{"id":20110,"date":"2020-01-17T16:35:55","date_gmt":"2020-01-17T11:05:55","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=20110"},"modified":"2021-07-02T15:28:51","modified_gmt":"2021-07-02T09:58:51","slug":"1st-puc-physics-model-question-paper-4","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-physics-model-question-paper-4\/","title":{"rendered":"1st PUC Physics Model Question Paper 4 with Answers"},"content":{"rendered":"

Students can Download 1st PUC Physics Model Question Paper 4 with Answers, 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 Model Question Paper 4 with Answers<\/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 the following questions: ( 10 \u00d7 1 = 10)<\/span><\/p>\n

    Question 1.
    \nName the experiment which established the nuclear model of atom.
    \nAnswer:
    \nGeiger-Marsdon experiment on alpha-particle scattering by gold foil.<\/p>\n

    Question 2.
    \nWhat is S.I. unit of luminous intensity?
    \nAnswer:
    \nThe SI unit luminous intensity is candela (cd).<\/p>\n

    Question 3.
    \nGive an example for two dimensional motion.
    \nAnswer:
    \nProjectile motion and circular motion.<\/p>\n

    Question 4.
    \nWhy don\u2019t action and reaction forces cancel each other?
    \nAnswer:
    \nAction and reaction forces act on two interacting bodies, so the forces don\u2019t cancel each other.<\/p>\n

    \"KSEEB<\/p>\n

    Question 5.
    \nA light body and a heavy body have the same momentum. Which one will have greater kinetic energy?
    \nAnswer:
    \nSince K.E. \u221d \\(\\frac{1}{m}\\) for same linear momentum, lighter body has more kinetic energy.<\/p>\n

    Question 6.
    \nGive the expression for moment of inertia of a circular disc of radius R about its diameter.
    \nAnswer:
    \nI = \\(\\frac{M R^{2}}{4}\\)<\/p>\n

    Question 7.
    \nWhat are geostationary satellites?
    \nAnswer:
    \nGeo-stationary satellites orbit around the Earth with the speed of the satellite equal to the rotational speed of the Earth. Hence, these satellites appear to be fixed with respect to a place on the surface of the Earth.<\/p>\n

    Question 8.
    \nWhat is shear deformation?
    \nAnswer:
    \nShear deformation is the ratio of lateral displacement of one edge to the vertical distance between the opposite faces of the body.<\/p>\n

    Question 9.
    \nHow does melting point of ice changes with increase of pressure?
    \nAnswer:
    \nThe ice will go to a lower melting point.<\/p>\n

    \"KSEEB<\/p>\n

    Question 10.
    \nGive an example for a wave which can travel through vacuum.
    \nAnswer:
    \nElectromagnetic wave.<\/p>\n

    Part – B<\/span><\/p>\n

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

    Question 11.
    \nGiven the relative error in the measurement of the radius of a circle is 0.02, whal is the percentage error in the measurement of its area?
    \nAnswer:
    \nA = \u03c0r2<\/sup>
    \n\\(\\frac{\\Delta \\mathrm{A}}{\\mathrm{A}}=2 \\mathrm{r} \\frac{\\Delta r}{r}\\)
    \n= 2 \u00d7 r \u00d7 0.02 = 0.04r
    \n(\\(\\frac{\\Delta \\mathrm{A}}{\\mathrm{A}}\\)) = 0.04r<\/p>\n

    Question 12.
    \nWhat are the significance of velocity – time graph?
    \nAnswer:
    \nThe slope of the line on a v-t graph gives the acceleration of the particle.<\/p>\n

    Question 13.
    \nWhat is resolution of a vector? What is the x-component of a vector A, that makes an angle 300 with x-axis.
    \nAnswer:
    \nA single vector can be resolved in two or more number of directions each of which is known as the component of a vector. This splitting is known as resolution of vector.
    \n\\(\\overrightarrow{\\mathrm{A}}_{x}=\\left(A \\cos 30^{\\circ} \\hat{i}\\right)=\\frac{\\sqrt{3}}{2} \\hat{i}\\)<\/p>\n

