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# ICSE Set Qa1 Year Icse Physics98 Qa1.php Physics Exam Paper for students online

 Physics - 1998 ( I.C.S.E) You are on Answer - 1 SECTION-I Compulsory:To be attempted by all candidates Q1(a)(i)What is the S.I unit of work? Ans:The S.I unit of work is Joule (J).One joule of work is said to be done when one Newton of force displaces a body through a displacement of one meter in the direction of force. Therefore 1 J = 1 Nm Q1(a)(ii)A body of mass 1 Kg falls from a height of 5 m. How much energy does it possess at any instant (take g = 9.8m/s2) ? Ans: E = mgh  = 1Kg   x 9.8m/s2  x 5m  = 49 J Q1(a)(iii)When a body moves in a circular path , how much work does it do? Ans:During circular motion the centripetal force is perpendicular to the displacement. Therefore W = F.S = F S cos90o = 0 Q1(b)The diagram given below shows the use of a lever. (i)State the principle of moments as applied to the above lever. (ii)Give an example of this class of lever. (iii)If FA=10 cm,AB=500 cm, calculate the minimum effort required to lift the load. Ans:(i)The vector sum of moments of all coplanar forces acting on a body in the equilibrium state about any point is zero.In the above case moment of effort and moment of load about F is zero. (ii)This is class III type of the lever, as the effort lies in between the fulcrum and load. ex. Fire Tongs (iii)Taking moments about F moment of Effort about F = moment of load about F E x 10 =  50 x 510 Therefore, E = 2550 N Q1(c)(i)State Archimedes' Principle. Ans:Archimedes' Principle states," When a solid body is partially or completely immersed in a liquid, it apparently loses its weight equal to the weight of the displaced liquid ". Q1(c)(ii)A man first swims in sea water and then in river water. 1. Compare the weight of sea water and river water displaced by him. 2. Where does he find it easier to swim and why? Ans:1. The weights of displaced water in both the cases is the same because in each case the weight of swimming man is the same so the buoyant force i.e the weight of the displaced liquid is the same. 2. It is easier for man to swim in sea water than in river water because volume of sea water displaced is less comparatively. Q1(d)(i) The graph below represents a cooling curve for a substance being cooled from a higher temperature to a lower temperature. 1. What is the boiling point of the substance? 2. What happens in the region DE? 3. Why is the region DE shorter than the region BC? Ans:1. Boiling point of the substance is 150oC. As BC portion of the graph represents the change of state and from B to A  is the vapour state.So at B constant temperature is the boiling point = 150oC 2. DE is the region which represents the solidification of the liquid  or fusion of the solid state . 3. BC represents the change of state from liquid to vapour or vapour to liquid so this is the region during which latent heat of vaporisation is required . During DE the change takes place either from solid to liquid or liquid to solid where latent heat of fusion is required but latent heat of fusion is less than the latent heat of vaporisation. Therefore DE is shorter than BC. Q1(d)(ii)At what approximate temperature will water boil in a pressure cooker? Ans:The approximate temperature inside the pressure cooker is 120oC to 125oC Q1(e)The diagram below shows a ray of white light AO incident on a rectangular glass block, which is silvered at one surface. This ray is partly refracted. (i) Copy the diagram and trace the path of the refracted and reflected ray. Show at least two rays emerging from the surface PQ. (ii)How many images are formed in the above case? Which image is the brightest? Ans:(i) (ii)Multiple images will be formed due to weak reflection at O and strong reflection at B and so on as shown in the figure.A1 is the first image due to weak reflection at O and A2 is the brightest image due to strong reflection at B Q1(f) The diagram below shows a students vision from his near point. (i) What defect is he suffering from? (ii) State one cause for the occurrence of this defect. (iii) Copy the diagram and complete to show how this defect is corrected. Ans:(i) He is suffering from long-sightedness i.e hypermetropia as he cannot see clearly the objects lying near the eye and only objects lying at longer distances can be seen. (ii)The cause of this defect is that the eye loses