Solution. Class 10 - Science. Revision Test. Section A

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Christine Holt
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1 Solution Class 10 - Science Revision Test Section A 1. (a) since resistances are in parallel R (b) Total current I 4.8 A (c) If I 1, I 2 and I 3 be the current through 2 respectively. Therefore, I 1 3 A I A I A 2. In n resistors each of resistance R are connected in parallel, then net resistance R p is: times Now 44 but R 176 Ω 44 or n 4 4 resistors each 176 Ω connected in parallel with result in net resistance of 44 Ω causing a current of 5A to flow through when connected to 220V. 3. Here, R 1 100, R 2 50, Rs500 Let the equivalent resistance R Resistors are connected in parallel. Mob ALL THE BEST-- 1 / 14

2 R Current through all the three appliances, I 7.04 A Since the electric iron connected to the same source that takes as much current as all the three appliances. So, Resistance of the electric iron Current through the electric iron 7.04 A 4. Here I 2.5 ma A; V 12 volt; R? V IR or R R 5. (i) Since (length of the conductor) Length of wire C is more than A and B. Therefore wire C has higher resistance. (ii) Resistivity of all wires is same as material of all the wires is same.electrical Resistivity of a wire depends on the nature of the material and not on the dimensions of a wire. 6. All of them are made of copper,the resisvity remains same. Let the resistivity be (i) (ii) (iii) Conductor(b) has the highest resistance. It has the largest length and the smallest area of cross-section compared to other two. The following conclusion is also proved mathematically. Mob ALL THE BEST-- 2 / 14

3 7. Resistance of heater coil, R As the coil is cut into two identical parts, resistance of each part When connected in parallel, the net resistance is given by Energy liberated 400J Energy liberated per second in the new combination is 400J 8. (i) As the total resistance (equivalent resistance) is 4, the 6 resistor cannot be in series as in series combination the equivalent resistance is greater than the largest individual resistance. So, it must be in parallel with the other resistors. In parallel connection, the equivalent resistance (4 ) has to be less than all the resistances. So, the resistors of 2 and 3 cannot be in parallel at one time with 6. So, the resistors have to be in a mixed combination. Let us consider the combination shown in the figure. The equivalent resistance between B and C (which are in parallel). The resistance between A and D So, the combination shown in the figure is true. Mob ALL THE BEST-- 3 / 14

4 (ii) Here, R 1 2, R 2 3, R 3 6, and R 1 Since the equivalent resistance of the combination is of lesser value than any of the resistors of the combination, it is clear that the resistors should be connected in parallel. It can be further confirmed by using the formula 1 i.e, R1 Therefore, resistors should be connected in parallel. 9. Resistance R of a conductor depends upon length and area A of conductor. It has been experimentally confirmed that: R l, the lenth of the wire R, where A is the area cross-section of the wire Combining above, we have: R or R where is a constant called specific resistance or electric resistivity of the material of the conductor. If l 1, A 1, R. i.e. specific resistance (or electric resistivity) of a conductor is the resistance of a wire of unit length and a unit area cross-section. It may be defined as: Specific resistance of a conductor is the resistance of unit cube of the conductor. Unit of : 10. It is given that potential difference (V) 220 V. Resistance of coil A (R A ) Resistance of coil B (R B ) 24 (i)when either coil A or B is used separately, the current (I) V/R 220 V/ A. Mob ALL THE BEST-- 4 / 14

5 (ii)when two coils are used in series, total resistance, R A + R B Current flowing (I) V/ R S 220 V/ A (iii)when the two coils are joined in parallel,total resistance (R p ) 1/24 + 1/242/24 R p 12. Current (I) V/R p 220V / A 11. Modified circuit is as shown. Since 5Ω, 8 Ω and 12 Ω are in series, therefore the total resistance in series. R s R 1 + R 2 + R Ω Current through circuit I Reading of ammeter 0.24A I. R Ω 12. The resistance of a heater coil is less than that of electric bulb filament. When the heater is switched on in parallel, more current start flowing through the heater coil and current through the bulb filament decreases, making it dim. After some time, when the heater coil becomes hot its resistance increases. As a result, current through the heater coil decreases and hence,the current through the bulb filament increases and thus dimness of the bulb decreases. 13. (i) V 6V R 1 1 R 2 2 Mob ALL THE BEST-- 5 / 14

