{"id":20096,"date":"2020-01-17T15:25:55","date_gmt":"2020-01-17T09:55:55","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=20096"},"modified":"2021-07-02T15:28:51","modified_gmt":"2021-07-02T09:58:51","slug":"1st-puc-physics-model-question-paper-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-physics-model-question-paper-3\/","title":{"rendered":"1st PUC Physics Model Question Paper 3 with Answers"},"content":{"rendered":"

Students can Download 1st PUC Physics Model Question Paper 3 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 3 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.
    \nWhat is physics?
    \nAnswer:
    \nPhysics deals with the laws of nature and natural phenomena associated with matter and energy.<\/p>\n

    Question 2.
    \nGive the number of significant figures in 5.300 \u00d7 103<\/sup>.
    \nAnswer:
    \nFour<\/p>\n

    Question 3.
    \nWhat is the angle between velocity and acceleration at the peak point of a projectile projected for maximum range?
    \nAnswer:
    \n– 90\u00b0 or 270\u00b0
    \n\"1st<\/p>\n

    Question 4.
    \nWhat is the use of mechanical advantage of a lever?
    \nAnswer:
    \nThe mechanical advantage of the lever is the ratio of output force to input force. By using lever with a little effect more work can be derived.<\/p>\n

    \"KSEEB<\/p>\n

    Question 5.
    \nName the experiment to measure the value of gravitational constant.
    \nAnswer:
    \nCavendish experiment.<\/p>\n

    Question 6.
    \nWhich property of a body is responsible for regaining its shape and size when deforming force acting on it is removed?
    \nAnswer:
    \nElasticity.<\/p>\n

    Question 7.
    \nName the instrument used to measure atmospheric pressure.
    \nAnswer:
    \nBarometer.<\/p>\n

    Question 8.
    \nWhat is absolute zero temperature?
    \nAnswer:
    \nAbsolute zero of temperature is that temperature of a gas at which its volume and pressure become zero.
    \nZero kelvin = -273.15\u00b0C \u2248 -273\u00b0C.<\/p>\n

    Question 9.
    \nDefine mean free path.
    \nAnswer:
    \nThe average distance a molecule can travel without colliding is called the mean free path.<\/p>\n

    \"KSEEB<\/p>\n

    Question 10.
    \nWhat is the basis of the phenomenon of interference?
    \nAnswer:
    \nThe two interfering waves should be identical and superpose at a point at small angles (travelling in the same direction) so that modification of energy takes place at the point of superposition of waves.<\/p>\n

    Part – B<\/span><\/p>\n

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

    Question 11.
    \nWhat is parallax? Mention its use.
    \nAnswer:
    \nParallax is the change in the position of an object with respect to it background, when the object is seen from two different positions.
    \nParallax is used to determine long distance.<\/p>\n

    Question 12.
    \nA car is traveling with a uniform velocity of 30ms-1<\/sup>. The driver applies the breaks and the car comes to rest in 10 seconds. Calculate the retardation.
    \nAnswer:
    \nu = 300 ms-1<\/sup>
    \nt = 10s
    \nv = 0
    \n\u2234a = \\(\\frac{v-u}{t}=\\frac{0-30}{10} = -3 \\mathrm{ms}^{-2}\\)
    \n\u2234 Retardation is 3 ms-2<\/sup>.<\/p>\n

    \"KSEEB<\/p>\n

    Question 13.
    \nDistinguish between scalars and vectors.
    \nAnswer:
    \nScalars:<\/p>\n

      \n
    • Scalars require only magnitude to represent them.<\/li>\n
    • Scalars require simple algebraic rules for addition or subtraction.<\/li>\n<\/ul>\n

      Vectors:<\/p>\n

        \n
      • Vectors require both magnitude and direction to represent them.<\/li>\n
      • Vectors require law of polygon of addition of vectors.<\/li>\n<\/ul>\n

        Question 14.
        \nWhen is the work done negative? Give an example.
        \nAnswer:
        \nIf the force and the displacement are in opposite directions, then the work is said to be negative. Work done by a force against the gravity is negative.<\/p>\n

