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Answer page-Ph. A to K.

posted Jan 29, 2015, 2:14 AM by Ranmini Perera   [ updated Sep 18, 2019, 7:17 PM by Upali Salpadoru ]

How..zat....?   



      A

      B

 

        C

Circular Motion.

1.1

Period is the time for 1 revolution.  

16 rev. in 5 second

Therefore 1 rev in  5/ 16 s.  =  0.31 s.

1.2 

Velocity = distance / Time

       V =Circumference  / T   ms-1

        V = (2 x 3.14x 1.0)/ 0.31  ms-1

             = 20.26   ms-1

1.3 .


Centripetal acceleration = V2 / R

                                    =20.26 x20.26 /  1

                                       = 410.47 


1.4  

Centripetal force = mass x acceleration.

Fc =  0.4 x 410.47 N.

= 164.2 N.

Option

Fc = mv2/ r

  = 0.4 x 20.26 x20.26  / 1

  =164.2  N.


Q 2.0

2.1

 Friction provided the Centripetal force.

Fc = MV2 / R

Fc = 0.2 x V2 / 0.25 N

=  0.2 x V2 / 0.25

V2 =3x 0.25 / 0.2

   V  = √3.75  = 1.94 m. s-1


 2.2  Circumference of record = 2 π R

Rev / s =  2x.3.14 x 0.25/1.94   = 0.8

In 1 min   0.8 x 60 = 48 rpm


3.1 

 Vertical force is the weight.

   = mg

   = 800 x 9.8   N.    =  7840 N.

3.2 

 If  N is the Normal, vertical becomes N cos θ

     N cos θ  = N

                 N = 7840 cos 20 ˚.

          N = 7840 x 0.94  =  7367 N

3.3 

The Horizontal component is =  N sine θ

                                      = 7840 x 0.34

                                        =  2681 N  


3.4 

 Maximum speed will depend on the Centripetal force (Fc)

       This is the horizontal component of N.  

      Fc = mv2 / r

   2681 = 800 x v2 /  50

       V2 = 2681 x50 /  800


    D

                               E        


       Electricity

  Q.1.0

Conductors:

Resistors:-

Insulators.

Copper,   Salt solution,  Gold.

Nichrome. Fuse wire,

Nitrogen. Leather.


Q.2.0

  1. Left  2. Ammeter. 3. Resistor. 4. Voltmeter,  5. Bulb.


Q.3.0

1.  AA type torch cell -    1.5 V, . DC.        2.  Household current-  UK and Commonwealth - 230 / 240 V,    AC,          3.   Car battery - (normal) 12 V, DC,

      4. Car battery charger-  12 V  DC,         5.  Car alternator-    12 V 0 AC.


Q.4.0


Question

No.


Condition of

Fuse

Lights or not.

Switch

1.

2

3

4

5

Red bulb

Blue bulb

.1.

Closed

Open

Closed

Closed

Good.

Lights

Lights.

2.

Closed

Open

Open

Closed

Good.

No.

Lights.

3.

Closed

Open

Closed

Off

Good.

Lights

No.

4.

Open

closed

Closed

Closed

Good.

No

No

5.

Closed

Closed

Closed

Closed

Melts

No.

No.

Q.5.0

  1. V=ir    R1 = V/i

                       R1 = 12/ 3  ………….R1 =  4

2.              R2 = V/i

                  R2 = 12/2   ……………..R2 = 6 Ω.

3.    V = ir

      12= (3 +2)  x R……………...R= 12/5 =  2.4Ω

4.     5 A.                       

5.  V = ir

   12 = ix4……………….i=  12/4 ……...i =  3 A

      Electromagnetic Induction

Q. 1.0

1.1 As the copper tuns in the coil cut the horizontal  magnetic lines each turn adds an emf. So the EMF will slowly increase until the movement stops.

1.2 Emf will be high to begin with and slowly decline as the turns come out of the magnetic field. The current will be in the opposite direction.

1.3  The effect will be as in 1.1 but the current will be in the other direction.

1.4 Similar to what happened in 1.2.

Q. 2.0

2.1 clock-wise.

2.2 Counter clock-wise.

Q.3.0

3.1  No current.

3.2  No current

3.3   B to A  (Right hand rule)


Q . 4.0

4,1  N to S

4.2  B to A. (Use right hand rule)

4.3  V = B. l. v

   V  = 5x 0.25 x 5

        = 6.25V.

