Subject Materials for Physics

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A concave lens of focal length 15 cm , forms an image 10 cm from the lens. How far is the object. Solution : Given: F = -15cm. V = -10cm. U = ? 1/f = 1/v - 1/u.
1 LIGHT: REFLECTION AND REFRACTION EXERCISES Choose the correct alternative in Question 1-5 1. The image formed by a concave mirror is observed to be virtual, erect and larger than the object Then the position of the object should be [A] Between the focus and center of curvature. [b] At the centre of curvature. [c] Beyond the centre of curvature. [d] Between the pole of the mirror and its focus. Ans: [ d] 2.

Where should an object be placed, so that a real and inverted image of the same size is obtained by a Convex lens? [a] at the focus of the lens. [b] at twice the focal length. [c] at infinity. [d] between the optical centre of the lens and its focus.

Ans: [ b ] 3. A spherical mirror and a thin spherical lens have each a focal length of -15cm. [a]both are concave. [b] both are convex. [c] the mirror is concave, but the lens is convex. [d] the mirror is convex , but the lens is concave. Ans: [ a ] 4. Nomatter how far you stand from a spherical mirror , your image appears erect. The mirror may be [a] plane [b] concave [c] convex [d] either plane or convex 5.

6.

Ans: [c ]. The beams of light , one red and other green, fall on the same spot on a white screen. The colour on the screen will appear to be [a] magenta [b] blue [c] cyan [d] yellow Ans: [ d ] An object 5cm high is held 25 cm away from a converging lens of focal length 10 cm . Draw the ray diagram and find the position , size and the nature of the image formed. Sol: Given, O = 5 cm U = -25 cm F = + 10 cm V = ?

2 I

= ?

Ray Diagram:

7.

A concave lens of focal length 15 cm , forms an image 10 cm from the lens. How far is the object Solution : Given: F = -15cm V = -10cm U = ? 1/f = 1/v - 1/u 1/-15 = 1/-10 1/u 1/u = - 1/10 + 1/15 1/u = -1/30 U = -30cm The object should be placed at a distance of 30cm in front of the mirror RAY DIAGRAM:

8.

An object is placed at a distance of 10cm from a convex mirror of focal length of 15cm . Find the position and nature of the image Solution: Given : f = +15cm u = -10cm v = ?’ 1/ f = 1/u +1/v 1/15 = -1/10 + 1/v 1/15 + 1/10 = 1/v 1/v = 1/6 v = 6cm Position of the image: Since V is formed at a distance of 6cm away on right side of the mirror . Nature of the image : Since the mirror is convex , image is virtual , erect and small sized.

3 9. An object 5.0cm in length is pl aced at a distance of 20cm in front of a convex mirror of radius of Curvature 30cm . F ind the position of the image , its nature and size. Solution: Given: Size of the object [O] = +15 cm R = +30 cm F = R/2 = +15cm U = -20cm V=? 1/f = 1/v + 1/u 1/15 = 1/v-1/20 1/v = 1/15 +1/20 1/v = 7/60 v = 60/7= 8.6cm Position of the image : Since v is +ve and image formed at a distance of 8.6 cm back side of the mirror [right side of the mirror] Size of the image: m = I/O = -v/u = I/5 = -60 / 7x 1/20 I = 15/7= 2.2cm I [ size of the image] 2.2cm Nature of the image: Since the mirror is convex image is virtual , erect and small sized.

10.

An object of size 7cm is placed at 27cm in front of a concave mirror of focal length 18cm. At what distance from the mirror should a screen be placed so that a sharp focused image can Be obtained? Find the size and nature of image. Solution: Given Size of the object [o] = +7cm F = -18cm U= -27cm V=? 1/f = 1/v+1/u 1/-18cm = 1/v-1/27 1/v= -1/54 v = -54cm Since Vis – ve screen should be placed at 54cm in front of the mirror. Size of the image : m= I/O= -v/u I/7=-[54/27] I= -14cm Nature of the image : image is real , inverted and magnified.

4 11.

Light of wavelength of 500nm in air, enters galas plate of refractive index 1.5. Find [i] speed [ii] frequency and [iii] wave length of light in glass Assume that the frequency of light remains same in both media Solution: Given Wave length of light in air λa = 500 nm = 500x10-9 m Refractive index of glass μg = 1.5 Speed of light in air C = 3x108 m/s C = νλ ν = 3 x 108 / 500 x 10-9 = 6 x 1014 Hz (i) Speed : μg = C / speed of light in glass speed of light glass = 3 x 10 8 / 1.5 = 2 x 108 m/s (ii) Frequency

since frequency does not change during refraction, it remains constant = 6 x 10 14 Hz

(iii) Wave length

λg = Vg / νg = 2 x 108 / 6 x 10 14 = 330 nm

12

The refractive index of dense flint glass is 1.65, and for alcohol it is 1.36, with respect To air. What is refractive index of dense flint glass with respect to alcohol ? So: Given

aμg

= 1.65 aμal = 1.36 alμg = aμg / aμal = 1.65 / 1.36 = 1.21

13.

Find the focal length of concave lens of power -2.D Given P = -2D F=? P = 1/f f = 1/-2 = - 0.5m = 50cm Focal length of concave lens = -0.5 mt

14.