    \"KSEEB<\/p>\n

    Question 14.
    \nGive the general conditions of equilibrium of a rigid body.
    \nAnswer:
    \n(1) The vector sum of the forces on the rigid body is zero for translatory equilibrium
    \n\\(\\sum_{i=1}^{n} \\vec{F}_{i}=0\\) and (acceleration a = 0 )
    \n(2) The vector sum of the torques on the rigid body is zero for rotatory equilibrium (a=0) i.e.
    \n\"1st
    \ni.e., the components of X, Y, and Z independently vanish to zero for linear equilibrium.
    \n\"1st
    \nSum of X components, Y components and Z components of torque on the particles, vanish for rotational equilibrium.<\/p>\n

    Question 15.
    \nWrite Stoke\u2019s formula for viscous drag force. Explain the terms.
    \nAnswer:
    \nF = 6\u03c0\u03b7av where v – terminal speed of liquid, \u03b7 – coefficient of viscosity, a – radius of the spherical object and F – viscous drag force.<\/p>\n

    Question 16.
    \nWhat is meant by anomalous behavior of water. What is its significance?
    \nAnswer:
    \nThe density of water increases from 0\u00b0C to 4\u00b0C and thereafter decreases with increase of temperature.
    \n\"1st
    \nThis behaviour of water from 0\u00b0C to 4\u00b0C is known as anomalous expansion of water. Water at 4\u00b0C has the maximum density and sinks down. However, water below the surface, relatively having lower density, rises above. Water becomes lighter as the temperature falls below 4\u00b0C. Ice at 0\u00b0C has the minimum density and floats on water. Hence marine animals can survive below ice. It is water everywhere below ice.<\/p>\n

    \"KSEEB<\/p>\n

    Question 17.
    \nState and explain first law of thermodynamics.
    \nAnswer:
    \nThe energy (\u2206Q) supplied to the system goes in partly to increase the internal energy of the system (\u2206U) and the rest in doing work on the environment (\u2206W).
    \n\u2206Q = \u2206U + \u2206W
    \nwhere \u2206Q \u2192 heat supplied, \u2206U \u2192 change in internal energy, \u2206W \u2192 work done<\/p>\n

    Question 18.
    \nWhat is periodic motion. Give an example.
    \nAnswer:
    \nA particle event that repeats itself in regular intervals of time is known as the periodic motion. Both circular and elliptical motions of electrons around the atomic nucleus are examples of periodic motions.<\/p>\n

    Part – C<\/span><\/p>\n

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

    Question 19.
    \nDerive the expression for time of flight of a projectile.
    \nAnswer:
    \nExpression for time of flight:
    \n\"1st
    \nConsider v = (v0<\/sub> sin\u03b8) – gt
    \nPut v = 0, fort = time of ascent = ta
    \n<\/sub>ta<\/sub> = \\(\\frac{v_{0} \\sin \\theta}{g}\\)
    \nbut time of ascent = time of descent
    \nand time of flight T = ta <\/sub>+ td<\/sub>
    \ni.e., T = \\(\\frac{2 v_{0} \\sin \\theta}{g}\\) and T \u221d v0<\/sub> for a given angle of projection.<\/p>\n

    \"KSEEB<\/p>\n

    Question 20.
    \nArrive at the statement of principle of conservation of linear momentum from Newton\u2019s laws of motion.
    \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 21.
    \nProve that potential energy stored in a spring is, where k is the force constant of the spring and x is the change in length Of the spring.
    \nAnswer:
    \nWork done by the spring force = Ws= \\(-\\int_{0}^{x_{\\mathrm{m}}} \\mathrm{F}_{\\mathrm{s}} \\mathrm{d} \\mathrm{x}\\)
    \nW = \\(=-\\int_{0}^{x_{m}} \\mathrm{k} x \\mathrm{d} \\mathrm{x}\\)
    \nW = \\(-\\frac{1}{2} \\mathrm{k} x_{\\mathrm{m}}^{2}\\)
    \nWork done by the external pulling force = W= \\(\\frac{1}{2} \\mathrm{k} x_{\\mathrm{m}}^{2}\\)<\/p>\n