6 Total resistance, R R 1 + R I 2A Therefore, power (P 1 ) I 2 R 8W (ii) V 4V, R 1 12, R 2 2 Potential difference is constant in a parallel circuit. Hence power dissipated across the 2 resistor is directly given as: 8W Ratio 1:1 As an interesting observation might be that power dissipated in parallel circuit does not depend on the resistor connected in parallel,while in a series circuit,as the current flow is determined by all the resistors,power dissipation is dependent on the other resistors in the circuit. 14. Death may be caused if a heavy current flows through the body. Current flows not due to potential of the body but due to potential difference. When a person standing on earth touches a 220 V line, the potential difference of two points is 220 V [Since the potential of earth on which he stands is at 0 V potential] A bird however is at 11,000 V potential but no current flows through it. 15. I 30 ma A, R 50 Ω; V? V IR , or V 1.5 volt Section B 16. Focal length, f -10cm Image distance,v +20cm Object distance Vcm Mob ALL THE BEST-- 6 / 14

7 V -20cm m +1 As m+1 therefore image is of same size as the object and the image formed is real. Object at 2F 1 image formed at 2F 2 Same size & Real & inverted 17. Here u - 10 cm (u is always negative), v? f and r for convex mirror are always positive. Using the formula we have or v +6 cm or Image is 6 cm behind the mirror (right or mirror). It is virtual and errect. Mob ALL THE BEST-- 7 / 14

8 Magnification Hence image is 0.6 times the size of the object. Again where h 2 is the size of the image and h 1 the size of the object. h 2 mh 1 0.6(5) 3 cm. Hence image is 3 cm. high. 18. There are mainly two laws of refraction of light (i) The incident ray, the normal and the refracted ray, all lie in the same plane. (ii) If the ray goes from an optically rarer medium to an optically denser medium, it bends towards normal. When the light goes from air to glass, it bends towards the normal. However if the ray goes from an optically denser medium (glass) to an optically rarer medium (air), it bends away from the normal. (iii) Ratio of sine of angle of incidence to sine of angle of refraction is constant. i.e. constant, where is the refractive index of second medium with respect to first medium. 19. Radius of curvature, R m; Object distance, u m; Image distance, v? Height of the image, h'? Focal length, f 1.50 m Since, or v m (approximately) The image is 1.15 m at the back of the mirror. Magnification m Mob ALL THE BEST-- 8 / 14

9 +0.23 Image is virtual, effect and smaller in size by a factor of Since the lens is concave, hence f is negative Given: u - 30 cm; f - 10 cm; h 2.5 cm; v?; h'? The lens formula for concave lens is v -7.5 cm The negative sign indicates the virtual nature of the image. The image is at a distance of 7.5 cm from lens (in front of lens). The magnification m The positive sign with the magnification indicates that the image formed erect. The size of the image is determined by h'. m h' m Thus the image formed is virtual and erect. It is at a distance of 7.5 cm from lens and its size is cm. 21. Here 2.5, 1.5, V g ms -1 ; V D? Mob ALL THE BEST-- 9 / 14

10 , But Therefore V D 22. f +15 cm, u -10 cm. 1/f 1/v +1/u 1/v 1/15 +1/10 1/v 5/30 v + 6 cm. Magnification -v/u 0.6. The image is formed 6 cm behind the mirror, it is a virtual and erect image and diminished. 23. Image is located behind the mirror. It is highly magnified and virtual and erect. 24. h 1 7 cm, u - 27 cm, v?, f - 18 cm. (concave mirror) Now Negative sign of h 2 indicates that image is on the same side as that of the object. It is real, inverted and 14 cm in size. 25. (a) for Mob ALL THE BEST / 14