Two thin lenses of power + 3.5.D and -2.5.D are placed in contact . Find the power of focal Length of lens combination. Given power of concave lens P1 = +3.5 D Power of convex lens P2 = -2.5D P = P1+P2 = 3.5 – 2.5 =+1D Power of lens combination = + 1D P = 1/ f F = 1/P = 1/1 = 1 Focal length of lens combination = 1m

5 OPTICAL

INSTRUMENTS

Choose the correct alternative in Questions 1-5 1. The human eye can focus objects at different distances by adjusting the focal length of the eye lens. This due to [i] persistence of vision [iii] near sightedness [ii] accommodation [iv] far sightedness iii 2. Cinematography makes use of [i] accommodation [ii] persistence of vision [iii] least distance of distinct vision [iv] bi focal lens system ii 3. The human eye forms the image of an object at its [i] cornea [ii] pupil [iii] iris [iv] retina iv 4. The least distance of distinct vision for ayoung adult with normal vision is [i] 25m [ii] 25cm [iii] 2.5cm [iv] 2.5m ii 5. The change in focal length of an eye lens to focus the image of objects at varying distances is Done by the action of the [i] pupil [ii] retina [iii] ciliarymuscles [iv] blind spot iii 5. The change in focal length of an eye lens to focus the image of objects at varying distances is done by The action of the [i] pupil [ii] retina [iii] ciliary muscles [iv] blind spot 6. A lens used as a simple magnifier gives magnification of 6. What is its focal length ? m=6 D = 25cm f=? m = D/f f = 25/6 = 4.2cm focal length of given lens = 4.2cm 7. A compound microscope has an objective of focal length 1.0cm and an eyepiece of focal length 4.0cm If the tube – length is 20cm , fins the magnification of the compound microscope. Focal length of object lens= 1cm Focal length o feye lens = 4cm L = 20 cm M= ? M= LxD/F 0f eye and object lens = 125 magnification of compound microscope =125 8. A telescope has an objective of focal length 140 cm and the eyepiece of focal length 5cm .Find [i] the magnification of the telescope for viewing distant objects for normal adjustment , [ii] separation between the objective lens and the eye piece. F of O = 140 cm

6 F of E = 5cm M= ? L=? M= f of O / f of E = 140/5 =28 magnification of the telescope = 28 L = f of O + fof E = 145.cm Tube length = 145 cm 9. A 52 – year old near sighted person wears eye glass with a power of -5.5 dioptres for distance Viewing . His doctor prescribes a correction of + 1.5 dioptres for near vision section of his Bi focals . This is measured relative to the main part of the lens. [i] What is the focal length of his distance viewing part of the lens ? [ii] What is the focal length of the near vision section of the lens? Solution: (a). For distant-viewing part Power P = - 5.5 D focal length f =1/P= - 1/5.5 m = - 0.18 m. (b) For near vision section. For far point of defective eye Pf = - 5.5 D given power of accommodation of the eyePa=+4 D Hence, for near point of defective eye, Pn = - 5.5 + 4 = - 1.5 D This power comes after correction If x be the original power of near vision section of the eye lens, then x + 1.5 = -1.5 x = = 1.5 – 1.5 = - 3D original focal length of near vision section of the lens, x = - 1/3 m = - 0.33 m 10. The far [point of a myopic person is 80cm in front of the eye . What is the nature and power Of the lens required to enable him to see very distant objects distinctly? Far point of the myopic person = 80 cm For the lens to be used U = infinite V = - 80 cm P = 1/f = 1/v- 1/u = 100/ 80 – 1/infinite = -10 / 8 =- 1-25 power of th lens = -1.25D

ELECTRICITY ITS HEATING AND CHEMICAL EFFECTS 1. (i) (ii) (iii) (iv) (v) (vi)

Read the Following Statements and write True or False against each. The quantity of charge flowing past a point multiplied by the time gives the current. The Flow of charge through a conducting were connected to a cell is due to the chemical reaction inside the cell. The resistively all pure metals increase with rising temperature. Ohm’s law is a relation between the powers used the current and the potential difference in a circuit. A Series circuit has only one connecting bath for the electrons that moves through it. A Parallel circuit has multiple conducting paths. A conducting wire offers resistance to the flow of electrons because electrons repet each other in the wire.

Ans : (i) False (ii) True (iii) True

(iv) False (v) True (vi) False

7 2. (i) (ii) (iii) (iv)

Explain the following : Why is tungsten used almost exclusively for filament of incandescent lamps ? Why are the conductors of electric beating devices, such as toasters and electric irons, made of an alloy rather than a pure metal ? Why is the series arrangement not used for domestic circuit ? How does the resistance of a wire vary with its cross-sectional area ?