    Question 22.
    \nState Kepler\u2019s laws of planetary motion.
    \nAnswer:
    \nKepler\u2019s I law (Law of orbit): All planets revolve in elliptical orbits with Sun as one of its foci.
    \n\"1st
    \n2a – Major axis length
    \n2b – Minor axis length
    \nS-Sun at one focus P
    \nS1<\/sup> – The other focus of the ellipse
    \nP – Perihelion position of the planet.
    \nA – Apehelion position of the planet.<\/p>\n

    Kepler\u2019s II law (Law of areas) : The line joining the planet and the Sun sweeps out equal areas in equal intervals of time.
    \n\\(\\frac{\\Delta \\overrightarrow{\\mathrm{A}}}{\\Delta \\mathrm{t}}=\\frac{1}{2 \\mathrm{m}} \\overrightarrow{\\mathrm{L}}\\)<\/p>\n

    \"1st
    \nKepler\u2019s III law (Law of periods) : The square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi major axis of the ellipse.
    \ni.e., T2<\/sup> \u221d a3<\/sup> so that for two planets
    \n\\(\\left(\\frac{T_{1}}{T_{2}}\\right)^{2}=\\left(\\frac{a_{1}}{a_{2}}\\right)^{3}\\)<\/p>\n

    Question 23.
    \nProve that the centre of mass of a system moves with constant velocity in the absence of external force on the system.
    \nAnswer:
    \nWhen there is no external force the acceleration of the body is zero.
    \nForce = rate of change of momentum.
    \n\"1st
    \n\u2234M is not zero
    \nso this proves that velocity will always be constant in such scenario.<\/p>\n

    Question 24.
    \nState and explain Hooke\u2019s law. Define modulus of elasticity.
    \nAnswer:
    \nStatement: The ratio of stress to strain is a constant for a material within the elastic limit.
    \nModulus of elasticity = \\(\\frac{\\text { Stress }}{\\text { Strain }}\\)<\/p>\n

    Within the elastic limit, stress v\/s strain is a straight line \u2018A\u2019 is the elastic limit upto which Hooke\u2019s law is applicable. Beyond \u2018B\u2019 the yielding point, the wire extends but does not return to the initial state when the deforming force is removed. \u2018F\u2019 is the breaking point. \u2018EF\u2019 allows the material to be malleable and \u2018DE\u2019, ductile.
    \n\"1st<\/p>\n

    \"KSEEB<\/p>\n

    Question 25.
    \nState Bernoulli\u2019s theorem. What are its applications?
    \nAnswer:
    \nAlong a steamline, in a steady flow of non viscous fluid, potential energy, kinetic energy and pressure energy remain constant.
    \nApplications<\/p>\n

      \n
    • Uplift of air craft and<\/li>\n
    • Atomisers work on the principle of Bernoulli\u2019s theorem.<\/li>\n<\/ul>\n

      Question 26.
      \nWhat is Doppler effect. Write any two applications of Doppler\u2019s effect.
      \nAnswer:
      \nThe phenomenon of apparent change in frequency due to relative motion between the source of sound and the listener (observer) is known as Doppler effect.
      \nApplications of Doppler effect:<\/p>\n

        \n
      1. It is used in radar.<\/li>\n
      2. It is used by meteorologists to track storms.<\/li>\n
      3. Doppler effect of sound is used to diagnose heart problems in the hospital.<\/li>\n<\/ol>\n