        Question 15.
        \nState the law of conservation of angular momentum. Illustrate with an example.
        \nAnswer:
        \nIf the total external torque on a system is zero then the total angular momentum of the system is conserved (remains constant)
        \n\"1st<\/p>\n

        Ballet dancer : Ballet dancers bring their arms closer to their body in order to increase the angular speed and decrease the moment of inertia. In order to decrease the angular speed, they stretch their arms out, thereby increasing the moment of inertia.
        \nThus I1<\/sub>\u03c91<\/sub> = I2<\/sub>\u03c92<\/sub> = L = constant.<\/p>\n

        An acrobat: In the coiled position of the body, the moment of inertia decreases and this helps the acrobat perform 2 or 3 somersaults before gravity takes over. In the stretched position moment of inertia increases and almost the turning effect becomes zero.<\/p>\n

        Question 16.
        \nMention the uses of polar satellites?
        \nAnswer:<\/p>\n

          \n
        • Polar satellites can view polar and equatorial regions at close distances with good resolutions.<\/li>\n
        • Polar satellites are used for remote sensing, meteorology and environmental studies of the Earth.<\/li>\n<\/ul>\n

          Question 17.
          \nState and explain zeroth law of thermodynamics.
          \nAnswer:
          \nIf two thermodynamic systems are in thermal equilibrium with third individually, then the systems are said to be in mutual thermal equilibrium with each other.
          \nIf TA<\/sub> = TC <\/sub>, TB<\/sub> = TC<\/sub> then TA<\/sub> = TB<\/sub>. where ‘T’ represents the temperatures and A, B, C represent the thermodynamic systems.<\/p>\n

          \"KSEEB<\/p>\n

          Question 18.
          \nOn an average a human heart is found to beat 75 times in a minute. Calculate its frequency and time period.
          \nAnswer:
          \nNo. of beats = 75
          \nTime taken = 1 minute = 60 second
          \n\"1st<\/p>\n

          Part – C<\/span><\/p>\n

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

          Question 19.
          \nState and explain triangle law of vector addition.
          \nAnswer:
          \nIf two vectors are represented by the two sides of a triangle taken in order then their resultant is represented by the closing side in the opposite order.<\/p>\n

          Question 20.
          \nProve that for a particle in rectilinear motion under constant acceleration, the change in kinetic energy of the particle is equal to the work done on it by the net force?
          \nAnswer:
          \nThe difference of PE. or K..E. required by a body at two different positions is equal to the amount of work done.
          \n\\(\\int_{k_{i}}^{k_{i}} d(K E)=\\frac{1}{2} m\\left(v^{2}-u^{2}\\right)\\)
          \n= mas = Fs
          \n= work done
          \ni. e. W = Kf<\/sub> – Ki<\/sub>.<\/p>\n

          Question 21.
          \nDraw the stress verses strain graph for a metallic wire stretched up to fracture point. Define the terms proportional limit and fracture point.
          \nAnswer:
          \nStatement: The ratio of stress to strain is a constant for a material within the elastic limit.Modulus 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 22.
          \nFind the torque of a force \\(7 \\hat{i}+3 \\hat{j}-5 \\hat{k}\\) about the origin. The force acts on a particle whose position vector is \\(\\hat{i}-\\hat{j}+\\hat{k}\\).
          \nAnswer:
          \n\\(\\overrightarrow{\\mathrm{F}}=7 \\hat{i}+3 \\hat{j}-5 \\hat{k} ; \\vec{r}=\\hat{i}-\\hat{j}+\\hat{k}\\)
          \n\"1st
          \n\u2234\u03c4 = 15.74 Nm<\/p>\n

          Question 23.
          \nWhat is capillary rise? Write an expression for it. Explain the terms.
          \nAnswer:
          \nA rise in a liquid above the level of zero pressure due to a net upward force produced by the attraction of the water molecules to a solid surface.<\/p>\n

          Question 24.
          \nWrite any three 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

