= 115V. = = = C/m 2
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1 SPHS Class th Physics Solution. parallel-plate air capacitor has a plate area of cm and separation 5mm. potential difference of V is established between its plates by a battery. fter disconnecting a battery, the space between the plates is filled by ebonite (K =.6). Find out final surface-density of charge on the plates. (). -7 C/m () 6. 7 C/m 5. 7 C/m (D) 9. 7 C/m Solution: Capacity of the parallel plate air capacitor 4 o 8.86 = =.77 F d 5 Final capacity of the capacitor with dielectric between the plates is C = KC =.6.77, C = 4.6 F Initial charge on the capacitor.77 = 5. 9 C Since, the battery has been disconnected, the charge remains the same, therefore the new potential difference is 9 5. V = = 5V C 4.6 The surface density of charge remains the same in both the cases, i.e., 9 5. = = = 5. 7 C/m 4. spherical condenser has cm and cm as the radii of inner and outer spheres. The space between the two is filled with a dielectric of dielectric constant. Find the capacity when the inner sphere is earthed. r R () 5 F () F 5 9 F (D) F 5 5 Solution:(D) y choice of reference potential at infinity and the potential of earthed conductor zero, the given system can be visualised as combination of two spherical capacitors, both being at same potential difference. Connection wise these may be considered to be in parallel connection. C = 4ob + ab 4k b a = F 5
2 SPHS Class th Physics Solution. charge Q is distributed over two concentric hollow spheres of radii r and R (R >r) such that the surface densities are eual. Find the potential at the common centre. () 4 Q R r R r Q 4 R r R r () (D) 4 4 R r Q R r R r Q R r Solution: () + = Q... (i) =... (ii) 4r 4R from (i) and(ii) Qr QR = = r R r R Vcentre= V + V = = 4 Q R r R r 4 r R 4. n electric dipole of dipole moment P is placed in a uniform electric field E is stable euilibrium position. Its moment of inertia about the centroidal axis is I. If it is displaced slightly from its mean position find the period of small oscillation. () PE T () I I T 4 (D) PE T T 4 I PE I PE Solution:4 () When displaced at an angle, from its mean position the mean position the magnitude of restoring torue is PE P + E + For small angular displacement sin PE The angular acceleration is, pe cos I I
3 SPHS Class th Physics Solution Where T PE I I PE 5. Figure shows two conducting thin concentric shells of radii r and r. The outer shell carries charge. Inner shell is neutral. Find the charge that will flow from inner shell to earth after the switch S is closed. s r R r () () (D) Solution:5 (D) Let be the charge on inner shell when it is earthed. Vinner = 4 r r o / i.e. charge will flow from inner shell to earth. 6. Two infinitely large sheets S and S having surface charge densities and (>) respectively are placed at a distance d. Find the work done by the electric field when a charge particle Q is displaced by a distance a (a < d) at an angle 45 with the normal of the sheets. It is assume that the charge does not affect the surface charge densities of the two plates. () ( ) ( ) Solution : 6 () Work done by the charge d d () ( ) (D) None of these d P W E. dr PQ Q
4 SPHS Class th Physics Solution ( ) d 7. soap bubble has radius R, charge Q, surface tension T. Find the excess pressure in it. π R εt - 64π R εt - () () 4 4 π R ε π R ε 8π R ε T - π R ε 4 (D) none of these Solution:7 P exces 4T R 4 R 8. charge of C is brought from to C along the path as shown by arrow in the figure. The work done is ().75 J ().6J.6J (D).75J Solution:8 (D) w 4 rf ri =.75 J 8 cm F 9. The euivalent capacitance between and is (each of the capacitors obtained is of capacitance eual to C) 6 cm C () C 5 C () C 5 (D) C 5
5 .mm SPHS Class th Physics Solution Solution: D E 5 =E D C C 5C C C C. ring has charge Q and radius R. If a charge is placed at its centre then the increase in tension in the ring is Q () () zero 4πεR Q Q (D) 4π ε R 8π ε R Solution: (D)Consider a small element, is very small. Then = R() Q Q Change on is dq = R R dq Qθ T sinθ = = 4π R 4π R Tθ = Q Q or T= 4π R 8π R T cos T cos T T sin Tcos T. Two plane mirrors and are aligned parallel to each other, as shown in the figure. light ray is incident at an angle of at a point just inside one end of. The plane of incidence coincides with the plane of the figure. The maximum number of time the ray undergoes reflections (including the first one) before it emerges out is: m () 8 () (D) 4