        Part – D<\/span><\/p>\n

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

        Question 27.
        \nFind the magnitude and direction of the resultant of two vectors A and B in terms of their magnitudes and angle, between them.
        \nAnswer:
        \nTo find the magnitude of the resultant:
        \nFrom the OAC, \\(\\vec{P}+\\vec{Q}=\\vec{R}\\)
        \nCD is \u22a5r<\/sup>\u00a0to OD. From the pythagoras theorem,
        \nOC2<\/sup> = OD2<\/sup> + DC2<\/sup> ……….(1)
        \nbut DC2<\/sup> = AC2<\/sup> + AD2<\/sup> ……..(2)
        \nFrom the right angled triangle ADC,
        \nsin\u03b8 = CD\/AC
        \n\u2234AC sin\u03b8 = CD …….(3)
        \ncos\u03b8 = AD \/AC
        \nie., AD = AC cos\u03b8 ……..(4)
        \nEqn. (1) can be written as
        \nOC2<\/sup> = (OA + AD)2<\/sup> + DC2<\/sup>
        \nsin OC2<\/sup> = OA2<\/sup> + AD2<\/sup> + 20A.AD + DC2<\/sup>
        \nusing (2) and (4) we get
        \nOC2<\/sup> = OA2<\/sup> + AC2<\/sup> + 20A. AC cos\u03b8
        \ni.e. R2<\/sup> + P2<\/sup> + Q2<\/sup> + 2PQ cos\u03b8
        \nHence, R = (P2<\/sup> + Q2<\/sup> + 2PQ cos\u03b8)1\/2<\/sup>
        \nTo find the direction of the resultant:
        \n\"1st<\/p>\n

        \"KSEEB<\/p>\n

        Question 28.
        \nState Newton\u2019s second law of motion and hence derive F= ma.
        \nAnswer:
        \nNewton\u2019s II law of motion:
        \nStatement: \u201cThe rate of change in the linear momentum of a body is directly proportional to the impressed force and takes place in the direction of force applied.
        \nTo show that \\(\\vec{F}=m \\vec{a}\\).
        \nLet m be the mass of the body. Let be the initial linear momentum. Let be the final linear momentum as a result of the impressed force.
        \nBy definition \\(\\lim _{\\Delta t \\rightarrow 0} \\frac{\\Delta \\vec{p}}{\\Delta t}=\\frac{d \\vec{p}}{d t}\\) where \\(\\Delta \\vec{p}=\\vec{p}_{f}-\\vec{p}_{i}\\), and \\(\\frac{d \\vec{p}}{d t}\\) is instantaneous acceleration.
        \nFrom Newton\u2019s II law of motion,
        \n\"1st
        \nBut the magnitude of force \\(|\\overrightarrow{\\mathrm{F}}|\\) is so defined that k = 1 : We denote \\(|\\vec{F}|=m|\\vec{a}|\\)
        \nThus F=ma and \\(\\vec{F}=m \\vec{a}\\).<\/p>\n

        Question 29.
        \nDefine torque. Show that torque on a particle is equal to rate of change of its angular momentum.
        \nAnswer:
        \nThe rotating effect of force is called torque and is represented by \\(\\vec{\\tau}=\\overrightarrow{\\mathrm{r}} \\times \\overrightarrow{\\mathrm{F}}\\).
        \ni.e. t=rFsin8
        \nwhere O is the angle between r and F.
        \nFora system of n particles,
        \nNet angular momentum = \\(\\overrightarrow{\\mathrm{L}}=\\overrightarrow{\\mathrm{L}_{1}}+\\overrightarrow{\\mathrm{L}_{2}}+\\ldots \\ldots . \\overrightarrow{\\mathrm{L}_{n}}=\\sum \\overrightarrow{\\mathrm{r}_{1}} \\times \\overrightarrow{\\mathrm{p}_{1}}\\)
        \n\"1st<\/p>\n