            Question 25.
            \nState and explain the law of thermal conduction.
            \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 26.
            \nMention any three differences between standing waves and progressive waves.
            \nAnswer:<\/p>\n

              \n
            1. Progressive waves transfer energy, whereas stationary waves do not.\n
                \n
              1. No particles of the medium are in a state of rest in a PW but in a SW, the particles at nodes will be at rest.<\/li>\n<\/ol>\n<\/li>\n
              2. The particle velocities will remain the same for a given displacement in a PW, whereas in a SW, particles at nodes will have zero speed and at antinodes, particles have maximum speed.<\/li>\n
              3. y = f(x, t) for a PW and y = f(x) g(t) for a SW.<\/li>\n
              4. Distance between two points at which particles have same state of vibration is called wavelength in a PW whereas distance between two consecutive nodes or antinodes is half the wavelength.<\/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.
                \nWhat is velocity – time graph. Derive the equation of motion: v2<\/sup> – v0<\/sub>2 <\/sup>= 2ax.
                \nAnswer:
                \nLet \u2018v0<\/sub>\u2019 be the initial velocity of a particle describing uniform accelerated motion \u2018a’.
                \nSlope of the line AB = a = tan\u03b8 = where BC = v – v0 <\/sub>and AC = t
                \ni.e., a = \\(\\frac{v-v_{0}}{t}\\) ……….(1)
                \nBut area of trapezium = x = 1\/2 \u00d7 OD \u00d7 (OA + BD)
                \ni.e., x = 1\/2 t(v0 +<\/sub> v)
                \n\"1st
                \nFrom(1), t = \\(\\frac{v-v_{0}}{a}\\)
                \nHence (2) may be written as 2s = \\(\\left(\\frac{v-v_{0}}{a}\\right)\\)(v + v0<\/sub>)
                \n\u2234v2<\/sup> – v0<\/sub>2 <\/sup>= 2ax.<\/p>\n

                \"KSEEB<\/p>\n

                Question 28.
                \nDerive the expression for maximum safe speed of a vehicle on a banked road in circular motion.
                \nAnswer:
                \nFrom the figure, \u2018\u03b8\u2019 is the angle of elevation of outer edge over the inner edge. \u2018R\u2019 is the normal reaction force.
                \n\"1st
                \nR sin\u03b8 = \\(\\frac{m v^{2}}{r}\\)………(1)
                \nRcos\u03b8 = mg\u00a0 ………(2)
                \n(1) \u00f7 (2) gives
                \ntan\u03b8 = \\(\\frac{v^{2}}{r g}\\)
                \n\u2234vmax<\/sub> = \\(=\\sqrt{r g \\tan \\theta}\\)<\/p>\n

                Question 29.
                \nWhat is elastic collision? Obtain the expression for final velocities of two bodies undergoing elastic collision in one dimension.
                \nAnswer:
                \nA collision in which linear momentum and kinetic energy of the interacting system are conserved, is known as an elastic collision.
                \nFrom the law of conservation of linear momentum,
                \nm1<\/sub> v1i<\/sub> + m2<\/sub> v2i<\/sub> = m1<\/sub>v1f<\/sub> + m2<\/sub>v2f<\/sub> ……….(1)
                \nFrom the law of conservation of total kinetic energy,
                \n\\(\\frac{1}{2} m_{1} v_{1 i}+\\frac{1}{2} m_{2} v_{2 i}=\\frac{1}{2} m_{1} v_{1 f}^{2}+\\frac{1}{2} m_{2} v_{2 f}^{2}\\)….(2)
                \n\"1st<\/p>\n

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

                Question 30.
                \nWhat is escape speed? Obtain the expression for escape speed. What is the value of escape speed for earth?
                \nAnswer:
                \nIt is the minimum speed needed for a free object to escape from the gravitational influence of a massive body.
                \nFor a satellite to escape into infinite orbit,
                \n\u2206PE = PE\u221e<\/sub> – PEr\u00a0 <\/sub>\u2235PE\u221e<\/sub> = 0
                \nBut K.E = PE.
                \n\u2206PE = \\(-\\left(-\\frac{\\mathrm{GMm}}{\\mathrm{R}+\\mathrm{h}}\\right)\\)
                \n\"1st
                \nIt is assumed that at infinite distance PE = 0, v1<\/sub> = 0.
                \nIt is a free fall of an object from the point of infinity to a point on the surface in the absence of any frictional forces.<\/p>\n

