KVPY-SA 2018 Physics Paper with Solutions

The Kishore Vaigyanik Protsahan Yojana (KVPY) is a national level examination conducted by the Department of Science and Technology. The objective of the program is to motivate students to pursue a career in science research. The test is conducted at three different levels namely SA, SX and SB. The applicants are advised to practise previous years’ question papers in order to clear the aptitude test. KVPY- SA 2018 Physics paper is answered on this page. PDF download of the question paper is also available.

KVPY SA 2018 - Physics

Question 1: A block of wood is floating on water at 0°C with volume V0 above water. When the temperature of water increases from 0 to 10°C, the change in the volume of the block that is above water is best described schematically by the graph

KVPY SA 2018 Physics Solutions


    1. Answer: a

      Change in volume can occur due to following reason:

      (i) Density/volume of fluid changes due to increase or decrease in temperature.

      (ii) The volume of material which is immersed changes due to temperature change [This argument need not to be considered as change in volume of the material is infinitesimal as co-efficient of volume thermal expansion has a very low value]

      • For floating condition we can write [T = O°C] [Archimedes – principle]

      Weight of the block = Buoyant force on the block

      W = FB

      W = ρwatergVimmersed ……(1)

      • As buoyant force depend on the density of water if it changes due to temperature change [ρwater is maximum at 4°C]

      KVPY SA 2018 Physics Solved Paper

      From (1)

      mg =ρwaterg Vimmersed [m = mass of block]

      Vimmersed= m/ρwater

      Volume outside water:-

      Vo = Vtotal – Vimmersed

      0 – 4°C Density ↑ Vimmersed ↓ Vo

      4°C – 10°C Density ↓ Vimmersed ↑ Vo

      Hence volume outside Block changes according to immersed volume which depend on density of water.

      Above statement written to understand that density variation curve & volume outside water of the block with temperature change is same.

      V01ViV_{0}\propto \frac{1}{V_{i}}

      Vi1ρwaterV_{i}\propto \frac{1}{\rho _{water}}

      V0ρwaterV_{0}\propto \rho _{water}

    Question 2: A very large block of ice of the size of a volleyball court and of uniform thickness of 8m is floating on water. A person standing near its edge wishes to fetch a bucketful of water using a rope. The smallest length of rope required for this is about

    1. a. 3.6 m
    2. b. 1.8 m
    3. c. 0.9 m
    4. d. 0.4 m


    1. Answer: c

      From Archimedes principle:-

      Weight of ice-Block = Buoyant force on block

      KVPY SA 2018 Physics Paper Solved

      W = FB

      ρi. gV = ρw g Vimmersed

      Vimmersed[Vi]=[ρiρw]×VV_{immersed}\left [ V_{i} \right ]=\left [ \frac{\rho _{i}}{\rho _{w}} \right ]\times V

      LBHimmersed=[ρiρw]×LBHLBH_{immersed}=\left [ \frac{\rho _{i}}{\rho _{w}} \right ]\times LBH

      Himmersed=[ρiρw]×HH_{immersed}=\left [ \frac{\rho _{i}}{\rho _{w}} \right ]\times H

      We know that

      Density of ice, ρi = 0.9 gm/cm3

      Density of water

      ρw = 1 gm/cm3

      Hi = 0.9H

      Hi = 7.2 m

      So Height outside water:

      Ho=8 – 7.2

      Ho = 0.8m

      Nearest value is 0.9m

      Hence, the smallest length of rope required is 0.9m.

    Question 3: A box filled with water has a small hole on its side near the bottom. It is dropped from the top of a tower. As it falls, a camera attached on the side of the box records the shape of the water stream coming out of the hole. The resulting video will show

    1. a. the water coming down forming a parabolic stream
    2. b. the water going up forming a parabolic stream
    3. c. the water coming out in a straight line
    4. d. no water coming out


    1. Answer: d

      When a filled box with water with small hole falls freely, then pressure is same at inside and outside of hole i.e. atmospheric pressure. No water will come out of hole as there is no pressure difference.