4.4

4.5


Q. 5.0

5.1  Step down Transformer capable getting several voltages.

5.2  Vp /Vs = Np/Ns

230/ Vs = 1000/ 400

Vs=   230x400/1000

      =  92V

5.3

 230 V      1000

 V                800

V  =  230x800 / 1000

         204


Q. 6.0

6.1  B  ( use the left hand rule)

6.2  Current coming into the coil will change as the brushes will contact opposite ends. ( B will get connected to positive and so on)

5x2= 10 marks)  

Energy Cycle


             Q.1.0

   1  D,    2 B,   3  A   4. C, 5 A,

6  C, 7. B, 8 .A, 9. D. 10-A


2.0 Give the energy made use of and the main form of energy released.


Device / Action

Starting Energy

Final energy formed or released.

Green plant exposed to sun.

Light/Radiant

Chemical

Turning a bicycle dynamo .

Kinetic

Electric

Flashing a torch light.

Chemical

Electric- radiant/light

Wood burner.

Chemical

Thermal- radiant/light.

Stopping a moving car.

Kinetic

Thermal - vibrational/sound.

2x10 = 20 marks.



         Q.3.0

                             3-1  500 x10 x60  =300,000 J

 


3-2

Bread…...990x2   ……= 1980 J

Egg  ………                =   600 J

Butter…. 5 x31.2…    =   156 J

Sugar…...3x16…….    =     48 J

Total content……..=      2784  J   25% of this ….=   25 x 2784 ÷  100

                                                                    =    696 J.



3-3   Kinetic energy = ½ x mv2

                             = 1200 x100 x100 J

                             = 12000000 J…..12,000.kJ.

3.4   Capacity ……………………...=  3x 1000 kW.

       No. of houses to serve  =   3000x 1000 ÷ 500  …= 6000.Houses



Q.4.0.  

                      4-1 Both bulbs use equal amounts.

                      4-2  Fluorescent bulb.

             4-3  A. All the UV rays cannot be converted to light By the fluorescent chemical.  Some other rays may also can be produced.

                             B. Filament has to get heated to be incandescent. Lot of heat is wasted.

            4-4  A- Mercury vapour ionises  allowing electrons in the gas to emit photons at UV frequencies. The UV light is converted into visible light by the chemicals applied inside the bulb.

                   B When the current pass through the thin long tungsten filament it gets heated up and glows. The original red turns to white light as the temperature rises.

        


                                                                      F

Forces - Extensions.

  1.0

1.1  

For 5 N   the extension is  0.03m

For 10 N the extension is  =0.03 x2  = 0.06

Therefore the new length is  = 0.10 + 0.06 

 = 0.16 m

1.2 

 Spring constant = Force / extension

                        =  5 / .03  =    166.67 nm-1 

2.0

2.1  11 cm.

2.2  12 cm.

2.3  

The weight of the lower spring gets added

to the load.


3.0

3.1   Downward force  = + mg

         Tension       = - kx

At equilibrium they are balanced

        mg = -kx

m x 9.8  = - 6 x 0.1

         m  = - 0.6 / 9.8  =  6.1 x 10-2   kg.

3.2     P.E = 0.5 x kx2

                   =  0.5 x 6 x 0.1 x0.1 = 0.03 J

 
 


4.0

4.1    This is the length at zero force  .   = 10 cm.

4.2     This is where the graph changes = 20 N

4.3    K =  Force  /  Extension =   1.

H ans.jpg





5.0

               F = kx

     k = F / x  =  2450 / 0.08   

        =


 Forces - Friction   

1.1 A –. Fluid friction,   B – Unwanted   C -   Streamlined to reduce

1.2 A - Fluid friction,   B – Unwanted   C -   Streamlined to reduce.

1.3  A – Sliding friction,  B – Unwanted,   C = Lubricated to reduce.

1.4 – Rolling friction,  B – unwanted   C -   Axle is greased to reduce.

        OR A- Sliding friction  B – Un wanted  C Reduced by wheels.

1.5 – Fluid friction.  B – Wanted.    C – Large area of parachute.

 2.0

Material or Method.

How or where used?

Effect.

2.1. Using rubber

On sliding surfaces.

Increases friction

2. 2 Using ball bearings.

Around axles of rotating wheels.

Reduces rubbing friction.

2.3  Treads.

Shoe soles, tyres.

Increases depending on material and depth of tread.

2.4  Magnetic force.

Trains lifted over rails.

Reduces sliding friction.

2.5  Ice

Skating.

Ice melts due to pressure reducing frictiongreatly.

 

 

 

3.0

3.1   R  = W cos 35˚.

             =  250 x 0.82

             =204.8 N.

 

3..2  F = W sin 35˚.

      = 250 x 0.54

           =134

 

3.3 Fr  = F

       = 134 N

 

3.4  Coefficient of Fr =  F/ R

=   134 /  204.8                          =  0.65.

