{"id":21736,"date":"2020-01-28T16:10:25","date_gmt":"2020-01-28T10:40:25","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=21736"},"modified":"2021-07-02T15:26:24","modified_gmt":"2021-07-02T09:56:24","slug":"1st-puc-physics-previous-year-question-paper-march-2019-north","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-physics-previous-year-question-paper-march-2019-north\/","title":{"rendered":"1st PUC Physics Previous Year Question Paper March 2019 (North)"},"content":{"rendered":"

Students can Download 1st PUC Physics Previous Year Question Paper March 2019 (North), Karnataka 1st PUC Physics Model Question Papers with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Physics Previous Year Question Paper March 2019 (North)<\/h2>\n

Time: 3.15 Hours
\nMax Marks: 70<\/p>\n

General Instructions:<\/p>\n

    \n
  1. All parts are compulsory.<\/li>\n
  2. Draw relevant figure \/ diagram wherever necessary.<\/li>\n
  3. Numerical problems should be solved with relevant formulae.<\/li>\n<\/ol>\n

    Part – A<\/span><\/p>\n

    I. Answer ALL the following questions: ( 10 \u00d7 1 = 10 )<\/span><\/p>\n

    Question 1.
    \nHow many kilograms is in one unified atomic mass unit?
    \nAnswer:
    \n1 amu = 1.66 \u00d7 10-27 <\/sup>kg.<\/p>\n

    Question 2.
    \nState the law of triangle of addition of two vectors.
    \nAnswer:
    \nIf two vectors \\(\\overrightarrow{\\mathrm{A}}\\) and \\(\\overrightarrow{\\mathrm{B}}\\) acting at a point are represented by the two sides of a triangle taken in an order then the closing side of the triangle represented in opposite order with respect to the other vectors, gives their resultant vector.<\/p>\n

    Question 3.
    \nState Aristotle’s fallacy.
    \nAnswer:
    \nAristotle assumed that force is required in order to set bodies in motion. However if a body is already in an uniform motion along a straight line, no external forces are required.<\/p>\n

    Question 4.
    \nName the type of energy stored in a stretched or compressed spring.
    \nAnswer:
    \nPotential Energy.<\/p>\n

    \"KSEEB<\/p>\n

    Question 5.
    \nGive an example for torque or moment of couple.
    \nAnswer:
    \nWhen two equal and opposite forces are applied at the two ends of the lid of the bottle, the lid experience the torque or rotational effect.<\/p>\n

    Question 6.
    \nWrite the aim of Cavendish experiment in gravitation.
    \nAnswer:
    \nCavendish experiment determines the gravitational force of attraction between two ‘1 kg’ mass separated by ‘1 m’ apart.<\/p>\n

    Question 7.
    \nHow does strain depend on stress?
    \nAnswer:
    \nStrain is directly proportional to stress.
    \nStrain \u221d Stress.<\/p>\n

    Question 8.
    \nMention the value of steam point of water in Fahrenheit scale.
    \nAnswer:
    \nSteam point = 100\u00b0C = 212 \u00b0F.<\/p>\n

    Question 9.
    \nWhich quantity is kept constant in adiabatic process?
    \nAnswer:
    \nIn an adiabatic process, entropy oft he system will remain constant.<\/p>\n

    \"KSEEB<\/p>\n

    Question 10.
    \nThe function y = log (\u03c9t). The displacement y increases monotonically with time t, is it periodic function or non-periodic function.
    \nAnswer:
    \nNon-periodic function.<\/p>\n

    Part – B<\/span><\/p>\n

    II. Answer any FIVE of the following questions: ( 5 \u00d7 2 = 10 )<\/span><\/p>\n

    Question 11.
    \nName any two fundamental forces in nature.
    \nAnswer:<\/p>\n

      \n
    1. Gravitational force<\/li>\n
    2. Electro – weak force<\/li>\n
    3. Strong nuclear force<\/li>\n<\/ol>\n

      Question 12.
      \nDefine accuracy in the measurement. How does accuracy depend on precision in the measurement.
      \nAnswer:
      \nAccuracy refers to finding of the result very close to expected result whereas precision is limited to the error in the instrument.<\/p>\n

      \"KSEEB<\/p>\n

      Question 13.
      \nDistinguish between speed and velocity.
      \nAnswer:
      \nAverage Speed:<\/p>\n

