CUET Physics

Optics   87 96. Yellow light is refracted through a prism (a) dispersion (b) reflection producing minimum deviation. If i1 and i2 denote the angle of incidence and emergence, (c) refraction (d) diffraction then 101. What is the relation between the refractive (a) i1 = i2 (b) i1 > i2 index (n) and the wavelength of light (l)? (c) i1 < i2 (d) i1 + i2 = 90° 97. A ray falls on a prism ABC (AB = BC) and (a) n ∝ l (b) n ∝ 1 λ travels as shown in figure. The minimum refractive index of the prism material should (c) n ∝ l2 (d) n ∝ 1 be λ2 A 102. The refractive index n and the wavelength (l) of radiations are related as 90° (a) n = a+ b λ2 (b) n = al + b 90° (c) n = a+ b + c B λ2 λ4 C 4 (d) n = a+ b + c 3 λ λ4 (a) (b) 2 (c) 1.5 (d) 3 103. The relation 1 = (µ − 1)  1 − 1  is called f  R1 R2    98. An astronaut can see the outer space as (a) mirror formula (a) white (b) black (b) lens formula (c) blue (d) red (c) lens maker’s formula 99. Two convex lenses L1 and L2 are coaxially (d) Cauchy’s formula placed with respect to each other as shown 104. The length of a compound microscope is in figure. F1 and F2 are the positions of their focal points. An object OP is placed in front of (a) f0 (b) fe (c) f0 – fe (d) f0 + fe lens L1 and forms an image AB close to F2 and 105. The magnifying power of an astronomical between F2C2. The final image as seen by the eye will be telescope at normal adjustment is P L1 L2 (a) f0fe (b) f0 O F1 C2 fe A C1 F2 (c) fe (d) fe fe f0 f0 + fe B (a) inverted and will form beyond the object 106. A convex glass lens (mg = 1.5) has a focal at a distance greater than C1 C2 length of 8 cm when placed in air. What is the (b) inverted and will form between C1 and focal length of the lens when it is immersed in C2 water (mw = 4/3)? (c) inverted and will form at infinity (a) 2 m (b) 4 cm (d) none of these is correct (c) 16 cm (d) 32 cm 100. The sun may appear to be elliptical just before setting. This happens due to 107. Huygen’s wave theory fails to explain (a) reflection of light (b) refraction of light

88 PHYSICS (c) polarization of light 115. The phenomenon of diffraction can be treated as interference phenomenon if the number of (d) all the above coherent resources is 108. In Young’s double slit experiment, for which (a) one (b) two colour, is the fringe width least? (a) red (b) green (c) three (d) infinity (c) yellow (d) blue 116. The limit of resolution of eye is approximately 109. The magnifying power of a simple microscope (a) 1 mm (b) 1 cm is given by (symbols have their usual meanings). (c) 10 (d) 1′ (a) M = 1+ D (b) M = 1− D 117. Light waves can be polarized because they fe fe (a) have high frequencies (c) M = 1+ fe (d) M = 1− fe (b) have short wavelength D D (c) are transverse 110. The magnifying power of a compound (d) can be reflected microscope is given by (symbols have their usual meanings). 118. The frequency of a light wave in a material is 4 × 1014 Hz and wavelength is 5000 Å. The L  D  L  D  refractive index of material will be fe  f0  f0  fe  (a) M = 1 +  (b) M = 1 +  (a) 1.4 (b) 1.33   (c) 1.0 (d) 1.5 (c) M = L 1 + fe  L  1 + D  119. Which of the following colours is scattered D  f0  (d) M = f0  fe  minimum?    111. The refractive angle of prism is A and (a) violet (b) red refractive index of the prism is cot A/2. Then the angle of minimum deviation is (c) blue (d) yellow (a) 180° – 3A (b) 180° – 2A 120. A liquid is placed in a hollow prism of angle 60°. If the angle of minimum deviation is 30°, (c) 90° – A (d) 180° + 2A what is the refractive index of the liquid? 112. A plano-concave lens is made of glass of (a) 1.41 (b) 1.50 refractive index 1.5 and the radius of curvature of the curved face is 50 cm. The power of the (c) 1.65 (d) 1.95 lens is 121. What is the minimum distance between an (a) – 1.0 D (b) – 0.2 D object and its real image formed by a convex lens? (c) + 1.0 D (d) + 0.2 D (a) 0 (b) f 113. A person can see clearly only up to a distance of 30 cm. He wants to read a book placed (c) 2f (d) 4f at a distance of 50 cm from his eyes. What is the power of the lens he requires for his 122. What is the minimum distance between an spectacles? object and its virtual image in a concave lens? (a) – 2.0 D (b) – 1.33 D (a) 0 (b) f (c) – 1.67 D (d) + 2.0 D (c) 2f (d) 4f 114. Which of the following undergoes maximum 123. A plano-convex lens is made of a glass of diffraction? refractive index 1.5. The local length f of the lens and radius of curvature R of its curved (a) a-rays (b) g-rays face are related as (c) radio waves (d) light waves (a) f = R/2 (b) = R/3 (c) f = 2R (d) f = 3R/2

Optics   89 124. A thin convergent glass lens (mg = 1.5) has a 131. A convex lens of focal length f1 is put in contact power of + 5.0 D. When this lens is immersed with a concave lens of focal length f2. The combination will be have like a converging in a liquid of refractive index m1 it acts as a lens if divergent lens of focal length 100 cm. The value of m1 must be (a) f1 > f2 (b) f1 = f2 (a) 4/3 (b) 5/3 1 > 1 1 f1 f2 f2 (c) 4/5 (d) 3/5 (c) (d) f1 = 125. The dispersion of light in a medium implies 132. A convex lens of focal length f is put in contact that with a concave lens of the same focal length. The equivalent focal length of the combination (a) lights of different wavelengths travel is with different speeds in the medium (b) lights of different frequencies travel with (a) 0 (b) f different speeds in the medium (c) 2f (d) ∞ (c) the refractive index of the medium is 133. Dispersive power is zero for a different for different wavelengths of light (a) slab (b) lens (d) all of these (c) prism (d) none of the above 126. What should be the minimum distance 134. A beam of light consisting of red, green and between a convex lens and an object to blue colours is incident on a right angled produce a real image? prism as shown in figure. (a) 0 (b) f The refractive indices of the prism for red, green and blue colours respectively ae 1.39, (c) 2f (d) 4f 1.44, 1.47. The prism will 127. In which case is the image formed by a concave lens real? (a) 0 < u < f (b) f < u < 2f (c) 2f < u < ∝ (d) none of the above 45° 128. In which case is image formed by a convex (a) separate a part of the red colour from the lens virtual? green and blue colours (a) 0 < u < f (b) f < u < 2f (b) separate a part of the blue colour from the red and green colours (c) 2f < u < a (d) none of the above (c) separate all the three colours from one 129. A convex lens when placed in a certain another medium does not act as a lens. The refractive index of lens is n1 and that of the medium is (d) not separate any colour at all n2. Then 135. The correct curve between energy of photon (a) n1 > n2 (b) n1 = n2 (E) and its wavelength (l) is (c) n1 < n2 (d) none of these 130. A lens is made from a material of absolute E E refractive index n1 and is placed in a medium (a) (b) of absolute refractive index n2. The focal length f of the lens is related to n1 and n2 as l l E (a) f ∝ (n1 – n2) (b) f ∝ (n1 1 E − n2 ) (d) (c) (c) f ∝ (n1 + n2) (d) f ∝ (n1 1 ll + n2 )

90 PHYSICS 136. Fraunhoffer spectrum is a ________ spectrum. 143. The angle of prism is A if the angle of minimum deviation is (180° – 2A), then the (a) line emission (b) line absorption refractive index of the material of the prism is (c) band emission (d) band absorption (a) cos A (b) sin A 2 2 137. Which of the following is a correct representation of deviation and dispersion of (c) tan A (d) cot A light by a prism? 2 2 (a) R (b) R 144. A ray of light is incident normally on one of (c) V the faces of a prism of apex angle 30° and V m = 2 . The angle of deviation of the ray is R about R V (a) 12° (b) 30° (d) (c) 45° (d) 60° V 145. When the length of a microscope tube 138. Figure below shows a ray passing through a increases, its magnifying power prism. The case corresponding to minimum deviation is (a) (b) (a) increases (b) decreases (c) remains same (c) (d) (d) nothing can be decided 139. The colour of a star is an indication of its 146. The distance travelled by light in glass (m = 1.5) in a nanosecond will be (a) distance (b) weight (a) 20 cm (b) 30 cm (c) temperature (d) size (c) 40 cm (d) 50 cm 140. A ray of light travels in optical fibre due to 147. The splitting of a white light into several colours on passing through a glass prism is (a) refraction due to (b) total internal reflection (c) reflection (a) refraction (b) reflection (d) polarization (c) interference (d) diffraction 141. A convex lens has a focal length f. It is cut into 148. The resolving power of a telescope depends two parts along a line perpendicular to the on principal axis. Then the focal length of each (a) focal length of eye piece part is (b) focal length of objective lens (a) f (b) 2f (c) length of the telescope f (d) diameter of the objective lens 2 (c) (d) none of these 149. A light ray is incident normally on a plane mirror. The angle of reflection will be 142. The phenomenon of ‘mirage’ formation is (a) 0° (b) 30° due to (c) 45° (d) 90° (a) reflection of light (b) refraction of light 150. A given ray of light suffers minimum (c) diffraction of light deviation in an equilateral prism A. (d) total internal reflection of light Additional prisms B and C of identical shape and of the same material as A are now added as shown in figure. The ray will now suffer

Optics   91 B C 157. A converging lens is that which A (a) collects rays (a) no deviation (b) spreads rays (b) greater deviation (c) forms real image (c) same deviation as before (d) forms virtual image (d) total internal reflection 158. A convex lens of focal length f produces a real 151. A ray of light travels through four transparent image n times the size of the object, then the media with refractive indices m1, m2, m3, and distance of the object from the lens is m4, are shown in figure. The surfaces of all media are parallel. If the emergent ray ST is (a) nf (b)  n  f parallel to the incident ray PQ, we have  +   n 1  (c)  n + 1  f (d) (n + 1)f  n  m3 S m4 T 159. The reciprocal of wavelength is called m1 m2 (a) wave number (b) wave velocity S (c) phase difference (d) none of these Q R 160. How will the interference pattern in Young’s PP double slit experiment be affected if yellow light is replaced by blue light? (a) m1 = m2 (b) m2 = m3 (a) narrower (b) wider (c) m3 = m4 (d) m4 = m1 (c) fainter (d) brighter 152. A ray of light is incident at an angle of 60° on 161. Two waves originating from sources S1 and one face of a prism of angle 30°. The emergent S2 having zero phase difference and common ray of light makes an angle of 30° with the wavelength l will show complete destructive incident ray. Then the angle made by the interference at a point P is S1P – S2P = emergent ray with second face of the prism is (a) 30° (b) 60° (a) 2l (b) 3l 4 (c) 75° (d) 90° 153. Refer to Question 152, the angle of refraction (c) 5l (d) 11l 2 (r2) at another face of the prism is (a) 0° (b) 30° 162. Light is an electromagnetic wave. It speed in vacuum is given by (symbols have their usual (c) 45° (d) 60° meanings) 154. Refer to Question 152, the angle of refraction (r1) at first face of the prism is µ0 ∈0 (a) 0° (b) 30° (a) µ0 ∈0 (b) (c) 45° (d) 60° 155. Refer to Question 152, the refractive index of (c) ∈0 (d) 1 the material of prism is µ0 µ0 ∈0 (a) 1.33 (b) 1.414 163. When light passes from one medium into another medium, then the physical property (c) 1.52 (d) 1.732 which does not change, is 156. The nature of sun’s spectrum is (a) velocity (b) wavelength (a) line spectrum (c) frequency (d) refractive index (b) band spectrum (c) continuous spectrum (d) continuous spectrum with absorption lines

92 PHYSICS 164. The velocity of light is maximum in 172. A ray of light from air is incident on water, then which property of light will not change (a) diamond (b) water in water? (c) glass (d) vacuum (a) velocity (b) frequency 165. A single slit of width d is placed in the path of (c) amplitude (d) colour beam of wavelength l. The angular width of the principal maxima obtained si 173. When a red glass is heated in a dark room, it will appear d l (a) l (b) ± d (a) green (b) black (c) yellow (d) purple (c) 2l (d) 2d 174. Focal length can’t be measured in d l (a) m (b) mm 166. Which of the following phenomena is not (c) rad (d) km explained by Huygen’s construction of wavefront? 175. Magnifying power of a telescope can be increased by increasing (a) reflection (b) refraction (a) the focal length of the eyepiece (c) diffraction (d) origin of spectra (b) the length of telescope (c) the focal length of the objective 167. Huygen’s wave theory of light cannot explain (a) diffraction (b) interference (d) the diameter of the objective (c) polarization (d) photoelectric effect 176. A compound microscope has two lenses. The magnifying power of one is 5 and the 168. Interference is possible in combined magnifying power is 100. The magnifying power of the other lens is (a) light waves only (b) sound waves only (a) 15 (b) 20 (c) both light and sound waves (c) 50 (d) 95 (d) neither light nor sound waves 177. Which of the following prisms is used to see infrared spectrum of light? 169. In Young’s double slit experiment, the (a) Rock salt (b) Nicol intensity on screen at a point where path difference is l is K. What will be the intensity (c) Flint (d) Crown l 178. A light ray is incident normally on a plane at the point where path difference is 4 ? mirror. The angle of reflection will be (a) K (b) K (a) 45° (b) 60° 2 (c) 90° (d) 0° (c) K (d) zero 179. The minimum distance between an object 4 and its real image formed by a convex lens of focal length ‘f ’ is 170. Which of the following is conserved when light waves interference? (a) 3 f (b) 5 f 2 2 (a) energy (b) amplitude (c) 2f (d) 4f (c) intensity (d) momentum 180. The astronaut in the spaceship sees the sky as 171. If a water drop is kept between two glass plates, then its correct shape is shown by (a) White (b) Blue (c) Red (d) Black (a) (b) 181. Which of the following colours sulfers maximum deviation in a prism? (c) (d) (a) Orange (b) Yellow (c) Green (d) Blue

