Solved question paper for PHY May-2019 (DIPLOMA automobile engineering 1st-2nd)

Physics

Previous year question paper with solutions for Physics May-2019

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Question paper 1

  1. SECTION-A

    Q1. a) Fill in the blanks.      15x1=15

    i. Heat flows from a body at ______temperature to a body at ______temperature.

    Answer:

    High, low

  2. ii. The moment of momentum of a body is called______

    Answer:

    Angular momentum

  3. iii. Sounds waves are a _______wave.

    Answer:

    Mechanical wave

  4. iv. Dimensional formula of potential energy is_______

    Answer:

    M1L2T-2

  5. v. Speed is a ______Quantity and velocity is a ______quantity.

    Answer:

    Scalar, vector

  6. b) State True or False.
    vi. One Fermi is equal to 10-15 m.

    Answer:

    True

  7. vii. Rolling is a combination of rotational and translational motion.

    Answer:

    True

  8. viii. Centripetal force and Centrifugal force are equal in magnitude and opposite in direction.

    Answer:

    True

  9. ix. Moment of force is called Torque.

    Answer:

    True

  10. x. Small insects can walk on the surface of still water due to viscosity.

    Answer:

    False

  11. c) Multiple choice questions.
    xi. The significant figures in 0.09 are
    a) 1                    b) 2     

    c) 3                    d) 4

    Answer:

    (B) 2

  12. xii. Which of the following relation is not correct?
    a) v = rω               b) a =rα     

    c) l= rθ                  d) ω = 2π T

    Answer:

    (c) L=rΘ

  13. xiii. A flying bird possesses
    a) K.E. only                        b) P.E. Only

    c) both K.E. and  AP.E.      d) Wind energy

    Answer:

    (c) both KE and PE

  14. xiv. What causes reverberation?
    a) Reflection                b) refraction

    c) diffraction                 d) interference

    Answer:

    (a) reflection

  15. xv. To controls temperature one uses

    a) Thermocouple               b) Thermometer

    c) Thermostat                    d) Pyrometer

    Answer:

    (c) thermostat

  16. SECTION-B

    Q2. Attempt any six questions.                  6x5=30

    a. Convert 1 Newton of force into dyne using dimensional analysis.

    Answer:

    Dimensional formula of force= [MLT-2]

    So, let a=1, b=1, c=-2

    So, we have

    Newton à M1=1kg, L1=1m, T1=1sec

    Dyne à M2=1g, L2= 1cm, T2=1sec

     

     N1= 1Newton

    N2= ?

    Using the formula,

    N2 =  N1 [M1/M2]a [L1/L2]b [T1/T2]c

                    = 1[1kg/1g]1 [ 1m/1cm]1 [1s/1s]-2

                    = 1000 g X 100 cm X 1

                        1g X 1 cm

                    = 100000   = 105 dynes

  17. b. Show that Newton’s second law of motion is a real law of motion.

    Answer:

    Since Newton’s first & third law is contained in second law, thus it may also called as real law of motion.

    Derivation of 1st law: According to Newton’s 2nd law

    F= ma

    Where, F is applied force, m is mass of the body & a is acceleration of body. So, in absence of force

    i.e. F=0

    => ma =0

    a =0 [ since m cannot equal to 0]

    Zero acceleration means body at rest will remain at rest & a body in uniform motion will continue its uniform motion, which is Newton’s 1st law of motion.

    Derivation of 3rd law: Consider an isolated system of two bodies A & B. An isolated system is such that no force acts on the system.

    This is Newton’s 3rd law.

    Thus Newton’s 2nd law is real law of motion.

  18. c. Differentiate between Echo and reverberation.

    Answer:

    Echo is the phenomenon of repetition of sound of a source by reflection from an obstacle. It can be heard in open as well as closed spaces. Echo can only be heard when distance of source of sound & reflecting body is at least 17.2 m. Reverberation is the phenomenon of persistence or prolongation of audible sound after the source has stopped emitting sound. It is usually experienced in closed spaces.

  19. d. Define pressure and give its units.

    Answer:

    Pressure is defined as physical force exerted on an object. The force applied perpendicular to the surface of objects per unit area. Basic formula of pressure is Force per unit area. (F/A). Unit of pressure is Pascal (Pa).

  20. e. What do you mean by term viscosity and coefficient of viscosity? Give its units.