its power of accomodation. The focal length of the lens increases. It can be corrected using a convex lnes of suitable focal length. (iii) Q1(g)(i)The following diagram shows two straight wires carrying current.Copy the diagram and draw the pattern of lines of force around them and mark their directions. (ii)How is galvanometer converted to 1. an ammeter 2. a voltmeter Ans:(i) (ii)1. A galvanometer can be converted into an ammeter  by joining a suitable shunt S ( a small resistance) connected in parallel to the galvanometer of resistance G. The value of shunt S depends on the range of the ammeter and is given by S =     G/  ( I /Ig)-1                                                                                                                                                                Where, I is the range of the ammeter and Ig is the full scale deflection current for the galvanometer. 2. A galvanometer can be converted into a voltmeter by connecting a suitable high resistance X  in series with the galvanometer of resistance G and for which Ig is the full scale deflection current . If range of voltmeter is V than, V = Ig(G + X) Hence X = V/Ig - G Q1(h)(i) A refrigerator is marked 80W and 220V 1. How much energy does it consume in one day if on an average it is used for 20 hrs. a day? 2. What is likely to happen if the voltage drops to 50V? Ans: (i)1. Given P = 80W   , V= 220 V , t = 20 hrs.=72000 sec. Energy consumed per day =p x t = 80 x 72000 = 5760000 J. = 5760000/ 3600000 =  1.6 KWh.   2. If the voltage V` = 50 V and Resistance of the appliance = R = V2/P = 220 x 220/ 80  =  605 Ohm. Therefore power consumed in second case p`= v`2 /R  = 50 x 50/605   = 4.132 W Q1(h)(ii)Score off the incorrect words and write correctly the following: In wiring a three-pinplug , the brown sleeved wire is connected to the neutral/live pin and the green/blue sleeved wire is connected to the metal body of the appliance. Ans: In wiring a three pin plug, the brown sleeved wire is connected to live pin and green sleeved wire is connected to the metal body of the appliance. Q1(i)X-rays are produced in a coolidge tube when cathode rays are made to strike a target. (1).Why is the anode of the tube heated  up at the the time of emission of X-rays? (2). How is the strength and penetrating power controlled? (3) Which of the soft and hard X-rays has a longer wavelength and which has higher penetrating power? Ans:(1) In coolidge tube X-rays are emitted from the target (anode). when the cathode rays strike the target. The kinetic energy of fast moving electrons , is given to the anode(target in this case)and thus anode gets heated when X-rays are emitted. (2)  This strength (intensity) of X-rays is controlled by varying the filament (heater) current. Strength  of X-rays is increased on increasing the heater current and intensity is lowered by decreasing the current. The penetrating power (the quality) of X-rays is controlled by varying the potential difference between the anode and the filament. When the potential difference is increased, the penetrating power of the X-rays also increases and on decreasing the potential difference the penetrating power also   decreases. (3) The hard X-rays have high penetrating power and shorter wavelengths while the soft X-rays have low penetrating power but longer wavelengths. Q1(j)(i)State the transformation of energy taking place in a 1. Loud speaker 2. Washing machine Ans:1. In loud speakers when the sound is amplified, in microphone sound (mechanical)energy is converted into electrical energy and then after amplification electrical energy is converted into mechanical sound energy. 2. In Washing Machine electrical  energy is converted into mechanical energy. Q1(j)(ii)An a particle absorbs an electron. What does it change to? Ans:           He++   + - 1e0 -----He+ when a particle absorbs an electron, a singly ionised helium is formed. Q1(j)(iii)A fusion reaction is represented as follows 1H2 + 1H2----- 2He4 + x. Identify x. Ans: X is energy = 24MeV.\