Optics   93 182. Stars are not visible in the day time because (a) always bright (b) always dark (a) stars vanish during the day (c) either dark or bright depending on the position of S (b) stars hide behind the sun (d) neither dark nor bright (c) stars do not reflect sun rays during the day 188. The correct curve between the energy of photon (E) and its wavelength (l) is (d) atmosphere scatters sun light into a blanket of extreme brightness through E E which faint stars cannot be visible (a) (b) 183. Rainbow is due to (a) dispersion of light O lO l (b) reflection of light E E (c) refraction of light (d) total internal reflection of light (c) (d) 184. A ray of light is incident on an equilateral O lO l glass prism placed on a horizontal table. For minimum deviation which of the following is 189. The velocity of light is maximum in true? (a) water (b) glass Q R (c) vacuum (d) diamond P S 190. In Young’s double slit experiment, the fringe (a) PQ is horizontal width is found to be 0.8 mm. If the whole (b) QS is horizontal apparatus is immersed in water of refractive (c) RS is horizontal (d) either PQ or RS is horizontal index 4 , without disturbing the geometrical 185. When light travels from an optically rarer 3 medium to an optically denser medium, the arrangement, the new fringe width will be velocity decreases because of change in (a) frequency (b) wavelength (a) 0.60 mm (b) 0.30 mm (c) amplitude (d) phase 186. The light waves from two coherent sources (c) 10.7 mm (d) none of these of same intensity (I) produce interference. At minimum intensity of light is zero. What is 191. In Young’s double slit experiment, the the intensity of light at the maximum? (a) I (b) 4I distance between two slits is d and the (c) I2 (d) 4I2 3 187. In the set up shown in figure, the two slits S1 and S2 are not equidistant from the slit S. The distance between the screen and the slits is central fringe at O is then 3D. The number of fringes in 4m on the S1 5 SO screen, formed by monochromatic light of wavelength 3l, will be given by (a) d (b) d 9Dl 27 Dl (c) d (d) none of these Dl 192. In Young’s double slit experiment, one slit is covered with green filter and the other is covered with red filter. Then the interference pattern will be S2 (a) green (b) red (c) yellow (d) invisible

94 PHYSICS 193. If the distance between the slits is increased, 200. Which term associated with an optical the angular separation of interference fringes instrument indicates its capability to magnify in Young’s experiment would the dimensions of the object so that it appears enlarged? (a) increase (b) decrease (a) resolving power (c) remain the same (b) angular spreading (d) first increase then decrease (c) magnifying power 194. How does the fringe width of interference (d) efficiency fringes change when the whole apparatus of Young’s experiment is kept in a liquid of 201. What is the shape of interference fringes in a Young’s double–slit experiment? refractive index 2 ? (a) spherical fringes (a) fringe width would increase (b) concentric circular fringes (b) fringe width would decrease (c) fringe width would remain the same (c) cylindrical fringes (d) fringes would disappear 195. What causes colours as seen in soap bubbles? (d) straight lines parallel to the slits (a) polarization (b) total internal reflection 202. Yellow light of wavelength 600 Å produces (c) dispersion fringes of width 0.8 mm in Young’s double (d) none of these slit experiment. What will be the fringe width if the light source is replaced by another 196. When two coherent sources are far apart, the monochromatic source of wavelength 7500 Å fringe width would and the separation between the slits is doubled? (a) 2 mm (b) 1 mm (a) increase exponentially (c) 0.5 mm (d) 0.25 mm (b) remain the same 203. The Brewster’s angle for glass is about (c) be too small to be detected (a) 37° (b) 45° (d) first increase then decrease (c) 58° (d) 90° 197. If the two point coherent sources are infinitely 204. Two slits in Young’s interference experiment close to each other, the fringe width would have widths in the ratio 1 : 16. Then, the ratio become of intensity at the maxima and minima in the interference pattern would be (a) unity (b) zero (c) infinity (d) none of these (a) 16:1 (b) 4:1 198. Which of the following phenomena would (c) 25:9 (d) 25:16 be prominently observed when size of the obstacle is comparable to the wavelength of 205. Two coherent sources of intensities 100:1 the waves? interfere with each other. What is the ratio of intensities between the maxima and minima (a) polarization (b) interference in the interference pattern? (c) diffraction (d) dispersion (a) 5:3 (b) 3:2 199. Blue light is used to form a single–slit (c) 2:5 (d) 7:5 diffraction pattern on a screen. When the slit is made wider, the separation of the bright 206. A telescope has a diameter of 4 m. What is bands on the screen would limiting angle of resolution for 600 nm light? (a) increase (a) 1.22 × 10–7 rad (b) decrease (b) 1.63 × 10–8 rad (c) remain the same (c) 2.4 × 10–7 rad (d) first increase then decrease (d) 1.83 × 10–7 rad

Optics   95 207. The spectral line for a given element in the (c) (d) light received from a distant star is shifted towards the longer wavelength by 0.03%. 213. Two coherent waves are represented by Then, the velocity of star in the line of sight is y1 = a1 cos wt (a) 3 × 104 m/s (b) 6 × 104 m/s y2 = b1 sin wt (c) 9 × 104 m/s (d) 105 m/s The resultant intensity due to interference 208. The time of coherence is of the order of will be (a) 10–6 s (b) 10–8 s (c) 10–10 s (d) 106 s 209. Which of the following ratios defines fringe (a) (a1 + b1) (b) (a1 – b1) visibility? (c) (a12 – b12) (d) (a12 + b12) (a) Imax – Imin (b) Imax + Imin 214. A 100 W source is emitting radiations of Imax + Imin Imax – Imin wavelength 5000 Å. Then, the rate of emission (c) Imax (d) Imin of photons is of the order of Imain Imax (a) 1010 (b) 1015 210. The correct curve between width fringe (β) (c) 1020 (d) 1025 and distance (D) between the slits and the screen for a given light, is correctly shown by: 215. A monochromatic visible light comprises (a) a single ray of light bb (b) light of many colours with a single wavelength (a) (b) (c) light of many wavelengths with a single D D colour b b (d) none of these (d) (c) 216. When monochromatic light is replaced by white light in Fresnel’s biprism arrangements, the central fringe is DD (a) dark (b) white 211. In Young’s double slit experiment, the (c) yellow (d) red separation between the slits is doubled and the distance between the slits and screen is 217. In a Young’s double–slit experiment, if the halved. Then, the ratio of new fringe width to slits are of unequal width, the original fringe width is (a) bright fringe will not be formed at the centre (a) 1 (b) 1 (b) fringes will not be formed 8 4 (c) the positions of minimum intensity will (c) 4 (d) 8 not be formed 212. The correct variation between fringe width (d) nothing can be decided (β) and distance between the slits (d) is shown by 218. In Young’s double slit experiment, if two slightly different wavelengths are present in (a) (b) the light used, then (a) the central fringe will be white (b) there will be no fringes at all (c) the sharpness of fringes will increase (d) None of these

96 PHYSICS 219. The fringe width at a distance 40 cm from the 228. Which property of electrons is employed in slits in Young’s double experiment for light an electron microscope? of wavelength 6000Å is 0.036 cm. The fringe width at the same distance for λ = 5000Å will (a) negative charge (b) small mass be (c) wave nature (d) spin (a) 0.03 cm (b) 0.04 cm 229. The wave number is reciprocal of (c) 0.24 cm (d) 0.62 cm (a) frequency (b) wavelength 220. In a biprism experiment, the wavelength of (c) intensity (d) Avogadro’s number monochromatic light is 6000 Å. The distance between the two virtual images is 6 mm. Then 230. In order to get sustained interference of light, the number of fringes formed per cm on a the necessary condition is that two sources screen at a distance of 1 m is should (a) 50 (b) 100 (a) be narrow (c) 150 (d) 200 (b) have same amplitude 221. Which of the following waves are diffracted (c) be close to each other by an obstacle of size 10–2 m? (d) have constant phase difference (a) light waves (b) sound waves 231. Soap bubbles look coloured because of (c) X-rays (d) ultrasonic waves (a) dispersion (b) reflection 222. What changes on polarization of light? (c) interference (d) diffraction (a) phase (b) frequency 232. A rocket is giving away from the earth at a speed of 106 ms–1. If wavelength of light wave (c) wave length (d) intensity emitted by it be 5400 Å, its Doppler’s shift 223. In order to be invisible in vacuum, what should be the refractive index of a transparent would be medium? (a) 18 Å (b) 20 Å (a) greater than 1 (b) less than 1 (c) 180 Å (d) 1.8 Å (c) equal to 1 (d) infinity 233. Light travels through a glass slab of thickness d and having refractive index m. If c is velocity 224. The limit of resolution of an optical instrument of light in vacuum, then the time taken by arises on account of light to travel this thickness of glass is (a) polarization (b) scattering (a) µc (b) µd (c) interference (d) diffraction d c 225. Light waves can be polarized because they (a) have high frequencies (c) cd (d) µ µ dc (b) have short wavelength (c) are transverse 234. A plane mirror is approaching you at 8.5 cm (d) can be reflected and refracted per second. You can see your image in it. At what speed will your image approach you? 226. Light emitted by the Nicole prism is (a) 8.5 cm s–1 (b) 4.75 cm s–1 (a) unpolarized (c) 17 cm s–1 (d) none of these 235. Velocity of electromagnetic waves in vacuum (b) plane polarized (c) circularly polarized depends upon (d) elliptically polarized (a) velocity of source 227. Polaroid glass is used in sun glasses because (b) amplitude of waves (a) it has a good colour (c) frequency of waves (b) it is fashionable (d) none of the above (c) it is cheaper (d) none of these

Optics   97 236. The frequency of electromagnetic wave which 244. The wavelength range of visible light is is best suited to observe a particle of radius approximately from 3 × 10–6 m is of the order of (a) 4500 Å to 8500 Å (a) 1010 Hz (b) 1012 Hz (b) 4000 Å to 8000 Å (c) 3000 Å to 7500 Å (c) 1014 Hz (d) 1016 Hz (d) none of these 237. A convex mirror of focal length f produces an 245. When temperature of a medium increases, its image  1 th of the size of the object. Then the refractive index  n  (a) increases distance of the object from the mirror is (b) decreases (c) remains same (a) (n + 1)f (b) (n − 1) (d) first increases then decreases f 246. The variation of refractive index (m) with (c) (n – 1)f (d) n f wavelength (l) is correctly shown by 238. If a ray of light is incident on a plane mirror mm at an angle of 35°, then the angle of deviation produced by the mirror is: (a) 35° (b) 70° (c) 90° (d) 110° (a) (b) 239. A fish 12 cm long is 4 cm under the water l l level. Its length when viewed vertically above m will be (d) (a) 12 cm (b) 3 cm m (c) 4 cm (d) 8 cm (c) 240. A completely transparent material will ll become invisible in vacuum when its refractive index (m) is (a) 1 (b) > 1 247. The minimum value of m is for (c) < 1 (d) ≥ 1 (a) water (b) glass 241. The frequency of incident light is 3 × 1016 Hz. (c) diamond (d) air The frequency after reflection 248. In terms of wavelength of light (l), the intensity (I) of scattered light is given by (a) remains same (b) increases (a) I ∝ l2 (b) I ∝ l–2 (c) decreases (c) I ∝ l4 (d) I ∝ l–4 (d) first increases then decreases 242. When light suffers reflection, at the interface 249. What happens to the refractive index when between air and glass, the change of phase in frequency of light increases? the reflected wave equals (a) increase (a) p radian (b) p radian (b) decreases (c) 2p radian 2 (c) remains same (d) none of these (d) first increases then decreases 243. When x is distance of the object from the focus 250. Angle of minimum deviation (dm) through a and y is distance of image from the focus of prism depends on the the mirror (of focal length f), as per Newton’s formula: (a) material of the prism only (b) angle of prism only (a) f = x + y (b) f = xy (c) angle of incidence only (c) 1 = 1 (d) f 2= xy (d) all the above f + x y

98 PHYSICS 251. If a mirror is lying in a denser medium 257. Two thin lenses of focal lengths 20 cm and (refractive index m) and viewed from a rarer 25  cm are placed in contact. The effective medium, then the ratio of real radius of curvature to apparent radius of curvature power of the combination is equals (a) 1 dioptre (b) 9 dioptres 9 (d) none of these 1 (a) m (b) µ (c) 4.5 dioptres (c) µ (d) 1 258. When a thin convex lens (focal length f) is µ put in contact with a thin concave lens of the same focal length, the resultant combination 252. If atmosphere were not there, the sky would has a focal length equal to appear (a) f/2 (b) 2f (a) white (b) blue (c) 0 (d) ∞ (c) red (d) black 259. If the refractive index of water is 4/3 and that 253. A glass slab is placed in the path of a beam of of glass slab is 5/3, then the critical angle of convergent light. The point of convergence of light incidence for which a light ray tending to go from glass to water is totally reflected, is (a) remains at the same point given by (b) moves towards the glass slab (a) sin–1 (3/4) (b) sin–1 (3/5) (c) moves away from the glass slab (c) sin–1 (1/3) (d) sin–1 (4/5) (d) will undergo some lateral shift 260. A ray of light passes from vacuum into a medium of refractive index m, and the angle 254. Focal length of a glass lens is 2.4 cm. Its focal of incidence is found to be twice the angle of length when immersed in water would be refraction. Then the angle of incidence is (a) 1.2 cm (b) 4.8 cm (a) cos–1 (m/2) (b) 2 cos–1 (m/2) (c) 1.6 cm (d) 9.6 cm (c) 2 sin–1 (m) (d) 2 sin–1 (m/2) 255. It is possible to observe total internal reflection 261. A convex lens of focal length 0.5 m and when a ray of light travels from concave lens of focal length 1 m are combined. The power of the resulting lens will be (a) air into water (b) air into glass (a) 1 D (b) – 1 D (c) glass into water (c) – 0.5 D (d) 0.5 D (d) water into glass 262. If the critical angle for total internal reflection from a medium to vacuum is 30°, the velocity 256. Figure shows two parallel rays incident on of light in the medium is a mirror. They are reflected as parallel rays as shown in figure. What is the nature of the (a) 3 × 108 m/s (b) 1.5 × 108 m/s mirror? (c) 6 × 108 m/s (d) 3 × 108 m/s Incident rays 263. A fish looking up through the water sees the outside world contained in a circular horizon. Reflected rays If the refractive index of water is 4/3 and fish is 12 cm below the surface, the radius of the circle (in cm) is (a) Plane (b) Concave (a) 12 × 3 × 5 (b) 4 × 5 (c) 12 × 3 × 7 (d) 12 × 3 / 7 (c) Convex (d) Parabolic

