where the gravitational field is zero.[2]
24.What is mean by state variable? Give example.[2]
25.A sound source and listener are both stationary and a strong wind is blowing. Is there a Doppler effect?[2]
26.Define center of mass.[2]
27.A simple pendulum is suspended from the roof of a school bus which moves in a horizontal direction with an acceleration a, then the time period is: (a) T ∝ \(\frac{1}{g^{2}+a^{2}}\) (b) T ∝ \(\frac{1}{\sqrt{g^{2}+a^{2}}}\) (c) T ∝ \(\sqrt{g^{2}+a^{2}}\) (d) T ∝ (g² + a²)[2]
28.Give the equation of state for an adiabatic process.[2]
29.An ideal spring of spring constant k, is suspended from the ceiling of a room and a block of mass m is fastened to its lower end. If the block is released when the spring is un-stretched, then the maximum extension in the spring is: (a) 4\(\frac { mg }{ k }\) (b) \(\frac { mg }{ k }\) (c) 2\(\frac { mg }{ k }\) (d) \(\frac { mg }{ 2k }\)[2]
Part III — Long Answer Questions 10 × 5 = 50

Answer in detail. (Answer all questions.)

30.Calculate the temperature at which the rms velocity of a gas triples its value at S.T.P.[5]
31.Calculate the work done by a force of 30N in lifting a load of 2kg to a height of 10m (g = 10ms -2 )[5]
32.Explain the horizontal oscillations of a spring.[5]
33.Explain Horizontal projection. Derive the equation for its motion, horizontal range & time of flight.[5]
34.Derive the expression for mean free path of the gas.[5]
35.What is Reynold’s number? Give its significance.[5]
36.Can the coefficient of friction be more than one?[5]
37.Explain the variation of ‘g’ with altitude.[5]
38.State the principle of homogeneity and explain with an example.[5]
39.Justify that a uniform motion is an accelerated motion.[5]
🔑 Show Answer Key — Set 2
  1. 1. b) 78.4m
  2. 2. a) pure rotation
  3. 3. (a) 12.6 x 10 12 N/M 2
  4. 4. (d) Elliptical orbits around the Sun with Sun at one of its foci.
  5. 5. b) zero
  6. 6. b) g tan α
  7. 7. d) \(\hat{d}\) + \(\hat{e}\) = \(\hat{f}\)
  8. 8. a) 1
  9. 9. d) – 600J
  10. 10. b) 0.5J
  11. 11. a) 1
  12. 12. b) only in rotating frames
  13. 13. a) 1ms -2
  14. 14. (d) ML 2 T -2
  15. 15. (c) can be positive or negative
  16. 16. Heat engine is a device which takes heat as input and converts this heat in to work by undergoing a cyclic process.
  17. 17. The rotational equivalents for (i) mass and (ii) force are moments of inertia and torque respectively.
  18. 18. Comparison of simple harmonic motion and angular simple harmonic motion.
  19. 19. Force constant is defined as force per unit length.
  20. 20. It states that for small deformation, the stress is directly proportional to strain.
  21. 21. Formula:
  22. 22. Incase of obliging projection
  23. 23. Given: m 1 = m m 2 = 4m distance = r To find: V = ??
  24. 24. The quantities that are used to describe the equilibrium states of a Thermodynamic system.. Example: Pressure, Volume, Temperature.
  25. 25. When both source and listener are stationary, there is no relative motion between the source and the observer. Hence there is no Doppler effect.
  26. 26. The center of mass of a body is defined as a point where the entire mass of the body appears to be concentrated.
