Score Booster 150: 10th Board Science Exam - 2025 | PDF

Score Booster 150 SCIENCE 10th Board Exam -2025 Page 1 of 30 Every question solved is a step closer to success!  Score Booster 150 SCIENCE 10th Board Exam -2025...

Score Booster 150 SCIENCE 10th Board Exam -2025 Page 1 of 30 Every question solved is a step closer to success!  Score Booster 150 SCIENCE 10th Board Exam -2025 Score Booster 150 (Must-Solve Questions for 10th Boards) PHYSICS 4. When you stand in front of a plane mirror, your image Multiple choice questions- is: (a) real, erect, and smaller than you 1. A resistance wire is stretched so as to double its length. (b) real, erect, and the same size as you Its new resistivity will have a magnitude (c) virtual, erect, and smaller than you (a) 2 times its original value (d) virtual, erect, and the same size as you (b) 4 times its original value Sol. (d) virtual, erect, and the same size as you (c) 8 times its original value (d) Same as its original 5. Assertion (A): Alloys are commonly used in electrical Sol. (d) [resistivity depends on the nature of material] heating devices like electric iron and Heater. Reason (R): Resistivity of an alloy is generally higher 2. Three V-I graphs are drawn individually for two than that of its constituent metals but the alloys have resistors and their series combination. Out of A, B, and low melting points than their constituent metals. C, which one represents the graph for a series (a) Both assertion (A) and reason (R) are true and combination of the other two? reason (R) is the correct explanation of assertion (A). (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (c) Assertion (A) is true but reason (R) is false. (d) Assertion (A) is false but reason (R) is true Sol. (c) Alloys are held for electrical heating devices due to their light resistivity and high melting point compared to constituent metals. (a) A (b) B (c) C (d) None of the above 6. A light ray enters from medium A to medium B as Sol. (c) [More slope of V-I graph means more resistance. shown in the figure. The refractive index of medium B The slope of C is maximum, so its resistance is relative to A will be : maximum. 3. The laws of reflection hold good for : (a) plane mirror only (b) concave mirror only (c) all mirrors irrespective of their shape (d) convex mirror only Sol. (c) Explanation-The laws of reflection hold good for all reflecting surfaces irrespective of their shapes whether (a) greater than unity (b) less than unity plane or curved. (c) equal to unity (d) zero Sol. (a) Greater than unity In this case, since ray of light bends towards the normal, when it goes from medium A Page 2 of 30 Every question solved is a step closer to success!  Score Booster 150 SCIENCE 10th Board Exam -2025 to medium B, the medium B is denser medium. Speed Power in first case, P1  I 2 R of light in rarer medium ( v1 ) is greater than the speed 100% increase in current means that current becomes 2I of light in denser medium ( v2 ) i.e., v1  v2 Power in second case, P2   2 I  R  4 I 2 R 2 v1 Now, increase in dissipated power  P2  P1 Thus, refractive index of B relative to A n  v2  4 I 2 R  I 2 R  3I 2 R  n 1 Percentage increase in dissipated power. 3P 7. Assertion (A): The sunrise occurs a few minutes earlier  1 100  300% P1 and the sunset occurs a few minutes later than the actual times due to atmospheric refraction. Reason (R): The atmosphere bends the light rays from 9. Assertion (A): Electric appliances with a metallic body the Sun, causing the Sun to be visible even before it have three connections, whereas an electric bulb has a actually reaches the horizon in the morning, and to two pin connection. remain visible a little after it has actually set in the Reason (R): Three pin connections reduce heating of evening. connecting wires. (a) Both A and R are true and R is the correct (a) Both assertion (A) and reason (R) are true and explanation of A. reason (R) is the correct explanation of assertion (A). (b) Both A and R are true but R is not the correct (b) Both assertion (A) and reason (R) are true but explanation of A. reason (R) is not the correct explanation of assertion (c) A is true but R is false. (A). (d) A is false but R is true (c) Assertion (A) is true but reason (R) is false. Sol. (a) Both A and R are true, and R is the correct (d) Assertion (A) is false but reason (R) is true. explanation of A Sol. (c) The metallic body of an electrical appliances are Assertion (A): This statement is true. Due to connected