      Option (d) is correct

    Question 4: An earthen pitcher used in summer cools water in it essentially by evaporation of water from its porous surface. If a pitcher carries 4 kg of water and the rate of evaporation is 20 g per hour, temperature of water in it decreases by ΔT in two hours. The value of ΔT is close to (ratio of latent of evaporation to specific heat of water is 540 °C)

    1. a. 2.7 °C
    2. b. 4.2 °C
    3. c. 5.4 °C
    4. d. 10.8°C


    1. Answer: c

      As evaporation occur the temperature of the water decreases as it is a cooling process

      Rate of evaporation of water = 20 g per hour

      KVPY SA 2018 Physics Question Paper Solved

    Question 5: Two plane mirrors are kept on a horizontal table making an angle θ with each other as shown schematically in the figure. The angle θ is such that any ray of light reflected after striking both the mirrors return parallel to its incident path. For this to happen, the value of θ should be

    KVPY SA 2018 Physics Question Paper with Solutions

    1. a. 30°
    2. b. 45°
    3. c. 60°
    4. d. 90°


    1. Answer: d

      Question Paper with Solutions of KVPY SA 2018 Physics

      Angle of deviation [δ]:-

      δ = 360°– 2θ

      given:- δ = 180°

      180° = 360° – 2θ

      θ = 90°

      Hence; option (d) is correct

    Question 6: A certain liquid has a melting point of –50°C and a boiling point of 150°C. A thermometer is designed with this liquid and it’s melting and boiling points are designated as 0° L and 100°L. The melting and boiling points of water on this scale are

    1. a. 25° L and 75° L, respectively
    2. b. 0° L and 100° L, respectively
    3. c. 20° L and 70° L, respectively
    4. d. 30° L and 80° L, respectively


    1. Answer: a

      This question is based on calibration of thermometer on various temperature scale in which the value of two standard reference point are given i.e. melting and boiling points.

      Sample Paper with Solutions of KVPY SA 2018 Physics




      Now for water:-

      Sample Paper of f KVPY SA 2018 Physics

      Alternate solution:-

      Sample Paper of f KVPY SA 2018 Physics

      On Comparison

      1°L = 2°C

    Question 7: One can define an alpha–Volt (α V) to be the energy acquired by an α particle when it is accelerated by a potential of 1 Volt. For this problem you may take a proton to 2000 times heavier than an electron. Then

    1. a. 1 ∝ V = 1 eV/4000
    2. b. 1 ∝ V = 2eV
    3. c. 1 ∝ V = 8000 eV
    4. d. 1 ∝ V = 1 eV


    1. Answer: b

      From work kinetic - energy theorem:

      Work done= Change in kinetic energy

      W = ΔK

      Work done =q [ΔV]

      q = charge

      ΔV = potential difference

      qα = 2qe

      For same –potential:

      ΔKα = 2ΔKe

      1αV = 2eV

      Hence, option (B) is correct

    Question 8: In a particle accelerator, a current of 500 μA is carried by a proton beam in which each proton has a speed of 3 × 107 m/s. The cross sectional area of the beam is 1.50 mm2. The charge density in this beam in coulomb/m3 is close to–

    1. a. 10–8
    2. b. 10–7
    3. c. 10–6
    4. d. 10–5


    1. Answer: d

      From current Electricity:-

      i = neAVd

      n = no. of electrons per unit volume

      A = cross-sectional area

      Vd = Drift speed

      Charge-density = q/V

      q/V= ne = charge on particle per unit volume [charge density]

      ne = i/AVd

      =500×1061.5×106×3×107=\frac{500\times 10^{-6}}{1.5\times 10^{-6}\times 3\times 10^{7}}

      = 10–5 c/m3

      Hence option (D) is correct.

    Question 9: Which of the following is NOT true about the total lunar eclipse?

    1. a. A lunar eclipse can occur on a new moon and full moon day.
    2. b. The lunar eclipse would occur roughly every month if the orbits of earth and moon were perfectly coplanar
    3. c. The moon appears red during the eclipse because the blue light is absorbed in earth’s atmosphere and red is transmitted
    4. d. A lunar eclipse can occur only on a full moon day


    1. Answer: a

      A lunar eclipse can occur only on full moon day.

      Hence option (a) is not correct and this is the required answer.