  27. 27. (b) T ∝ \(\frac{1}{\sqrt{g^{2}+a^{2}}}\) Hint: T = 2π\(\sqrt{\frac{l}{g}}\) When a bus is moving g’ = \(\sqrt{g^{2}+a^{2}}\) ∴ T ∝ \(\frac{1}{\sqrt{g^{2}+a^{2}}}\)
  28. 28. The equation of state for an adiabatic proc ess is given by, PV γ = Constant
  29. 29. (c) 2\(\frac { mg }{ k }\) Hint:
  30. 30. Let, T = T 1 = 273K RMS velocity, C = \(\sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}}}\) … (1) When the RMS velocity is tripled, 3C = \(\sqrt{\frac{3 \mathrm{RT}_{2}}{\mathrm{M}}}\) … (2) Dividing equation (2) by (1) we get \(\frac { 3C }{ C }\) = \(\sqrt{\frac{3 \mathrm{RT}_{2} / \mathrm{M}}{3 \mathrm{R} \times 273 / \mathrm{M}}}\) 3 = \(\sqrt{\frac{\mathrm{T}_{2}}{273}}\) Squaring on both side 9 = \(\frac{\mathrm{T}_{2}}{273}\) ∴ T 2 = 273 x 9 = 2457K Temperature T 2 = 2457K
  31. 31. F = 30N m = 2kg s = 10 m g = 10 ms -2 θ = 0 W.D = ? W.D = \(\overline{F}\).\(\overline{S}\) = FS cos θ W.D = 30 x 10 = 300 J
  32. 32. Let us consider a system containing a block of mass m fastened to a massless spring with stiffness constant or force constant or spring constant k placed on a smooth horizontal surface (frictionless surface) as shown in Figure. Let x 0 be the equilibrium position or mean position of mass m when it is left undisturbed. When the mass is displaced through a small displacement x towards right from its equilibrium position and then released, it will oscillate back and forth about its mean position x 0. Let F be the restoring force (due to stretching of the spring) that is proportional to the amount of displacement of block: For one-dimensional motion, we get F ∝ x F = – kx where negative sign implies that the restoring force will always act opposite to the direction of the displacement. This equation is called Hooke’s law. It is noticed that, the restoring force is linear with the displacement (i.e., the exponent of force and displacement are unity). This is not always true. If we apply a very large stretching force, then the amplitude of oscillations becomes very large. m\(\frac{d^{2} x}{d t^{2}}\) = – kx \(\frac{d^{2} x}{d t^{2}}\) = – \(\frac { k }{ m }\)x … (1) Comparing the equation (1) with simple harmonic motion equation a = \(\frac{d^{2} y}{d t^{2}}\) = – ω²y, we get ω² = \(\frac { k }{ m }\) which means the angular frequency or natural frequency of the oscillator is ω = \(\sqrt{\frac{k}{m}}\)rad s -1 … (2) The frequency of the oscillation is f = \(\frac { ω }{ 2π }\) = \(\frac { 1 }{ 2π }\)\(\frac { k }{ m }\) Hertz … (3) and the time period of the oscillation is T = \(\frac { 1 }{ f }\) = 2π[/latex]\(\frac { m }{ k }\) seconds … (4)
  33. 33. Consider an object thrown horizontally with an initial velocity u, from atop of a tower of height h. The horizontal velocity remains constant throughout its motion and the vertical component of velocity go on increases. The constant acceleration acting along the downward direction is g. The horizontal distance travelled is x(t) = x and the vertical distance travelled is y(t)=y. since the motion is two-dimensional the velocity will have both horizontal (u x ) and vertical (u y ) components. Motion along horizontal direction: The particle has zero acceleration along the x-direction and so initial velocity ux remains constant throughout its motion. The distance travelled by projectile in a time’t’ is given by x = ut+1/2 at² x = u x t → (1) Motion along vertical direction Here uy =0, a = g, s = y S = ut + \(\frac { 1 }{ 2 }\) at² y = \(\frac { 1 }{ 2 }\) gt² → (2) from (1) t = x/u x sub in equation (2) y = \(\frac { 1 }{ 2 }\) g (x/ux)² y = k x² Where k = \(\frac{g}{2 u_{x}^{2}}\).x² This equation resemble the equation of a parabola. Thus the path followed by the projectile is a parabola. Expression for time of flight: The time taken for the projectile to complete its trajectory is called the time of flight. Let h be the height of the tower or the vertical distance traversed. Let T be the time of flight w.k. S = ut + 1/2 at² here s = y = h, u = u y, t = T, a = g T depends on height of tower or vertical distance & independent of Horizontal velocity. Expression for Horizontal Range: The horizontal distance covered by the projectile from the foot of the tower to the point where the projectile hits the ground it called horizontal range. w. k. t, S = ut + \(\frac { 1 }{ 2 }\) at² Here, t = T, a = 0, S = x = R, u = u x Hence R ∝ u ∝ & R ∝ \(\frac{1}{\sqrt{g}}\)