to the third pin which is connected to the atmospheric refraction, the light from the Sun is bent as earth. This is a safety precaution and avoids eventual it passes through the Earth's atmosphere. This bending electric shock. By doing this the extra charge flowing causes the Sun to appear above the horizon before it through the metallic body is passed to earth and avoids actually reaches it (early sunrise) and after it has shocks. There is nothing such as reducing the heating of actually set below the horizon (delayed sunset). connecting wires by three pin connections. Reason (R): This statement is also true. The refractive index of the Earth's atmosphere increases with altitude, 10. A constant current flowing in a horizontal wire in the causing the Sun's rays to bend slightly downwards as plane of the paper from East of West is shown in they pass through the different layers of the Figure. The direction of magnetic field at a point will atmosphere. As a result, when the Sun is below the be from North to South: horizon, its rays are still visible due to this bending, making the Sun appear earlier in the morning and stay visible longer in the evening. 8. If the current I through a resistor is increased by 100% (assume that temperature remains unchanged), the (a) directly above the wire. increase in power dissipated will be (b) directly below the wire. (a) 100% (b) 200% (c) at a point located in the plane of the paper, on the (c) 300% (d) 400% north side of the wire. Sol. (c) If I is current and R is resistance then, (d) at a point located in the plane of the paper, on the Power, P  I 2 R south side of the wire. Page 3 of 30 Every question solved is a step closer to success!  Score Booster 150 SCIENCE 10th Board Exam -2025 Sol. (b) Line WE shows a straight conductor through which Sol. If the refractive index of a glass lens is equal to the current is moving from E to W. When seen from east, refractive index of liquid then the glass lens placed in a the magnetic field lines appear in clockwise direction, transparent liquid will become invisible. i.e. S to N above the wire and N to S below the wire. This is in accordance with right hand thumb rule. 16. State one function of the iris and crystalline lens in the human eye. 11. The strength of magnetic field inside a long current Sol. Iris - It is a dark muscular diaphragm that controls the carrying straight solenoid is : size of the pupil. (a) more at the ends than at the centre crystalline lens - The crystalline lens of the human eye (b) minimum in the middle focuses the light that enters the eye and forms the (c) same at all points image on the retina. (d) found to increase from one end to the other Sol. (c) Magnetic field lines are straight and parallel inside 17. An alpha particle is placed in a magnetic field. Will it the solenoid. This indicates the same magnetic field. experience any force, if: Hence, inside the solenoid, the magnetic field is the (i) It moves in the magnetic field parallel to field lines. same throughout. (ii) It moves in the magnetic field perpendicular to field lines. Very short answer type questions- Sol. (i) No, because the force is zero if the current and field are in the same direction. 12. Activity based - A student determines the focal length (ii) Yes, because the force is maximum when current of a device X' by focussing the image of a distant object and magnetic field are perpendicular. on a screen as shown in the figure below then what type of mirror is X and give the correct explanation to Short answer type questions- support your answer. 18. Give reason for the following- (i) In solar furnaces concave mirrors are used (ii) Convex mirrors are used as rear view mirrors in vehicles. Sol. (i) Concave mirrors are used in solar furnaces because they have the ability to converge parallel rays of sunlight from the sun to a single focal point, creating a concentrated beam of heat at that point, which allows Sol. ‘X’ is a concave mirror because it focuses the image of for extremely high temperatures to be reached within a distant object at its focus and a real image will form the furnace. on the screen provided. (ii) A convex mirror is used as a rearview mirror in vehicles because it provides a wider field of view by 13. Should the resistance of a voltmeter be low or high? creating a diminished, virtual, and erect image of Give a reason. objects behind the car, allowing the driver to see a Sol. High. In parallel connection, less current passes larger area of traffic even if the objects are far away -sa through high resistance. 