    Question 10: Many exoplanets have been discovered by the transit method, wherein one monitors a dip in the intensity of the parent star as the exoplanet moves in front of it. The exoplanet has a radius R and the parent star has radius 100R. If I0 is the intensity observed on earth due to the parent star, then as the exoplanet transits

    1. a. the minimum observed intensity of the parent star is 0.9 I0
    2. b. the minimum observed intensity of the parent star is 0.99 I0
    3. c. the minimum observed intensity of the parent star is 0.999 I0
    4. d. the minimum observed intensity of the parent star is 0.9999 I0


    1. Answer: d

      Practice Paper of  KVPY SA 2018 Physics

      Intensity ∝ [Area] ∝ [Radius]2

      Ap = effective area of exoplanet

      As = effective Area of star


      Ip=[AsAp]×I0I_{p}=\left [ \frac{A_{s}}{A_{p}} \right ]\times I_{0}

      I=I0[AsAp]×I0I=I_{0}-\left [ \frac{A_{s}}{A_{p}} \right ]\times I_{0}

      I=I0[1R2(100R)2]I=I_{0}\left [ 1-\frac{R^{2}}{(100R)^{2}} \right ]


      I = 0.9999I0

    Question 11: A steady current I is set up in a wire whose cross-sectional area decreases in the direction of the flow of the current. Then, as we examine the narrowing region

    1. a. the current density decreases in value
    2. b. the magnitude of the electric field increases
    3. c. the current density remains constant
    4. d. the average speed of the moving charges remains constant


    1. Answer: b

      Given :-

      I = constant

      Current Density [J]=IA\left [ \vec{J} \right ]=\frac{I}{A}

       KVPY SA 2018 Physics Practice Paper

      J1AJ\propto \frac{1}{A}

      Hence option (A) & (C) are wrong due to above expression


      J=σE\vec{J}=\sigma \vec{E}

      σ\sigma = conductivity of the material

      EJ1AE\propto J\propto \frac{1}{A}

      magnitude of Electric-Field increases. [Option B is correct]

      I = neAVd

      Vd1AV_{d}\propto \frac{1}{A}

      Hence Drift speed/average speed increases along the length.

      Option (D) is incorrect

    Question 12: Select the correct statement about rainbow

    1. a. We can see a rainbow in the western sky in the late afternoon
    2. b. The double rainbow has red on the inside and violet on the outside
    3. c. A rainbow has an arc shape since the earth is round
    4. d. A rainbow on the moon is violet on the inside and red on the outside


    1. Answer: b

      KVPY SA 2018 Physics Solved Sample Paper

    Question 13: Remote sensing satellites move in an orbit that is at an average height of about 500 km from the surface of the earth. The camera onboard one such satellite has a screen on area A on which the images captured by it are formed. If the focal length of the camera lens is 50 cm, then the terrestrial area that can be observed from the satellite is close to

    1. a. 2 × 103A
    2. b. 106A
    3. c. 1012A
    4. d. 4 × 1012A


    1. Answer: c

      Area of Image = m2 × area of object

      Ai = m2Ao

      m = magnification = Height of image/Height of object

      = =v2u2A0=\frac{v^{2}}{u^{2}}A_{0}


      V= 50 cm = 50 x 10-2 m

      u = 500 Km = 500 × 103m

      A0=A[500×10350×102]2A_{0}=A\left [ \frac{500\times 10^{3}}{50\times 10^{-2}} \right ]^{2}

      = 1012 A

      Solved Sample Paper of KVPY SA 2018 Physics

    Question 14: Letters A, B, C and D are written on a cardboard as shown in the picture.

    Solutions of Sample Paper KVPY SA 2018 Physics

    The cardboard is kept at a suitable distance behind a transparent empty glass of cylindrical shape. If the glass is now filled with water, one sees an inverted image of the pattern on the cardboard when looking through the glass. Ignoring magnification effects, the image would appear as

    KVPY SA 2018 Physics Sample Paper Solutions


      1. Answer: d


        Lens is of cylindrical shape and we know the image formed by cylindrical lens is Erect & Laterally inverted.

        Hence option (d) is correct Answer

      Question 15: If a ball is thrown at a velocity of 45 m/s in vertical upward direction, then what would be the velocity profile as function of height? Assume g = 10 m/s2.