  34. 34. Let us consider a system of molecules each with diameter d. Let n be the number of molecules per unit volume. It is assumed that only one molecule is in motion and all others are at rest as shown in the figure. If a molecule moves with average speed v in a time t, the distance travelled is vt. In this time t, let us consider the molecule to move in an imaginary cylinder of volume itd2vt. It collides with any molecule whose center is within this cylinder. Hence, the number of collisions is equal to the number of molecules in the volume of the imaginary cylinder. It is equal to πd²vt. The total path length divided by the number of collisions in time t is the mean free path. It is assumed that only one molecule is moving at a time and other molecules are at rest. But in actual practice all the molecules are in random motion. Hence the average relative speed of one molecule with respect to other molecules has to be taken into account. After some detailed calculations the correct expression for mean free path, λ = \(\frac{1}{\sqrt{2} n \pi d^{2}}\) It is implied from the equation that the mean free path is inversely proportional to number density. When the number density increases the molecular collisions increases. Hence it decreases the distance travelled by the molecule before collisions. Rearranging the equation (2) using ‘m’ (mass of the molecule) ∴ λ = \(\frac{m}{\sqrt{2} \pi d^{2} m n}\) But mn – mass per unit volume = ρ (density of the gas)
  35. 35. Reynold’s number R c is a critical variable, which decides whether the flow of a fluid through a cylindrical pipe is streamlined or turbulent. R c = \(\frac { ρVD }{ η }\)
  36. 36. No, it cannot be more than 1 for normal plane surfaces. But when surfaces are so irregular that they have sharp minute projections and cavities on them. Then the coefficient of friction may be more than one.
  37. 37. Let us consider an object of mass m at a height h from the surface of the Earth. Acceleration experienced by the object due to Earth is If h << Re Using Binomial expansion and taking the terms upto first order. We find that g’
  38. 38. The principle of homogeneity of dimensions states that the dimensions of all the term in the physical expression should be same. This principle is used to check the correctness of the equation For example V 2 = U 2 + 2as Writing dimensions on both sides [LT -1 ]² = [LT -1 ]² + [LT -1 ]² [L 2 T 2 ] = [L 2 T 2 ] + [L 2 T -2 ] Here the dimensions of all the terms in the expression are same and equal to[L 2 T -2 ] So the equation is dimensionally correct.
  39. 39. In a uniform circular motion, the speed of the body remains the same but the direction of motion changes at every point. Fig. shows the different velocity vectors at different positions of the particle. At each position, the velocity vector V is perpendicular to the radius vector. Thus the velocity of the body changes continuously due to the continuous change in the direction of motion of the body. As the rate of change is of velocity is acceleration a uniform circular motion is an accelerated motion.
Brain Grain · braingrain.in
Physics — Practice Paper · Set 3
Class: 11Samacheer KalviMax Marks: 93
Name: ____________________Reg No: ____________
Part I — Multiple Choice Questions 15 × 1 = 15

Choose the correct answer. (Answer all questions.)

1.A student performs an experiment for determination of g = \(\frac{4 \pi^{2} l}{T^{2}}\) an error of ∆l. For that, he takes the time of n oscillations with the stopwatch of least count ∆T and he commits a human error of 0.1s. For which of the following data, the measurement of g will be most accurate? ∆l, ∆T, n (a) 5m, 0.2s, 10 (b) 5mm, 0.2s, 20 (c) 5mm, 0.1s, 20 (d) 1mm, 0.1s, 50[1]
2.Consider the x-axis as representing east, the y-axis as north, and the z-axis as vertically upwards. Give the vector representing each of the following points. (a) 5m northeast and 2m up. (b) 4m southeast and 3m up. (c) 2m northwest and 4m up.[1]
3.Two identically sized rooms A and B are connected by an open door. If the room A is air conditioned such that its temperature is 4° lesser than room B, which room has more air in it?(a) Room A(b) Room B.(c) Both room has same air(d) Cannot be determined[1]
4.If the distance between the Earth and Sun were to be doubled from its present value, the number of days in a year would be:(a) 64.5(b) 1032(c) 182.5(d) 730[1]
5.Which one of the following is not a conservative force? a) gravitational force b) electrostatic force between the charges c) Magnetic force between two magnetic dipoles d) Frictional force[1]
6.The magnitude of a vector can not be- (a) positive (b) negative (e) zero (cl) 90[1]