19. The current flowing through a resistor connected in a 14. A spherical mirror has a real focus. Identify the mirror. circuit and the potential difference developed across its Sol. A concave mirror has a real principal focus. ends are as shown in the diagram by milliammeter and voltmeter readings respectively. 15. Under what condition with a glass lens placed in a (a) What are the least counts of these meters? (b) What transparent liquid becomes invisible. is the resistance of the resistor ? Page 4 of 30 Every question solved is a step closer to success!  Score Booster 150 SCIENCE 10th Board Exam -2025 Potential drop across 10 Ω resistor is. V = IR = 0.2 A x 10 Ω =2V Sol. (a) least count of ammeter = 10 mA 23. “A ray of light incident on a rectangular glass slab Least coount of voltmeter = 0.1 V immersed in any medium emerges parallel to itself.” 2.4 Draw labelled ray diagram to justify the statement”. (b) = 9.6 ohm (250 mA = 0.25 A) 0.25 Sol. 20. Calculate the resistance of a 1 km long copper wire of area of cross section 2 × 10–2 cm2. The resistivity of copper is 1.623 × 10–8 ohm-meter. 1 Sol. R ρ A 1.623  104  1000  2  10 2  10 4 m 2  0.8110  81 21. How many 132 Ω resistors in parallel are required to carry 5 A on a 220 V line ? Sol. Given, V = 220 V, I = 5 A V = IR 24. The absolute refractive indices of glass and water are V 1.5 and 1.33 respectively. In which medium does light or R  I travel faster? Calculate the ratio of speeds of light in the In parallel combination, let the no. of resistors = x. two media. 132 220 Sol. Given : refractive index of glass, ng  1.5  x 5 Refractive index of water, nw  1.33 132 Since, refractive index of medium, Or,  44 x speed of light in air  c  n Or, x  132 speed of light in medium  v  44 c x  3 For glass ng  …..(i) vg The number of resistors = 3 c For water nw  …..(ii) 22. Three resistors of 10 Ω, 15 Ω and 20 Ω are connected vw in series in a circuit. If the potential drop across the 15 Since velocity of light in medium is inversely Ω resistor is 3 V, find the current in the circuit and proportional to its refractive index, the light will travel potential drop across the 10 Ω resistor. faster in optically rarer medium i.e., water. Sol. In series circuit same current flows through all the Dividing (i) by (ii), resistors. Current through 15 Ω resistor. ng vw vg nw V 3V 1  or  1   = 0.2 A nw vg vw ng R 15 5 Current in the circuit = 0.2 A Page 5 of 30 Every question solved is a step closer to success!  Score Booster 150 SCIENCE 10th Board Exam -2025 vg 1.33  vw 1.5 So, the ratio of vg and vw is 1.33 : 1.5. 25. Describe and illustrate with a diagram how we should arrange two converging lenses so that a parallel beam 28. A student is unable to see clearly the words written on of light entering one lens emerges as a parallel beam the black board placed at a distance of approximately 3 after passing through the second lens. m from him. Name the defect of vision the boy is Sol. Suppose we have two converging lens of focal lengths suffering from. State the possible causes of this defect f1 and f 2. We will keep the two converging lens at a and explain the method of correcting it. distance of f1  f 2 so that a parallel beam of light Sol. Student is suffering from myopia. entering one lens emerges as a parallel beam after The two possible reasons due to which the defect of passing through the second lens. vision arises are : excessive curvature of the eye lens and elongation of the eye ball. A student with myopia has the far point nearer than infinity, thus, the image of a distant object is formed in front of the retina. Here the focus of the two lenses should coincide. 26. Define the term power of accommodation. Write the modification in the curvature of the eye lens which Correction of myopia: This defect can be corrected by enables us to see the nearby objects clearly? using a concave lens of suitable power as it brings the Sol. The ability of the eye lens to adjust its focal length is image back on to the retina, thus the defect is corrected. called power of accommodation. The ciliary muscles modifies the curvature to some extent. The change in the curvature of the eye lens can thus change its focal length. When the ciliary muscles contract, the lens becomes thick and its focal length decreases, thus enables us to see nearby objects clearly. 29. What is ‘dispersion of white light’? State its cause. 