6 SPHS Class th Physics Solution Solution: () No. of ref = =.. glass prism of refractive index.5 is immersed in water (R.I. = 4/). The beam of light incident normally on the face is totally reflected to reach the face C, if () sin 8/9 () sin < / / < sin < 8/9 (D) None of these C Solution: () The light ray is totally reflected internally at the point P (say). Therefore the critical angle C. lso, sin c nw 4 / 8 sin c = sin 9 n / 9 g sin 8/9 C p. screen beaming a real image of magnification m formed by a convex lens is moved through a distance x. The object is the moved until a new image of magnification m is formed on the screen. The focal length of the lens is: () x m m () x m m x m m (D) None of these Solution: () In first case, +m = f and =m p f p () In the second case x nd m p () and () x p f x m = f ()
7 SPHS Class th Physics Solution m m = x/f x f = m m 4. small mirror of area and mass m is suspended in a vertical plane by means of a weightless string. beam of light of intensity I falls normally on the mirror and the string is deflected form the vertical through a very small angle. ssuming the mirror to be perfectly reflecting, obtain an expression for. I/ C () I c mg I c mg mg () (D) I c mg none of the above Solution:4 () When the light is incident on the mirror is totally reflected and there is no loss of energy. Change in momentum = incident momentum Energy Intensity = Time rea Intensity Momentum = Velocity of light Pressure, P = Force Rate of change in momentum = rea = I c = tan = P = mg (where c = speed of light) I c mg 5. n observer can see, through a pin-hole, the top end of a thin rod of height h, placed as shown in the figure. The beaker height is h and its radius is h. When the beaker is filled with a liuid up to a height h, he can see the lower end of the rod. Then the refractive index of the liuid is: h h h
8 SPHS Class th Physics Solution () 5 / () 5/ / (D) / Solution:5 () sin i sin r = m r h Eye x x + 4h = h - x m ( ) h - x + h h - x x sin r = = h ( ) h - x + h x h-x h...(i)...(ii) Using (i) and (ii), m= 5 6. The sun (diameter d) subtends an angle radians at the pole of a concave mirror of focal length f. What is the diameter of the image of the sun formed by the mirror? () f () f/ f C F M Solution:6 () Since the sun is at a very large distance, u is M very large and so (I/u) is practically zero. So, i.e., v = f v f i.e., the image of sun will be formed at the focus and will be real, inverted and diminished. Now as the rays from the sun subtend an angle radians at the pole, then in accordance with figure. d = [where d is the diameter of the image of the sun] FP f i.e. d = f 7. ray of light is incident on the left vertical face of a glass cube of refractive index, as shown in figure. The plane of incident is the plane of the page, and the cube P
9 SPHS Class th Physics Solution is surrounded by liuid (). What is the largest angle of incidence for which total internal reflection occurs at the top surface? c c () sin = μ μ - () sin = μ μ + sin = μ μ + (D) sin = μ μ - Solution:7 () Consider Point pplying Snell s law, sin = sin () ut = 9 c cos = sinc = () Elimination of between () and (), we get sin = 8. ray incident at a point as an angle of incidence of 6 enters a glass sphere of R.I. n = and is reflected and refracted at the farther surface of the sphere. The angle between the reflected and refracted rays at this surface is: () 5 () 6 9 (D) 4 Solution:8 Refraction at P. Sin 6 Sin r Sinr r= 6 P r Since r = r r = Refraction at Q Sin r Sin i Putting R= we obtain i = 6 Reflection at Q o r r 8 (r i) r r Q i
10 SPHS Class th Physics Solution = 8 - ( +6 ) = 9 9. What is the number of images of an object formed by two mirrors if the angle between the mirrors is in between /4 and /5? () () 8 6 (D) 4 Solution:9 (D) s we know, if angle between two mirrors is between and n n then the no. of images formed is n So here in the uestion it is and 4 4 no. of images will be 4.. If one face of a prism of prism angle and = is silvered, the incident ray retraces its initial path. What is the angle of incident? () 9 () 8 45 (D) 55 i D r E C Solution: s in ED D = 8 D = 6 Now as by construction D + r = 9 r = 9 6 = pplying Snell s law at surface C sini = ( )sin = = i = sin = 45. n achromatic combination of two prisms - one of uartz and other of flint glass is to be formed. If the refracting angle of the uartz prism is 4, find the refracting angle of the flint glass prism. lso find the net mean deviation produced by the combination. Given that for the flint glass : v =.694, r =.65 and for the uartz glass : v =.557, r =.54. ().8 () (D) 8.8 Solution.() For an achromatic combination, net dispersion =