        \n
      • The ratio of total distance traveled to the total time gives the average speed.<\/li>\n
      • Average speed is a scalar physical quantity.<\/li>\n
      • Average speed of a particle is finite; cannot be zero and always positive.<\/li>\n
      • Average speed is greater than the magnitude of average velocity.<\/li>\n<\/ul>\n

        Average Velocity:<\/p>\n

          \n
        • The ratio of net displacement of the particle to the total time taken gives the average velocity.<\/li>\n
        • Average velocity is a vector physical quantity.<\/li>\n
        • Average velocity of a particle may be zero, negative and positive.<\/li>\n
        • Average velocity (magnitude) is less than the speed of the particle.<\/li>\n<\/ul>\n

          Question 14.
          \nAnswer:
          \nStatement: Everybody continues to be in a state of rest or uniform motion along a straight line until an unbalanced force acts on it.
          \nExplanation:<\/p>\n

            \n
          • Consider the mouth of a glass tumbler covered by a thin sheet of paper with a heavy coin on it. When the sheet of paper is pulled suddenly, the coin falls vertically down-wards into the glass tumbler due to inertia of position.<\/li>\n
          • Consider a book lying on a table. A force is required in order to cause any linear motion in it. In the absence of an external force, the book continues to occupy that position..<\/li>\n<\/ul>\n

            Question 15.
            \nMention the relation between linear momentum and angular momentum with usual meanings.
            \nAnswer:
            \n\\(\\vec{L}=\\vec{r} \\times \\vec{p}\\) where \\(\\vec{p}\\) = linear momentum, \\(\\vec{L}\\) = angular momentum vector, \\(\\vec{r}\\) = radius or distance vector.<\/p>\n

            Question 16.
            \nWrite any two practical applications of Pascal’s law.
            \nAnswer:
            \nHydraulic brakes, hydraulic press, hydraulic lifts, hydraulic jack, injection syringes.<\/p>\n

            Question 17.
            \nGive any two assumptions of kinetic theory of gases.
            \nAnswer:<\/p>\n

              \n
            1. Collisions of molecules on the walls are perfectly elastic.<\/li>\n
            2. The size of the molecules are neglected when compared to the volume of the container.<\/li>\n
            3. The contribution to the net pressure due to inter-molecular collisions is neglected.<\/li>\n
            4. The root mean square value of speed is the average of velocities taken in all the three directions.<\/li>\n<\/ol>\n

              \"KSEEB<\/p>\n

              Question 18.
              \nHow does the period of oscillation of a pendulum depend on mass of the bob and length of the pendulum?
              \nAnswer:
              \n\u2018T\u2019 is independent of mass and directly proportional to the square root of the length of the pendulum.
              \n\\(\\mathrm{T} \\propto \\sqrt{\\mathrm{L}}\\)<\/p>\n

              Part – C<\/span><\/p>\n

              III. Answer any FIVE of the following questions : ( 5 \u00d7 3 = 15 )<\/span><\/p>\n

              Question 19. Deduce the expression for horizontal range of a projectile. For what angle of pro jection does horizontal range become maximum?
              \nAnswer:
              \nExpression for range of the projectile
              \nWe know that (v0 <\/sub>cos\u03b8) remains constant.
              \nInstantaneous distance along the horizontal is given by x = (v0 <\/sub>cos\u03b8)t
              \nfor x = R (range of the projectile),
              \nt = T= time of flight.
              \n\"1st
              \nHence R= \\(\\left(v_{0} \\cos \\theta\\right)\\left(\\frac{2 v_{0} \\sin \\theta}{g}\\right)\\)
              \nNote: R = \\(=\\frac{2}{\\mathrm{g}} v_{0 \\mathrm{x}} v_{0 \\mathrm{y}}\\)
              \ni.e., R = \\(\\frac{v_{0}^{2} \\sin 2 \\theta}{g}\\) where sin2\u03b8 = 2sin\u03b8 cos\u03b8
              \nHence R \u221d v0<\/sub>2 <\/sup>for the given angle of projection.<\/p>\n

              \"KSEEB<\/p>\n

              Question 20.
              \nWrite any three laws of friction.
              \nAnswer:<\/p>\n