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

        Question 30.
        \nUsing kinetic theory of gases derive an expression for finding the pressure of an ideal gas in terms of mean squared speeds of the molecules.
        \nAnswer:
        \nLet an ideal gas be enclosed in a cubical vessel of side l. Area of cross wall will be A =l2<\/sup>. Let Vx<\/sub> be the velocity of the molecule hitting the plane of wall (of the vessel) YZ. After collision, it rebounds with the same speed but in the opposite direction along a straight line. The change in ‘ velocity of the molecule along the X direction is vx<\/sub> – (-vx<\/sub>) = 2vx<\/sub>.
        \nThe change in linear momentum imparted to the wall in the collision process = 2.mvx where m is the mass of the molecule.
        \nBut force exerted on the wall = rate of change in momentum = \\(\\frac{2 m v_{x}}{\\Delta t}\\)
        \nPressure exerted on the wall due to 1 molecule = \\(\\frac{\\text { Force }}{\\text { Area }}\\)
        \n\u2234Pressure due to single molecule = \\(\\frac{2 m v_{x}}{\\left(A^{2}\\right) \\Delta t}\\) where A = l2<\/sup>.
        \nTotal number of molecules hitting the wall and returning back = (Volume) (number density) Distance travelled by the molecule = (vx<\/sub>\u2206t)
        \nVolume covered = (vx<\/sub>\u2206t)(A)
        \nTotal number of molecules hitting the wall on an average = \\(\\frac{1}{2}\\) (vx\u2206tA)(n)
        \nHence pressure exerted on the wall = P
        \nP = \\(\\left(\\frac{2 m v_{x}}{A \\Delta t}\\right)\\left(\\frac{1}{2} v_{x} \\Delta t A n\\right)\\)
        \ni.e., P = mnvx<\/sub>2<\/sup>
        \n\"1st
        \nBy symmetry (isotropic condition of speed)
        \n\\(\\overline{\\mathbf{v}}_{x}^{2}=\\overline{\\mathbf{v}}_{y}^{2}=\\overline{\\mathbf{v}}_{z}^{2}\\)
        \nSo, the mean of the squred speed along any one axis
        \n= (\\(\\frac{1}{3}\\)) \\(\\overline{\\mathbf{v}^{2}}\\)
        \nThus P = mn (\\(\\frac{1}{3}\\)\\(\\overline{\\mathbf{v}^{2}}\\)).
        \ni.e., P = \\(\\frac{1}{3}\\)mn\\(\\overline{\\mathbf{v}^{2}}\\)<\/p>\n

        \"KSEEB<\/p>\n

        Question 31.
        \nDescribe Carnot\u2019s cycle and write the expression for its efficiency.
        \nAnswer:
        \nWorking of Carnot\u2019s engine:
        \nStep 1 : The working substance (ideal gas) is enclosed \u00a1n a non-conducting wall and conducting bottom of a cylinder fitted with air tight non conducting piston. This is placed on the source having an infinite thermal capacity at a steady temperature.
        \nThe top surface is conducting and the rest non conducting. As a result, the gas expands isothermally. The work done
        \nby the system.
        \nQ = W1<\/sub> = \u03bcRT1<\/sub> log\\(\\frac{V_{2}}{V_{1}}\\)(curve A B)
        \n\"1st<\/p>\n

        Step 2 : The working substance is now placed on a non L conducting platform. as a result ofwhich no heat exchange takes place between the system and the surroundings. [he system expands adiabatically at the expense of its internal energy. The gas cools. The work done by the system.
        \n\\(\\mathrm{W}_{2}=\\frac{\\mu \\mathrm{R}}{\\gamma-1}\\left(\\mathrm{T}_{1}-\\mathrm{T}_{2}\\right)\\)(curve B C)<\/p>\n

        Step 3 : The working substance is now placed on the sink maintained at a steady low temperature T,K. The system undergoes isothermal compression at this temperature. The pressure of gas increases and volume decreases without any change in the internal energy and specific heat of gas remains at infinity.
        \n\\(\\mathrm{W}_{3}=\\mu \\mathrm{RT}_{2} \\log \\left(\\frac{\\mathrm{V}_{4}}{\\mathrm{V}_{3}}\\right)\\) (curve C D)
        \n\"1st<\/p>\n

        Step 4: The working substance is placed on a non-conducting platform, Under thermal isolation. the system undergoes change in its internal energy and its specific heat remains at zero. Adiabatic compression results in increase in the pressure and temperature at the expense of work being done on the system. The system is allowed to reach its initial slate. This completes one cycle of operation. The area hounded by the curves gives the amount of heat converted into work.
        \n\\(\\mathrm{W}_{4}=\\frac{-\\mu \\mathrm{R}}{\\gamma-1}\\left(\\mathrm{T}_{1}-\\mathrm{T}_{2}\\right)\\) (curve D A)
        \ni.e., Net work done = W =W1<\/sub> + W2<\/sub> +W3<\/sub> + W4<\/sub>
        \nWork done = \u03bcR(T – T2<\/sub>) \\(\\log _{e}\\left(\\frac{V_{2}}{V_{1}}\\right)\\)
        \n\u2234Efficiency \u03b7 = \\(\\frac{\\mathrm{w}}{\\mathrm{Q}}=1-\\frac{\\mathrm{T}_{2}}{\\mathrm{T}_{1}}\\)<\/p>\n