Optics   99 264. To a bird in air, a fish in water appears to be at 271. In an experiment to find the focal length of a concave mirror, a graph is drawn between the 24 cm from the surface. If amw = 4 , then the magnitude of u and v. The correct graph will 3 look like true depth of the fish is uu (a) 24 cm (b) 30 cm (c) 32 cm (d) 18 cm (a) (b) 265. Figure shows three cases of a ray passing v v through a prism of refracting edge A. The u case corresponding to minimum deviation is u (d) AA (c) (a) (b) vv A A 272. A convex lens is dipped in a liquid whose refractive index equals the refractive index of (c) (d) the lens. Then its focal length will be (a) 1 (b) – 1 (c) 0 (d) ∞ 266. An air bubble in glass slab (m = 1.5) from one 273. The nature of sun’s spectrum is side is 7 cm and from the other side is 3 cm. The thickness of the glass slab is (a) line spectrum (a) 15 cm (b) 6 cm (b) band spectrum (c) 20 cm (d) 12 cm (c) continuous spectrum 267. A ray of light enters from a denser medium (d) continuous spectrum with a few into rarer medium. The speed of light in rarer absorption lines medium is twice that in denser medium. What is the critical angle for total internal 274. Which of the following colours is scattered reflection (TIR) to take place? minimum? (a) red (b) blue (a) 30° (b) 45° (c) violet (d) yellow (c) 60° (d) 90° 275. A ray of light is incident normally on one face of an equilateral prism of refractive index 1.5. 268. Two thin lenses are in contact and the focal Then the angle of deviation is length of the combination is 80 cm. If the focal length of one lens is 20 cm, then the power of (a) 30° (b) 45° the other lens would be (c) 60° (d) 80° (a) – 2.75 D (b) 3.75 D 276. Dispersion is explained on the basis of (c) – 3.75 D (d) 1.66 D (a) Rayleigh’s scattering law 269. The colour of a star is an indication of its (b) Snell’s law (a) mass (b) size (c) Cauchy’s formula (c) distance (d) temperature (d) Total internal reflection 270. If the angle of a prism is 60° and the angle of 277. Rainbow is due to minimum deviation is 40°, then the angle of refraction would be (a) dispersion of light (b) reflection of light (a) 30° (b) 40° (c) refraction of light (c) 20° (d) 10° (d) total internal reflection of light

100 PHYSICS 278. The focal length of a convex lens is 2.5 cm. Its (c) (d) magnifying power will be (a) 4 (b) 5 (c) 6 (d) 11 281. An astronomical telescope of ten–fold angular magnification has a length of 55 cm. The focal 279. The focal lengths of objective and the eye- length of the objective is piece of a compound microscope are f0 and fe respectively. Then (a) 4 cm (b) 5 cm (a) f0 > fe (b) f0 < fe (c) 40 cm (d) 50 cm (c) f0 = fe (d) f0 ≥ fe 280. If a drop of water is placed between two glass 282. A convex lens is dipped in a liquid whose refractive index is equal to the refractive plates, then its shape is correctly shown by index of the lens. Then its focal length will (a) increase (a) (b) (b) decrease (c) remain same (d) become infinite CASE BASED QUESTIONS Case Study–1 and the corresponding time difference between actual sunset and apparent sunset is about 2 minutes. The Read the following passage and answer the apparent flattening (oval shape) of the sun at sunset questions that follow: and sunrise is also due to the same phenomenon. The refraction of light through the atmosphere 1. The sun may appear to be an elliptical just is responsible for many interesting phenomena. For before setting. This happens due to example, the sun is visible a little before the actual sunrise and until a little after the actual sunset due (a) dispersion to refraction of light through the atmosphere. By actual sunrise we mean the actual crossing of the (b) reflection horizon by the sun. Figure below shows the actual and apparent positions of the sun with respect to the (c) refraction horizon. The figure is highly exaggerated to show the effect. The refractive index of air with respect to (d) differaction vacuum is 1.00029. Due to this, the apparent shift in the direction of the sun is by about half a degree 2. The earth takes 24 h to rotate once about its axis. How much time does the sun take to Apparent position shift by 1° when viewed from the earth? of the sun (a) 2 minutes (b) 3 minutes (c) 4 minutes (d) 5 minutes Horizon 3. The maximum angle of refraction when a ray of light is refracted from glass into air Observer (a) 0° (b) 30° (c) 45° (d) 90° Earth 4. Velocity of light in glass is 2 × 108 ms–1 and that in air is 3 × 108 ms–1. By how much would Actual position of the sun an ink dot appear to be raised when covered by a glass plate 6 cm think? Atmosphere (a) 1 cm (b) 2 cm Advance sunrise and delayed sunset due to atmospheric refraction (c) 3 cm (d) 4 cm

Optics   101 Case Study–2 1. While moving in a bus or a car during a Read the following passage and answer the hot summer day, a distant patch of road, questions that follow: especially on a highway, appears to be wet. But we do not find any evidence of wetness On hot summer days, the air near the ground when we reach that spot. This is due to becomes hotter than the air at higher levels. The refractive index of air increases with its density. Hotter (a) looming (b) mirage air is less dense, and has smaller refractive index than the cooler air. If the air currents are small, that is, the (c) refraction (d) none of these air is still, the optical density at different layers of air increases with height. As a result, light from a tall 2. Which of the following statement in incorrect? object such as a tree, passes through a medium whose refractive index decreases towards the ground. Thus, (a) Total internal reflection takes place a ray of light from such an object successively bends away from the normal and undergoes total internal when light travels from rarer to denser reflection, if the angle of incidence for the air near medium. the ground exceeds the critical angle. This is shown in Fig. (b). To a distant observer, the light appears (b) The bending of light as it travels from to be coming from somewhere below the ground. one medium to another medium need The observer naturally assumes that light is being not occur all in a single step. reflected from the ground, say, by a pool of water near the tall object. Such inverted images of distant (c) If the medium gradually changes tall objects cause an optical illusion to the observer. its index of refraction, a light ray is This phenomenon is called mirage. gradually bent towards the medium with the higher index. Light (d) None of these. Air at a uniform temperature 3. The phenomenon of ‘mirage’ formation is due to (a) A tree is seen by an observer when the air above the ground is at uniform temperature. (a) reflection of light Refractive index (b) refraction of light constant (c) diffraction of light (d) total internal reflection of light. 4. Brilliance of diamond is due to (a) interference of light (b) diffraction of light (c) refraction of light Refractive index (d) none of these decreasing Air warmer Case Study–3 near ground Read the following passage and answer the questions that follow: (b) When the air close to the ground is hot, light bends gradually as undergoing total internal reflection, Optical fibre medical instruments may contain and the apparent image of the tree may mislead bundles of optical fibres. An optical fibre instrument the observer into thinking that there is a pool of used to see the internal parts of human body is water in front of the tree! endoscope. The endoscope facilitates the physicians to see the internal parts of body without performing surgery. The main part in endoscope is fibrescope. Figure shows what it would see at the end of endoscope. Two of the main fibre optic features used in an endoscope are cables that light up (illuminate) the

102 PHYSICS inside the stomach. The other main feature is a fibre 4. Figure shows a cross section of a ‘light-pipe’ optic viewing cable that allows to see physician what made of a glass fibre of refractive index 1.68. the problem is. The outer covering of the pipe is made up of material refractive index 1.44. What is the Air/water nozzle range of the angles of the incident rays with the axis of the pipe for which the total internal Illumination Viewing reflection inside the pipe takes place as shown fibre optic fibre optic in the figure? Biopsy/Suction channel Illumination Cladding fibre optic i¢ i¢ Illumination Endoscope Viewing r t Core (a) ≈ 59° (b) ≈ 31° (c) ≈ 60° (d) ≈ 53.5° Stomach Case Study–4 Both of these fibre optic cables rely on total Read the following passage and answer the internal reflection. questions that follow: 1. A ray of light travels in optical fibre due to Microscopes have opened up a whole new dimension in science. By using microscopes (a) refraction scientists were able to discover the existence of the microorganisms, study the structure of cells, and see (b) total internal reflection the smallest parts of plants, animals, and fungi. (c) reflection Microscopes use a visible light and magnifying lenses to detect small objects that are not visible by (d) polarization the naked eye. The higher the resolution, the better is our ability to detect two distinct objects without 2. An optical fibre (n = 1.72) is surrounded by any blurring of the image. The resolution of the best a glass coating (n = 1.50). The critical angle modern microscope is determined by the diffraction for total internal reflection at the fibre-glass properties of light, which limit the resolution to about interface will be: half the wavelength of light. In other words, with a good modern microscope, one can observe objects as (a) 59.69° (b) 62.61° small as half the wavelength of the illuminating light. (c) 60.69° (d) 58.76° 1. Which of the following statements is true for a simple microscope? 3. Which of the following is incorrect statement? (a) When a small object is placed between (a) Light undergoes successive total internal the principal focus and optical centre reflections as it moves through an optical of a convex lens, its virtual, erect and fibre. magnified image is formed. (b) Optical fibres consist of thousands of (b) When an object is placed at the principal extremely thin and long strands of high focus of a covex lens, its magnified image quality glass/quartz. is formed at infinity. (c) Optical fibres are used for optical signal (c) Under practical conditions the transmission. magnification of a simple microscope is ≤ 9. (d) None of these (d) None of these.

Optics   103 2. A convex lens of focal length 6.25 cm is used (d) It requires a material medium for its as a magnifying glass. If the near point of the propagation. observer is 25 cm from the eye and the lens is held close to the eye then distance of the 2. The branch of optics dealing with the object from the lens will be formation of images using the concept of straight line propagation of light is called (a) – 5 cm (b) – 6 cm (a) geometrical optics (c) – 3 cm (d) – 4 cm (b) physical optics 3. The magnifying power (M) of a simple microscope is given by (c) corpuscular optics (a) 1 + f (b) 1 − D (d) quantum optics D f 3. In the set-up shown in figure, the two slits (c) 1 − f (d) 1 + D S1 and S2 are not equidistant from the slits S. D f Then the central fringe at O is 4. A woman uses +1.5 D glasses to have normal S1 O vision from 25 cm onwards. She uses a 20 D S lens as a simple microscope to see an object. The maximum magnifying power if she uses S2 the microscope together with her glass is (a) 4 (b) 5 (c) 6 (d) 7 Case Study–5 (a) always dark Read the following passage and answer the (b) always bright questions that follow: (c) either bright or dark depending on the The wave theory was not readily accepted position of S primarily because of Newton’s authority and also because light could travel through vacuum and it was (d) neither dark nor bright felt that a wave would always require a medium to propagate from one point to the other. However, when 4. A beam of light AO incident on a glass slab Thomas Young performed his famous interference (m = 1.54) in a direction as shown in figure experiment in 1801, it was firmly established that given below. The reflected ray OB is passed light is indeed a wave phenomenon. The wavelength through a Nicol prism (N). On rotating the of visible light was measured and found to be Nicol prism, we find that extremely small; for example, the wavelength of yellow light is about 0.5 mm. Because of the smallness AB of the wavelength of visible light (in comparison to the dimensions of typical mirrors and lenses), light 33° 33° can be assumed to approximately travel in straight O lines. (a) there is no change in intensity 1. Which of the following is not a property of (b) the intensity decreases somewhat but light? rises again (a) It can travel through vacuum. (c) the intensity is reduced to zero and (b) It has a finite speed. remains zero (d) the intensity gradually reduces to zero (c) It involves transportation of energy. and then again increases

104 PHYSICS Case Study–6 (b) Rays (c) Plane wavefront Read the following passage and answer the questions that follow: In the propagation of sound waves in a gas, these waves travel in all directions. If the medium is isotropic, the velocity of propagation of sound waves is the same in all directions. So, in this case, the wavefronts are spherical (Fig. (a)). The rays are normal to the wavefronts. S (a) Spherical wavefronts. (d) none of these (b) Plane wavefronts. 3. Which of the following statement is incorrect? As the spherical wavefronts advance further, (a) Each point on a wavefront acts as a fresh their curvature goes on decreasing. As a result of this, source of disturbance. a small portion of a spherical wave observed from a long distance is a plane wave. In this case, the rays are (b) The rays of light are always normal to unidirectional. the wavefront. To sum up, the wavefronts of waves travelling in (c) The shape of the wavefront emitted by a a single direction are plane (Fig. (b)). light source in the form of a narrow slit is cylindrical Plane waves can be produced in water by moving a straight rod up and down on the surface of water. (d) None of these 1. Which of the following wavefront is produced 4. Huygen’s concept of secondary waves is by the point source of light? useful in (a) Spherical wavefront (a) explaining polarisation (b) Cylindrical wavefront (c) Plane wavefront (b) determining the focal length of a lens (d) Circular wavefront 2. Which of the following wavefronts (c) geometrical reconstruction of a wavefront corresponds to a beam of white light coming from a very far off source? (d) none of these (a) Case Study–7 Read the following passage and answer the questions that follow: The figure shows a surface XY separating two transparent media, medium-1 and medium‑2. The lines ab and cd represent wavefronts of a light wave Rays Spherical wavefronts