 Q2(a)A bullet of mass 50 g is moving with a velocity of 500 m/s. It penetrates 10 cm into a still target and comes to rest. 1. Calculate the K.E possessed by the bullet. 2. The average retarding force offered by the target. Ans:1.Mass of the bullet = m = 50 g   = 50  /1000     Kg                                                          velocity of the bullet = v = 500 m/s Therefore, K.E of the bullet = 1 /2  mv2 = (1/2)  x (50 x 500 x 500)/1000   = 6250 Joules 2.Suppose the distance moved by the bullet in the target = S m and, the average force of resistance by the target           = F  N Therefore by law of conservation of energy       F.S      = 1/2 mv2 Hence,  the average retarding force                          F   = (1 /2.  mv2)/S    =  (6250 )/(10/100)   N F =  62500 N Q2(b)The diagram alongside shows a system of 5 pulleys. 1.Copy the diagram and complete it by drawing a string around the pulleys. Mark the position of load and effort. 2. If the load is raised by 1 m, through what distance will the effort move? Ans: 1 The complete diagram of a string around the pulleys is shown as follows  with the marked positions of effort and load. 2. Let the distance moved by the effort = D Distance moved by the load = d = 1 m (given) Let the load = W and the effort applied = E  By the principle of machines E. D = W . d Therefore, D/d = W/E But W/E = 5 Hence D = (W X d)/E =  5 x  1 m                                \D = 5 m Q2(c)A flat - bottomed test-tube floats upright  with some lead shots, on the surface of water 10 cm of its length immersed. The tube and the lead shots were found to weigh 20 g, when weighed in a beam balance.The same test-tube when floated in Kerosene was found to float with 12 cm immersed. 1. What mass of water does it displace when floated on water? 2. What is the area of cross-section of the tube? 3. calculate the R.D of Kerosene. Ans:1.  By law of floatation the weight of the body = weight of the displaced liquid But weight of the body (tube and lead shots) = 20 g Therefore, weight of the displaced water = 20 g Hence, mass of water displaced = 20 g 2. Let the area of cross-section of the flat bottom tube = A  cm2  Therefore , volume of displaced water = h x A  where , h = 10 cm   (given) But volume of 20 g water = 20 cc Therefore h x A = 20 10A = 20 Hence A = 2 cm2  3. Weight of water displaced by the body (tube + lead shots)   =  weight of Kerosene displaced by the body  (tube + lead shots) h x A  x d = h` x A x d` Therefore, h/h`  = d`/d                        Hence, d`/d  =  The related density of Kerosene  = h/h` = 10 /12 = 0.833 R.D  of Kerosene = 0.833
 Q 3(a)Explain why bottled soft drinks are more effectively cooled by cubes of ice than by iced water. Ans:Ice blocks absorb more heat from the bottle which is used to melt the ice blocks where as iced water does not require this heat from the bottle. So in contact with the ice blocks bottles lose more heat and so there is more cooling effect. Q 3(b) 10 g of ice at 0oC absorbed 5460 J of heat to melt and change to water at 50oC.Calculate the specific latent heat of fusion of ice. [Specific heat capacity of water is 4200 J/Kg/oC] Ans: In the conversion of 10 g ice at 0oC to 10 g of water at 50oC. There are two steps:- Firstly, Conversion of 10 g of ice at 0oC into 10 g of water at 0oC. Let heat required in this case = Q1  J Q1= 10 x L  J  where, L = Latent specific heat of fusion of ice in J/g Secondly, 10 g of water at 0oC is converted into 10 g of water at 50oC Let heat required in this case = Q2  J Q2 = 10 x 4200 x 50 / 1000 =  2100 J                                                            But  total heat required = Q J  =  5460 J (given) Therefore Q1 + Q2  = Q Hence, Q1= Q - Q2 10 L = 5460 - 2100 = 3360  J L = 3360 /10 = 336 J/g = 336000 J/Kg Q3(c) In an experiment to determine the specific latent heat of vaporisation of steam 'L' , the following measurement were taken : Mass of calorimeter + stirrer = x Kg Mass of water = y Kg Initial temperature of water = t1oC Final  temperature of mixture = t2oC (Given S.H.C. of calorimeter and water are S1 and S2 respectively) Express 'L' in terms of the above data. Ans: Suppose the mass of the steam condensed   =   m Kg The specific latent heat of vaporisation of steam = L J/Kg Therefore heat lost by the steam and hot water formed from steam Q1 = mL + mS2(100 -  t2)  J  ---------(1) Heat gained by calorimeter + stirrer and cold water  contained in calorimeter Q2 =  x S1  ( t2 - t1) + y  S2 ( t2 - t1) Q2 =  (x S1+ y  S2)( t2 - t1)---------(2) But by heat equation ,(Heat lost by steam  and hot water )  =  (Heat gained by cold calorimeter + stirrer and cold water) i.e , Q1 = Q2 From (1) and (2)     mL + mS2(100 -  t2) = (x S1+ y  S2)( t2 - t1) mL = (x S1+ y  S2)( t2 - t1) - mS2(100 -  