7.A solid sphere of mass m and radius R rolls without slipping an the horizontal surface such that V cm = V 0 a) The K.E of rotation is 1/5 mV 0 b) The total K.E in 7/10 mV 0 ² c) The mechanical energy is mgh + 7/10 mV 0 ² d) All are correct[1]
8.A body of mass 4 m is lying in xy-plane at rest. It suddenly explodes into three pieces. Two pieces each of mass m move perpendicular to each other with equal speed v. The total kinetic energy generated due to explosion is _______. (AIPMT 2014) a) mv² b) \(\frac { 3 }{ 2 }\)mv² c) 2 mv² d) 4 mv²[1]
9.Infinitesimal quantity means –(a) collective particles(b) extremely small(c) nothing(d) extremely larger[1]
10.Which of the following gases will have least rms speed at a given temperature?(a) Hydrogen(b) Nitrogen(c) Oxygen(d) Carbon dioxide[1]
11.The work done by Sun on Earth in one year will be:(a) zero(b) non-zero(c) positive(d) negative[1]
12.Two forces each of magnitude ‘F’ have a resultant of the same magnitude. The angle between two forces a) 45° b) 120° c) 150° d) 60°[1]
13.Choose the correct statement _______. a) The frictional forces are dependent on the roughness of the surface. b) The kinetic friction is proportional to normal reaction c) The friction is independent of area of contact d) All statements are correct[1]
14.A force vector applied on a mass is represented as \(\vec { f }\) = 6\(\vec { i }\) – 8\(\vec { j }\) + 10\(\vec { k }\) and accelerates with 1ms -2, what will be the mass of the body in kg _______. a) 10\(\sqrt{2}\) b) 20 c) 2\(\sqrt{10}\) d) 10[1]
15.Two blocks of mass m 1 6 kg and m 2 = 3 kg as in figure coefficient of friction between m 1, and m 2 and between m 1 and surface is 0.5 and 0.4 respectively. The maximum horizontal force to can be applied to the mass m 1 so that they move without separation is _______. a) 41 N b) 61 N c) 81 N d) 101 N[1]
Part II — Short Answer Questions 14 × 2 = 28

Answer briefly. (Answer all questions.)

16.What are longitudinal waves? Give one example.[2]
17.What are processes involves in a Carnot engine?[2]
18.Define the gravitational field. Give its unit.[2]
19.State Newton’s Universal law of gravitation.[2]
20.Define one mole.[2]
21.Give the expression for work done by the gas.[2]
22.Let the source propagate a sound wave whose intensity at a point (initially) be I. Suppose we consider a case when the amplitude of the sound wave is doubled and the frequency is reduced to one-fourth. Calculate now the new intensity of sound at the same point?[2]
23.Imagine that the gravitational force between Earth and Moon is provided by an invisible string that exists between the Moon and Earth. What is the tension that, exists in this invisible string due to earth’s centripetal force?[2]
24.People often say “for every action there is an equivalent opposite reaction”. Here they meant ‘action of a human’. Is it correct to apply Newton’s third law to human actions? What is mean by ‘action’ in Newton’s third law? Give your arguments based on Newton’s laws.[2]
25.An object is thrown with initial speed of 5ms -1 with an angle of projection of 30°. What is the height and range reached by the particle?[2]
26.Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. Calculate the speed of each particle.[2]
27.Two streamlines cannot cross each other. Why?[2]
28.Define orbital velocity.[2]
29.Define the time period of simple harmonic motion.[2]
Part III — Long Answer Questions 10 × 5 = 50

Answer in detail. (Answer all questions.)

30.Explain Doppler Effect.[5]
31.Briefly explain centrifugal force with suitable examples.[5]
32.Explain Joule’s Experiment of the mechanical equivalent of heat.[5]
33.Discuss various modes of heat transfer.[5]
34.State Newton’s second law:[5]
35.The moon is orbiting the earth approximately once in 27 days. What is the angle transversed by the moon per day?[5]
36.Explain the need for banking of tracks.[5]
37.Explain the method to find the center of gravity of an irregularly shaped lamina.[5]
38.Explain how the interference of waves is formed.[5]
39.Derive the expression for total acceleration in the non-uniform circular motion.[5]
🔑 Show Answer Key — Set 3
  1. 1. (d) 1mm, 0.1s, 50
  2. 2. 5m northeast and 2m up. (a) The vector representation of 5m N-E and 2m up is (5i + 5j) Cos 45° + 2\(\hat{k}\) (b) 4m south east and 3m up. The vector representing 4m south east and 3m up is (4i – 4j) cos 45 + 3\(\hat{k}\) \(\frac{4(i-j)}{\sqrt{2}}\) + 3\(\hat{k}\) (c) 2m north west and 4m up. The vector representing 2m northwest and 4m up
  3. 3. (a) Room A Hint: As temperature of room A is less than that of room B evidently, Room A has more air in it.