Optics   105 travelling in medium-1 and incident on XY. The lines Case Study–8 ef and gh represent wavefronts of the light wave in medium-2 after refraction. Read the following passage and answer the questions that follow: 1. Light travels as a There are various applications of the interference (a) parallel beam in each medium which is defined as the phenomenon of which two or more coherent waves will superimpose on each (b) convergent beam in each medium other and form a single resultant wave which will have greater or lower or of the same amplitude. The (c) divergent beam in each medium interference can be constructive or destructive which is resultant from the interaction of the waves which (d) divergent beam in one medium and are coherent with each other. Some of the applications convergent beam in the other medium. of the interference are: 2. The phases of the light wave at c, d, e and f 1. Optical testing               2. Space applications are fc, fd, fe and ff respectively. It is given that fc ≠ ff . In optical testing, interference is used in testing surface quality like: flat surface, spherical surface, (a) fc cannot be equal to fd roughness of surface etc. Whereas in space applications (b) fd can be equal to fe include radio astronomy, measuring light intensity, in (c) (fd – ff) is equal to (fc – fe) retrieving images from the telescopes. (d) (fd – fc) is not equal to (ff – fe). 3. Speed of light is 1. For sustained interference, we need two sources which emit radiations (a) the same in medium-1 and medium-2 (a) of the same intensity (b) larger in medium-1 than in medium-2 (b) of the same amplitude (c) larger in medium-2 than in medium-1 (c) having a constant phase difference (d) different at b and d. (d) none of these 4. Direction: The question has a paragraph followed by two statements, Statement‑1 and 2. The interference fringes formed by a thin oil Statement-2. Of the given four alternatives film on water, when seen in the yellow light after the statements, choose the one that of sodium lamp will be describes the statements. (a) coloured (b) yellow and black A thin air film is formed by putting the convex surface of a plane-convex lens over a plane (c) green (d) red glass plate. With monochromatic light, this film gives an interference pattern due to light 3. In Young’s double slit experiment, the slits reflected from the top (convex) surface and are separated by 0.28 mm and the secreen the bottom (glass plate) surface of the film. is placed 1.4 m away. The distance between the central bright fringe and the fourth bright Statement-1: When light reflects from the fringe is measured to be 1.2 cm, then the air-glass plate interface, the reflected wave wavelength of light used in the experiment suffers a phase change of p. will be: Statement-2: The centre of the interference (a) 4 × 10–6 m pattern is dark. (b) 5 × 10–6 m (c) 6 × 10–7 m (a) Statement-1 is true, Statement-2 is false. (d) 6.5 × 10–7 m (b) Statement-1 is true, Statement-2 is true, 4. Light waves from two coherent sources of intensity ratio 81 : 1 produce interference. Statement-2 is the correct explanation of The ratio of the maxima and minima in the Statement-1. interference pattern is (c) Statement-1 is true, Statement-2 is true, (a) 9 : 1 (b) 1 : 9 Statement-2 is not the correct explanation (c) 16 : 25 (d) 25 : 16 of Statement-1. (d) Statement-1 is false, Statement-2 is true.

106 PHYSICS Case Study–9 (c) Diffraction fringes become broader and farther apart. Read the following passage and answer the questions that follow: (d) The diffraction pattern disappears. As required by wave theory of light, light bends 2. A bright spot at the centre of the geometrical round corners. Since wavelength of light is extremely shadow of a small circular disc placed in the small (of the order of 6 × 10–5 cm), the amount path of light is due to the phenomenon of of bending is too small to be detected without careful examination. When light passes through an (a) diffraction aperture whose dimensions are comparable with the wavelength of light, the light is found to diverge. (b) interference There is encroachment of light within the geometrical shadow of the opaque obstacle. (c) dispersion The encroachment of light in the region of (d) polarisation geometrical shadow is due to the bending of light around the corners of the aperture or obstacle. 3. The fringe width b of a diffraction pattern and The diffraction effects can be observed by the slit width d are related as viewing a distant source of light through a fine cloth handkerchief. The luminous border surrounding the 1 profile of a mountain just before the sun rises is also (a) b ∝ d (b) b ∝ d a diffraction effect. (c) b ∝ d (d) b ∝ 1 1. A single slit diffraction pattern is obtained by d2 using a beam of red light. What happens if the red light is replaced by a blue light? 4. Determine the angular spread between central maximum and first order maximum (a) There is no change in diffraction pattern. of the diffraction pattern due to a single slit of width 0.25 mm, when light of wavelength (b) Diffraction fringes become narrower 5890 Å is incident on it normally. and crowded. (a) ± 2.390 × 10–3 rad (b) ± 3.028 × 10–3 rad (c) 2.90 × 10–3 rad (d) ± 3.530 × 10–3 rad

Unit 10 Communication Systems MULTIPLE CHOICE QUESTIONS Tick (3) the most appropriate choice amongst (c) Electromagnetic waves of frequencies the following: higher than 30 MHz penetrate ionosphere. 1. Which range of frequencies is suitable for sky wave propagation ? (d) Satellite communication uses sky wave mode of propagation. (a) 1 kHz to 500 kHz (b) 1 MHz to 2 MHz 5. The maximum line-of-sight distance dM between two antennas having heights hT and (c) 2 MHz to 20 MHz hR above the earth is (d) above 30 MHz. (a) R (hT + hR ) (b) 2R/(hT + hR ) 2. A TV tower has a height of 80 m. The (c) RhT + 2RhR (d) 2RhT + 2RhR maximum distance upto which TV 6. The frequency band used in the downlink of satellite communication is transmission can be received is equal to (radius of earth = 6.4 × 106 m) (a) 16 km (b) 32 km (c) 80 km (d) 160 km. (a) 9.5 to 2.5 GHz (b) 896 to 901 MHz 3. Which of the following frequencies (c) 3.7 to 4.2 GHz (d) 840 to 935 MHz will be suitable for beyond the horizon 7. In amplitude modulation, the bandwidth is communication ? (a) twice the audio signal frequency. (b) thrice the audio signal frequency. (a) 10 kHz (b) 10 MHz (c) 1 GHz (d) 1000 GHz. (c) thrice the carrier wave frequency. 4. Pick out the correct statement in the (d) twice the carrier wave frequency. propagation of electromagnetic waves for communication purposes. 8. If both the length of an antenna and the wavelength of the signal to be transmitted are (a) Space wave propagation is achieved by doubled, the power radiated by the antenna ionospheric reflection. (a) is doubled. (b) Sky wave propagation is used for line- (b) is halved. of-sight communication. (c) remains constant. (d) is quadrupled. 144

Communication Systems   145 9. If the maximum amplitude of an amplitude (a) 8 kHz (b) 4 kHz modulated wave is 25 V and the minimum amplitude is 5 V, the modulation index is (c) 7.6 kHz (d) 3.8 kHz. (a) 1 (b) 1 15. The TV telecast is to cover a radius of 120 km 5 3 (given the radius of the earth = 6400 km), the height of the transmitting antenna is (c) 3 (d) 2 . (a) 1280 m (b) 1125 m 2 3 (c) 1560 m (d) 79 m. 10. A modem is a 16. Identify the INCORRECT statement from the following. (a) modulating device only. (b) demodulating device only. (a) AM detection is carried out using a rectifier and an envelope detector. (c) modulating and demodulating device. (b) Pulse position denotes the time of rise or (d) transmitting device. fall of the pulse amplitude. 11. Which of the following four alternatives is (c) Modulation index m is kept ≥ 1, to avoid not correct ? We need modulation distortion. (a) to increase the selectivity. (d) Facsimile (FAX) scans the contents of the document to create electronic signals. (b) to reduce the time lag between transmission and reception of the 17. The mobile telephones operate typically in information signal. the range of (c) to reduce the size of antenna. (a) 1 – 100 MHz (d) to reduce the fractional bandwidth, that (b) 100 – 200 MHz is the ratio of the signal bandwidth to the centre frequency. (c) 1000 – 2000 MHz 12. The distance of coverage of a transmitting (d) 800 – 950 MHz. antenna is 12.8 km. Then, the height of the antenna is 18. In television transmission, what type of modulation is used ? (Given that radius of earth = 6400 km) (a) Only amplitude modulation (a) 6.4 m (b) 12.8 m (b) Only frequency modulation (c) 3.2 m (d) 16 m. (c) Both amplitude and frequency modulation 13. Which one of the following is INCORRECT statement in the transmission of (d) TV signal does not need any kind of electromagnetic waves ? modulation. (a) Ground wave propagation is for high 19. Why do we need carrier wave of high frequency transmission. frequency to transmit audio signal over long distances ? (b) Sky wave propagation is facilitated by ionospheric layers. (a) High frequency carrier wave can propagate with a faster speed. (c) Space wave is of high frequency and is suitable for line of sight communication. (b) High frequency carrier waves offer availability of higher transmission (d) Space wave is used for satellite bandwidth. cummunication. 14. 1000 kHz carrier wave is amplitude (c) High frequency carrier waves offer modulated by the signal frequency availability of lower transmission 200–4000 Hz. The channel width of this case bandwidth. is (d) High frequency carrier waves is easy to produce.

146 PHYSICS 20. In an amplitude modulation with modulation 28. Find the mismatch index 0.5, the ratio of the amplitude of the carrier wave to that of the side band in the (a) Sky wave communication : modulated wave is  Frequency upto 30 MHz (a) 4 : 1 (b) 1 : 1 (b) Line-of-sight communication : (c) 1 : 2 (d) 2 : 1.  Frequency greater than 40 MHz 21. A ground receiver in line-of-sight (c) Mobile telephony : communication cannot receive direct waves due to  Frequency range 800-950 kHz (d) Facsimile : Static document. (a) its low frequency 29. In a video signal for transmission of picture, what value of bandwidth is used in (b) curvature of earth communication system ? (c) its high intensity (a) 2.4 MHz (b) 4.2 MHz (d) smaller antenna. (c) 24 MHz (d) 42 MHz. 22. The waves that are bent down by the 30. A tuned amplifier circuit is used to generate a ionosphere are carrier frequency of 2 MHz for the amplitude modulation. The value of LC is (a) ground waves (b) surface waves (c) direct waves (d) sky waves. 1 1 × 106 × 106 23. A television tower of height 140 m can (a) 3π (b) 2π broadcast its signal upto a maximum area of (Radius of earth = 6.4 × 106 m) (c) 1 (d) 2 1 . 4π × 106 × 106 (a) 1.56 × 106 km2 (b) 5.6 × 103 km2 (c) 5.6 × 1010 km2 (d) 1.56 × 109 km2. 31. A point-to-point communication mode is seen in 24. The range of frequency bands used for television VHF service is (a) satellite cable communication. (b) television transmission. (a) 540 – 1600 kHz (b) 88 – 108 MHz (c) FM radio transmission. (d) fax transmission. (c) 3.7 – 4.2 GHz (d) 54 – 72 MHz. 32. If the heights of transmitting and the receiving 25. An example of point to point mode of antennas are each equal to h, the maximum communication is line-of-sight distance between them is (R is the radius of earth) (a) FM radio (b) standard FM radio (a) 2Rh (b) 4Rh (c) television (d) telephony. (c) 6Rh (d) 8Rh . 26. The radiating power of a linear antenna of 33. The ionospheric layer acts as a reflector for length l for a wavelength l is proportional to the frequency range (a) l (b) l2 (a) 1 kHz to 10 kHz (b) 3 to 30 MHz λ λ2 (c) 3  to 30 kHz (d) 100 kHz to 1 MHz. 34. An AM radio station operating at 630 kHz is (c) l (d) l2 λ2 λ permitted to broadcast audio frequencies up to 6 kHz. The band pass filter in its modulation 27. A repeater is a combination of circuit can retain the frequencies (a) receiver and modulator. (b) receiver and transducer. (c) receiver and transmitter. (d) receiver and amplifier.

Communication Systems   147 (a) 636 kHz, 630 kHz (a) Sky waves are not used in long distance communication. (b) 12 kHz, 6 kHz (c) 1260 kHz, 6 kHz (b) Their propagation takes place by total internal reflection. (d) 1260 kHz, 630 kHz. (c) Sky waves support the so-called AM 35. A transducer, in communication system, is a band. device that (d) The frequency of sky waves ranges (a) is a part of the antenna. typically from 3 MHz to 30 MHz. (b) is a combination of a receiver and a 41. The length of antenna to transmit waves of 1 transmitter. MHz will be (c) converts audio signals into video signals. (a) 3 m (b) 15 m (d) converts physical variable into (c) 30 m (d) 300 m. corresponding variations in the electrical signal. 42. Amplitude modulation has 36. A TV tower is 120 m high. How much more (a) one carrier. height is to be added to it, if its coverage range is to become double ? (b) one carrier with high frequency. (a) 120 m (b) 240 m (c) one carrier with two side band frequencies. (c) 360 m (d) 480 m. (d) one carrier with infinite frequencies. 37. A message signal of frequency 10 kHz and 43. The waves used for line-of-sight (LOS) peak value of 10 volts is used to moderate a communication are carrier of frequency 1 MHz and peak voltage 20 volts. The modulation index and side (a) space waves (b) sky waves bands produced are (c) ground waves (d) sound waves. (a) 0.4 and 1200 kHz, 990 kHz 44. The frequencies that are reflected and transmitted at ionospheric layer respectively (b) 0.5 and 1010 kHz, 990 kHz are (c) 0.2 and 1010 kHz, 1000 kHz (a) 3 kHz and 10 MHz (d) 0.5 and 1500 kHz, 1000 Hz. (b) 10 MHz and 40 MHz 38. The wireless communication frequency (c) 10 MHz and 20 MHz bands for Cellular Mobile Radio (mobile to base station) are in range of (d) 35 MHz and 70 MHz. (a) 76–88 MHz (b) 900–1000 MHz 45. The gap between the frequency of the side bands in an amplitude modulated wave is (c) 896–901 MHz (d) 896–1000 MHz. (a) twice that of the carrier signal. 39. A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency (b) twice that of the message signal. 2 MHz. The frequencies of the resultant signal is/are (c) the same as that of the message signal. (d) the same as that of the carrier signal. (a) 2005 kHz, 2000 kHz and 1995 kHz 46. A transmitting antenna at the top of a tower has a height of 20 m. For obtaining 40 km as the (b) 2000 kHz and 1995 kHz maximum distance between the transmitter and receiver for satisfactory communication (c) 2 MHz only in LOS mode, the height of receiving antenna should be (radius of the earth R = 64 × 105 m) (d) 2005 kHz, and 1995 kHz. 40. Which of the following is incorrect about sky (a) 30 m (b) 35 m waves ? (c) 40 m (d) 45 m.