Ans : (i) Tungsten is used almost exclusively for filament of in candescent lamps for the following reasons. It has high resistance among metals. It has high melting point. So it can be operated at a temperature of 2700oc. (ii) Conductors of Electric heating devices such as toasters and electric irons are made up of an alloy rather than a pure metals because. • The resistivity of an alloy is generally higher than that of pure metals which form that alloy. • Resistiavity of alloys changes less rapidly with change in temperature • They do not oxidise (burn) readily at high temperature unlike pure metals. (iiii) Series arrangement is not used in domestic circuit because. • Any Fault produced in one appliances will cut down Flow of current to other circuits or appliances. • Resultant resistance will become very large which may well affect the flow of current. • Potential difference across an appliance may not be sufficient since variable potential difference produced at each appliance. (iv) Resistance of a wire increases with decrease of cross sectional area and decreases with increase of cross sectional area. It means resistance of a wire is indirectly proportional to its cross sectional area.(A) 2. A Piece of wire having resistance ‘R’ is cut into Four equal parts. (i) How will the resistance of each part compare with the original resistance. (ii) If four parts are placed in parallel How will the resistance of the combination compare with the original value. Sol:

4. Sol.

Initial resistance of the wire=R. Let initial length of the wire be L (i) After cutting four equal parts each part has ¼ th length. i.e L/4 AS R is directly proportional to L resistance of each part will become ¼ th of the original distance. (ii) Resistance of each part = R/4 In parallel 1/R = 1/R1 + 1/R2+ !/R3 …………. = 4/R+4/R+4/R+ 4/R = 16/R Rp=R/16 A piece of wire is drawn by pulling it until its length is doubled. Compare new resistance with original value. before redrawing the wire, its length = L1, Cross Sectional area = A1, and its resistance = R1 after redrawing the wire its length = L2 = 2L1, Cross Sectional area= A2 and Resistance = R2 since volume is same, L1 x A1 = L2 x A2 A1/A2 = L2/L1=2 since the material of the wire is same, ρ1 = ρ2 R1A1/ L1 = R2 A2/L2 R2=A1/A2 x L2/L1 x R1 = 2 x 2 x R1 = 4 R1 The new resistance will be Four times the original.

8 Sol.

5. A copper wire has diameter 0.5 mm and a resistivity of 1.6 10-6 ohm cm. How much of the wire should be required to make a 10 ohm coil. ρ=1.6 x 10-6 ohm.cm = 1.6 x 10-8 ohm.m d=0.5 mm = 0.5 x 10-3 m R=10 ohm L= ? L= R A/ρ = R π r 2/ρ = 10 x 3.14 x 0.25 x 10-6/ 1.6 x 10-8 x 4 = 122.6 metres.

6. Draw a schematic diagram of circuit consisting of a battery of Four 2v cells. A 5 ohm resistor, 8 ohm resistor, 12 ohm resistor, and a plug key all connected in series. Circuit diagram

7. Redraw the circuit of above problem 6, putting an ammeter to measure the current through the resistors and Voltameter to measure the Voltage across the 12 ohm resistor. What would be reading in ammeters and Dolt meter? Sol.

V= 4 x 2 = 8v R1= 5 ohm, R2=8 ohm, R3=12 ohm voltage across R3=? Current I = ?

Diagram:In series R= R1+R2+R3 = 5+8+12 = 25 ohm I = V/R = 8/25 ampere Reading of ammetre = 8/25 ampere Voltage across R3 = I x R3 = (8/25) x 12 = 3.84 v reading of volt meter = 3.84 v 8.Two lamps, one rated 100 w at 220 v, and the other 60w at 220w, are connected in parallel to a 220v supply. What current is drawn from the supply line. Sol.

Given

Supply voltage For first bulb 100 = 484 ohms

P1 = 100w, V1=220 v, P2 = 60 w, V2=220 v V = 220 v, R1= V21/ P1

= (220 x 220 )/

9 For Second bulb In parallel

Current in the circuit

R2= V22/ P2 = (220 x 220) /60 = 484/6 ohms 1/R=1/R1 + 1/R2 = 160/( 220 x 220) R=220 x 22/16 I = V/R = (220 x 16)/ (220 x 22) = 16/22 = 0.727 A = 0.727 A

9. A heater coil connected to a 220v has a resistance of 150 ohm. How long will it take for this coil to heat 1 kg of Water from 20oC to 60oC . Assume that all heat is taken up water? Sol. V = 220 v R=150 ohm m=1 kg θ= 60-20 = 40oC s= 4180 J/KgoC t= ? Heat produced by the Coil = H= V2t/R, = (220 x 220 x t) / 150 Heat gain = Q = msθ= 1 x 4180 x 40 we know that H=Q ( 220 x 220 x t)/150 = 1 x 41180 x 40 t=518.18 s time required = 518 sec. 10. In the circuit show in figure calculate (i) The current flowing through arm AB, AC and CDE. (ii) Potential difference across AB, CD and DE. Sol :-

R1=6 ohm, R2= 4 ohm, R3=12 ohm, V=1.5 volts from diagram , Arm AB and CDE are connected in

parallel. Arm CD and DE are connected in series. Resultant Resistance of arm CE = 6+4 = 10 ohm Resultant Resistance of the circuit R = (10 x 12)/22 = 60/11 ohm I = V/R = 1.5 x ( 11/60 ) = 0.275 A (i) Current across arm AB .125 A Current across arm AC Current across arm CDE

= VAB / R3 = 1.5/12 =

(ii) Potential difference across AB across CE = 1.5 v Potential difference across CD 6 = 0.9 v Potential difference across DE 4 = 0.6 v

=

Potential difference

=

0.15 x R1 = 0.15 x

=

0.15 x R2 = 0.15 x

= 0.275-0.125 = 0.15A = 0.15 A

11. A 40 w lamp require 0.182 A of current at 220 v while 60w lamp require 0.272 A of current at 220 v . If 40w lamp and 60w lamp are connected in series with a 220 v lines, How many amperes of current will Flow through each lamp ?