        Question 32.
        \nDerive the Newton\u2019s formula to find the speed of a longitudinal wave in an ideal gas. What is the Laplace correction to obtain the speed of sound in air?
        \nAnswer:
        \nv = \\(\\sqrt{\\frac{p}{\\rho}}\\) where \u2018p\u2019is pressure on gaseous medium. \u03c1 – density of gaseous medium.
        \nConsider adiabatic variation in the pressure and volume of gaseous medium.
        \nPV = constant.
        \nSince no change of heat takes place between the system and the surrounding.
        \nA(PV\u03b3<\/sup>) = 0
        \ni.e., \u03b3PV\u03b3-1\u00a0\u00a0<\/sup> \u2206V+V\u03b3\u00a0 \u00a0<\/sup> \u2206P = 0
        \ni.e., Bulk modulus = \\(-\\frac{\\Delta \\mathrm{P}}{\\frac{\\Delta \\mathrm{V}}{\\mathrm{V}}}\\) = \u03b3P
        \nHence in the expression V = \\(\\sqrt{\\frac{\\mathrm{E}}{\\rho}}\\), Elasticity \u2018E is replaced by \u03b3P.
        \nLaplace equation is written as v = \\(=\\sqrt{\\frac{\\gamma \\mathrm{P}}{\\rho}}\\).
        \nWhere \u03b3 = 1 + 2\/f and \u2018f\u2019 is number of degrees of freedom of gaseous molecule.<\/p>\n

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

        Question 33.
        \nTwo towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?
        \nAnswer:
        \nThe distance between two buses just when one takes is vT.
        \nu = 20 km\/hr
        \nWhen they move in the same direction, relative velocity = v – u
        \nThe time to cover the distance,
        \nt= 18 mm = \\(\\frac{18}{60}\\)hrs
        \ndistance = t (v-u)
        \nt (v-u) = vT
        \nvT = \\(\\frac{3}{10}\\)(v-u)……..(1)
        \nWhen they move in the opposite direction, relative velocity = v + u
        \n\u2234Time to cover the distance, t = 6 minutes = \\(\\frac{6}{60}\\)hrs
        \n\u2234t (v + u) = vT
        \nvT = \\(\\frac{1}{10}\\)(v + u)……..(2)
        \nEquating vT from the eqn (1) and (2)
        \n\\(\\frac{3}{10}\\)(v-u) = \\(\\frac{1}{10}\\)(v + u)
        \n3v – 3u = v + u
        \n\u2234v = 2u
        \nv = 40 km\/hr …….(3)
        \nSubstituting in eqn (2)
        \n40 xT = \\(\\frac{1}{10}\\)(40 + 20)
        \nT = \\(\\frac{3}{20}\\)hrs
        \nT = 9 minutes.<\/p>\n

        \"KSEEB<\/p>\n

        Question 34.
        \nA person trying to lose weight(dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 in each time. Assume that the potential energy lost each time she lowers the mass is dissipated, (a) How much work does she do against the gravitational force? (b) Fat supplies 3.8 \u00d7 107<\/sup> J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?
        \nAnswer:
        \nGiven m = 10 kg, g = 9.8 ms-2<\/sup> , h = 0.5m, n = 1000
        \nEnergy supplied by fat is 3.8 \u00d7 107<\/sup> J \/kg, efficiency = 20%
        \n\u223420 % of energy supplied by fat = 3.8 \u00d7 107<\/sup> \u00d7 0.20 = 0.76 \u00d7 10-7<\/sup> Jkg-1<\/sup>
        \nWork done against the gravity = n mgh =103<\/sup> \u00d7 10 \u00d7 9.8 \u00d7 0.5 = 4.9 \u00d7 104<\/sup>J
        \nMass of fat used = \\(\\frac{4.9 \\times 10^{4}}{0.76 \\times 10^{7}}\\) = 6.45 \u00d7 10-3<\/sup> kg<\/p>\n