  4. 4. (b) 1032
  5. 5. d) Frictional force
  6. 6. (b) negative
  7. 7. d) All are correct
  8. 8. b) \(\frac { 3 }{ 2 }\)mv²
  9. 9. (b) extremely small
  10. 10. (d) Carbon dioxide Hint: v rms = 1.73\(\sqrt{\frac{k \mathrm{~T}}{m}}\)
  11. 11. (d) negative
  12. 12. b) 120°
  13. 13. d) All statements are correct
  14. 14. a) 10\(\sqrt{2}\)
  15. 15. c) 81 N
  16. 16. The direction of vibration of particles in a medium is parallel to the direction of propagation of the wave. Example: Sound waves traveling in air.
  17. 17. Isothermal expansion Adiabatic expansion Isothermal compression Adiabatic compression.
  18. 18. The gravitational field intensity \(\overrightarrow{\mathrm{E}}_{1}\) at a point is defined as the gravitational force experienced by unit mass at that point. It’s unit N kg -1.
  19. 19. The gravitational force between two masses is directly proportional to the product of masses and inversely proportional to square of the distance between the masses.
  20. 20. One mole of any substance is the amount of that substance which contains Avogadro number (NA) of particles (such as atoms or molecules).
  21. 21. In general, the work done by the gas by increasing the volume from V i to V f is given by W = \(\int_{V_{i}}^{V_{f}} \mathrm{P} d \mathrm{~V}\)
  22. 22. Given:
  23. 23. Mass of moon = 7.34 x 10 22 kg Distance between the moon and earth = 3.84 x 10 8 m
  24. 24. Newton’s third law is applicable to only humans actions that involve physical force. Third law is not applicable to humans’ psychological actions or thoughts.
  25. 25. u = 5 m/s θ = 30° h max = ? R = ? Height reached Range:
  26. 26. Force acting on a particle Since particle, moving circular path experience centripetal force,
  27. 27. If two streamlines cross each other, there will be two directions of flow at the point of intersection which is impossible.
  28. 28. Orbital velocity is the velocity required to put the satellite into its orbit around the earth.
  29. 29. The time period is defined as the time taken by a particle to complete one oscillation. It is usually denoted by T.
  30. 30. When the source and the observer are in relative motion with respect to each other and to the medium in which sound propagates, the frequency of the sound wave observed is different from the frequency of the source. This phenomenon is called Doppler Effect.
  31. 31. Circular motion can be analyzed from two different frames of reference. One is the Inertial frame where Newton’s laws are obeyed. The other is the rotating frame of reference which is noninertial as it is accelerating. To use Newton’s I and II law in the rotational frame of reference the pseudo force called as centrifugal force is needed. The centrifugal forces appear to act on objects with respect to rotating frames. To explain consider an example, In the case of a whirling motion of a stone tied to a string, assume the stone has angular velocity ω in an inertial frame. If the motion of the stone is observed from a frame which is also rotating along with the stone with the same angular velocity ω then the stone appears to be at rest. This implies that in addition to the inward centripetal force (-mrω²) there must be an equal and opposite force that acts on outwards equal to (+ mrω 2 ). So the total force acting on the stone in the rotating frame is equal to zero (- mrω² + mrω² = 0) This outward force acting on the stone + mrω² is called centrifugal force.
  32. 32. Joule showed that mechanical energy can be converted into internal energy and vice versa. In his experiment, two masses were tied with a rope and a paddle wheel as shown in Figure. When these masses fall through a distance h due to gravity, both the masses lose potential energy which is equal to 2mgh. When the masses fall, the paddle wheel turns. Due to the turning of wheel inside water, frictional force comes in between the water and the paddle wheel. This causes a rise in temperature of the water. This confirms that gravitational potential energy is converted to internal energy of water. The temperature of water increases due to the work done by the masses. In fact, Joule was able to show that the mechanical work has the same effect as giving heat energy. He found that to raise lg of an object by 1°C, 4.186J of energy is required. In earlier days the heat was measured in calorie. 1 cal = 4.186 J This is called Joule’s mechanical equivalent of heat.