10 Sol. For 1st Lamp,

P1=40w, V1=220 v R1=V12/P1=(220 x 220)/ 40 = 4840/4 ohm For 2nd Lamp, R2 =V22/P2= (220 x 220)/60 = 4840/6 ohm In series R=R1 + R2= (5 x 4840)/12 I = V/R = 220 x 12/( 5 x 4840) = 0.109 A

Five dry cells each of 1.5 v have internal resistance of 0.2, 0.3, 0.4, 0.5 & 12 ohms. when connect in series. 12. What current will these five cells furnish through 10ohm resistance ? Sol.

Total emf of the circuit E Total internal resistance r

= R = I = I = =

5 x 1.5 = 7.5 v 0.2+0.3+0.4+0.5+12=13.4 ohm 10 ohm ? E/(R+r) = 7.5 /(10+13.4) 0.32 A

The values of Current I, Flowing in a given resistor for corresponding values of potential difference V, across 13. the resistor are given below. 2.0 3.0 4.0 I 0.5 1.0 V 0.5 2.5 6.75 11.0 15.0 between v and I and Plot a graph Calculate resistance of the resistor ?

y

5.0 4.5 4.0 3.5 V 3.0 2.5 2.0 1.5 1.0 0.5

x 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5

I

11 MAGNETIC EFFETS OF ELECTRIC CURRENT 1.

Check the following statements. Write true or False against each. (i) Like magnetic poles attract each other ; unlike poles repel (ii) If you strike a sharp edge of a metallic knife against the N-Pole of a bar magnet, it will induce a N-Pole (iii) The Magnetic field produced by a current in straight conductor has no poles. (iv) An electric generator is a device that converts electrical energy into mechanical energy Ans : (i) False (ii) False (iii) True (iv) False 2.

The phenomenon of electromagnetic Induction is (i) The process of charging a body (ii) Process of generating magnetic field due to current passing through a coil (iii) Producing induced current in a coil by relative motion between a magnet and the Coil (iv) The process of rotating a coil of an electric Motor.

Ans : The Correct alternative is (iii) 3.

The device used for producing current is called a (i) Generator (ii) Dolt Meter (ii) Ammeter

(iv) Galvanometer

4. What are Magnetic field lines? How is the direction of a magnetic field at a point determined? Mention two imp properties of the magnetic field lines? Ans : Magnetic Field Line :A Magnetic lines of Force, is a line straight or curved, in the Magnetic field of a magnetic. Pole or magnetic dipole, such that the tangent at any point of this line gives the direction of the magnetic field at that point. Direction of Magnetic Field at a point :Important properties of Field Lines. (i) Magnetic Field lines start from a N-pole and end at a S-pole. (ii) Two lines of Force do not intersect each other (iii) They tend to contract longitudinally. (iv) They tend to expand late rally (v) The number of magnetic field lines of force passing normally per unit area about a point, gives the intensity of the magnetic field at that point. 5.

Draw a rough sketch of pattern of field lines due to a Current flowing through a circular coil (i) (ii) Solenoid carrying current Ans : Pattern of Field lines for a current flowing (i) Through a circular coil. (ii) Pattern of Field lines for solenoid carrying current

12

6.

State the rule to determine the direction of (i) Magnetic Field produced around a current carrying conductor (ii) Force experienced by a straight conductor carrying current placed in a Magnetic Field which is perpendicular to it. (iii) Current induced in a circuit by the changing magnetic flux due to the motion of magnet.

Ans : (i) (ii) (iii)

The rule is « Right hand thumb rule » The rule is « Fleming’s left hand rule » The rule is « Fleming’s Right hand rule »

On what factors does the force experienced by a current carrying conductor placed in a uniform magnetic field 7. depend ? Ans : Fore experienced by a current carrying conductor placed in a uniform magnetic field depend upon following factors. (i) Strength of the current (I) (ii) Strength of the magnetic Field(B) (iv) length of the conductor(L) 8. Explain the principle and working of an electric motor with the help of a diagram. What is the function of split-ring commutator ?

N

S

Ans : Principle : It works on the principle of motion of a current carrying conduction in a magnetic field according to Fleming’s right hand rule. Diagram : Working

:

A direct current source is connected between metallic brushes B1 and B2 K when current passes through the it flows in arms CB and AD in a direction perpendicular to the magnetic field equivalent opposite force act on these arms.(in a direction according to Fleming’s left hand rule) and they form a couple. The coil rotates in clock wise direction. After half rotation split parts of ring change brushes(I) B1 touches R2

13 and B2 touches R1. Hence current become reverse the arms but couple acts in the same direction as before. The continuous rotating the shaft and which it is wrapped. Thus rotate the motion become available. Function of split ring Commutator : Due to split ring commutators during every half rotation current become reverse which in tern make the couple to act in the same direction all time. Hence function of split-ring commutator is to provide continuous clock wise rotation of armature. 9.

(A)

A coil of copper wire is connected to a Galvanometer. What would happened if a bar magnet is (i) (ii) (iii)

Pushed into the coil which it’s N Pole entering first. Pulled out of the bar magnet. Held stationary inside the coil.

(i) (ii)

When N Pole of a magnet pushed into the coil galvanometer shows a deflection as if current is induced. When N Pole of a magnet pulled out of the coil galvanometer shows a deflection which is opposite to

(iv)

When a Bar magnet was held stationary inside the coil there is no deflection in galvanometer.

above.