 

 

4.0 Answers.

4.1 

           240 up.  

4.2 

        120 West.

 

4.3  

         C =  Fr/R

            =  120 / 240

            = 0.5

 

4.4

New reaction force  240 + 200 = 440 N

Using the formula

                C =  Fr/ R

We get

        0.5 =  Fr / 440

        Fr   =   440 x 0 .5                     

              =  220 N                               

 

 


  G

   H

Heat energy

Q.1.0  Answers

1.1 Temparature.   1.2  Conduction,  1.3 Joule  ,  1.4   Latent heat,  1.5 Specific heat.

Q.2.0 Answers

2.1 Heat =  m.c.t,    That is  = 25 x 4.18 x 80.

                                              = 8360. J

2.2.         Heat gained by Al =  150 x0.4 x 80.  

                                                =   4800 J

2.3.    Yes.  When heat is produced outside vessel can waste a lot by convection and radiation

2.4.    Silver.   It reduces neat loss by radiation.

Q.3.0   Answer.

Let C be the specific heat of the object.

Heat gained by water =  40x 4.18 x 8 J

Heat lost by object  = 20 x c x (100- 28).

That is  40x4.18x8 = 20x xc x72

                   C= 1337 / 1440  =0 ..93 J/g per ˚C

Q. 4.0 Answer

Let L be the latent heat of melting.

Then heat gained by ice = 41 x L.  

Electric energy given = Vit.

41 x L  = 1.5 x 0.03 x300.     

                            L   =  (1.5 x0.03 x300)/ 42

                           L =  0.33J 

Q. 5.0

 1. Protects the glass bottle. Prevents heat loss to some extent.
 2. Prevents convection and evaporation.
3. Contains the liquid. Double walled flask holds the vacuum and offers a surface foe silver coating.
4. Prevents radiation.
5. Prevents mainly conduction.


    I

    J

    K

Kirchhoff's Law


 Q. 1.0

  1.1.1   

         V= ir,

    V = 1.5 x 4 = 6 V

1.    

 1. According to Kirchhof first law Incoming current and leaving current should add up to zero..

     Lets name   Current from A to B as X .

               Then    X + 1-3 = 0

                                 X = 2 A.

2

2. Let V be the voltage of the bulb

    V – 6   +6 – 12 = 0

    V = 12 .

 2.        2. Finding V1.

        V = ir.    

             V = 2 x 6     = 12 V.

3.            3     Finding V2.

               V = ir.    

                V = 1 x 6  =  6 V.

       4    Finding V3.

               Use Kirchhoff’s voltage Law.

              Going clockwise in outer loop.

             + 6  - V 2 + V1 + V3 = 05

                 + 6 – 6 + 12 + v3 = 0

                                   V3  = - 12 V.

    5    Finding V 4

     Going clockwise in outer loop

                   + 6 – V4 – V3 = 0

              +6 –V4 – 12 = 0

                    V4 =  6 V.

.  Q 3.0.

     Loop 1

     1.Find  A1.

              V= ir

             12= ix 6

              I = 12/6 =  2 A

. 3.0.

     Loop 1

           1.  Find  A1.

                      V= ir

                     12= ix 6

            I = 12/6 =2

    At Loop 2

          2. Find A 2.

                 V=ir

                 6= ix6

                I = 1 A


     Find A3.

    Incoming current = leaving current.

                    2 + 1 = 3 A.

4.  Find V1

    V1 = 2x 6   = 12 V.

5 . Find V2.

     V2 = 1x6 = 6 V.

    Outer loop

 6. Find V3.

    Using Kirchhoff law:-

    6 + V-12 = 0

          V = 6 V

Q. 4.0

 W

1.       

I/r1 + !/r2 = 1?R

     1/3 +1/3  =1/ R

              1/R = 2/3

        R = 3/2  =1.5 Ω .                                                                                

2.      

     r1 + r2 =R

    R = 3 + 1.5 Ω.

    R = 4.5 Ω.

3.        Total voltage in the loop 

            =  6 v + 6 v = 12 v

     As V = ir    

        12 = i x 4.5

      I = 12/4.5  = 2.67  A

4.             

Current in ‘b’ has to be the same as in ‘a’.

5.                 C     

          V =ir

          V =2.7 x 3

           V=8,1 V

6.         d.    

 As the current gets split into 2 parallel resistors current in d is half of 2.67           = 1.3 A

7    e.    

Using Kirchhoff law

               8.1 – 6 +e  -6 = 0

               e      = 12- 8.1

                       = 3.9 V


                           


 

 

 

  

 

  •     

Q.1.0    1.1  ….At  45°

           


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