    Answer:

    Viscosity is defined as the degree up to which fluid resists the flow under an applied force. It is measured by tangential friction force acting per unit area divided by the velocity gradient under condition of streamline flow.

    Viscosity = Shear stress / Shear rate

    SI unit of viscosity is m2/sec.

    Coefficient of viscosity is the term to measure viscosity of given fluid. It is constant for a liquid and depends on its nature. If ‘f’ is the force of friction between two layers of fluid having area in square cm and represented by distance will have viscosity given by fµA(dv/dx)

    or

    f = ɳA(dv/dx)

    ɳ = coefficient of viscosity

    SI unit of coefficient of viscosity is Ns/m2.

  21. f. Convert 90˚ F into Kelvin scale

    Answer:

    We have given temperature 90°F

    1st we have to convert 90°F to Celcius scale

    T(°C) = (T(°F)-32) X 5/9

                    =             (90-32) X 5/9

                    =             58 X 5/9

                    =             32.22°C

    Now add 273 to 32.22°C to convert it into Kelvin scale

                    =             32.22 + 273

                    =             305.22 K

     

    So, 90°F = 305.22 K

  22. g. Define free, forced and resonant vibrations.

    Answer:

    Free vibrations are the oscillations where the total energy remains the same over time. This means that the amplitude of vibrations stays the same.

    Forced vibrations occur when an object is forced to vibrate by the means of external periodic input of force.

    Resonance vibration is the forced vibration where the frequency of the vibration is very close or same to the natural frequencies of the object.

  23. h. Frivtion is a necessary evil. Explain.

    Answer:

    Friction is said to be natural evil because it is useful as well as harmful. Friction helps us to walk on the surface of the earth & holding the objects. Without friction many essential processes cannot be done.

    On the other hand, it causes wearing & tearing in the machinery parts. It causes heat produced in moving parts due to which it also damages thing. So, it is an essential evil.

  24. i. State different modulus of elasticity.

    Answer:

    The three different modules of elasticity are

    Young’s Modulus

    Shear Modulus

    Bulk Modulus

        Young modulus is a mechanical property that measures the stiffness of a solid material. It defines the relationship between the stress and strain in a material in the linear elasticity regime of a deformation. 

        The shear modulus (or modulus of rigidity) describes an objects tendency to shear (deform) when acted upon by opposing forces. It is defined as shear stress over shear strain.

     Shear Modulus =  

    StrainThe bulk modulus of a substance is a measure of how resistant to compression that substance is. It is defined as the ratio of the infinitesimal pressure increase to the resulting relative decrease of volume.

  25. SECTION-C

    Attempt any three questions.             3x10=30
    Q3. Explain conservation of mechanical energy of a freely falling body.

    Answer:

    Consider a body of mass m placed at point A at height h from ground.

    H= height from the ground

    U = 0 is initial velocity

    V1 = velocity of body at point B

    V = velocity of body at point C

     

           (i) At point A  (PE  =max, KE =0)

    PEA = mgh

    KEA = 0

    Total Mechanical Energy = PEA + KEA

                                                                              = mgh + 0

                                                      = mgh

    (ii)           AT point B

                    V12 -0 = 2gs

                    V12  = 2gs

    KEB = 1/2m V12 = 1/2m(2gs) = mgs

    PEB = mg(h-s)

    Total Mechanical Energy = PEB + KE= mg(h-s) + mgs = mgh

     

    (iii)          AT point C

                    V2  - 02 = 2gh

                    V2  = 2gh

    KEC = 1/2mv2 = 1/2m(2gh) = mgh

    PEC = 0

    Total Mechanical Energy = KEC + PEC

                                    = mgh + 0

                                    = mgh

    So, total mechanical energy of the body at point A, B, C is same. So, total mechanical energy of the free falling body is conserved.

  26. Q4. a) It is easier to pull a lawn roller than to push it. Explain.

    Answer:

    Pushing friction force is always greater than that of in case of pulling force. The force applied to move the body depends upon the effective weight. As effective weight in case of pulling is less than in case of pushing, so less force is required. Thus, it is easier to pull a lawn roller rather than to push it.

  27. b) What is banking of roads? Explain.

    Answer:

    The phenomenon of raising the outer edge of the curved road above the inner edgeso as to provide the necessary centripetal force to the vehicles to take a safer turn & curved road is called Banking of the road. When the vehicle moves along the horizontal curved road, necessary centripetal force is supplied by the force of the friction between the wheels and surface of the road. To increase the centripetal force at the sharp edges the road is banked to avoid the skidding of the vehicles from road.

