260. From what height a block of ice is dropped in order that it may completely melt. It is assumed that 20% of energy of fall is retained by ice. [168000m] Hints: Actually the ice doesn’t melt completely when it is dropped from any height. But just in the view of examination, take 20% PE = mL and determine height.
261. Find the rms speed of Nitrogen at NTP. Density of nitrogen is 1.29 kg/m3 at NTP. [ 484.65 m/s ]
262. How much heat is needed to change 10 g ice at – 10 0C to steam at 100 0C. Given, specific heat capacity of ice is 0.5 cal/gK, Latent heat of ice 80 cal/gK and that of steam 540 cal/gK. [30450 J]
263. Helium gas occupies a volume of 0.04 m3 at a pressure of 2×105 N/m2 and temperature 300K. Calculate the mass of the helium and rms speed of its molecule. Relative molecular mass of helium = 4 , molar gas constant = 8.3 J/ mol K. [ 1366.57 m/s]
264. What is the result of mixing 100g of ice at 0 0C and 100 g water at 100 0C ? Latent heat of ice 3.36 x 105 J/K. [ 10 0C ]
265. Estimate the rate of heat loss through a glass window of area 2 m2 and thickness 4 mm when the temperature of the room is 300 K and that of outside is 5 0C? [ 11000K watt ]
266. A cylinder of gas has mass of 10.0 kg and pressure of 8.0 atmosphere at 27 0C. When some gas is used in a cooled room at – 3 0C, the gas remaining in the cylinder at this temperature has a pressure 6.4 atm. Calculate the mass of the gas used. [ 1.1 kg ]
267. Assuming that the thermal conductivity of woolen glove is equivalent to a layer of quiescent air 3 mm thick, determine the heat lost per minute from the man’s hand of surface area 200 cm2 0n a winter’s day when atmospheric temperature is 3 0C. The skin temperature is taken 34 0C and thermal conductivity of air is 24 x 10-3 W/mK. [ 354 J/min]
268. Density of a gas is 1.775 kg/m3 at 27 0C and 1 atm pressure. If the specific heat capacity at constant pressure is 846 J/kgK, find the ratio of the specific heat capacity at constant pressure to that at constant volume. [ 1.29]
269. In a gas at standard condition, what is the length of the side of a cube that contains a number of molecules equal to the population of the earth, 6 x 109 ?
OPTICS
261. Find the rms speed of Nitrogen at NTP. Density of nitrogen is 1.29 kg/m3 at NTP. [ 484.65 m/s ]
262. How much heat is needed to change 10 g ice at – 10 0C to steam at 100 0C. Given, specific heat capacity of ice is 0.5 cal/gK, Latent heat of ice 80 cal/gK and that of steam 540 cal/gK. [30450 J]
263. Helium gas occupies a volume of 0.04 m3 at a pressure of 2×105 N/m2 and temperature 300K. Calculate the mass of the helium and rms speed of its molecule. Relative molecular mass of helium = 4 , molar gas constant = 8.3 J/ mol K. [ 1366.57 m/s]
264. What is the result of mixing 100g of ice at 0 0C and 100 g water at 100 0C ? Latent heat of ice 3.36 x 105 J/K. [ 10 0C ]
265. Estimate the rate of heat loss through a glass window of area 2 m2 and thickness 4 mm when the temperature of the room is 300 K and that of outside is 5 0C? [ 11000K watt ]
266. A cylinder of gas has mass of 10.0 kg and pressure of 8.0 atmosphere at 27 0C. When some gas is used in a cooled room at – 3 0C, the gas remaining in the cylinder at this temperature has a pressure 6.4 atm. Calculate the mass of the gas used. [ 1.1 kg ]
267. Assuming that the thermal conductivity of woolen glove is equivalent to a layer of quiescent air 3 mm thick, determine the heat lost per minute from the man’s hand of surface area 200 cm2 0n a winter’s day when atmospheric temperature is 3 0C. The skin temperature is taken 34 0C and thermal conductivity of air is 24 x 10-3 W/mK. [ 354 J/min]
268. Density of a gas is 1.775 kg/m3 at 27 0C and 1 atm pressure. If the specific heat capacity at constant pressure is 846 J/kgK, find the ratio of the specific heat capacity at constant pressure to that at constant volume. [ 1.29]
269. In a gas at standard condition, what is the length of the side of a cube that contains a number of molecules equal to the population of the earth, 6 x 109 ?
OPTICS
Reflection of Light
270. A ray of light falls normally on a plane mirror. What are (i) angle of incidence, (ii) angle of reflection, (iii) glancing angle, (iv) angle of deviation? [0, 0, 900, 1800] Hints: (i) Angle of incidence is the angle made by incident ray with normal. (ii) Angle of incidence and angle of reflection are equal. (iii) Glancing angle is the angle made by incident ray with the plane of surface. (iv) Angle of deviation is the angle between incident and angle of reflected ray.
271. A man is walking towards a plane mirror with velocity 5 m/s (a) normally, (b) making angle 300 to the normal. What is the relative velocity between the man and his image? [ 10 m/s, 8.7 m/s] Hints: (a) The image also moves with same velocity as object moves but the motion of image is in opposite direction. So, relative velocity between them is 2v. (b) In second case, component of velocity along the normal is v Cos. So, relative velocity is 2v Cos.
272. An object is placed 4 cm from a concave mirror of radius of curvature 12 cm. Calculate the image position and respective magnification. [ 12 cm, 3 ] Hints: Focal length is half of the radius of the curvature. Use mirror formula to obtain image position.
273. An object is placed 16 cm from a convex mirror of focal length 10 cm. Calculate the image distance and magnification. [ 6 cm, 2/5 ] Hints: Take focal length of convex mirror negative and apply mirror formula.