148 PHYSICS 47. Choose the correct statement : 51. For commercial telephonic communication, the frequency range for speech signals is (a) In amplitude modulation the amplitude of the high frequency carrier wave (a) 50 Hz to 1000 Hz is made to vary in proportion to the amplitude of the audio signal. (b) 3000 Hz to 4500 Hz (b) In amplitude modulation the frequency (c) 1000 Hz to 2000 Hz of the high frequency carrier wave is made to vary in proportion to the (d) 300 Hz to 3100 Hz. amplitude of the audio signal. 52. The role of envelope detector in an AM receiver is to (c) In frequency modulation the amplitude (a) retrieve the message signal. of the high frequency carrier wave is made to vary in proportion to the (b) rectify the AM signal. amplitude of the audio signal. (c) modify the AM signal. (d) modulate the message signal. (d) In frequency modulation the amplitude 53. Speech signal is in the range of of the high frequency carrier wave is made to vary in proportion to the (a) 3700 to 7000Å wavelength frequency of the audio signal. (b) 20 Hz to 20 kHz frequency 48. The process of superimposing message signal (c) 300 to 3100 Hz frequency on high frequency carrier wave is called (d) 540 to 1600 kHz frequency. (a) Amplification (b) Demodulation 54. Wavelength of the wave with 30 MHz frequency is (c) Transmission (d) Modulation. 49. If the height of TV tower is increased by 21%, (a) 1 cm (b) 10 cm the transmission range is enhanced by (c) 100 cm (d) 1000 cm. (a) 10% (b) 5% 55. To transmit a signal of frequency, wm, with a carrier frequency, wc , in AM transmission, the (c) 15% (d) 25%. bandwidth of the filter and amplifier is 50. The range of a communication system can be (a) wm (b) 2wm extended by a (c) wc (d) wc – wm. (a) modulator (b) transmitter (c) demodulator (d) repeater. CASE BASED QUESTIONS Case Study–1 involvement at every stage of communication was Read the following passage and answer the necessary. However, in modern communication questions that follow: systems, the information is first converted into electrical signals and then sent electronically. This Communication of information is a basic human has the advantage of speed, reliability and possibility activity. A person wishes to tell something or give of communicating over long distances. You are using a message to another person sitting near him. Then these everyday such as telephones, TV and radio he speaks and transmits sound waves through air transmission, satellite communication etc. medium or channel. On the other side, the other person receives the message by listening through his/ 1. Communication is the process of her ears. This method will fail if distance between the two persons is large. In early days, long distance (a) keepting in touch messages were carried by a messenger on foot/horse/ cart etc. People also used coded signalling methods (b) exchange information through smoke or flags or beating of drums. Human (c) broad casting (d) entertainment by electronics.

Communication Systems   149 2. Which of the following is the element of a 4. In space communication, the sound waves communication system? can be sent from one place to another: (a) Transmitter (a) through space (b) Channel (c) Receiver (b) through wires (d) All of these. (c) by superimposing it on undamped electromagnetic waves 3. The term used “to collect the information about an object and a place without physical (d) by superimposing it on damped contact” is called: electromagnetic waves. (a) modulation 5. Television signals are (b) communication (c) amplification (a) Amplitude modulated (d) remote sensing. (b) Frequency modulated (c) Phase modulated (d) Both frequency and amplitude modulated.

150 PHYSICS Answers UNIT 1:  ELECTROSTATICS 1. (c) 2. (c) 3. (d) 4. (a) 5. (c) 6. (b) 7. (b) 8. (d) 9. (a) 10. (a) 11. (b) 12. (b) 13. (b) 14. (c) 15. (a) 16. (b) 17. (b) 18. (a) 19. (d) 20. (d) 21. (c) 22. (c) 23. (c) 24. (d) 25. (d) 26. (a) 27. (a) 28. (b) 29. (a) 30. (b) 31. (d) 32. (c) 33. (a) 34. (b) 35. (a) 36. (c) 37. (b) 38. (a) 39. (b) 40. (a) 41. (b) 42. (b) 43. (c) 44. (d) 45. (a) 46. (d) 47. (a) 48. (b) 49. (c) 50. (c) 51. (d) 52. (a) 53. (c) 54. (a) 55. (b) 56. (a) 57. (a) 58. (b) 59. (c) 60. (d) 61. (a) 62. (a) 63. (c) 64. (c) 65. (c) 66. (b) 67. (a) 68. (a) 69. (b) 70. (d) 71. (b) 72. (b) 73. (b) 74. (d) 75. (c) 76. (a) 77. (b) 78. (b) 79. (a) 80. (d) 81. (a) 82. (d) 83. (c) 84. (d) 85. (d) 86. (a) 87. (d) 88. (a) 89. (d) 90. (d) 91. (c) 92. (b) 93. (d) 94. (a) 95. (b) 96. (c) 97. (c) 98. (b) 99. (b) 100. (d) 101. (c) 102. (c) 103. (d) 104. (a) 105. (c) 106. (c) 107. (a) 108. (b) 109. (c) 110. (d) 111. (a) 112. (b) 113. (a) 114. (c) 115. (d) 116. (a) 117. (b) 118. (b) 119. (d) 120. (d) 121. (b) 122. (b) 123. (a) 124. (c) 125. (c) 126. (c) 127. (a) 128. (c) 129. (c) 130. (a) 131. (b) 132. (b) 133. (c) 134. (b) 135. (a) 136. (a) 137. (b) 138. (a) 139. (d) 140. (d) 141. (a) 142. (a) 143. (a) 144. (b) 145. (d) 146. (c) 147. (a) 148. (b) 149. (a) 150. (a) 151. (b) 152. (b) 153. (d) 154. (d) 155. (c) 156. (d) 157. (c) 158. (c) 159. (d) 160. (b) 161. (d) 162. (c) 163. (c) 164. (a) 165. (d) 166. (b) 167. (c) 168. (d) 169. (b) 170 (c) 171. (a) 172. (a) 173. (d) 174. (b) 175. (b) 176. (a) 177. (b) 178. (c) 4. (a) 5. (c) Case Study–1 4. (a) 5. (d) 1. (b) 2. (c) 3. (b) 4. (c) 5. (b) Case Study–2 1. (a) 2. (c) 3. (b) Case Study–3 1. (b) 2. (d) 3. (b) 150

ANSWERS   151 Case Study–4 2. (d) 3. (b) 4. (d) 5. (c) 2. (d) 3. (a) 4. (c) 5. (c) 1. (c) 2. (a) 3. (d) 4. (d) 5. (a) Case Study–5 1. (d) Case Study–6 1. (c) HINTS/SOLUTIONS 1. Electric charge can exist only as an integral 9. When the charges are present in air or multiple of charge on an electron (– e) vacuum, i.e., q = ± ne, where n is an integer.  1  q1q2 2. Charge less than the charge on an electron F =  4pε0  r2 , (i.e., e = 1.6 × 10–19 C) is not possible.   3. The net force is zero as the three forces are When the same charges are placed at the same represented by the three sides of a triangle distance in a medium of relative permittivity taken in the same order. er or K, → →→ force between the charges is given by, 4. Torque,  t = p × E .  1  q1q2  4pε0εr  r2 → 1   q1q2  rˆ F′ =   ,   r2  5. F =  4pε0   where er = 5 e0 =  1  q1q2 F = 5N = 1 N. 4pF  r2 F′ = εr 5 SI units of e0 will be N–1 m–2 C2. 10.        q = ne 6. F =  1  q1q2 = 106 × 1.6 × 10–19 C  4pε0  r2   = 1.6 × 10–13 C or e0 =  1  q1q2 12. Fm = F0  4pF  r2 K = (ML T–2)–1 ( AT )2 = [M–1 L–3 T4 A2]. In case of metals, K = ∞, therefore Fm → 0 in L2 the thickness of sheet. So, force between the → charges will decrease. 7. When an electric dipole of moment p is held q1q2 r2 at an angle q with the direction of a uniform 13. F∝   or  F ∝ r–2 →→ It is of the form, y ∝ x, and represents a straight line passing through the origin, as electric field E , a torque t acts on the dipole, given as shown in (b). → →→ 1 14. F ∝ r2 t = p × E = pE sin q, in magnitude. 1 8. The circuit being incomplete, no current It is of the form, y ∝ x , and represents a flows through its body. Hence, the parrot is not affected practically. rectangular hyperbola, as shown in (c).

152 PHYSICS 15. On introducing a dielectric slab between the 26. Capacitance of a parallel plate Capacitor with charges, the electrostatic force between them decreases. a conducting slab of thickness t is given by, 17. Positive charge implies loss of electrons and C′ = ε0 A negative charge implies gain of electrons. d−t Theoretically speaking, mass of sphere with charge – q coulomb will be more. Also, the Here, d variation in their masses depends on the t = 2 magnitude of q. \\ ⇒ C′ = ε0 A = 2C d 18. Both the charges exert equal and opposite C′/C = 2 : 1. forces on each other in accordance with 28. Charge resides on the surface of a sphere. →→ 30. For the system to be in equilibrium, net force Newton’s third law of motion (F12 = − F21) . on Q = 0. 19. F0 =ε QqQ Fm ε0 So, the required ratio is e : e0. dd 20. We have, QV = Q × Work done \\ Qq + QQ = 0 4pε0 d 2 4pε0 (2d)2 Charge = work done or energy Q ⇒ q + 4 = 0 21. Electron gains K.E. as it jumps across a pot. Q difference of V volts. ⇒ q = – 4 . ⇒ eV = 1 Mv2 31. The electric field or electric intensity (E) due 2 to two oppositely charged infinite planes is: 2eV (a) s at any point between the plates and ⇒ v = M . ε0 22. Charge always resides on the outer surface of (b) zero for all external points. a sphere. Fe ke2 r2 So, the force on P is same as that on Q. Fg r2 Gm 23.    = × 32. (a), (c), (d) Effective capacitance across the 9 × 109 × (1.6 × 10−19 )2 combination (5 capacitors in series, = 6.67 × 10−11 × (1.67 × 10−27 )2 each 2 mF) is given by, 1 = 5  ⇒ C = 2     C 2 5 mF ; 1036 11 24. F = 9 × 109 × (3 + 3)µC × (3 − 3)µC (b) C2 = 7 ⇒ C2 = 7 mF (first row) (second row) r2 1 3 = 9 × 109 × 6µC × 0 = 0. Also, C3 = µF  r2 Effective capacitance(C) is given by (two rows 25. Capacity of a parallel plate capacitor with air in parallel), as dielectric is, 1 1 =1201 3 7 ε0 A C = + d C = 10 21 where A is area of each plate and d is distance ⇒ C = µF . between the plates.

ANSWERS   153 (c) C1, C2 and C3 are effective capacitors of → each row, and then the three rows are in parallel. 39. To orient the dipole at an angle with E , work has to be done on the dipole which is stored 1 1 µF as P.E. of the dipole, given by C1 4 = 4 ⇒ C1 = →→ P.E = – p . E . Also, C2 = 1 µF  and C3 = 1 µF 40. Potential at the centre = potential on the 4 2 surface. \\ C = C1 + C2 + C3 41. Radius of the big drop, R = 4r =  1 + 1 + 1  µF = 1 mF   4 pR3 = 64 × 4 pr 3   4 4 2  3 3  33. Inside the sphere at any point, E = 0. As R becomes four times and C ∝ r, therefore, 34. Force, F = qE capacity becomes four times. C + 3/2 This force due to the electric field acts 42. P Q upwards. 3 mF 3 mF 35. E1 > E2. Effective capacitance between P and Q is 1 mF. 36. \\ 1 = 1 + 1 + 1 3 + 3 3 2 mF 4 mF C 2 PQ 2 2 = 1 2C + 3 = 3 3 3 mF ⇒ 1– 2 mF 4 mF ⇒ 2C + 3 = 6 P Q 3 4 mF 2 3 mF 8 mF ⇒ C = µF = 1.5 mF. \\ 1 = 1 + 1 + 1 1 C 4 3 8 43. Due to an electric dipole, E ∝ r3 = 6 +8+ 3 = 17 ⇒ E ∝ r–3. 24 24 44. At the centre of a charged ring of radius r, or C = 24 µF . 17 E = 0. 37. → 45. Electric field intensity E acts along the dipole axis. So, q = 0°. 46. On the surface of an irregularly shaped charged conductor, electric intensity may be different at different points, but electric potential is same at every point. 47. For a soap bubble, ⇒ C = (1 + 1)mF = 2mF. C ∝ r 38. Q = VC Q1 = VVCC=21 CC=21 r1 . i.e., capacitance of a sphere is directly \\ Q2 r2 proportional to its radius. 48. Charge remains conserved.