                \"KSEEB<\/p>\n

                Question 31.
                \nWhat is isothermal process? Obtain the expression for work done in an isothermal process.
                \nAnswer:
                \nThe process in which changes in pressure and volume of a gas system takes place at constant temperature of the system, is known as isothermal process.
                \nFor an isothermal process, PV = constant.
                \nAt any intermediate stage with pressure P and change in volume from V to V + dV,
                \nWork done dW = P dV. However for the entire process W = \\(=\\int_{V_{1}}^{V_{2}}\\)PdV.
                \n\"1st
                \nFor V2<\/sub> > V1<\/sub>, W > 0 and for V2<\/sub> < V1<\/sub>, W < 0.
                \nIn an isothermal expansion, the gas absorbs heat and does work, while in an isothermal compression, work is done on the gas by the environment and heat is released.<\/p>\n

                Question 32.
                \nWhat is simple harmonic motion? Derive an expression for velocity and acceleration in simple harmonic motion.
                \nAnswer:
                \nA SHM is the simplest form of oscillatory motion in which the force on the oscillating particle or body is directly proportional to the displacement from the mean position and the force directed towards the mean position.
                \nWe know that, the displacement of the particle
                \ny = A sin \u03c9t
                \nBut v = \\(\\frac{d y}{d t}\\)
                \n\u2234v = \\(\\frac{d}{d t}\\)(A sin \u03c9t)
                \nand acceleration,
                \na = \\(\\frac{d v}{d t}\\)
                \n\u2234a = \\(\\frac{d}{d t}\\)(A\u03c9 cos \u03c9t)
                \na = -\u03c92<\/sup> A sin \u03c9t<\/p>\n

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

                Question 33.
                \nA cricketer can throw a ball to a maximum horizontal distance of 100 m..How much high above the ground can the cricketer throw the same ball? And for how long the bail remains in the air in this case?
                \nAnswer:
                \nR = 100 m
                \nRmax<\/sub> = \\(\\frac{v^{2}}{g}\\)\u00a0 \u00a0 \u03b8 = 45\u00b0
                \n\u2234v2<\/sup> = Rmax<\/sub> x g
                \nv2<\/sup> = 100 \u00d7 9.8
                \nv2<\/sup> = 980 m2<\/sup>s-2<\/sup> and v = 31.3 ms-1<\/sup>
                \n\u2234the maximum height can be calculated, by
                \n\"1st
                \nT = 4.52s<\/p>\n

                \"KSEEB<\/p>\n

                Question 34.
                \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 rev.\/min in a horizontal plane. What is the tension in the string? W hat is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
                \nAnswer:
                \n\"1st
                \nTmax<\/sub> = \\(\\frac{m v^{2}}{r}\\)
                \nTmax<\/sub> = mr\u03c92<\/sup> [v = r\u03c9]
                \n= 0.25 x 1.5 x \\(\\left(\\frac{4 \\pi}{3}\\right)^{2}\\)
                \nAs the string can withstand a maximum tension of 200 N.
                \nTmax<\/sub> = \\(\\frac{m v_{\\max }^{2}}{r}\\)
                \nTmax<\/sub> = \\(\\frac{T_{\\max } \\times r}{m}\\)
                \n\"1st<\/p>\n