      Physics Sample Paper Solutions  KVPY SA 2018


        1. Answer: a

          From Newton’s Third equation of motion

          V2 = u2 + 2as

          Physics Practice Paper KVPY SA 2018

          Here:- a = – g

          At ,S = Hmax = h

          V = 0

          h=V22g=(45)22×10h=\frac{V^{2}}{2g}=\frac{(45)^{2}}{2\times 10}

          h = 101.25 m = 101 m

          As h = V2/2g

          at s = 0 ; V = 45m/s

          at s = h ; V = 0

          hV2h\propto V^{2} – Parabolic- Variation

          Physics Question Paper KVPY SA 2018

          Hence; option (A) is correct

        Question 16: A coffee maker makes coffee by passing steam through a mixture of coffee powder, milk and water. If the steam is mixed at the rate of 50 g per minute in a mug containing 500 g of mixture, then it takes about t0 seconds to make coffee at 70°C when the initial temperature of the mixture is 25°C. The value of t0 is close to (ratio of latent heat of evaporation to specific heat of water is 540°C and specific heat of the mixture can be taken to be the same as that of water)

        1. a. 30
        2. b. 45
        3. c. 60
        4. d. 90


        1. From principle of calorimetry:

          Heat provided by steam = Heat gain by mixture

          For Steam

          Physics Question Paper KVPY SA 2018

          Qsteam = Qmixture

          MsL + MsS Δ θs = Mm SΔ θm

          From the definition of Latent Heat = MsL

          From the definition of Specific Heat = MsSΔ θs

          From the definition of Specific Heat = Mm SΔ θm

          ΔMsΔt[L+SΔθs]=MmSΔθmΔt\frac{\Delta M_{s}}{\Delta t}[L+S\Delta \theta _{s}]=M_{m}\frac{S\Delta \theta _{m}}{\Delta t}

          ΔθmΔt=ΔMs/ΔtMm[LS+Δθs]\frac{\Delta \theta _{m}}{\Delta t}=\frac{\Delta M_{s}/\Delta t}{M_{m}}\left [ \frac{L}{S}+\Delta \theta _{s} \right ]

          Substituting the values

          45Δt=50500[540+30]\frac{45}{\Delta t}=\frac{50}{500}[540+30]

          Δt=4557×60second\Delta t=\frac{45}{57}\times 60 second

          Δt=45second\Delta t=45 second(approx)

        Question 17: A person in front of a mountain is beating a drum at the rate of 40 per minute and hears no distinct echo. If the person moves 90 m closer to the mountain, he has to beat the drum at 60 per minute to not hear any distinct echo. The speed of sound is

        1. a. 320 ms–1
        2. b. 340 ms–1
        3. c. 360 ms–1
        4. d. 380 ms–1


        1. Answer: c

          For not hearing Echo, time interval between the beats of drum must be equal to the time of Echo.

          Physics Paper KVPY SA 2018 Solved

          t1= 2x/V

          [f1 =1/t1]




          f2 = 60 per minute

          f2 = 2[x-90]/V

          6060=2xV180V\frac{60}{60}=\frac{2x}{V}-\frac{180}{V} [Substituting the value of 2x/V from equation (1)]



          V= 360 m/s

        Question 18: A glass beaker is filled with water up to 5 cm. It is kept on top of a 2 cm thick glass slab. When a coin at the bottom of the glass slab is viewed at the normal incidence from above the beaker, it’s apparent depth from the water surface is d cm. Value of d is close to (the refractive indices of water and glass are 1.33 and 1.50, respectively)

        1. a. 2.5
        2. b. 5.1
        3. c. 3.7
        4. d. 6.0


        1. Answer. b

          Physics Questions of  KVPY SA 2018 with Solutions

          Apparentdepth[d]=d1μ1+d2μ2Apparent-depth[d]=\frac{d_{1}}{\mu _{1}}+\frac{d_{2}}{\mu _{2}}

          Apparentdepth[d]=d1μw+d2μgApparent-depth[d]=\frac{d_{1}}{\mu _{w}}+\frac{d_{2}}{\mu _{g}}

          Substituting the values of µw and µg

          = 5[43]+2[32]\frac{5}{\left [ \frac{4}{3} \right ]}+\frac{2}{\left [ \frac{3}{2} \right ]}