  28. Q5. State and prove Bernoulli’s theorem.

    Answer:

    Bernoulli’s Theorm

    It states that an increase in the speed of a fluid occurs simultaneously with the decrease in the pressure or decrease in fluid potential energy.

    Let the viscosity, pressure and area of a fluid column at a point A be V1, P1, A1  and at another point B is V2, P2, A2.

    Let the volume be moved from A & B to M & N.

    Let AM = L1 & BN = L2

    Now if we can compress the fluid then we have A1L1 = A2L2

    Net work done per volume = P2 – P1

    KE per volume = 1/2ρv2

    KE gained per volume = ½ρ

    PE gained per volume = ρ g (h2-h1)

    Now, P1-P2 = 1/2P( V22 – V12) + Pg (h2-h1)

    P1-P2 =1/2PV22 -1/2P1+ Pgh2 – Pgh1

    P1+1/2PV12+Pgh1 =P2 +1/2 PV22+ Pgh2

     P + 1/2PV  + Pgh =const

    This is called Bernoulli’s theorem.

  29. Q6. What are different modes of transfer of heat? Explain with examples.

    Answer:

    The three basic modes of transfer of heat are:

    1. Conduction
    2. Convection
    3. Radiation

     

    1. Conduction of heat :-  It is a process in which heat is transferred from hotter part of the body to the colder part of the body without involving any actual movement of molecules in the body. Heat transfer through the process of conduction occurs in substances which are in direct contact with each other. E.g. heating one side of the needle other side also get hot. Or heat will transfer from the burner to the cooker.
    2. Convection :- in this process heat will transferred in the liquid & gaseous from a region of higher temperature to the region of lower temperature due to the movement of molecules between the regions. E.g. heating of milk or boiling of milk in pan.
    3. Radiation :- it is the process in which heat is transferred from one body to another body without involving molecules or medium and gets transferred in th form of radiation. E.g. heat from sun or heat from bulb.

  30. Q7. a) Give difference between scalar and vector quantities.

    Answer:

    S. No

    Scalar

    Vector

     

     

     

    1

    Scalar quantities have only magnitude

    Vector quantities have magnitude as well as direction

    2

    These have no directions

    These have directions associated with them

    3

    These are represented by symbol A

    These are represented by            

                              

    4

    Examples Mass, Temperature, Speed

    Examples Velocity, Acceleration, Force

  31. b) Check the correctness of relation t = 2π √ l/g where l is length and g is acceleration due to gravity.

    Answer:

    1. We have

    t =

    Let us prove the correctness of the above relation using dimensional analysis

    t = M°L°T1

           = (M°L1T°)/(M°L1T-2)1/2

    = (M°L°T2)1/2

    = M°L°T1

    Now we have dimensional formula for LHS & RHS

    So, we get

    t =

    M°L°T1 = M°L°T1

    So, the equation is correct.

  32. Q8. Define three coefficient of thermal expansion. Establish relation between them.

    Answer:

    Thermal expansion is the tendency of the matter to change in shape, area & volume in response to a change in temperature. The three coefficients are

    Linear expansion (α) = it means change in only one direction (length).

    Areal Expansion (β) = it gives change in area of a object due to change in temperature.

    Volume Expansion (γ) =  it occurs in case of gases, liquids, solids. It gives change in volume w.r.t. change in temperature.

    Relationship between α:β:γ = 1:2:3

  33. Q9. State and explain Newton’s law of motion.

    Answer:

    Newton gave three laws of motion and are as under

    1. First Law = It states that every object persists to stay at rest or in motion in a straight line if there is no external force acting on it. It is also called law of inertia.
    2. Second law = It states that force is equal to change in momentum per change in time.

    i.e. F = dp/dt

    But p = mv [m= mass, v= velocity]

    F = d(mv)/dt

       = m dv/dt  [dv/dt = a]

       = ma

    i.e for constant mass m, force F equals to the mass time acceleration.

    1. Third law = It states that for every action there is equal and opposite reaction.

  34. Q10. State and prove conservation of linear momentum.

    Answer:

    Law of conservation of linear momentum states that the total momentum of the system is always conserved if no external force acts on an object or system of objects.

    Consider two bodies A & B of masses m1 & m2 which are moving with velocity u1 & u2. After collision, velocity changes to v1 & v2 respectively.