10. Draw a labeled diagram to explain the principle underlying the working of electric generator. Ans : Principle : It works on the principle of electro magnetic induction. Working : Armature The shaft is rotated by some mechanical means. As the shaft rotates magnetic Flux introduce the coil changes hence induced E.M.F in the coil is produced in the N circuit. According to Faraday’s laws of electro magnetic induction. Magnitude of induced e.m.f changes from ‘O’ to maximum and then maximum to O. Through half rotation of the coil. During next half rotation direction of the current in both the arms changes giving rise to net induced current in the direction DCBA. Thus after every half rotation the polarity of the current in the respective arms changes. Thus magnitude and direction of induced e.m.f changes after equal intervals of time as shown in the fig.

S

brushes

diagram : What is the function of an earth wire? Why it is necessary at the mechanical appliances? 11. Ans : Function of earth wire : Due to long use some covered wires inside the appliance may become bare and may contact with metallic body of the appliance. In such case the appliance gives a shock. The function of the earth wire is to keep the potential of the appliance ‘o’ and shock is avoided. Necessity to earth the metallic appliances: It is necessary to earth the metallic appliances as the safety measure to avoid shocks. 12. What is short circuit and over load in electric supply? Ans : Short circuiting :

14 It means that the two wires live and neutral have come in contact with each other either due to their insulations having been damaged or fault in appliance. In such case the resistance of circuit decreases to very small value which makes current increase enormously it result in heating the line wires and producing spark at the place of short circuit. Over loading : When high powered appliances like refrigerator, air conditions and electric iron etc.. are switched on simontaneously total current through may exceeds the tolerant limit causing damage and Fire This situation causes. 13. Describe an experiment to illustrate the action of an electric switch. Ans : Material required : Two iron nails a thin strip of aluminum foil a Battery Plug Key Torch Bulb Procedure : (1) (2) (3) (4)

Fix two iron nails vertically on a table. Fix the two ends of a thin strip of aluminum foil of length about 5 cm to each iron nail. Connect the nails to a battery through a plug key and a small bulb. When the key is plugged bulb will glow then the strip is burnt the bulb becomes off. The circuit is broken.

SOURCES OF ENERGY EXERCISES 1. Explain why only apart of the solar energy that strikes the upper regions of atmosphere reaches the Surface of the earth . Ans: Only a part of solar energy reaches the surface of the earth because [i] some parts of solar radiation gets reflected back to the space by clouds and etc; [ii] the atmosphere absorbs significant portion of solar energy such as ultra violet radiation, X-ray s etc; 2. Exposure to which component of solar radiation could be a health hazard? Ultra violet radiation 3. With the help of a diagram explain the construction and working of a box type of solar cooker. Construction: It consists of an insulated and no n conducting wooden or metallic box. Those box are painted in black .The walls of the box are made thick with a lining of non Conducting material to prevent lose of heat. Plane mirror reflector is attached to the box an dits position is adjustable . It increases the affective area at the collection of solar energy. These provided with containers whose outer sides are black painted. A thick glass sheet covers The container. Working: The food to be cooked is taken in the containers which are then placed inside the box .The Containers are covered with glass sheet. The cooker is placed in sun light and the position of reflector is so adjusted that a strong beam of Sun light is reflected on the top covered with glass sheet .Sun light passes through transparent Glass sheet and is observed by the black painted walls of the container and the box . The Infrared radiation in the sun light heat the box and food inside the container .Temperature may Go up to 100 c to 140 c within two to three hours and the food get s cooked. Diagram:

15 sun rays

reflector

glass lid

box containers 4. Mention any two differences between a box type and a concentrator type solar heater. A ns : BOX TYPE SOLAR COOKER : [i] It use a plane mirror which only reflects but it does not concentrate solar energy. [ii] It attain relatively low temperature [iii] It ca n not provide baking and frying CONCENTRATED TYPE SOLAR COOKER: [i] It uses parabolic mirror or concave mirror which concentrates solar energy. [ii] It can attain high temperature. [iii] It can provide baking and frying. 5.

The surface area of a concentrator type solar heater is 5m It can reflect 80% of solar radiation Incident on it while it absorbs the rest. Calculate the energy concentrated by the heater at its Focus in 2 hours if solar energy were delivered to it at the rate of 0.4 kw /m Given: Area of the surface [A] = 5 m2 Time [t] = 2 h = 7200s Efficiency[n] = 80%= 0.8 Rate of reception of solar energy = 0.4kw/ m2 Energy focused [E] = nPAt = 0.8x0.4x5x7200 = 11520kj energy focused = 11520kj