        Question 35.
        \nPhobos is a satellite of the planet mars. Phobos has a period 7 hours, 39 minutes and an orbital radius of 9.4 \u00d7 103<\/sup> km. calculate the mass of mars. Assume that earth and mars move in circular orbits around the sun, with martian orbit being 1.52 times the orbital radius of earth. What 1 the length of the martian year in days?
        \nAnswer:
        \nT = \\(\\frac{4 \\pi^{2}}{\\mathrm{GM}_{m}} \\mathrm{R}^{3}\\) M = \\(\\frac{4 \\pi^{2}}{\\mathrm{G}} \\frac{\\mathrm{R}^{3}}{\\mathrm{T}^{2}}\\)
        \nMm<\/sub> = \\(=\\frac{4 \\times(3.14)^{2} \\times(9.4)^{3} \\times 10^{18}}{6.67 \\times 10^{-11} \\times(459 \\times 60)^{2}}\\)
        \nMm<\/sub> = 6.48 \u00d7 10 kg
        \nUsing Kepler\u2019s III law,
        \n\\(\\frac{\\mathrm{T}_{\\mathrm{M}}^{2}}{\\mathrm{T}_{\\mathrm{E}}^{2}}=\\frac{\\mathrm{R}_{\\mathrm{MS}}^{3}}{\\mathrm{R}_{\\mathrm{ES}}^{3}}\\)
        \nWhere RMS<\/sub> \u2192 Mars sun distance
        \nRES<\/sub> \u2192 Earth – sun distance
        \n\u2234TM<\/sub> = \\((1.52)^{\\frac{3}{2}}\\) \u00d7 365
        \nTM<\/sub> = 684 days.<\/p>\n

        Question 36.
        \n\u2018Thermacole\u2019 icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6h. The outside temperature is 45\u00b0 C, and co-efficient of thermal conductivity of thermacole is 0.01 J s-1<\/sup>m-1<\/sup>K-1<\/sup>. [ Heat of fusion of water = 335 \u00d7 10 J kg].
        \nAnswer:
        \nGiven, L = 3.36 \u00d7 105 Jkg-1\u00a0 \u00a0\u00a0<\/sup> K = 0.01 Wm-1<\/sup>K-1<\/sup>
        \nd = 0.05 m, t = 6 \u00d7 3600 = 21600s
        \nA = 0.3 \u00d7 0.3 = 0.09 m2<\/sup>, \u03b81<\/sub> \u2248 \u03b82<\/sub>,= 45\u00b0C
        \nWe know that Q = \\(\\frac{\\mathrm{KA}\\left(\\theta_{1}-\\theta_{2}\\right) \\mathrm{t}}{\\mathrm{d}}\\) = mL
        \n\"1st<\/p>\n

        \"KSEEB<\/p>\n

        Question 37.
        \nA spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?
        \nAnswer:
        \nSpring constant = \\(\\frac{50 \\mathrm{kg}}{0.2 \\mathrm{m}}\\) =250 kgm-1<\/sup>
        \nWe know that T = \\(=2 \\pi \\sqrt{\\frac{m}{k}}\\) ; T = 0.6s (given)
        \ni.e., m = \\(\\frac{\\mathrm{kT}^{2}}{4 \\pi^{2}}=\\frac{250 \\times(0.6)^{2}}{4 \\times(3.142)^{2}}\\)
        \nm = 2.279 kg.<\/p>\n","protected":false},"excerpt":{"rendered":"

        Students can Download 1st PUC Physics Model Question Paper 4 with Answers, 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 Model Question Paper 4 with Answers Time: 3.15 Hours Max Marks: 70 General …<\/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\/20110"}],"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=20110"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20110\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=20110"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=20110"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=20110"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}