27. (a) List two causes of hypermetropia. Draw a ray diagram to show the dispersion of white (b) Draw ray diagrams showing (i) a hypermetropic eye light by a glass prism. and (ii) its correction using a suitable optical device. Sol. Splitting of white light into its seven constituent colours Sol. (a) Hypermetropia is caused due to following reasons: due to refraction is known as dispersion of white light. (i) Shortening of the eyeball Cause of dispersion : When a beam of white light enters (ii) The focal length of crystalline lenses is too long. a prism, it gets refracted and splits into seven constituent colours. The splitting of the light ray occurs due to the different bending angle for each colour. Thus, each colour ray when passing through the prism bends at different angles with respect to the incident beam, thus giving rise to a spectrum. Page 6 of 30 Every question solved is a step closer to success!  Score Booster 150 SCIENCE 10th Board Exam -2025 Two 9 ohm resistors in parallel connected to one 9 ohm resistor in series. 1 1 1 2    RP 9 9 9 30. How will you use two identical prisms so that a narrow beam of white light incident on one prism emerges out 9  RP   of the second prism as white light? 2 Sol. We can place one prism straight and the other identical 9 R  9    13.5 prism upside down, so that the light that emerges from 2 the second prism is white light. The light that emerges (ii) from the first prism is incident on the second prism in such a way that the one emerging from the second one gives a white light on the screen that is kept perpendicular to the emerging ray. Two 9 ohm resistors in series connected to one 9 ohm resistor is parallel. Rs  9  9  18 1 1 1 3    R 18 9 18  R  6 31. Find the power of a convex lens which forms a real and inverted image of magnification -1 of an object placed 33. State any two factors on which the magnetic field at a distance of 20 cm from its optical centre. produced by a current carrying straight conductor Sol. Magnification of the lens is given by depends. Mention the rule which helps to find the v v direction of its magnetic field m   1  [u = –20 cm] u 20 Sol. Factors on which the magnetic field produced by a ∴ v = 20 cm current carrying conductor depends: As v = u then (i) Strength of current passing through the conductor. 20 (ii) Distance of the point of measurement from the f  cm = 10 cm = 0.1 m 2 conductor. 1 1 Right Hand Thumb Rule gives the direction of Power of the lens, P  D D  10 D magnetic field. f  inm  0.1 34. What is solenoid? Draw the field lines of the magnetic 32. Show how would you join three resistors, each of field produced on passing current through and around a resistance 9 so that the equivalent resistance of the current carrying solenoid. combination is (i) 13.5 , (ii) 6 ? Sol. Definition: A coil of many circular turns of insulated Sol. (i) copper wire wrapped closely in the shape of a cylinder is called solenoid. Page 7 of 30 Every question solved is a step closer to success!  Score Booster 150 SCIENCE 10th Board Exam -2025 So, E = 3 x 100 W x 10 h x 30 = 90,000 Wh = 90 kWh. (c) Total cost of electricity = Total unit of energy consumed x Cost per unit = 90 kWh x 6.50 = Rs. 585. 37. A small candle, 2.5 cm in size is placed at 27cm in front of a concave mirror of radius of curvature 36cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved? Magnetic field lines through and around a current Sol. Given: R = 36 cm carrying solenoid. 36cm So, f   18cm 2 35. A household uses the following electrical appliances: (i) Refrigerator of rating 400 W for ten hours each day. 1 1 1   (ii) Two electric fans of rating 80W each for twelve f v u hours each day. Given: (iii) Six electric tubes of rating 18 W each for 6 hours –Object distance (u) = –27 cm (negative because the each day. object is in front of the mirror) Predict the amount of electricity bill of the household Substituting the values into the mirror formula: for the month of June if the cost is Rs. 3.00/unit. 1 1 1 Sol. June month is of 30 days.   18 v 27 We know, E = P x t 1 3 2 1 Electricity consumption in one day    v 54 54 54 = 400 x 10 + 2 x 80 x 12 + 6 x 18 x 6 Thus, v  54cm = 4000 + 1920 + 648 = 6568 Wh = 6568/1000 kWh = 6.568 kWh = 6.568 units. 