16 6. Why is not possible to make use of solar cells to meet all our energy needs? State at least two reasons to support your answer Ans: Reasons for not using solar cells to meet our energy needs 1.They are very expensive because the semi conductor material needed for the solar cells must be very pure and expensive to make. 2.The energy stored in batteries provide only direct current for devices which require A.C.curret, D.C. is converted into A.C. conversion reduce efficiency and increase it’s cost. 7. Mention any four areas where solar cells are being used as a source of energy? ANS:Solar cells are being used in the areas such as 1. To provide electrical power for spacecrafts. 2. Street lighting. 3.To provide electricity to those houses which were situated in the seas and off- shore drilling platform. 4.To run calculator, watches, clocks etc. 8. Why wind energy farms can be established only at the specific locations. Give reasons to support your answer Ans: Wind energy forms can be establish at specific location because [1]. Wind is not available at all time. [2]. Wind is not available at all places. [3]. Wind should strong and steady to maintain desired level of production. Minimum wind velocity required for function of wind mill is about 15 km/h. Hence only few areas can satisfy above conditions. Hence wind mills can be established at specific locations which satisfy above conditions. 9. It is difficult to use hydrogen as a source of energy although its calorific value is quite high. Explain? Ans: Hydrogen is not used as domestic fuel in spite of it’s high calorific value. The reasons are [1]. It is highly combustible gas and explode when combined with oxygen. [2]. It is difficult to handle hydrogen. [3]. Production cost of hydrogen is very high. [4]. It was not easily available. [5]. Not easily liquefiable. Hence it is difficult to transport. 10. state the forms in which energy stored in the oceans manifests itself .Which one of these is utilized in OCET systems ? The forms in which energy manifests it self in ocean are [i] tidal energy [ii] Ocean wave energy Ocean thermal energy [ocean thermal energy is utilized in OCET system] 11.With the help of a labeled diagram explain the process of destructive distillation of wood . WORKING OR PROCEDURE: Heating the wood in horizontal tube is started initially at slow flame after some time heating is made intense And on strong heating volatile substance expel out of the wood and get condensed in water in tube B At the bottom in form of a black liquid. This dark black liquid is called wood tar Uncondensed volatile substance leave through this side tube in the form of gas. This combustible Gas is called wood gas. When gas is stop coming out heating is stopped. The black residue left in tube is collected which is charcoal.

17 tube A charcoal delivery tube Burner

wood gas cork tube B

12. Why is charcoal considered a better fuel than wood ? What are the disadvantages of converting wood into Charcoal Charcoal is considered a better fuel than wood because [i] charcoal ha s high calorific value [ 33kj/gm where as wood as low calorific value 17kj/gm] [ii] charcoal does not produce smoke on heating where as wood produces a lot of smoke . [iii] charcoal is compact where as wood is bulkish. [iv] charcoal is easy to handle where as handling wood is painful. DISADVATAGES :charcoal is a n expensive fuel [ii] on destructive distillation of 1 kg of wood yields 0. 25 kg charcoal which is very low 13. Describe the steps involved in obtaining biogas and explain what is meant by anaerobic decomposition steps involved in obtaining biogas are as follows [i] cattle dung or human excreta or domestic sewage and water are mixed in equal proportions in mixing tank to form a slurry [ii] the slurry goes into the inlet tank and then into the digester and fills about 1/3 rd of the digester [iii] biogas which is mixture of CH4 , H2S , H2, and CO2 is supplied to consumer through pipes thus we get biogas. ANOROBIC DECOMPOSITION : The process in which an organic matter under go fermentation by anaerobic bacteria in presence of water And absence of air is called anaerobic decomposition. 14. State the advantages of obtaining biogas from animal dung and bio wastes. 15. Why are fossil fuels classified as non renewable sources of energy? What steps should be taken to conserve These sources ? Fossil fuels are classified as non renewable sources of energy because [i] formation of fossil fuels such as coal , petroleum , natural gas takes billions of years which is very long time . [ii] formation of fossil fuels require certain very slow changes under special circumstances STEPS SHOULD BE TAKEN TO CONSERVE FOSSIL FUELS: [i] since fossil fuels are non renewable sources of energy , these should be used for very limited

18 purposes [ii] give main emphasize on use of non conventional sources of energy such as solar energy ,wind energy [iii] use power saving devices [iv] reduce wastage during extraction of coal , natural gas etc

16.The heat produced on complete combustion of a fuel could raise the temperature of 2 kg of water from 20 c To 70 c . Calculate the calorific value of the fuel , if the specific heat capacity of water were 44.2j/g c. Assume that the heat taken by the container is negligible . Mass of fuel [m] = 10g Mass of water [M] =2kg = 2000g T = 70 -20 = 50 S = 4.2 j /g c Clorific value =? Q= ms0 =2000x4.2x50 = 4.2x10 0000 10g of fuel produce 4.2x100000 j 1 g of fuel produce 4.2x 100000j/10 = 4.2x10000 calorific value of given fuel = 4.2x10000j =42 kj/g 17.How much energy will 1kg mass of wood yield on complete combustion if its calorific value were 15kj/g? calorific value of the fuel= 15 kj/g mass of the fuel = 1 kg = 1000g 1 g of fuel yield = 14 kj 1000g of fuel yield= 15x1000kj heat produced by 1kg of the fuel = 15x1000kj 18.Suppose the average solar energy incident on the green canopy [leaves ] of a tree were 10 j per day . The Calorific value of the wood obtained from this tree is 15 kj/g . If the tree could convert 1% of incident solar Energy as wood, how many days will it take to produce 10 kg of wood.? Heat received per day = 10 J Efficiency of conversion = 1% Mass of the wood to be produced m = 10kg = 10000g Calorific value of wood = 15 kj /g Number of days required [n] = ? Total energy required = 15x10000kj Since it is only 1% of total solar energy incident there fore total energy should be incident = 15x 106 No of days = 15x 106/105 No of days = 150 days

19.State the condition essential for combustion to take place . for combustion to takes place following conditions are essential [i] the presence of combustible substance [ii] the presence of a supporter of combustion such as air or oxygen [iii] attainment of ignition temperature

19 20. Explain how knowledge of the conditions of combustion could help in fire fighting . Knowledge of the conditions of combustion help us in what efforts should be made to put off a fire It tells us that to put off a fire we can either of the following [i] remove combustible substance from the surroundings [ii] pour dry sand to prevent oxygen and air

NUCLEAR FISSION AND FUSION EXERCISES 1. Two important nuclear reactions are described incompletely by the following equations: Fill in the gaps in these equations by employing the conservation of the masses and the atomic numbers of the reacting species and products. 27

Al13 + 4He2 → 30Si14 + ?