154 PHYSICS 1 57. 49. As V ∝ r and r is doubled, so potential C4 becomes half, equal to 8 volt. 50. C= ε0 A = 10 F AB d C1 C2 C3 When the insulating medium separating the C5 two conductors in a capacitor is other than air, its capacity becomes K times the capacity The given arrangement shows a balanced in air. Wheatstone bridge circuit. \\ C′ = KC = 6 × 10 F = 60 F. C1 = C4 = C =1 q 40 C C3 C5 C 52. C = VA − VB = {10 − (− 10)} V \\ 1 = 1 + 1 =C2 ⇒ CS1 = C CS1 C C 2 40 C = 20 V = 2F. 1 1 +1 =2 C CS2 C C C 2 53. Capacitance of a sphere is directly Similarly, =   ⇒  CS2 = proportional to its radius. So, effective capacitance 4 Given, 4 pR3 = m × 3 pr 3 Cp = CS1 + CS2 = C + C = C. ⇒ 3 2 2 R = m1/3r 58. A capacitor works in A.C. circuits only as it But C ∝ radius blocks D.C. \\ C′ = m1/3 C. 59. Both the spheres, hollow copper sphere and 54. Let R be the radius of the big drop and r is solid copper sphere, have the same radius, radius of the small drop. hence same capacity. \\ pR3 = m × 4 pr 3 Hence, charge q (= CV) must be same for the 3 spheres. or R = (m)1/3 r 60. The capacitance of a parallel plate capacitor is mq mq directly proportional to the dielectric constant V1 = 4pε0R = 4pε0 (m)1/3 r (er) of the insulating medium between the plates i.e. C ∝ er. = m2/3 V or V1 = m2/3 V. C = ε0dεr , er = 1 in case of air or vacuum 55. If V is p.d. between the plates, ε0 A V = Ed Also, C = d−t , which gives C > C0. ⇒ V = 20 V = 1000 NC–1. So, the presence of conducting slab between E = d 2 cm the plates of a capacitor also increases the capacitance. Hence, (a), (b) and (c) are 56. incorrect. CC C/2 61. We have, = 1 = 1 + 1 CS 6 4 CC C/2 2+3 5 = 12 = 12 We have, Cs = C/2 + C/2 = C.

ANSWERS   155 ⇒ Cs = 12 µF kq(Q − q) 5 69. F = r2 So, Q = VCs = 120 × 12 C = 288 µC   = k (qQ − q2 ) 5 r2 \\ V4 = Q = 288µC = 72 V. Differentiating F with respect to q we get, 4µF 4µF dF k [Q 2q] 62. The three capacitors are connected in parallel. dq = r2 − \\ Equivalent capacitance, C1 is given by, dF For maximum value of F, dq = 0 C1 = C + C + C or Q – 2q = 0 or C1 = 3C. Q 63. q = VC = 200 × 4 × 10–6 C or q = 2. = 8 × 10–4 C Energy produced gets stored as heat energy 70. Imagine a unit positive charge being kept at O. Net force (f1) due to Q and 3Q acts along 1 1 × 8 × 10–4 × 200 J OA. Similarly, net force (f2) due to 2Q and 4Q = 2 qV = 2 acts along OB. The resultant acts along the bisector of ∠AOC i.e., parallel to BC. = 8 × 10–2 J = 0.08 J. AC 64. Capacitance, C = Aε0 d f1 R f2 It is independent of the material of the plates. 65. g is earthed, hence has zero potential. O 66. Charge on the other plate is – 20 C. 67. It is a case of a balanced Wheatstone bridge. DB The capacitor, having 2 C capacitance, has no role to play. The circuit is equivalent to, 71. →→ = 3000 N = 1000 NC­–1 3C E = F/q CC C/2 \\ DV = Ed 1 P Q= P Q = 1000 × 100 V = 10 V. CC C/2 72. As one of the plates is removed, E becomes half. \\ Equivalent capacitance C1 is given by, C1 = C + C = C. So, F(= qE) becomes half. 2 2 73. Energy gained by an electron as it is 68. C = Aε0 , where t is thickness of the copper accelerated through a potential difference of d−t 1 volt = 1 eV foil. = 1.6 × 10–19 C × 1J Given, 1C \\ d – t  d = 1.6 × 10–19 J. C′ = A ε0 = C. 74. The bird is not affected practically as no d current flows through its body; it does not form a complete circuit.

156 PHYSICS 75. ne = Q = 6.4 × 10–9 C  q 1 6.4 × 10−9 C (ii) E =  4pε0  R2  (r = R, surface point) or n = 1.6 × 10−19 C   = 4 × 1010. (iii) E = 0 (r < R, inside point) 76. Mass of 1 electron = 9.1 × 10–31 kg Mass of 4 × 1010 electrons The correct variation is shown by (c). = 4 × 1010 × 9.1 × 10–31 kg 84. Electric intensity due to uniformly charged sphere, = 36.4 × 10–21 kg (i) E ∝ r(r < R, inside point) = 3.64 × 10–20 kg 1 77. DV = E D d (ii) E ∝ r2 (r > R, outside point) 1 (iii) E ∝ R2  (r = R, surface point) = 10 × 300 = 30 V The correct variation is shown by (d). 100 1 85.      C = 4pe0 R 78. E ∝ r2 . ⇒ C = 9 × 109 × 10–6 m R = 4pε0 1 79. V ∝ r . = 9000 m = 9 km 80. Electric potential at any point inside the 86.     1 = 1 + 1 + 1 + 1 4 CS C C C C =C conductor is constant and equals the potential 4 on the surface of the conductor. ⇒ CS = C = 1 F (Given) Sq ⇒ C = 4 F. 81. We know, fE = ε0 87. For equilibrium, net force on Q = 0 fES1 Q/ε0 As per given condition, fES2 = Q/ε0 =1 \\ kQQ + kqQ = 0 (2r)2 r2 So, the reqd. ratio is 1. kQ  Q q r2  4 l1 or    + = 0 82. E = 2pε0r  i.e. E ∝ r Q Where E is electric intensity due to an which gives, 4 + q = 0 infinitely long thin wire at a distance r, l being the linear charge density of the wire. Q or q = – 4 1 88. Lines of force start from positive charge and E ∝ r end on negative charge. It is of the form, y ∝ 1 . The graph is a DV x 89.    X = e0 L Dt rectangular hyperbola, as shown in (d). = (N–1C2m–2) × L × ML2T −2 Q×T 83. Electric field (E) due to a uniformly charged conducting spherical shell to radius R at a distance r from its centre is given by: = (ML–2)–1 Q2 × L–2 × L × ML2r −2 Q×T (i) E =  q  1  (r > R, outside point)  4pε0  r2 Q   = T or X = current.

ANSWERS   157 90. Since the plates have been connected 96. K = Cm 105 = 2.1 alternately, (+) plates of all (n – 1) capacitors C0 = 50 are joined to one point and (–) plates of (n – 1) capacitors are joined to the other point. This 97. Since the number of free electrons entering P per second equals the number of electrons means we have (n – 1) capacitors joined in leaving P, there is no net charge inside P. parallel. So, electric flux remains zero or remains \\ Cp = (n – 1) y. constants. 91. In series, V = V1 + V2 99. Effective capacitance of the circuit is given by,   = (3 + 3) Volt = 6 Volt C1′ = 1 + 1 = 2 = 2 Cp Cp Cp 2C 92. Electric intensity, ∑ ∑dQ ldl E = (4pε0 )r2 = (4p ε0 )r2 where Cp = 2C l(pr) ⇒ C′ = C = 25 µF (4p ε0 )r2 ∑ = So, charge on each plate of the capacitor is given by, ⇒ E = ∑ l Q = CV 93. 4ε0r . = ± 25 × 10–6 × 200 C = ± 5 × 10–3 C. P(+Q) 100. A cannot repel B. 101. K = Cm = 105 C0 50 r r Or ⇒ K = 2.1 S R 102. Initially the current flows continuously till (+Q) (+Q) such time the capacitor is fully charged. During this time, electric flux between the PSR is an equilateral ∆ and O is its centroid. plates is changing. However, when the Consider a unit positive charge being kept capacitor gets fully charged, electric flux at O. It will be repelled equally by three (+ Q reaches its maximum and current becomes each) charges at the three corners of ∆ PSR. zero. → 103. No current flows through the capacitor. Effective resistance in the circuit = 4 Ω + 1 Ω So, resultant E at O would be zero (by = 5 Ω. 1 10V symmetry). K given in the question is 4p ε0 . 104. I = 5Ω = 2A Q2 Pot. drop across 1 Ω = (2 × 1) V = 2V 94. We have, U = 2C , when the battery is Pot. drop across C and D = 8 V disconnected, total charge Q gets shared by two capacitors. However, net charge on a charged capacitor is zero. So, energy stored in each capacitor 105. Electric field lines start from (+) and end on = (Q2/C2=)2 Q8=C2 U2 (–). No two lines of force can ever intersect. 4. Electric lines of force are normal to the surface of the conductor. So, (a), (b), and (d) Work are incorrect. 95. Potential difference = Charge 2J = 20 C = 0.1 V.

158 PHYSICS 106. Particles, 1 and 2 are negatively charged and 116. Electric flux, f = E(dS) cos q particle 3 is positively charged. Acceleration of the charged particle in the electric field, Since area at a point is zero, f = 0. a = E(e/m) 118. f = E dS Since particle 3 is deflected the mostest = Force × dS = MLT −2 × L2 experiences highest acceleration. Also, E Charge Q being same for all particles, (e/m) is highest for particle 3. or    f = [ML3 T–2 Q–1] 107. Q = CV  or  Q ∝ V. 119. Electric flux depends only on charge enclosed and not on the radius of the sphere. 108. Electric field due to a point charge at its own 120. fE = 1 Sq location is zero. ∈0 109. Work done in charging a capacitor is stored in Here, Sq = 3mC – 3mC = 0 it in the form of electrostatic energy, given by E = 1 CV 2 = 1 QV = 1 Q2 So, fE = 0. 2 2 2 C 121. Potential gradient = Potential length Where Q is in coulomb, V is in volt and C is in = Vm–1. farad and energy is in joule. 1 = 1 + 1 + 1 = 1+ 2 + 3 110. Charge on the electron = 1.6 × 10–19 C 122.    CS 12 6 4 12 1 or CS = 2 mF ∴ 1 C = 1.6 × 10−19 Also, Cp = (12 + 6 + 4) mF = 22 mF = 6.25 × 1018 electrons. CS 2 1 111. If ball 1 is neutral, balls 2 and 3 may be \\ Cp = 22 = 11 . positive, balls 4 and 5 may be negative. This clearly satisfies the given condition. 123. Energy gained by the electron = eV 112. An electric dipole kept in a non-uniform where e is in coulomb and V is in joule per coulomb. electric field will experience force as well as joule torque. So, eV is in  coulomb × coulomb  joule.   113. Electric intensity = Force Charge 124. Under the influence of electric field, dielectric [MLT −2 ] gets polarised, therefore, the field lines are [E] = [AT] = [MLT–3 A–1]. correctly shown in (c). 114.   WPQ = – q (VQ – VP) 125.    1 = 1 + 1 + 1 = – q (– Er) C 2 3 6 = q Er  dV E  3+2+1 dr  = 6 = 1 mF = − \\ Charge on each capacitor = CV = 1 × 10 mC = 10 mC 115. For a line charge, E = l 1 CV2 linear charge density 2p ∈0 r , where l is 126. Energy stored by the capacitor = 2 1 1 × 4 × 10–6 × (400)2 J ⇒ E ∝ r . = 2 = 32 × 10–2 J = 0.32 J

ANSWERS   159 127. Cp = (2 + 2 + 2) mF = 6 mF. 137. 1 Volt = 1 joule per coulomb 128. The given arrangement is equivalent to three 138. Inside the hollow sphere, V = constant capacitors joined in parallel. = Potential on the surface of the sphere 1 \\ Cp = 3ε0 A d Outside the sphere, V ∝ r 1 So, the correct graph is (a). 2 129.    E= CV 2 139.   Vnet = V1 + V2 = 1 ×  1 × 10− 6  × (300 × 300)J = KQ + KQ , where K = 1 2 300 r r 4pε0 = 150 × 10– 6 J = K (20 − 20) = 0 = 1.5 × 10– 4 J r 130. Energy supplied by the battery = QV 140.    fE = 1 Sq ε0 = CV2 In case of a dipole, Sq = 0. = 1 × 10− 6 × 300 × 300 141. We know, U =  1  q1q2 , as distance 300  4pε0  r12 = 3 × 10– 4 J decreases, U increases. So, when an uncharged capacitor is connected 142. Charge remains constant to a battery, on charging the capacitor only half of energy supplied gets stored in the 143. Electric intensity is force per unit charge capacitor. 144. Charge = current × time 131. The arrangement shows a wheatstone bridge. [Q] = [M0 L0 AT] As C1 = C4 = 1, the bridge is balanced. 145. E.m.f. is energy per unit charge. C3 C5 146. As per Gauss Theorem in electrostatics, 1 = 1 + 1 = 1 =1 q CS1 2 2 1 fE = ε0 ⇒ CS1 = 1 mF 147.   Eeq = 1 Eax Also, CS2 = 1 mF 2 \\ CP = effective capacitance between P ⇒ E2 = 1 E1  ⇒  E1 =2 and Q 2 E2 = CS1 + CS2 = (1 + 1) = 2 mF 148. We have, 132. A capacitor blocks D.C. and works in A.C. dV circuit only E = – dx Charge AT = – d (4x2 + 10x − 6) = – 8x – 10 134. s = area = L2 dx [s] = [M0 L– 2 TA] \\ E(x = 1m) = – 8 × 1 – 10 = – 18 Vm– 1 135. It is a case of balanced wheatstone bridge 149. There are six equal faces in a cube. circuit, so effective capacitance between P and Q is 2 mF. \\SI unfiEts)co=m6i1εn0g(qo×u1t0−t6h)rough each face (in 136. It is also a case of balanced wheatstone bridge = q × 10− 6 circuit, so effective capacitance between A 6ε0 and B is 2 mF.