t2) Hence, L = { (x S1+ y  S2)( t2 - t1) - mS2(100 -  t2)}/  m . Q4(a)(i) Differentiate between "forced vibration" and "resonance". (ii) How does frequency and amplitude affect a musical sound? Ans:Forced Vibration   The vibrations which take place  under the influence of external periodic force , are called forced vibrations. In this case the body vibrates with a frequency other than its natural (free) frequency, but vibrates with a frequency of the externally applied periodic force. Resonance   When the frequency of the applied force is equal to the natural frequency of the body, the body readily takes up the vibrations and begins to vibrate with an increased amplitude. This phenomenon is known as Resonance. (ii)The loudness of musical sound depends upon its amplitude as Loudness is directly proportional to the square of the amplitude of vibrating surface.The frequency affects the pitch of the musical sound . The pitch is directly proportional to the frequency. Q4(b)Two musical notes of the same pitch and loudness are played on a violin and a piano. Their waveforms are as shown in the figures .  Explain why the wave patterns are different. Ans:A note played on the piano has a large number of tones  while the note played on violin contains only one or few Secondary tones. So the quality of musical sound of the same loudness and of the same pitch produced by two different instruments are distinguished by their different waveforms. Q4(c)(i)An electromagnetic wave has a frequency of 500 MHz and a wavelength 60 cm. Calculate the velocity of the wave. Name the medium through which it is travelling. (ii) A wave has a wavelength of 0.01 �. Name the wave. Ans:(i)We know that the velocity of light of em wave in vacuum or free space = c = 3 x 108  m/s If the medium in which em wave is travelling is other than vacuum then Let  the velocity of the wave = v But v = nl where n is the frequency = 500 MHz = 5 x 108 Hz and the wavelength l = 60 cm  = 0.6 m Therefore v= 5 x 108 x 0.6 = 3 x 108  m/s. As the velocity of the electromagnetic wave in the medium is the same as that in the free space. Hence, the medium is vacuum or free space. (ii) We know that the electromagnetic waves having wavelength range = 10-10 m to 6 x 10-15 m  are gamma rays. The given wavelength is 0.01� = 0.01 x 10-10 m  = 10-12 m .Hence the given wave is a Gamma ray. Q5(a)What is meant by (i)Critical angle (ii)Total internal reflection? Ans:(i) The Critical angle for a  given pair of two media of light , is that particular angle of incidence  in the denser medium corresponding to which the angle of refraction in the rarer medium is of 90o. If the angle of incidence in the denser medium 'b' is equal to the critical angle C and the corresponding angle of refraction 'r' is equal to 90o in the rarer medium 'a' then the refractive index amb = 1/ sin C (ii) If the angle of incidence in the denser medium is greater than the critical angle for the given pair of media then  there no refraction takes place but total internal reflection takes place and the incident ray is reflected back in the same denser medium obeying the laws of reflection. This phenomenon is known as total internal reflection. Q5(b)(i)Draw a diagram to show that white light can be split up into different colors. (ii)Draw another diagram to show how the colors can be combined to give the effect of white light. (iii) How would you show the presence of ultra-violet and infra-red rays in the spectrum? Ans:(i) (ii) (iii) Infra red spectrum is detected by its heating property, using a thermopile or a blackened bulb thermometer. Ultra Violet spectrum is detected by property of fluorescence by using a thin tube   containing a solution of quinine - sulphate is found to fluorescence with bluish light when placed in the path of ultra-violet radiation. Q5(c) When yellow paint and blue paint are mixed we get green color. When yellow light and blue light are mixed white light is obtained. Give reasons for the observations mentioned. Ans:Paints and pigments reflect more than one color and when they are mixed, the resultant color is the common color out of the reflected colors . Blue reflects blue, green, violet  and yellow reflects green , yellow and orange. So green is common. Hence, when yellow paint is mixed with blue paint we get green color. But on the other hand blue and yellow are the complementary colors in case of spectral-colors.