39

K19 + ? → 42Ca20 + 1H1

Ans;

27

Al13 + 4He2 → 30Si14 + 1H1 (Proton) K19 + 4He4 → 42Ca20 + 1H1

39

2. Nuclear reactions have been successfully demonstrated with the following energetic projectiles: Neutrons , α- particles , protons , and γ- rays .Complete the following reactions 27

Al13 + 2H1 → ? +1 0n Mg12 + ? → 23Na11 + 1H1

Ans :

24

sol :

27

Al13 + 2H1 → 28N14 +1 0n 24 Mg12 + γ- rays → 23Na11 + 1H1

3. Weknow that for thorium , Th , the atomic number Z=90 , for protactinium ,pa Z=91 and for U, Z=92.Suppose a proton is ejected from the nucleus of Pa , which of the following will be the product of this emission : [i] thorium [ii] uranium [iii] zinc; or [iv] none of the first three possibilities. Sol: the correct answer is (i) 4. We have learnt that in fission the nucleus of an element split into two nuclei will the fission products

20 Have [i] The same atomic number [ii] The same atomic mass [iii] The lower atomic number [iv] Higher atomic mass Give reason for the selection we have made Sol: The correct answer is (iii) Reason: The sum of the atomic numbers of two product nuclei is equal to the atomic number of fissioned nucleus 5. In fusion two nuclei merge into each other to form a new nucleus the product of the fusion is a nucleus Of [i] lower atomic number and lower atomic mass than the reacting nuclei [ii] same atomic number and same atomic mass has the reacting nuclei [iii] higher atomic number and lower atomic mass than the reacting nuclei [iv] higher atomic number and higher atomic mass than the reacting nuclei [v] lower atomic number and higher atomic mass than the reacting nuclei Give reason for your choice. Sol: The correct answer id (IV) Reason: The product nucleus has atomic mass is still less than sum atomic masses of reacting nuclei 6. The energy released during a fission of 1kg of U-235 is equivalent to ? [i] energy released by burning of 25 kg of coal [ii] energy released by burning of 2,500kg of coal [iii] energy rekeased by burning of 2,500,000kg of coal [iv] energy rekeased by burning of 250,000,000kg of coal Sol: The correct answer is (iii) 7. In fission and fussion reactions energy realeased due to [i] chemical reaction [ii] conversion of kinetic energy of the reactant [iii] conversion of mass into energy [iv] conversion of gravitational energy Sol: The correct answer is (iii) 8. Design a hypothetical experiment to create a nucleus of the precious metal platinum Pt 78 195 on the following ingredients ; 120Sn 80, 11 Na23 , 58Ni28 , 4He2 , 1H1 and 1n0 , You can use fusion reaction Ans: The hypothetical reaction may be 2 23Na11 + 2 58Ni28 + 331n0 → 195 Pt 78 9. In the chain reaction involving fission of U-235 what is the final stable fission product? What is It used for? Ans: The final stable product is plutonium [ 239 Pu94] it is used as 1. Fission fuel 2. For making atomic bombs 10. What are the roles of the moderator and the coolant fluids used in a nuclear reactor ? Name a few Moderators and coolants employed widely? Ans; Role of Moderator: Moderator slow down the speed of neutrons and it reduces the energy of neutrons from 2Mev ro 0.02 ev this is called thermal neutron. Moderators are made of (i) Graphite core (or) normal water (or) Heavy water Role of coolant : Liquid sodium is used as coolant. It carries the heat energy produced in nuclear reactor to heat exchanger.

21 THE UNIVERSE EXERCISES 1. Why are the images of astronomical objects obtained by Hubble Space Telescope better than those obtained By telescopes on the ground? Ans; Images of astronomical objects attaines by hubbles space telescope better than those obtained By telescopes on the ground because [i] hubble space telescope obtain the images in dust free space which is clear [ii] hubble telescope uses X-rays. 2. Which is the force that keeps the solar system together? Ans: Gravitation force of attraction 3. What is the size of the solar system? Ans: Size of the solar system is about 4. What is the colour of a star due to? Ans; Colour of a star is due to it s temperature 5. What is the source of energy of a star like the sun ? Ans: Nuclear fusion of hydrogen is the source of energy o fa star like the sun. 6. The radius of the sun is 7x 10 m. Its mean distance from the earth is 1.5x 10m.What is its angular diameter The radius of the moon is 1740 km . Its distance from the earth is 384000km . What isits angular diameter ? If the moon comes in between the earth and the sun can it cover the sun ? D oes this have any connection With the solar eclipse? (1)

d = diameter of sun D = distance between sun and earth θ = angular diameter of sun θ = d /D = 2x 7x 108 / 1.5 x 10 7 = 0.95x10-2 Angular diameter of sun = 0.95x10-2 radians (2) d = diameter of moon D = distance between moon and earth θ = angular diameter of moon = 0. 90625 x 10 -2 Angular diameter of moon= 0.90625x10-2 Since θm < θs , moon can’t cover whole sun. Moon covers the sun the eclipse is total. If only a part of moon covers the sun the eclipse id partial. 7. Why is it that we can see the solar corona only at the time of the total solar eclipse ? Ans: On sun‘s surface there are extremely hot gases but less dense. These layers are very faint and not Visible in presence of strong light. At time of total solar eclipse moon covers the central part of the sun. it reduces the intensity Of sun light. In this situation we can see the outer most layer which is called corona.