          = 154+43\frac{15}{4}+\frac{4}{3}


        Question 19: A proton of mass m and charge e is projected from a very larger distance towards an α particle with velocity v. Initially, α particle is at rest, but it is free to move. If gravity is neglected, then the minimum separation along the straight line of their motion will be

        1. a. e2/4 π ε0mv2
        2. b. 5e2/4 π ε0mv2
        3. c. 2e2/4 π ε0mv2
        4. d. 4e2/4 π ε0mv2


        1. Answer: d

           KVPY SA 2018 Physics Questions with Solutions

          From conservation of linear momentum [As ∑F = 0]

          Pinitial = Pfinal

          mv = [m + 4 m] V’

          v = 5V’

          V’ = v/5

          Now from conservation of mechanical energy:-

          Questions with Solutions of  KVPY SA 2018 Physics Paper

          Method II

          Applying conservation of mechanical Energy with respect to centre of mass frame:-

          Answers of  KVPY SA 2018 Physics Paper= potential energy at minimum separation dmin

          12μvrel2=K(2e)(e)dmin\frac{1}{2}\mu v_{rel}^{2}=\frac{K(2e)(e)}{d_{min}}

          Where, 2e=charge of α particle, e=charge of proton

          And K=Coulomb's constant=1/4πε0

          μ = Reduced Mass = m1m2/m1+m2

          Where, m1= mass of α-particle and m2= mass of proton

          μ=(4m)(m)(4m+m)=4m5\mu =\frac{(4m)(m)}{(4m+m)}=\frac{4m}{5}

          12[4m5]v2=2Ke2dmin\frac{1}{2}\left [ \frac{4m}{5} \right ]v^{2}=\frac{2Ke^{2}}{d_{min}}

          dmin=5e24πε0v2md_{min}=\frac{5e^{2}}{4\pi \varepsilon _{0}v^{2}m}

        Question 20: A potential is given by V(x) = k(x + a)2/2 for x < 0 and V(x) = k(x – a)2/2 for x > 0. The schematic variation of oscillation period (T) for a particle performing periodic motion in this potential as a function of its energy E is

        KVPY SA 2018 Physics Paper


          1. Answer: b

            KVPY SA 2018 Physics Paper with Answers

            For small energy; particle oscillates in one well either it will oscillate about x = +a OR x = –a & Time period of this oscillation is T. [Below energy Level (1/2)Ka2 ]

            When this particle gain energy equal to the [(1/2)Ka2 ] it will oscillate in both well so total time period now become 2T so there is a sudden jump in time period.

            Now as you increase energy more than [(1/2)Ka2 ] then:

            KVPY SA 2018 Sample Physics Paper

            [This curve indicates that it takes less than T/4 time period from x = – a to x = 0]

            Hence time period starts decreasing from 2T. Hence option B is correct.

            Now for high energy we can conclude that curve gets symmetrical and time period again asymptotically approaches T.

            Physics Question Paper of  KVPY SA 2018

          Video Lessons - KVPY SA 2018 - Physics

          KVPY-SA 2018 Physics Paper with Solutions

          KVPY-SA 2018 Physics Paper with Solutions
          KVPY-SA 2018 Physics Paper with Solutions
          KVPY-SA 2018 Physics Paper with Solutions
          KVPY-SA 2018 Physics Paper with Solutions
          KVPY-SA 2018 Physics Paper with Solutions
          KVPY-SA 2018 Physics Paper with Solutions
          KVPY-SA 2018 Physics Paper with Solutions
          KVPY-SA 2018 Physics Paper with Solutions
          KVPY-SA 2018 Physics Paper with Solutions
          KVPY-SA 2018 Physics Paper with Solutions
          KVPY-SA 2018 Physics Paper with Solutions
          KVPY-SA 2018 Physics Paper with Solutions
          KVPY-SA 2018 Physics Paper with Solutions
          KVPY-SA 2018 Physics Paper with Solutions
          KVPY-SA 2018 Physics Paper with Solutions
          KVPY-SA 2018 Physics Paper with Solutions
          KVPY-SA 2018 Physics Paper with Solutions
          KVPY-SA 2018 Physics Paper with Solutions
          KVPY-SA 2018 Physics Paper with Solutions
          KVPY-SA 2018 Physics Paper with Solutions
          KVPY-SA 2018 Physics Paper with Solutions