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    Physics – Ans

    Physics
    1DADBACBADC
    11ADBDDABDCC
    21DBAABAACAD
    31DCBBCCBABB
    41BDCDDACADC

    section B answer 3 only

    ============================

    10a )

    Diffraction is the ability of waves to bend around obstacles in their path

    10bi)

    Critical angle is the highest angle of incidence in a denser medium when angle of refraction in the less dense mediumis 90 degrees

    10bii )

    Critical angle = 90- 44= 46degrees

    The refractive index of the glass is obtained as follows

    The refractive index( n )=sini / sinr = sin 46/sin 90= 0 . 7193/ 1

    = 0 . 7193

    10ci )

    fo= 200 Hz

    f 1 = 3 v / 4 l => closed pipe

    f 1 = v / l => open pipe

    but 3 fo= f 1 => closed pipe

    3 * 200 = f 1 => f 1 = 600 Hz

    Also f 1 = 2 fo=> open pipe

    2 fo= 600

    fo= 600 / 2 = 300 Hz

    10cii )

    v = 330 m / s

    L =?

    fo= 200 Hz

    fo= v / 4 l

    => 200 = 330 / 4 l

    800 l = 330

    l = 330 / 800

    l = 0 . 4124m

    10ciii )

    fo= v / 2 l

    fo= 300 Hz

    v = 330 m / s

    300 = 330 / 2 l

    l = 330 / 600

    l = 0 . 55m

    ============================

    9 ai)

    i ) Nature of surface

    ii) medium of transmission

    9 aii )

    i ) state or phase of the substance ( i : e solid ,or gas )

    ii) temperature of the meduim

    9 b )

    i ) temperature

    ii) specific heat capacity of the body

    9 c )

    the statement means that the amount of heat energy required to change 1 kg of liquid mercury to gaseous mecury without change in temperature is 2 . 72* 10^ 5 jkg^ – 1

    9 di )

    Q= MCDSin

    V ^ 2 / R t = MCDsin

    ( 220 )^ 2 * 4 * 60/ 35= M * 4200* ( 100 – 28)

    331885. 7 = 302400m

    m = 331885. 7 / 302400 = 1 . 098 kg

    9 dii )

    V ^ 2 / R t = MLv

    ( 220 )^ 2 * 5 * 60/ 35= 0 . 3 * Lv

    414857. 14= 0 . 3 Lv

    Lv = 414857. 14/ 0 . 3

    Lv = 1382857 . 13JKg ^ – 1

    ============================

    12a )

    This is defined as the amount of energy that must be supplied to a nucleus to completely separate it ‘ s nuclear particles ( nucleons )

    12b )

    i ) They have short wavelength and high frequency.

    ii) They are highly penetrating .

    iii ) They travel in straight lines.

    iv ) They don’ t require material medium for their propagation .

    12c )

    i ) It is used in production of electricity .

    ii) It is used to study and detect charges in genetic engineering .

    iii ) It is used in agriculture .

    iv ) It is used in treatment of cancer .

    12di )

    E = hf – hfo

    but f = v / landa

    E = v / landa . h – wo

    Where wo = hfo = work function

    f = frequency

    landa = wavelength

    Hence

    hf = hfo – E

    f = hfo – E / h

    f = wo – E / h

    Recall ; that v = f landa

    Therefore f = v / landa = 3 × 10^ 8 / 4 . 5 × 10- 7

    = 3 / 4 . 5 × 10^ 8 + 7

    = 6 . 6 × 10^ 14Hz

    f = 6 . 6 × 10^ 14Hz

    12dii )

    E = hf

    = 6 . 6 × 10^ – 34 × 6 . 6 × 10^ 14Hz

    = 43. 56× 10^ – 20J

    12diii )

    Energy of the photoelectron E = hf – vo

    = Energy of incident electron – work function

    = 4 . 356 × 10^ – 19J – 3 . 0 × 10^ – 19J

    = 1 . 356 × 10^ – 19J

    ============================

    section A Answer 5 Only

    ============================

    1 a )

    Strain can be defined as the ratio of extension per unit length

    Strain = extension / length

    1 b )

    Strain = extension / length

    Let original length = L

    Final length = 2 L

    Extension = 2 L – L = L

    Strain = L / L = 1

    strain = change in length / original length

    ============================

    6 )

    Given constant = 2 . 9 × 10^ – 3 mk

    Temperature = 57degreeC = ( 57+ 273 ) k = 330 k

    Using landamaxT = constant

    landamaxT330 = 2 . 9 × 10^ – 3

    landamax = 2 . 9 × 10^ – 3 / 330

    landamax = 8 . 788 × 10^ – 6 m

    The speed of electromagnetic wave, v = 3 × 10^ 8 m / s

    Using V = f landa

    f = v / landa

    = 3 × 10^ 8 / 8 . 788 × 10^ – 6

    = 3 . 4 × 10^ 13Hz

    ============================

    5 )

    Range = u ² Sin2 tita / g

    At maximum range

    Sin2 tita = 1

    2 tita = sin ^ – 1 ( 1 )

    2 tita = 90dgrees

    Tita = 90/ 2 = 45degree

    Maximum height reached = u ² sin ² tita / 2 g

    = u ² ( sin 45) ² / 2 g

    = 200² ( sin 45) ² / 2 ( 10)

    = 40000 ( 1 / √2 ) 2 / 20

    = 40000 ( 1 / 2 ) / 20

    = 20000 / 20

    = 1000metres .

    ============================

    4 a )

    An intrinsic semiconductor is an

    undoped semiconductor that is a pure semiconductor without any significant dopant

    species present .

    4 b )

    The P Type semiconductor is a type of semiconductor that

    carries a positive charge , while the N type semiconductor carries a

    negative charge .

    ============================

    2 )

    i ) Silicon Dioxide

    ii) Silica Powder

    iii ) Germanium Tetrachloride

    ============================

    3 )

    i ) Ferromagnetic material

    ii) Diamgentic material

    iii ) Pamangentic material

    ============================

    7 a )

    LASER stands for Light Amplification by Stimulated Emission of Radiation .

    7 b )

    A laser is a device that emits a beam of coherent light through an optical amplification process .

    ============================

    Completed

    ============================



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    Master Solution April 20, 2018 Categories: WAEC 817


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    2 Comments

    1. Happiness says:
      April 4, 2018 at 7:20 pm

      How can i get the ans after paying

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    2. Ekwealor onyebuchi says:
      March 4, 2018 at 5:17 pm

      how am i sure if ur service is genure

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