160 PHYSICS 150. Energy per unit volume (u) = 1 ε0E2 =  1  Q1(Q − Q1 ) 2  4p∈0  r2 V For F to be minimum, But E = d dF 1 1 \\ u = 1 ε0E2 dQ1 = 0 =  4p∈0  r2 (Q − 2Q1 ) 2 d2   → → →→ →→ ⇒ Q – 2Q1 = 0 151. t = p × E, t is a vector product of p and E. Q1 1 152. We have, q ⇒ Q = 2 p= l pr = l 161. Metal is a good conductor, hence K = ∞. l 162. Substituting values in the following \\ r = p expression, Distance between the charges, 2r = 2l Fp =  1  × e2  1036 p Fg  4p∈0  Gmp2 \\ New dipole moment, It should be noted that in case of electrons, p′ = p × 2l = 2p Fe = (9 × 109) l p p Fg 153. Work done in charging the capacitor is stored (1.6 × 10− 19 )2 × (6.67 × 10− 11) × (9 × 10− 31)2 as electrostatic energy. 154. E = 21 CV2 = 1 QV = 1 Q2  1042 2 2 C Note: mp = 1.67 × 10– 27 kg 163. In the incomplete circuit, the current does not (a), (b) and (c) are all correct. flow through the body of the bird. 155. Electric force per unit area on the surface of a Kee K e2 charged conductor is, 164.        F = r2 = r2 2s∈20 , where s is surface density of charge. K(2e) (2e) Force is independent of separation between F′ = (2r)2 the plates but varies inversely as area of the plates. K e2 = r2 = F 156. Electric field does not exist inside the conductor. Note: Charge on an alpha particle is twice the Q Charge charge on a deuteron. 157. C = V = Potential 165. SI unit of ∈0 is given by, 158. Electric field due to a point charge at its own F =  1  q1q2 location is zero.  4p ∈0  r2 159. Coulomb’s law is valid for only point charges \\ ∈0 =  1  q1q2 → N–1 m– 2 C2 and for distances > 10–15 m.  4pF  r2 160. Given, Q = Q1 + Q2 F =  1  Q1Q2  4p∈0  r2

ANSWERS   161 166.   F∝ q1q2 171. Dielectric gets polarised, hence no field lines r2 inside the body of the dielectric. The electric lines are correctly shown in (a). F′ ∝ 2q1 × 2q2 ∝ 16F* 2 172. Toleration limit for each capacitor = 250 V  r 2 \\  No. of capacitors in series in each row ⇒ n = 16 1 kV 1000 = 250 = 250 = 4 167.   1 = 1 + 1 + 1 = 3+2+1 If there are n such rows, then total capacity CS 2 3 6 6 ⇒ CS = 1 mF = n × 2 = 16 (Given) \\ Charge flowing through 3 mF = V CS ⇒ n = 8 = 10 V × 1 mF So, total number of capacitors = 4 × 8 = 32. = 10 mC. 173. When the plates of a capacitor are joined by a 168. We have, C = 4p∈0 R, where R = 6400 km metal, both the plates of the capacitor acquire 1× 6400 × 103 the same potential. \\ C = 9 × 109 So, p.d. between the plates = 0  711 × 10– 6 F = 711 mF \\ Capacitance has a capacity given by, 169. Cp = 3 mF + 3 mF = 6 mF QQ Now, three capacitors each of capacitance V = 0 = infinity 6 mF are joined in series. 174. Because net force q1 is zero, we have \\ Equivalent capacitance between P and Q 2k q1q k q1q1 k q1q1 a2 ( 2 a)2 2a2 is given by, = = 1 = 1 + 1 + 1 1 CS 6 6 6 k = 4pε0 where 31 ⇒ q1 = 6 = 2 ⇒ q1 = 2 2 \\ CS = 2mF q1 q 170. Capacitance across BAD, C1 = 1 µF = 2 2 2 Capacitance across B and D, C2 = 1mF q1 q Capacitance across BCD, C3 = 1 µF a 2 These capacitances C1, C2 and C3 are in q a q1 parallel, so total capacitance across B and D = C1 + C2 + C3 175. We have, =  1 + 1 + 1  µF F2 = q′1 q′2 2 2 F1 q1q2 = 2mF (+ 2 − 4) (+ 6 − 4) − 4 = (+ 2) (+ 6) = 12

162 PHYSICS ⇒ F2 = −4 4. Given: q = –3.2 × 10–7 C 12 12 Charge on one electron, ⇒ F2 = – 4N e = –1.6 × 10–19 C i.e., 4N attractive force \\ Number of electrons transferred to 176. C = KC0, also K > 1 polythene piece from wool, ⇒ C increases   n = −3.2 × 10−7 = 2 × 1012 Q −1.6 × 10−19 Potential, V = C 5. Mass of each electron As Q is constant, V will decrease = 9 × 10–31 kg Q2 \\ Mass transferred to polythene = (2 × 1012) × (9 × 10–31) kg P.E. (U) = 2C = 1.8 × 10–18 kg As C increases, U decreases. 177. Nucleus of hydrogen atom has one proton Case Study–2 of charge (+ e). The revolting electron carries 2. there is a minimum permissible magnitude charge (– e). of charge As per Coulomb’s law, 3. Number of electronic charges in one coulomb, → = K(e) (− e)rˆ n = q = 1 coulomb r2 e × 10−19 coulomb F 1.6 = − Ke2 →r = 0.625 × 1019 r3 4. Number of molecules in 10 g of water 178. When two capacitors, each of 4mF are joined = 6.02 × 1023 × 10 in series, then 18 1 = 1 + 1 = 1 Note: Number of molecules in 1 g of water CS 4 4 2 Avogadro number = CS = 2mF = Atomic weight If third capacitor of 4mF is joined in parallel A molecule of water (H2O) is made up of two across the combination, then hydrogen atoms and one oxygen atom. Each hydrogen atom contains one electron and one Cp = (2 + 4) mF oxygen atom contains 8 electrons. Thus, each = 6 mF molecule of water has 10 electrons. Case Study–1 \\ Number of free electrons in 10 g 2. Given: n = 30, q = ?,  e = –1.6 × 10–19 C We know, of water q = ne = 6.02 × 1023 × 10 × 10 18 e = 30 × (–1.6 × 10–19 C) = –4.8 × 10–18 C = 3.344 × 1024 3. q = n|e| Negative charge, q = ne or n = q = 8.4 × 1018 = 52.50 = 3.344 × 1024 × 1.6 × 10–19 C |e| 1.6 × 10−19 = 5.35 × 105 C As n is not an integer, the given value of 5. Charge given out in 1 second charge is not possible. = 109 × 1.6 × 10–19 C = 1.6 × 10–10 C

ANSWERS   163 Time required to accumulate a charge of 1 C 1 d [q(Q − q)] = 0 4πε0 r 2 dq 1C or    = 1.6 × 10−10 C s−1 = 6.25 × 109 s 6.25 × 109 or     d [q(Q – q)] = 0 365 × 24 × 3600 dq = years = 198.2 years or    q(– 1) + (Q – q) = 0 Case Study–3 or Q = 2q  or  Q =2 2. q1 = q2 = 2 C, r = 2 × 103 m, F = ? q 2× 2 Case Study–4 (2 × 103 )2 F = 9 × 109 N 3. V= 1q 4πε0 r = 4.5 × 103 N = 4.5 kN 9 × 109 × 79 × 1.6 × 10−19 6.6 × 10−15 3. q1 = q2 = q (say), = V r = 5 Å = 5 × 10–10 m, = 1.7 × 107 V F = 3.7 × 10–9 N, n = ? 1 q 1 (q)(q) = 1 (ne)(ne) 4. V = 4πε0 r 4πε0 r2 4πε0 r2 F = = 1 n2e2 9 × 109 × 4 × 10−7 40 r2    = 0.09 V [\\ Quantisation of Charge: q = ne]    = 4 × 104 V 5. E = 20 N C –1, V = 10 J C–1, r = ?, q = ? 3 .7 × 10–9= 9 × 109 n2 × (1.6 × 10−19 )2 E = 1 q = 20 ...(i) (5 × 10−10 )2 4πε0 r2 ...(ii) or n2 = 3.7 × 10−9 × 25 × 10−20 =4 and V = 1 q = 10 9 × 1.6 × 1.6 × 10−29 4πε0 r or n = 2 Dividing (ii) by (i), we get 4.      F= 1 q1q2  ...(i) r = 10 m = 0.5 m 4πε0 d2 20 Again, F = 1 q1q2 ...(ii) Again, from equation (ii), we get 3 4πε0 r2 r2     9 × 109 × q = 10 d2 0.5 Dividing (i) by (ii), 3 = or q = 1 ×5C × 109 9 or r = 1.732 d = 5.56 × 10–10 C 5. F= 1 q(Q − q) Case Study–6 4πε0 r2 5. Q = CV Q and r are fixed. So, F is a function of q. For given V, Q ∝ C Q is more for A For F to be maximum, dF =0 \\  C is more for A. dq

164 PHYSICS UNIT 2:  CURRENT ELECTRICITY 8. (d) 16. (b) 1. (d) 2. (a) 3. (d) 4. (c) 5. (c) 6. (b) 7. (b) 24. (a) 9. (c) 10. (a) 11. (a) 12. (d) 13. (c) 14. (a) 15. (a) 32. (a) 17. (c) 18. (b) 19. (a) 20. (a) 21. (c) 22. (a) 23. (d) 40. (a) 25. (b) 26. (a) 27. (a) 28. (c) 29. (d) 30. (a) 31. (c) 48. (d) 33. (b) 34. (a) 35. (a) 36. (c) 37. (a) 38. (a) 39. (c) 56. (c) 41. (a) 42. (d) 43. (d) 44. (b) 45. (d) 46. (a) 47. (a) 64. (b) 49. (c) 50. (a) 51. (d) 52. (b) 53. (c) 54. (a) 55. (a) 72. (a) 57. (a) 58. (b) 59. (b) 60. (c) 61. (a) 62. (d) 63. (a) 80. (b) 65. (d) 66. (c) 67. (a) 68. (a) 69. (a) 70. (b) 71. (a) 88. (c) 73. (b) 74. (b) 75. (a) 76. (d) 77. (a) 78. (d) 79. (c) 96. (c) 81. (c) 82. (b) 83. (d) 84. (a) 85. (b) 86. (a) 87. (d) 104. (c) 89. (d) 90. (a) 91. (c) 92. (a) 93. (d) 94. (b) 95. (c) 112. (a) 97. (d) 98. (c) 99. (b) 100. (d) 101. (b) 102. (b) 103. (a) 120. (c) 105. (d) 106. (b) 107. (d) 108. (c) 109. (b) 110. (c) 111. (a) 128. (d) 113. (d) 114. (b) 115. (c) 116. (b) 117. (b) 118. (d) 119. (d) 136. (a) 121. (a) 122. (b) 123. (b) 124. (a) 125. (c) 126. (d) 127. (a) 144. (a) 129. (c) 130. (c) 131. (a) 132. (d) 133. (c) 134. (b) 135. (b) 152. (c) 137. (a) 138. (c) 139. (c) 140. (a) 141. (d) 142. (b) 143. (c) 160. (d) 145. (b) 146. (a) 147. (a) 148. (a) 149. (a) 150. (b) 151. (b) 168. (c) 153. (b) 154. (b) 155. (c) 156. (c) 157. (d) 158. (b) 159. (b) 176. (a) 161. (d) 162. (c) 163. (a) 164. (d) 165. (d) 166. (a) 167. (a) 184. (d) 169. (c) 170. (b) 171. (b) 172. (d) 173. (c) 174. (d) 175. (b) 192. (c) 177. (a) 178. (d) 179. (d) 180. (c) 181. (c) 182. (c) 183. (a) 200. (a) 185. (d) 186. (a) 187. (b) 188. (c) 189. (a) 190. (c) 191. (c) 208. (b) 193. (c) 194. (b) 195. (d) 196. (c) 197. (d) 198. (c) 199. (c) 216. (c) 201. (a) 202. (a) 203. (c) 204. (a) 205. (b) 206. (a) 207. (a) 209. (d) 210. (a) 211. (c) 212. (a) 213. (a) 214. (d) 215. (d) 217. (c) 218. (b) 219. (b) 220. (d) 221. (b) 222. (b) Case Study–1 3. (a) 4. (a) 5. (c) 1. (a) 2. (c) 3. (d) 4. (b) 5. (c) Case Study–2 3. (b) 4. (d) 5. (c) 1. (a) 2. (d) Case Study–3 1. (d) 2. (d)

ANSWERS   165 HINTS/SOLUTIONS 1. Charge is transported by both negative and Current Area positive ions within the cell. 8. Current density = 2. P = EI [Current density] = [M0 L– 2 T0 A] 3. Slope of the graph = I = 1 = tan q 9. Equivalent circuit is given by, V R 2W 2W ⇒ R = cot θ 2W 2V I 2W 2W q V 2W 2W O 2W 2W 4. Polarization is a major defect in Voltaic cell. ⇒  5. q = ne. Also, 1 A = 1C 2V 2W 1s ∴ n = q 4W e 4W ⇒  = 1C 1.6 × 10− 19 C 2V 2W = 6 .25 × 1018 6. Following three resistances are possible. ⇒  (i) r ≡r r 2V 2W 2W (ii) ≡r ∴ I = 2V = 0.5 A 2 10. A 4Ω r (iii) r r 2W 2W ≡ 2r 7. Four different combinations are possible. 2W 2W (i) rrr ≡ 3r B 2W 2W r (ii) r ≡r ⇒  A B r 3 2W 2W 4W r (iii) r r ≡ 3r ⇒  A 2W 2W B 2 ⇒  A r ∴ 2 W 4/3 W B rr (iv) ≡ 2r R = 2W + 4Ω = 10 Ω 3 3 3

166 PHYSICS 11. 2 W parallel to 2 W gives 1 W resistance which is = 4  ρl  in series with 1 W, hence gives 2 W resistance.  A  2W or R′ = 4r PQ 14. It satisfies the condition of balanced 2W Wheatstone bridge, i.e., 2 W and 2 W in parallel give an equivalent P = R resistance of 1 W. Q S 12. Volume = pr2l = Therefore, the given circuit is equivalent to, π RR 4 = × D2l AB π  D 2 RR 4  2  2R = × × l′ or l′ = 4l...(i) ≡ A B R = ρl = 4ρl ...(ii) 2R πD2 πD2 R ≡ A B 4 15. The given circuit is a balanced Wheatstone bridge circuit. R′ = 4ρl′ = 4ρ × 4l  D 2 πD2 π  2  4 rr P rQ [Using (i) and (ii)] rr = 16  4ρl   πD2  = 16 R So, the equivalent resistance between P and Q is r. ⇒ R′ = 16 W( R = 1 W) 16. The equivalent circuit of the given circuit is, In general, if radius of cross-section of the metallic wire becomes n times, its resistance becomes 1 times. In this case, n = 1. 1W 1W  n4  2 A 1W ρ l 1W 1W B A 1W 13.  r = Volume of wire = Area × length The circuit between A and B′ is a balanced = A × l Wheatstone bridge circuit and has a resistance of 1 W. This is in series with 1 W. = A × 2l 2 So, equivalent resistance between A and B is 2 W. ∴ R′ = ρ × 2l A 17. For a perfectly ohmic conductor, graph between I and V is a straight line passing 2 through the origin.