11 SPHS Class th Physics Solution dispersion by one prism = dispersion by other prism. Þ m - m = m - m ( ) ( ) v r v r ( ) ( ) Þ = Þ =.64 D - D = m - - m - ( ) ( ) m m m m v r = -.64 = -.8 Net mean deviation = Dm - Dm = -.8. Find the dispersion produced by a thin prism of 8 having refractive index for red light =.56 and for violet light =.68. () 4.6 () (D) 8.6 Solution We know that dispersion produced by a thin prism = m - m ( ) v R Here m v =.68, m R =.56 and = 8. = ( ) 8 =.6.. Two plane mirrors are inclined to each other such that a ray of light incident on the first mirror and parallel to the second is reflected from the second mirror parallel to the first mirror. Determine the angle between the two mirrors. () 6 () 9 (D) 8 Solution: () Let be the angle between the two mirrors OM and OM. The incident ray is parallel to mirror OM and strikes the mirror OM at an angle of incidence eual to. From figure we have M = OC = MOM = M D a a N E d N O Similarly for reflection at mirror OM, we have MCD = CO = MOM = C M
12 SPHS Class th Physics Solution Now in triangle OC, = 8, therefore = 6 4. small filament is at the centre of a hollow glass sphere of inner and outer radii 8 cm and 9 cm respectively. The refractive index of glass is.5. Calculate the position of the image of the filament when viewed from outside the sphere. () 9 cm () 9 cm 9 cm (D) +9 cm Solution: 4 () For refraction at the first surface, u = 8 cm, R = 8 cm, =, =.5 m m m - m - = v u R = v ' 8-8 v' = 8 cm Glass O = 8 cm O = 9 cm m=.5 O ir It means due to the first surface the image is formed at the centre. For the second surface u = 9 cm, =.5, R = 9 cm m m m - m - = v u R = v 5-9 v = 9 cm Thus, the final image is formed at the centre of the sphere. 5. thin rod of length f/ is placed along the optical axis of a concave mirror of focal length f. Such that its image which is real and elongated just touches the rod. Calculate the magnification. () / () 5/ / (D) / Solution:5 () Let be the length of the image. mf Then, m= Þ = f
13 SPHS Class th Physics Solution l rod u image f/ u v lso image of one end coincides with the object, Þ u' = f f f 5f u' = u + Þ f - = Putting in mirror formula, + = Þ + = u + f + mf u f 5f + f + mf 5f f Þ = Þ m = m n object of length.5 cm is placed at a.5 f from a concave mirror where f is the magnitude of the focal length of the mirror. The length of the object is perpendicular to the principal axis. Find the length of the image. Is the image erect or inverted? () 8 cm () cm 5 cm (D) cm Solution:6 f O F.5 f The focal length F = f and u =.5 f, we have, + = or + = u v F -.5f v f = - = - v.5f f f v = f f Now, m = - v /u = - = -.5f h or or h h 5 cm h = - = - = - The image is 5 cm long. The minus sign shows that it is inverted. 7. Select a graph for concave mirror between /v and /u.
14 SPHS Class th Physics Solution /v /v () () O /u O /u /v /v (D) O /u O /u Solution: 7 For a mirror u v f let y = v, x = u x + y = f, y = x + /f This is the euation of a straight line with slope = ( ) and intercept = f Hence the graph would be /v /f O /f /u 8. Convex lens of cm focal length is combined with a concave lens of 6 cm focal length. Find the focal length of the combination. () 5 cm () 5 cm cm (D) cm Solution:8 () Here f = cm, f = -6 cm, F =? Use the formula = + = - = - F f f 6 5 \ F = - 5 cm
15 SPHS Class th Physics Solution 9. n achromatic convergent doublet has power of +D. If power of convex lens is + 5D, then ratio of the dispersive powers of a convergent and divergent lenses will be () : 5 () : 5 : (D) None of these Solution 9 () P = +D P 5D P D P f w w s P f w 5 w 5. Eye piece of an astronomical telescope has focal length of 5 cm. If angular magnification in normal adjustment is, then distance between eye piece and objective should be () 5 cm () 5 cm 55 cm (D) 75 cm Solution: In normal adjustment, L = f + fe f f andm or f 5 cm fe 5 L = 5 cm + 5 cm = 55 cm.
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