                \n
              1. Limiting friction depends on the nature of the surfaces in contact but not on the surface area as long as normal reaction remains the same.<\/li>\n
              2. Force of friction always acts tangential to the surface and opposite the direction of motion.<\/li>\n
              3. The limiting friction is directly proportional to the normal reaction between two surfaces in contact.<\/li>\n
              4. The kinetic friction becomes constant, once there is a relative motion between the two surfaces.<\/li>\n
              5. \u03bcs<\/sub> >\u03bck<\/sub> > \u03bcr<\/sub> where \u03bcs<\/sub>, \u03bck<\/sub> and \u03bcr<\/sub> are coefficient of friction due to static friction, kinetic friction and rolling friction respectively.<\/li>\n<\/ol>\n

                Question 21.
                \nDeduce the work-energy theorem for a constant force.
                \nAnswer:
                \nThe difference of P.E. or K.E. required by a body at two different positions is equal to the amount of work done.
                \n\"1st
                \ni.e., W = Kf<\/sub> – Ki<\/sub>.<\/p>\n

                Question 22.
                \nDeduce the equations of motion of centre of mass.
                \nAnswer:
                \n\"1st<\/p>\n

                Question 23.
                \nName the three types of moduli of elasticity.
                \nAnswer:<\/p>\n

                  \n
                1. Young’s Modulus<\/li>\n
                2. Bulk Modulus<\/li>\n
                3. Rigidity Modulus.<\/li>\n<\/ol>\n

                  \"KSEEB<\/p>\n

                  Question 24.
                  \nWhat is capillary rise? Write the expression for height of capillary rise in the capillary tube with usual meanings.
                  \nAnswer:
                  \nThe phenomenon of rise or fall of a liquid in a capillary tube is known as capillarity.
                  \n\"1st<\/p>\n

                  Question 25.
                  \nObtain an expression for thermal stress.
                  \nAnswer:
                  \nY = \\(\\frac{\\text { stress }}{\\text { strain }}\\) so that stress = (Y) (strain)
                  \ni.e. stress = Y \u221d \u2206\u03b8
                  \nwhere stress represents thermal stress, Y – Young’s Modulus of elasticity.<\/p>\n

                  Question 26.
                  \nShow that specific heat capacity of a solid is equal to three times that of Gas constant (C = 3R).
                  \nAnswer:
                  \nFor solids U = 3RT.
                  \nAt constant pressure, \u2206Q = \u2206U
                  \n\u2234C = \\(\\frac{\\Delta \\mathrm{Q}}{\\Delta \\mathrm{T}}\\) = 3R .<\/p>\n

                  Part – D<\/span><\/p>\n

                  IV. Answer any TWO of the following questions : ( 2 \u00d7 5 = 10 )<\/span><\/p>\n

                  Question 27.
                  \nWhat is velocity – time graph. Deduce x = v0<\/sub>t +\\(\\frac{1}{2}\\)at2<\/sup> by using velocity-time graph.
                  \nAnswer:
                  \nA graph of velocity of a particle plotted along the Y-axis and time along the X-axis is known as v-t graph and the curve is known as velocity time curve.
                  \nLet \u2018v0<\/sub>\u2019 be the initial velocity of a particle. Let ‘a’ be the B uniform acceleration v – t graph gives a straight line with a constant slope, tan\u03b8 = m
                  \ni.e. a = \\(\\frac{v-v_{0}}{t}\\)……..(1)
                  \n\"1st
                  \nFrom the figure OABD is a trapezium.
                  \nArea of trapezium OABD = Area of rectangle OACD + Area of triangle ACB
                  \nThe second term on the right indicates the additional distance covered by the particle due to acceleration.
                  \ni.e., area of trapezium = (OA)(OD) + 1\/2 (AC)(BC) = v0<\/sub>t + 1\/2 t(v – v0<\/sub>) by using (1)
                  \nArea of trapezium = v0<\/sub>t + 1\/2 at2<\/sup>
                  \nFrom the dimensional analysis. the right hand terms indicate the distance travelled. From the principle of homogeneity, the left hand side term should indicate the distance covered.
                  \nHence area of the trapezium = x = v0<\/sub>t + 1\/2 at2<\/sup>
                  \nFor a uniform motion, a = 0 and x = v0<\/sub>t
                  \nFor a particle starting from rest v0 <\/sub>= 0, x = 1\/2 at2<\/sup>
                  \nin a vector form, \\(\\vec{x}=\\vec{v}_{0} t+\\frac{1}{2} \\overrightarrow{a t}^{2}\\)<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 28.
                  \nState and illustrate the law of conservation of linear momentum for any two colliding particles in a closed system.
                  \nAnswer:
                  \nStatement: In an isolated system of collision of bodies, the total linear momentum before impact is equal to the total linear momentum after impact.
                  \n\"1st
                  \nLet m1<\/sub> and m2<\/sub> be the masses of two bodies moving along \\(\\vec{v}_{1 i}\\) and \\(\\vec{v}_{2 i}\\). Let \\(\\vec{v}_{1 f}\\) and be the \\(\\vec{v}_{2 f}\\) be final velocities after the impact.
                  \nAt the time of impact the force of action acts on the body B and the force of reaction acts on A.
                  \nApplying Newton\u2019s III law of motion
                  \n\\( |Force of action on \\mathrm{B}|=-| Force of reaction on \\mathrm{A} |\\)
                  \n\"1st
                  \nThis shows that the total final linear momentum of the isolated system equals its total initial momentum.<\/p>\n