 Q6(a)(i) State Ohm's law. (ii) State the factors that alter the resistance of a conductor. Ans:(i) when an electric current is passed through a conductor,the potential difference across the ends of the conductor is directly proportional to the current passing through it, provided the physical conditions temperature etc. remain unchanged. If P.D =v and electric current = I then, v a I (ii) The resistance of a conductor 'R' depends upon the following factors :- 1. R  a l  , the length of the conductor 2. R a 1/A where, A is the area of cross-section of the conductor Combining 1 and 2, we get R a l/A  . Changing this proportionality into an equation we get R= l /A where is the constant of proportionality and is known as the resistivity of the material of the conductor.Hence, the third factor which affects the resistance R of the conductor is the resistivity of the conductor. It is different for different materials. Q6(b)The following circuit diagram shows three resistors 2W, 4W and RW connected to a battery of emf 2V and internal resistance 3W. A main current of 0.25 ampere flows through the circuit. (i) What is the P.D across the 4W resistor? (ii) Calculate the P.D across the internal resistance of the cell. (iii) What is the P.D across the RW or the 2W resistor? (iv) Calculate the value of R. Ans:(i)P.D between Q and M = (0.25) 4 = 1.00 V (ii)P.D across the internal resistance of the cell = 0.25 x 3 = 0.75 V (iii)Let the P.D across P and Q = VPQ But the total emf applied = E = VPQ + 1 V + 0.75 V = 2 V (given) Therefore, VPQ = 2 - 1 - 0.75 = 0.25 V (iv)Let the resultant of RW and 2W be XW Therefore, 1/x = 1/R + 1/2  Hence, x = 2R/ (2 + R) Therefore Ix =VPQ (0.25 )2R/ (2 + R) = 0.25 V On simplifying we get R = 2 W Q6(c) State two ways of increasing the speed of rotation of a D.C motor. Ans:The speed of rotation of a D.C motor can be increased in the following two ways:- 1. By increasing the number of turns 2. By increasing the magnetic field Q7(a)(i)Draw a neat labeled diagram of a hot cathode ray tube. (ii) What is the effect on the beam particles if 1. a hotter filament is used 2. the anode voltage is increased? (iii) Name one device in which the cathode ray tube is used. Ans:(i) (ii)1. If the temperature of the filament is high, than the number of electrons emitted per second will be more. As electrons are the beam particles so the beam particles will be more when a hotter filament is used. 2. When the anode voltage is increased the speed of the beam particles (electrons) will be more and hence the K.E of the beam particles will be increased. (iii)Cathode ray tube is used in various devices for example, in Television Q7(b) State the factors that affect the rate of emission of electrons in a photoelectric cell. Ans:1.The rate of emission of photoelectrons in a photoelectric cell constitutes the photoelectric current which increases as the intensity of the incident light is increased i.e the photoelectric current is directly proportional to the intensity of incident light. 2. The rate of emission of photoelectrons in a photoelectric cell depends upon the area of the cathode of the  photoelectric cell. The larger the area the more the electrons are emitted per second. Q7(c)(i)A thorium isotope 90Th223 undergoes an a-decay and changes into Radium. What is the atomic number and mass number of the radium produced? (ii)If the radium undergoes a further disintegration and emits two b particles, represent this reaction in the form of an equation. (iii)What is the source of energy released during the decay? Ans:(i) We know that a = 2He4 90Th 223 = 2He4 + 88Ra 219 Therefore, Atomic number of Ra = 88 Atomic weight of Ra = 219 (ii)We know that  b particles = -1e0 88Ra 219 = 2  -1e0  + 90Th 219 (iii) There is always some mass that disappears in the process of decay and an equivalent amount of energy is released. The mass disappeared in the process is called mass defect. Energy released is given by the expression E = m.c2   where m is the mass defect and c is the velocity of light.