22 8. Apart from scientific interest why should a constant watch is kept on asteroids? Ans: Though asteroids orbit around sun in gap between mars and Jupiter. There remains a small probability Of their colliding with the earth. It was also occurred 65billions years ago. Such a collision can cause A lot of damage to life and property on the earth. so apart form scientific interest, a constant watch should be kept on asteroids . 9. Give a brief description of the internal structure of the earth. How is this information derived? Internal structure of earth can be divided into crust, mantle and core. CRUST: The outer most layer of the earth is called crust. It does not have uniform thickness, thick Is about 35km to 60km under continents? It is only 10km under ocean. It contain light rich in aluminum And silica. It has low average density of 3g/cm3. MANTLE: The region in between the crust and the core of earth is called mantle. It extended up to 2900km Below the crust. It contains layers of iron silicate and magnesium silicate rocks. The density varies from 4g/cm3 to 6g/cm3 . CORE : The inner most part of the earth is called core. The radius of the core is about 3400 Km.It contains mostly iron and some nicker. At the centre of the earth the temperature is about 40000C and the pressure is about 3.7 million atmosphere. 10. Calculate the distance covered by the light in one year. Name the unit that is equivalent to this Distance. Speed of the light is 3x10 m/s. Ans: We know that the speed of light in vacuum = 3x108 m/s distance traveled by light in 1 second = 3x108 m. distance traveled by light in 1 year = 3x108 x 24 x 365 x 3600 = 9.5 x 1015m 11. How do we estimate the age of the universe ? Ans: Age of the universe can be astimated by nthe method of Radiometric dating. Principle: In case of Uranium, 1/3 of nuclei decay into stable nuclei lead every 4.5 billion years. By radiometric measurements, proportion of uranium and lead nuclei in the sample is found out.Using the principle stated above, the age of the meteorite can be determined. Since meteorites come from outer space of universe, age of universe is to age of meteorites. 12. List some applications of artificial satellites. Ans: i. artificial satellites are used in communication like telecommunication, T.V. and Radio broad casting, fax , internet etc., ii. artificial satellites are used to weather monitoring and weather forcasting. iii. artificial satellites are used to estimate ground resources such as forest survey, minerals,crop yield and ground water survey. iv. artificial satellites are used to collect information about outer space. v. artificial satellites are used in spying purpose. 13. Name the organization which looks after the space programme in India. List some of its achievements . Ans: ISRO: Indian Space Research Orgnisation. Achievements of ISRO. i. Isro launched Indian National Satellite series of satellites like , Insat-1 , Insat-2, Insat-3. ii. Isro launched Indian remote sensing satellite system, like IRS-1 and IRS-P series iii. Isro launched Rohini satellite series.

23 iv. v.

Isro developed and tested Polar Satellite Launch Vehicle (PSLO). It successfully developed Geo Synchronous Satellite launch vehicle (GSLV)

14. Distinguish between natural and artificial satellites. Natural satellites: i. The satellite which revolve around earth naturally in its own orbit. ii. Man doesn’t have control over natural satellites. iii. these cannot be used for telecommunication, weather monitoring etc,. example; moon. Artificial satellites. i. the satellite which revolve around earth and made by man are called artificial satellites. ii. man has control over artificial satellites. iii. these can be used for communication,weathermonitoring etc,. eg. INSAT-1 15. A satellite is orbiting the earth at a height of 2000km . Compare its time period and orbital velocity with Those of a geostationary satellite. [Refer to table 17.1] h = 2000km , R =6400km r =R + h = 6400 + 2000 = 8400km =84x105 M = 6 x1024kg ,G = 6.67 x10-11 T = 2π √(R=h)3/GM = 7642.76 sec. Velocity V = √GM/r =6.9 x 103 m/s. Its time period is less than geostationary satellite but its orbital velocity is greater than geostationary orbits. 16. What are the essential properties of a rocket fuel . What is the composition of the liquid fuel ?What Is the composition of solid fuel? i. it must burn very rapidly but not explode. ii. It mustg produce large volumes of gas at high pressure and high temperature. iii. It should not left any residue in the rocket. iv. It should have high calorific value Liquid fuels: Liquid hydrogen and liquid oxygen , and kerosene. Solid fuels: Powdered aluminium and ammonia percholarate. ii.Powdered aluminium and ammonia nitrate. 17. Distinguish between polar and equatorial orbits for artificial satellites. Which orbit is suitable for A geostationary satellite? Which orbit would be suitable for a satellite collecting data for weather Predction?` Ans: Polar Satellites: i. Orbits which are pass over poles of earth . ii. it is suited for remote sensing satellite. Equotorial orbit: i. Orbits which pass parallel to equator of earth. ii. it is suited for Geostationary satellite. Polar satellites are best suitable for satellite collecting data weather prediction.