                  Question 29.
                  \nState and explain parallel axes and perpendicular axes theorems of moment of inertia.
                  \nAnswer:
                  \nStatement: The moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through the centre of mass and the product of mass of the body and square of the distance between the two parallel axes.
                  \nI = I0<\/sub> + Md2<\/sup>.
                  \nLet \u2018d\u2019 be the distance between the two parallel axes. Let I be the moment of inertia about an axis passing through the centre of mass of a thin rod of length L.
                  \n\"1st
                  \nWe know that I0<\/sub> = \\(\\frac{\\mathrm{ML}^{2}}{12}\\) and d = \\(\\frac{\\mathrm{L}}{2}\\)
                  \nHence moment of inertia at one end = \\(\\frac{M L^{2}}{12}+M\\left(\\frac{L}{2}\\right)^{2}\\)
                  \nI = \\(\\frac{M L^{2}}{4}\\left[\\frac{1}{3}+1\\right]=\\frac{M L^{2}}{3}\\)<\/p>\n

                  Statement : The moment of inertia about an axis perpendicular to two other axes acting in the same plane with their point of intersection being a point on it and the (third) axis passing through the common point, is equal to the sum of moments of inertia about the two axes.
                  \ne.g: Iz<\/sub> = Ix<\/sub> + Iy<\/sub>.
                  \nLet M be the mass of the disk of radius R.
                  \nM.I. about a point passing through the centre and perpendicular to the plane containing X and Y is I = \\(\\frac{M R^{2}}{2}\\)
                  \nSince X and Y are in the same plane, Ix<\/sub> = Iy<\/sub>
                  \n\"1st
                  \n\u2234 Iz<\/sub> = Ix<\/sub> + Iy<\/sub> becomes Iz<\/sub> = 2Ix<\/sub>
                  \nhence Ix<\/sub> = \\(\\frac{I_{Z}}{2}=\\frac{M R^{2}}{4}\\)
                  \ni.e. moment of inertia of a circular disc about the diameter \\(\\frac{I_{Z}}{2}=\\frac{M R^{2}}{4}\\)<\/p>\n