                Question 35.
                \nA solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s-1<\/sup>. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? Calculate the magnitude of angular momentum of the cylinder about its axis.
                \nAnswer:
                \nm = 20 kg
                \nw =100 rad\/s
                \nr = 0.25 m
                \n(i) The moment of inertia of the solid cylinder,
                \nI = \\(\\frac{m r^{2}}{2}\\) = \\(\\frac{1}{2}\\) \u00d7 20 \u00d7 (0.25)
                \nI = 0.625 kgm2<\/sup>
                \n(ii) The kinetic energy of rotation, KE = \\(\\frac{1}{2} \\mathrm{I} \\omega^{2}\\)
                \nKE \\(=\\frac{1}{2} \\mathrm{I} \\omega^{2}=\\frac{1}{2} \\times 0.625 \\times(100)^{2}\\)
                \nK.E = 3125J
                \n(iii) Angular momentum,
                \nL = I\u03c9
                \ni.e.,L = 0.625 \u00d7 100
                \nL = 62.5 Js<\/p>\n

                \"KSEEB<\/p>\n

                Question 36.
                \nA brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250\u00b0 C if the original lengths arc at 40.0\u00b0 C? Is there a \u2018thermal stress\u2019 developed at the junction? The ends of the rod are free to expand.
                \n(Co-efficient of linear expansion of brass = 2.0 \u00d7 105<\/sup> K-1<\/sup>\u00a0, steel = 1.2 \u00d7 10-5<\/sup>\u00a0K-1<\/sup>).
                \nAnswer:
                \nT1<\/sub> = 40\u00b0C, T2<\/sub> = 250\u00b0C
                \n\u2234\u2206T = 210\u00b0C
                \nFor brass rod, initial length (L0<\/sub>) = 50 cm
                \nLet final length = L
                \n\u03b1b<\/sub>\u00a0= 2 \u00d7 10-5<\/sup> \/K
                \nL = L0<\/sub>\u00a0<\/sub>(1 + \u03b1b <\/sub>\u2206T) = 50 (1 + 2 \u00d7 10-5<\/sup> \u00d7 210) = 50(1 + 420 \u00d7 10-5<\/sup>)
                \nL = 50.21 cm
                \n\u2234Increase in length (\u2206L) = L – L0<\/sub> = 50.21 – 50
                \n\u2206L = 0.21 cm
                \nFor steel rod,
                \nInitial length (L0<\/sub>1<\/sup>) = 50 cm
                \nLet final length = L1<\/sup>
                \n\u03b1s<\/sub> =1.2 \u00d7 10-5<\/sup> K-1<\/sup>
                \n\u2234L1<\/sup> = L0<\/sub>1<\/sup> (1 + \u03b1s <\/sub>\u2206T) = 50 (1 + 1.2 \u00d7 10-5<\/sup> \u00d7 210) = 50 \u00d7 1.00252
                \nL1<\/sup> = 50.126 cm
                \n\u2234increase in length (\u2206L1<\/sup>) = 50.126 – 50 = 0.126 cm
                \n\u2234Total increase in length = \u2206L + \u2206L1<\/sup> = 0.21 + 0.126 = 0.336 cm<\/p>\n

                \"KSEEB<\/p>\n

                Question 37.
                \nA wave travelling along a string is described by, y (x, t) = 0.005 sin (80.0 \u00d7 – 3.0 t), in which the numerical constants are in St units (0.005 m, 80.0 rad m-1<\/sup>, and 3.0 rad s-1<\/sup>). Calculate (a) the amplitude, (b) the wavelength, and (c) the frequency of the wave. Also, calculate the displacement jr of the wave at a distance x = 30.0 cm and time t = 20 s?
                \nAnswer:
                \nThe given equation is in the form
                \ny = A sin (kx – wt)
                \n\u2234A = 0.005 m,\u00a0 \u03c9 = 3rads-1<\/sup>
                \n\"1st
                \ny (x, t) = -0.005 sin (3.0 t – 80.0 x),
                \n= -5 \u00d7 10-3<\/sup> sin (80 \u00d7 0.3 – 3 \u00d7 20)
                \n= 0.713 m<\/p>\n","protected":false},"excerpt":{"rendered":"

                Students can Download 1st PUC Physics Model Question Paper 3 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 3 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\/20096"}],"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=20096"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20096\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=20096"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=20096"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=20096"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}