1 v 1 54 Now, magnification m =  or   2 Electricity consumption in one month O u 2.5 27 = 6.568 x 30 = 197.04 units I = –2 x 2.5 = –5cm (inverted image). ∴ Total cost = Rs. 197.04 x 3 Thus, the image is real, inverted and enlarged. = Rs. 591.12 If the candle is moved closer to the mirror: -The image will move further away from the mirror Long answer type questions - (beyond 54 cm). -Therefore, the screen must be moved backward to 36. (a) An electric bulb is rated at 200 V-100 W. What is its maintain a sharp image. resistance? (b) Calculate the energy consumed by 3 such bulbs if 38. Study the circuit shown in which three identical bulbs they glow continuously for 10 hours for the complete B1, B2 and B3 are connected in parallel with a battery month of November. of 4.5 V. Answer any four questions (a) to (e). (c) Calculate the total cost if the rate is Rs 6.50 per unit. Sol. (a) Given, V = 200 volts and P = 100 watt V 2  200  2 V2 40000 As P  or R     400W R R 100W 100 (b) Electrical energy consumed, E = number of units x Power of each unit x time x total days. Here, n = 3, P = 100 W, t = 10 hours, Days = 30 Page 8 of 30 Every question solved is a step closer to success!  Score Booster 150 SCIENCE 10th Board Exam -2025 (a) What will happen to the other two bulbs if the bulb 88 RP  B3 gets fused ? n (i) They will also stop glowing V  IR (ii) Other bulbs will glow with same brightness 88 (iii) They will glow with low brightness R n (iv) They glow with more brightness 88 220  n 10 (b) If the wattage of each bulb is 1.5 W, how much n  4 resistors readings will the ammeter A show when all the three bulbs glow simultaneously. (e) (iv) There is a path of flow of electron change (i) 1.0A (ii) 2A through each resistor. So there are several paths. (iii) 1.5W (iv) None of the above 39. (a) State and explain the laws of refraction of light with (c) Find the total resistance of the circuit: the help of a labelled diagram. (i) 1.0Ω (ii) 4 (b) What is meant by the refractive index of a (iii) 1.5Ω (iv) 2 substance? Sol. (a) Laws of refraction (d) How many resistors of 88 Ωare connected in parallel to carry 10 A current on a 220 V line: (i) 2 resistors (ii) 1 resistors (iii) 3 resistors (iv) 4 resistors (e) Find the statement which does not justify parallel connection: (i) The voltage of each component is same (ii) Overall resistance is lower than the resistance of single component (iii) Automobile headlight is connected in parallel (iv) There is a single path for the flow of electrons/charge First law: According to the first law of refraction, the Sol. (a) (ii) Explanation: Other bulbs will glow with same incident ray, the refracted ray and the normal at the brightness point of incidence all lie in the same plane. (b) (i) 1.0 A Explanation: When the bulbs are in Second law: According to the second law of refraction, parallel, wattage will be added (4.5 W) and the the ratio of the sine of angle of incidence to the sine of ammeter reading would be, refraction is constant for a given pair of media. (c) (ii) 4.5 Ammeter reading = 1.0A (b) The refractive index is the measure of bending of a V  4.5V light ray when passing from one medium to another. It can also be defined as the ratio of the velocity of a V R  4.5 light ray in an empty space to the velocity of light in a I substance, n = c/v  4.5 (d) (iv) 4 resistors 40. It is desired to obtain an erect image of an object, using 1 n a concave mirror of focal length of 12 cm. Equivalent resistance,   RP 88 (i) What should be the range of distance of object when object placed in front of the mirror? Page 9 of 30 Every question solved is a step closer to success!  Score Booster 150 SCIENCE 10th Board Exam -2025 (ii) Will the image be smaller or larger than the object? Sol. (a) (i) Power is the degree of convergence or Draw a ray diagram to show the formation of the image divergence of light rays achieved by a lens. in this case. It is defined as the reciprocal of its focal length. (iii) Where will the image of this object be, if it is 1 i.e., P = placed 24 cm in front of the mirror? Draw a ray f diagram for this situation also to justify your answer. Show the positions of pole, principal focus and the (ii) Principal focus : Incident rays parallel to principal centre of curvature in the above ray diagrams. axis, after reflection either converge to as appear to Sol. Focal length of the concave mirror f = 12 cm diverge from a fixed point on the principal axis known (i) If the object is placed between the pole and focus of as principal focus of the spherical mirror. the concave mirror, then the image formed is virtual (b) (i) for a spherical lens, according to lens formula, and erect. Therefore, the range of distance of the object 1 1 1 should be 0 < u