                  V. Answer any TWO of the following questions: ( 2 \u00d7 5 = 10 )<\/span><\/p>\n

                  Question 30.
                  \nState and explain the laws of thermal conductivity and hence mention the SI unit of coefficient of thermal conductivity.
                  \nAnswer:
                  \nThe quantity of heat conducted during the steady state is directly proportional to the,
                  \n(i) Area of cross-section of a conductor.
                  \n(ii) Temperature difference between the two points or ends (\u03b81<\/sub> – \u03b82<\/sub>).
                  \n(iii) Time of passage of heat (t) and inversely proportional to the thickness or distance between the two ends of the conductor (d).
                  \ni.e., Q = \\(\\frac{\\mathrm{KA}\\left(\\theta_{1}-\\theta_{2}\\right) \\mathrm{t}}{\\mathrm{d}}\\)
                  \nwhere \u2019K’ is thermal conductivity of the material of the conductor.
                  \nThe SI unit of thermal conductivity is Wm-1 <\/sup>K-1<\/sup>.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 31.
                  \nDeduce the expression for energy stored in a body executing simple harmonic motion.
                  \nAnswer:
                  \nWe know that total energy of a particle executing SHM,
                  \nTE = KE + PE
                  \nwhere K.E= \\(\\frac{1}{2}\\) mv2<\/sup>
                  \nK.E = \\(\\frac{1}{2}\\)mA2<\/sup> cos2<\/sup> (\u03c9t + \u03a6)\u03c92<\/sup>
                  \nand P.E = \\(\\frac{1}{2}\\) ky2<\/sup>
                  \ni.e., P.E = \\(\\frac{1}{2}\\) m\u03c92<\/sup>. A2<\/sup>sin2<\/sup>(\u03c9t + \u03a6)
                  \nHence T.E = \\(\\frac{1}{2}\\) m\u03c92<\/sup>. A2 <\/sup>(cos2<\/sup>(cost + \u03a6) + sin2<\/sup> (\u03c9t + \u03a6))
                  \nor T.E = \\(\\frac{1}{2}\\) m\u03c92<\/sup>. A2<\/sup>
                  \nRepresenting m\u03c92 <\/sup>= k
                  \nwe get TE = \\(\\frac{1}{2}\\) kA2<\/sup>
                  \nK.E of the particle=\\(\\frac{1}{2}\\) m\u03c92<\/sup>. A2 <\/sup>– \\(\\frac{1}{2}\\) m\u03c92<\/sup>. x2<\/sup>
                  \n= \\(\\frac{1}{2}\\) m\u03c92<\/sup>(A2 <\/sup>-x2 <\/sup>)
                  \nwhere x = A cos (\u03c9t + \u03a6)
                  \nAlso K.E = \\(\\frac{1}{2}\\) m\u03c92<\/sup>(A2<\/sup>)(1 – cos2<\/sup>(\u03c9t + \u03a6))
                  \ni.e., K.E = \\(\\frac{1}{2}\\) m\u03c92 <\/sup>A2<\/sup>sin2<\/sup>(\u03c9t + \u03a6)
                  \nK(t) – K.E of a particle.
                  \n\"1st
                  \nU(t) of a particle.
                  \nT.E = K(t) + U(t) = \\(\\frac{1}{2}\\) kA2<\/sup>
                  \nwhere A – amplitude, k = m\u03c92<\/sup>, m – mass of the particle
                  \n\u03c9 = angular frequency = \\(\\frac{2 \\pi}{\\mathrm{T}}\\)
                  \nT = Period of SHM
                  \nTE = KE + PE = \\(\\frac{1}{2}\\)m\u03c92<\/sup> A2<\/sup> cos2<\/sup> (\u03c9t + \u03a6) + \\(\\frac{1}{2}\\) m\u03c92 <\/sup>A2<\/sup>sin2<\/sup>(\u03c9t + \u03a6) = \\(\\frac{1}{2}\\) m\u03c92 <\/sup>A2<\/sup><\/p>\n

                  Question 32.
                  \nDiscuss the mode of vibration of air columns in a closed pipe and hence define the fundamental frequency of vibration.
                  \nAnswer:
                  \nLet ‘L’ be the length of the closed pipe. A pipe with one end closed is known as a closed pipe system. Le V be the velocity of sound in air. Length of half segment = \\(\\frac{\\lambda_{0}}{4}\\)
                  \ni.e., L = \\(\\frac{\\lambda_{0}}{4}\\) i.e., \u03bb0<\/sub> = 4L.
                  \n\"1st
                  \nThe least mode of vibration is called fundamental mode i.e.,
                  \nfundamental frequency f0<\/sub> = \\(\\frac{\\mathbf{v}}{\\lambda}=\\frac{\\mathbf{v}}{4 L}\\)
                  \nIn the second mode of vibration.
                  \n\"1st
                  \ni.e., f2 <\/sub>= \\(\\frac{5 v}{4 L}\\)
                  \nHence f0<\/sub> : f1<\/sub> : f2<\/sub> :……..: fn<\/sub> : : 1 : 3 : 5 :………: (2n – 1)<\/p>\n