11 putting the values of u and f, we get the value of v - a. As v is -ve image is also formed on the same side as of the object. Position of the object Between O & F (b). Applying, v3a. Position of the object Between F and 2F Section C 26. Earth is surrounded by envelope of gases called atmosphere. At the time of sun-rise (or at sun-set), light has to travel greater thickness [AB at sun-rise and BC at sun-set] than what it covers, when sun is over-head at noon [it has to travel only DB]. The wavelength of blue colour is about half that of red, blue light is scattered nearly 16 times red. As a result, the sun appears red at sun rise or at sun-set due to negligible scattering of red in comparison with blue. Mob ALL THE BEST / 14

12 27. There is no change to the image distance in the eye when we increase the distance of an object from the eye. To see closer or distant object clearly, the eyes, due to its ability of accommodation, can increase or decrease focal length of its lens, so that the image is always formed at retina. The eye lens is composed of a fibrous, jelly-like material. To see closer or distant objects, focal length of its lenses can be increased or decreased to some extent by the contraction and relaxation of the ciliary muscles. 28. In normal eye, the far point is infinity. The lens used should be such that an object at infinity forms virtual image at 1.2 m. Now, v -1.2 m, u, f? We know or f -1.2 m Power, P D A concave (or divergent) lens of focal length 1.2 m and power D should be used to restore proper vision. 29. Here u - ; v - 40 cm., f? Using we have Mob ALL THE BEST / 14

13 or f - 40 cm m or P -2.5 D He should use concave or divergent lens of focal length -0.4 m and power -2.5 D. 30. (i) For distant vision f cm (ii) For near vision f 66.7 cm 31. The pupil of the eye acts as a variable aperture whose size can be varied with the help of the iris. When the light is very bright, the iris contracts the pupil to allow less light to enter the eye. However in dim light the iris expands the pupil to allow more light to enter the eye. When we enter from a bright light area to a dim light area, the pupil is not able to expand quickly to allow more light into the eye. We are therefore not able to see the objects in dim light room momentarily. 32. It is that defect in the eye due to which a person is unable to distinguish between certain colours due to insufficient or no cone shaped cell on retina. We know there are three types of cone cells - one responding to blue, second to red and third to green colour. The colour blind persons do not possess some cone cells that respond to certain colours. Colour blindness is a genetic disorder caused due to inheritance. There is no cure to this defect. A colour blind person can seen everything clearly but is not able to distinguish between different colours. The founder of atomic theory of matter, Dalton was colour blind. 33. The ability of the human eye to continue to see the image of an object for a very short duration even after the removal of that object is called persistence of vision. It is due to persistence of vision that we are able to see movie pictures in a cinema hall. The pictures in the form of a long film are projected on the screen at a rate of about 24 pictures per second. Under these conditions, the image of one picture Mob ALL THE BEST / 14

14 persists on the retina of the eye till the image of the next picture falls on the screen, and so on. Due to this, the slightly different images of the successive pictures present on the film merge smoothly with one another and give us the feeling of continuity and moving images. 34. The defect of the vision here is hypermetropia (Longsightedness) Given that, u -25 cm and v -75 cm Using the formula, Therefore, - + Or f cm Power of the lens D The + sign with f or the power indicates a convergent lens. 35. Presbyopia is due to decrease in elasticity of eye-lens. With age, the elasticity of eye lens decreases. The near point goes away from the eye and the far point comes near the eye. Due to this, a person suffering from this is not able to see near as well as far away objects clearly. It can be corrected by using bifocal lenses or using tow spects, one for reading purpose and one for seeing distant objects. Mob ALL THE BEST / 14

The Hyman Eye and the Colourful World

The Hyman Eye and the Colourful World In this chapter we will study Human eye that uses the light and enable us to see the objects. We will also use the idea of refraction of light in some optical phenomena