  33. 33. There are three modes of heat transfer: Conduction, Convection and Radiation. Conduction: Conduction is the process of direct transfer of heat through matter due to temperature difference. If two objects are placed in direct contact with one another, then heat will be transferred from the hotter object to the colder one. Convection: Convection is the process in which heat energy transfer is by actual movement of molecules in fluids such as liquids and gases. In convection, molecules move freely from one place to another. Example: Boiling of water in a pot. Radiation: Radiation is a form of energy transfer from one body to another by electromagnetic waves. Example: Solar energy from the Sun.
  34. 34. Newton second law state that” The force acting on an object is equal to the rate of change of its momentum: F = \(\frac { dp }{ dt }\) If m is the mass_of the object, and v its velocity of motion then \(\overline{p}\) = m\(\overline{v}\): The above equation can be written as F = \(\frac { dp }{ dt }\)(m\(\overline{v}\) ) = m\(\frac { dv }{ dt }\) ∴ F = ma
  35. 35. : Angle described in 27 days = 2π rad = 360° days Angie described in one day = 2π/27 radian = \(\frac { 360° }{ 27 }\) θ = 13.3°
  36. 36. In a leveled circular road skidding mainly depends on the co-efficient of static friction n s. The coefficient of static friction depends on the nature of surface which has a maximum limiting value. To avoid this usually “the outer edge of the road is slightly raised compared to inner edge”. This is called banking of roads or tracks. The angle of inclination called banking angle. Let the surface of the road make angle θ with horizontal surface. Then the normal force makes an angle θ with vertical. When the car takes a turn, two forces are acting on the car. (a) Gravitational force mg (downwards) (b) Normal force N (Perpendicular to surface). Normal force ‘N’ can be resolved into two components N cos θ and N sin θ and balances downward gravitational force. N sin θ provides necessary centripetal acceleration, According to II law N cos θ = mg N sin θ = \(\frac{m v^{2}}{r}\) Dividing the above equations, tan θ = \(\frac{v^{2}}{r g}\) V = \(\sqrt{r g \tan \theta}\) ∴ The banking angle θ and radius of curvature of the road or track determines the safe speed of car at the turning. If the speed exceeds this safe limit, then it starts to skid outward but the frictional force comes into effect and provides an additional centripetal force to prevent outward skidding. But at the same time if the speed is less than the safe limit it starts to skid inward and again frictional force come into effect which reduces centripetal force to prevent inward skidding However if the speed of the vehicle is sufficiently greater than the correct speed the frictional force cannot stop the car from skidding. So to avoid skidding in circular road or tracks they are banked.
  37. 37. The center of gravity of an irregularly shaped lamina by pivoting it at various points by trail and error. The lamina remains horizontal when pivoted at the point where the net gravitational force acts, which is the centre of gravity shown figure. When the body is supported at the centre for gravity, the sum of torques acting on all point masses of the rigid body becomes zero. Moreover the weight is compensated by the normal reaction force exerted by the pivot. The body in static equilibrium and hence it is horizontal.
  38. 38. Interference is a phenomenon in which two waves superimpose to form a resultant wave of greater, lower, or the same amplitude. Let us consider two harmonic waves having identical frequencies, constant phase difference φ, and same waveform (can be treated as coherent source), but having amplitudes A 1 and A 2, then y 1 = A 1 sin(kx – ωt) … (1) y 2 = A 2 sin(kx – ωt) … (2) Suppose they move simultaneously in a particular direction, then interference occurs (i.e., the overlap of these two waves). Mathematically y = y 1 + y 2 … (3) Hence by substituting equation (1) and equation (2) in equation (3), we get By squaring and adding equation (5) and (6), we get, A² = A 1 ² + A 2 ² + 2A 1 A 2 cosφ … (8) Since, intensity is square of the amplitude (I – A²), we get, I = I 1 + I 2 + 2\(\sqrt{\mathrm{I}_{1} \mathrm{I}_{2}} \cos \varphi\) … (9) This means the resultant intensity at any point depends on the phase difference at that point.
  39. 39. If the velocity changes both in speed and direction during circular motion, then we get non-uniform circular motion. Whenever the speed is not the same in a circular motion then the particle will have both centripetal and tangential acceleration. The resultant acceleration is obtained by the vector sum of centripetal and tangential acceleration Let the tangential acceleration be a t. Centripetal acceleration is v²/r. The magnitude of the resultant acceleration is a R = \(\sqrt{a_{t}^{2}+\left(\frac{v^{2}}{r}\right)^{2}}\) IV. Exercises:

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