                  VI. Answer any THREE of the following questions: ( 3 \u00d7 5 = 15 )<\/span><\/p>\n

                  Question 33.
                  \nA stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 revolutions per minute in a horizontal plane. What is the tension in the spring?
                  \nAnswer:
                  \nGiven m = 0. 25 kg, R = 1.5 m, co = 40 rev. min-1<\/sup>
                  \ni.e., \u03c9 = \\(\\frac{40 \\times 2 \\pi}{60}\\) rad s-1<\/sup>
                  \n\u03c9\u00a0= \\(\\frac{40 \\times 2 \\pi}{60}\\)
                  \n\u03c9 = \\(\\frac{4 \\times 3.14}{3}\\) = 4.19 rad\/s-1<\/sup>
                  \nLinear speed = v = rw = 1.5 \u00d7 4.19 = 6.285 ms-1<\/sup>
                  \n\"1st
                  \nT = 6.584 N<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 34.
                  \nA pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 minute. If the tank is 40 m above the ground and efficiency of the pump is 30%, how much of electrical power is consumed by the pump?
                  \nAnswer:
                  \nGiven V = 30 m3<\/sup>, t = 15 min = 15 \u00d7 60 = 900s
                  \nh = 40 m, \u03b7 = 30% , density \u03c1 = 103<\/sup> kg m-3<\/sup>, g = 10 ms-2<\/sup>
                  \nPotential Energy = mgh = V\u03c1gh = 30 \u00d7 103<\/sup> \u00d7 10 \u00d7 40
                  \ni.e. Work done = 12 \u00d7 106<\/sup>J.
                  \n\"1st
                  \nTrue power required = \\(\\frac{100}{30}\\) \u00d7 1.33 \u00d7 104<\/sup> = 4.433 \u00d7 104<\/sup>\u00a0W<\/p>\n

                  Question 35.
                  \nIf two spheres of equal masses with their centres 0.2 m apart attract each other with a force of 1 \u00d7 10-6<\/sup> kg wt. What wrould be the value of their masses?
                  \ng = 9.8ms-2<\/sup> and G = 6.67 \u00d7 10-11<\/sup> Nm2<\/sup>kg-2<\/sup>.
                  \nAnswer:
                  \nGiven r = 0.2 m,\u00a0 F = 1 \u00d7 10-6<\/sup> kgwt =1 \u00d7 10-6<\/sup> \u00d7 9.8 = 9.8 \u00d7 106 <\/sup>N
                  \nG = 6.67 \u00d7 10-11<\/sup> Nm2 <\/sup>kg-2<\/sup>, m1<\/sub> = m2<\/sub> = m
                  \n\"1st
                  \ni.e., mass used is 0.511 \u00d7 103<\/sup> kg<\/p>\n

                  Question 36.
                  \nThe sink in Carnot’s heat engine is at 300K and the engine works at an efficiency 0.4. If the efficiency of the engine is to be increased to 0.5. Find by how many Kelvin the temperature of the source should be increased.
                  \nAnswer:
                  \nGiven \u03b7 = 0.4,\u00a0 n1<\/sup>\u00a0= 0.5,\u00a0 T = 300K
                  \n\"1st
                  \nThe source temperature should be increased by 300 K.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 37.
                  \nA train is moving at a speed of 72 kmph towards a station sounding a whistle, of frequency 640 Hz. What is the apparent frequency of the whistle as heard by a man standing on the platform when train approaches towards him.
                  \nGiven : Speed of sound = 340 ms-1<\/sup>.
                  \nAnswer:
                  \nVs<\/sub> = 72 kmph = 20 ms-1\u00a0<\/sup> f = 640 Hz, V = 340 ms-1<\/sup>
                  \nW.K.T. apparent frequency
                  \n\"1st
                  \n\u2234f1<\/sup>\u00a0= 680 Hz
                  \nApparent frequency of sound as heard by the observer will be 680 Hz.<\/p>\n","protected":false},"excerpt":{"rendered":"

                  Students can Download 1st PUC Physics Previous Year Question Paper March 2019 (North), Karnataka 1st PUC Physics Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 1st PUC Physics Previous Year Question Paper March 2019 (North) Time: 3.15 Hours Max Marks: …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/21736"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=21736"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/21736\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=21736"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=21736"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=21736"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}