ANSWERS   167 18. Equivalent circuit is, 1W 1W 21. 2 W in series with 2 W gives 4 W, which in parallel with 4 W gives 2 W. Now, 2 W in series 1W 1W 1W with 2 W gives 4 W, which in parallel with 4 W gives 2 W. PQ Next, 2 W in series with 2 W gives 4 W, which 5W in parallel with 2 W gives 4 Ω. 5W 3 =P Q 1 = 1 + 1 ⇒ rp 4 2 ⇒ 5W 1 = 1+1 = 2 = 3 rp 55 5 4 rp = 5 = 2.5 W ⇒ rp = 4 Ω 2 3 19. 1 = 1+1 22. I = V = VG, where G is conductance R 84 R ∴ I = 10 × 2 = 1 + 2 = 3 = 20 A 8 8 23. Temp. of the conductor remains constant for ∴ R = 8 Ω Ohm’s law. 3 Also, I = 8 =3A 1W 1W 8/3 1W 24. 1 W 1 W  =  20. The given circuit is a balanced Wheatstone A B bridge circuit. AB 1W P = 24 = 1 2W Q 2 = R = 162 = 1 S 2 AB 1W ⇒ P = R 1 = 1 + 1 = 3 Q S rp 2 1 2 Equivalent circuit is given by, ⇒ rp = 2 Ω 3 2W 4W A B 25. Siemen (S) is the SI unit of conductance. 26. Equivalent circuit can be written as, 6 W 12 W 3W 1 1 1 1W 1W 1W 3W 8W rp 6 18 ∴ = + 3W = 3+1 = 2 A 18 9 12 V ⇒ rp = 4.5 W

168 PHYSICS 1 = 1 + 1 + 1 = 3 =1 1 = 1+ 1 + ... n times rp 3 3 3 3 rp r r ⇒ rp = 1 W = n Total resistance of the circuit r = 1 W + 1 W + 1 W + 1 W + 8 W ⇒ rp = r n = 12 W 33. Carbon has a negative temp. coeff. of ∴ I = 1122 VΩ = 1 A resistance 34. The equivalent resistance of the circuit can be found using condition of balanced 27. When closed, the whole of current flows Wheatstone bridge. through the first segment of the circuit and no ∴ Total resistance of the circuit = 2 W current flows through the rest of the circuit. It will lead to a short circuit. I = 22 V =1A Ω 28. The given circuit is equivalent to, 1W 1W 2W 2W AB 2W 4W 2W 2W 2W 35. ºA 2W 1W 2W ºA 5W B B 4V 29. Metal copper is an ohmic conductor. 1 = 1+ 1 = 1 ⇒ rp 4 4 2 30. Their e.m.f.s are added when cells are Also, connected in series. rp = 2 W ∴ E, r rs = 2 W + 2 W = 4 W 31. I = 4 V =1A 4 Ω R 36. Slope of the graph gives resistance of the conductor. R(T1) > R(T2) I = E r R+ Since resistance of the conductor increases with temperature, = E T1 > T2. [ R + r −2 R r]+ 2 R r 38. It is a case of balanced Wheatstone Bridge = E r circuit. ( R− r )2 + 2 R 1W 1W The current in the circuit will be maximum, if P 1W Q ( R − r) = 0 1W 1W ⇒ R = r. So, the equivalent resistance between P and Q 32. For smallest resistance, n resistances must be is 1 W. joined in parallel.

ANSWERS   169 39. The Wheatstone Bridge (principle) is used to 48. The wire used in heating appliances is measure unknown resistance. generally made of nichrome. 40. The given network is equivalent to Wheatstone 49. The resistance of an ideal voltmeter is infinity. Bridge circuit (balanced). Voltmeter is always connected (in parallel) across a resistor. 1W 1W P 1W → → 1W 1W Q 50. vd =– e E τ, where t is relaxation time. m ºP 2W Q 51. If n cells, each of emf E, are connected in ºP Q parallel, the effective emf of the combination 2W is E, and not n E. 1W 52. We know, I = A ne vd, where n is the electron density (number of electrons per unit volume) ⇒ A vd = const. ⇒ (pR2) V = p (2R)2 V′ 41. Resistance of a metallic conductor decreases ⇒ V′ = V with decrease in temp. 4 Also, resistance being inversely proportional 53. to relaxation time, the relaxation time increases. 42. R = ρl , R = 1 W A Volume of wire = Al = A′l′ = A (2l) 2 ∴ R′ = ρ 2l = 4ρl = 4R = 4 W A/2 A 43. The value of resistance is 41 × 103 W ± 20%. 44. It is a case of a balanced Wheatstone Bridge circuit. P = 2 = 1; R = 6 = 1 Q 4 2 S 12 2 So, no current flows through 5 W resistance. 54. Let each resistance be r. Hence, if 5 W is replaced by 10 W, the current through the main circuit remains the same as We have, 1 = n resistance of the circuit does not change. R r 45. The Wheatstone Bridge is balanced in case of ⇒ R = r (d). n 46. E/l = constant. 47. 1 = 1+1 = 3 When resistances are connected in series, rp 24 4 equivalent resistance (RS) is given by ∴ rp = 4 Ω RS = n r = n2 R 3

170 PHYSICS 55. Given, r1 + r2 = 10.5 W Total resistance of the circuit Total resistance in the circuit = (10.5 + 1.5) W = (1 + 2) W = 12 W = 3 W I = 12 V =1A ∴ I = 3 V =1A 12 Ω 2 Ω 56. Resistance of human body is about 12 kW. 62. The given arrangement can be written as: 57. Given, r1r2 =3 3W r1 + r2 Suppose r1 is burnt and gets broken such that A 2W 6W its resistance becomes 12 W. 8W B 4W So, 12 r2 = 3    r 12 + r2   ≡ A 3W ⇒ 12 r2 = 36 + 3 r2 ⇒ r2 = 4 W 8 W 12 W 5 5 B 58. 1 = 1+ 1 = 3 r rp r 2r 2r 3W ∴ rp = 2r   ≡ A 4 W B 3 r \\ Equivalent resistance between P and Q = r + 2r + r = 8r 3 3 4 W 59. Current through the circuit R= 3 8V−3V 5V   ≡ A B (6 + 4) Ω 10 Ω = = = 0.5 A R1 1+1+1 7r + 12 34r 12r The direction of current is from d to c. We have,   = = 60. Potential difference across each resistance is \\ R = 12r = 4 same. 7r + 12 3 Current through each resistor Solving we get, r = 6 W = 220 A = 1 A 2W 2200 10 Current in ammeter will be the current flowing through 4 resistances on left of 1W 2V ammeter which is, 63.  1 +1 +1 + 1  = 4 A 1W 10 10 10 10  10 61. Net emf in the circuit or e.m.f of the battery 2W = 3 V, as one cell has its terminals reversed i.e., (1.5 + 1.5 + 1.5 – 1.5) V 1W 2V ⇒ 2W

ANSWERS   171 1 1+ 1 1W 1W rp 2 2 = =1 ⇒ rp = 1 W 2V Also, rs = rp + 1 = 2 W \\ I = 2V =1A \\ I = 2 =1A 2Ω 2 64. A thermistor with negative temperature 68. 1 = 1+1 = 1 coefficient of resistance will show decrease rp 44 2 in resistance with heating due to increase in current. \\ rp = 2 W The correct variation is shown by (ii). I = 8 V =4A 2 Ω 65. In case of a thyristor, there are two values of 69. It is the case of a balanced Wheatstone voltages for one value of current on V–I graph. Bridge, hence no current flows through the galvanometer. Let current I flow through The correct variation is shown by (iv). ABC, then (2 – I) flows through ADC. 66. Total resistance of the circuit due to resistors = (6 + 7 + 4) W = 17 W \\ 2 I I = 45 =3 − 15 Internal resistance of the battery = 1 W ⇒ I = 6 – 3I Total effective resistance = 17 W + 1 W ⇒ I = 3 A 2 = 18 W I = E 70. Current through the potentiometer wire, 18 I = 4 = 1A 9 + 3 3 Pot. drop across 6 W = 6 I = 3 (Given) ⇒ I = 1 Resistance of 50 cm length = 3×1 = 3Ω 2 12 2 E 1 E = Potential drop across 3Ω = 3×1 = 1V 18 2 2 23 2 \\ = ⇒ E = 9 V 71. Potential gradient, K = 10 Vm– 1 67. The given arrangement is equivalent \\ Potential difference across the given length (l) = Kl 1W 1W = K × 20 = 10 × 20 =2V 100 100 1W 72. The Wheatstone Bridge will be most sensitive 1W 1W when it is balanced. The essential condition is, P = Q = R = S 2V 1W 73. We know, P = W = H = V2 t t R Applying condition for the balanced Also, P′ = V2 Wheatstone bridge, the given arrangement 2R can be written as,

172 PHYSICS P′ 1 81. Power, P = VI P 2 ⇒ = V2 R P = I2 R = 2 ⇒ P′ = ⇒ R1 = ( 250)2 and R2 = (250)2 100 200 So, the rate of generation of heat will be halved. V2 R V2 We have, electric energy per second = R 74. We know, P = H1 R2 H2 R1 V2 \\ = P ⇒ R1 = (250)2 200 R1 P2 200 = × 100 R2 P1 100 (250)2 So, = = =2 75. 1 W = 1 V × 1 A ⇒ H1 = 1 H2 2 = 1 A × 1 W × 1 A 82. Let r1, r2 be the resistances of the wires of = 1 A2 W different diameters. 76. Heat energy consumed by the heater in one In series combination, hour = 1000 W × 1 hour H1 = V 2t ...(i) = 1 KWh (r1 + r2 ) = 3.6 × 166 J In parallel combination, 77. H = (I2 r t) Joule V 2t r1r2 H = (I2 r) Joule H2 = ...(ii) t Rate of generation of heat, (r1 + r2 ) 78. Resistance of lamp, Dividing (ii) by (i) we get, R = V2 = 220 × 220 H2 = (r1 + r2 )2 >1 P 100 H1 r1r2 = 484 W ∴ Current through the lamp when used at ∴ H2 > H1 200 V, I= V 84.        P = 2200 W R V = 220 V 200 = 484 = 0.41 A V2 R We know, P = 79. Amount of heat produced (H) in a conductor, H = I2 R t ⇒ R = 220 × 220 = 22 W 2200 ⇒ H ∝ I2 When I is doubled, H becomes four times. 85. Same current flows through r1 and r2. Also, H ∝ Dq, Also, H = I2 r t where Dq is rise in temp. of the conductor. H1 r1 H2 r2 So, the rise in temp. becomes four times. ∴ = >1 80. Electric energy consumed = Power × time ⇒ H1 > H2 So, heating would be more across r1. = 100 × 6 × 30 Wh = 18000 kWh = 18 kWh

ANSWERS   173 87. If current flowing through 10 W is I, we have, V2 R I2R = 10 cal s– 1 or = 20 W ⇒ I2 = 10 In parallel combination, V2 =P 10 R/4 Same magnitude of current will flow through V2 R the other branch i.e., 4 W and 6 W. or 4 = P \\ Heat evolved per second across 4 W, is From (i) we get, P = 80 W given by, H = I2 × 4 V2 R = 1 × 4 94. Power, P = = 4 cal s– 1 P1 R2 V2 40 × 40 \\ P2 = R1 P 60 88. Resistance of a lamp = = 80 ⇒ R1 = 1 = P2 3 R2 2 P1 ⇒ R = Ω Current for each lamp, ⇒ P1 = 2 P2 1 I = P = 60 = 3A 95. Emf. of a car battery is about 12 V. V 40 2 If n lamps are connected in series across 240 V, 96. If E is the applied e.m.f. of the source of e.m.f. i.e., a battery of internal resistance r and R is I = 2n4R0 external resistance, then current in the circuit is given by, ⇒ n = 2430 × 2× 3 I = E × 80 (R + r) = 6 At the max. power, R = r 89. R = ρl = ρl =  4ρl  1 \\ I = E A πD2  π  D2 2r 4 \\ Max. output power R′ =  4ρl  ( 1 = R = I2 r =  E 2 × r = E2  π  2D)2 4  2r  4r 90. P = 100 W 98. R = ρ l A V = 250 V So, the bulb will fuse if pot. drop across the R′ = ρ  nl  =  ρl  n2 bulb becomes 300 V.   A  A / n  91. The safe current for a fuse wire varies directly = n2R as r3/2. 92. The safe current is independent of the length 100. Volume of the wire remains constant of the fuse wire in series with an electrical appliance. V = lA = l′ A′ 93. In series combination, the power dissipated ⇒ l (pr2) = l × π (pr)2 p2 = V =5W ∴ R′ = ρ l′ = ρ l  × 1 = R 4R A′ ρ2  ρ2 A ρ4