274. A 10 cm long stick is placed along an axis of a convex length of focal length 15 cm. If the nearer end of the stick is 5 cm away from the mirror, calculate the length of the image.
[ 3.75 cm]
Hints: Take two point object distances u1 and u2 5 cm and 15 cm and determine respective image position v1 and v2. The difference in image position v gives the length of image.
275. A mirror forms an erect image 30 cm from the object and twice its height. Where the object must be situated? What is the radius of curvature? Assuming the object is real, determine whether the mirror is concave or convex? [ 10 cm, 40 cm ] Hints: Draw ray diagram of the formation of magnified and virtual image and use mirror formula. The mirror must be concave because convex mirror always forms diminished image. Object distance = u, Image distance = v and given that u + v = 30 cm Magnification m = v / u So, v = 2u Use mirror formula to calculate focal length. Since image is virtual, image distance is negative.
276. A coin 2.54 cm in diameter held 254 cm from the eye just covers the full moon. What is the diameter of image of the moon formed by the concave mirror of radius of curvature 1.27 m? [ 6.35 x 10-3 m ]
277. A driving mirror consists of a cylindrical mirror of radius 10 cm and length over the curved surface is 10 cm. If the eye of the driver be assumed at a great distance from a mirror, find the angle of view. [ 1140 ]
278. A small convex mirror is placed 60 cm from the pole and on the axis of a large concave mirror, radius of curvature 200 cm. The position of the convex mirror is such that a real image of distance object is formed in the plane of the hole drilled through the concave mirror at its pole. Calculate (i) R of the convex mirror, (ii) the height of the real image if the distance object subtends an angle of 0.5 0 at the pole of the concave mirror. [ 240 cm, 1.3 cm ]
279. A concave mirror of focal length 40 cm and a plane mirror are placed at a distance of 60 cm. A point object is placed midway between the two mirrors. Find the final position of image if the ray first meets the plane mirror. [ 72 cm ] Hints : Object distance (u) = 60 + 30 = 90 cm and focal length (f) = 40 cm. Obtain image distance from the mirror formula.
280. An object is situated 40 cm from a convex mirror. When a plane mirror is inserted between object and convex mirror at a distance 32 cm from the object, the image in the two images coincide. What is the focal length of the convex mirror?
[ – 60 cm ]
Hints: Object distance (u) = 40 cm and image distance (v) = – (32-8) = – 24 cm. Then apply mirror formula.
270. A ray of light falls normally on a plane mirror. What are (i) angle of incidence, (ii) angle of reflection, (iii) glancing angle, (iv) angle of deviation? [0, 0, 900, 1800] Hints: (i) Angle of incidence is the angle made by incident ray with normal. (ii) Angle of incidence and angle of reflection are equal. (iii) Glancing angle is the angle made by incident ray with the plane of surface. (iv) Angle of deviation is the angle between incident and angle of reflected ray.
271. A man is walking towards a plane mirror with velocity 5 m/s (a) normally, (b) making angle 300 to the normal. What is the relative velocity between the man and his image? [ 10 m/s, 8.7 m/s] Hints: (a) The image also moves with same velocity as object moves but the motion of image is in opposite direction. So, relative velocity between them is 2v. (b) In second case, component of velocity along the normal is v Cos. So, relative velocity is 2v Cos.
272. An object is placed 4 cm from a concave mirror of radius of curvature 12 cm. Calculate the image position and respective magnification. [ 12 cm, 3 ] Hints: Focal length is half of the radius of the curvature. Use mirror formula to obtain image position.
273. An object is placed 16 cm from a convex mirror of focal length 10 cm. Calculate the image distance and magnification. [ 6 cm, 2/5 ] Hints: Take focal length of convex mirror negative and apply mirror formula.
274. A 10 cm long stick is placed along an axis of a convex length of focal length 15 cm. If the nearer end of the stick is 5 cm away from the mirror, calculate the length of the image.
[ 3.75 cm]
Hints: Take two point object distances u1 and u2 5 cm and 15 cm and determine respective image position v1 and v2. The difference in image position v gives the length of image.
275. A mirror forms an erect image 30 cm from the object and twice its height. Where the object must be situated? What is the radius of curvature? Assuming the object is real, determine whether the mirror is concave or convex? [ 10 cm, 40 cm ] Hints: Draw ray diagram of the formation of magnified and virtual image and use mirror formula. The mirror must be concave because convex mirror always forms diminished image. Object distance = u, Image distance = v and given that u + v = 30 cm Magnification m = v / u So, v = 2u Use mirror formula to calculate focal length. Since image is virtual, image distance is negative.
276. A coin 2.54 cm in diameter held 254 cm from the eye just covers the full moon. What is the diameter of image of the moon formed by the concave mirror of radius of curvature 1.27 m? [ 6.35 x 10-3 m ]
277. A driving mirror consists of a cylindrical mirror of radius 10 cm and length over the curved surface is 10 cm. If the eye of the driver be assumed at a great distance from a mirror, find the angle of view. [ 1140 ]
278. A small convex mirror is placed 60 cm from the pole and on the axis of a large concave mirror, radius of curvature 200 cm. The position of the convex mirror is such that a real image of distance object is formed in the plane of the hole drilled through the concave mirror at its pole. Calculate (i) R of the convex mirror, (ii) the height of the real image if the distance object subtends an angle of 0.5 0 at the pole of the concave mirror. [ 240 cm, 1.3 cm ]
279. A concave mirror of focal length 40 cm and a plane mirror are placed at a distance of 60 cm. A point object is placed midway between the two mirrors. Find the final position of image if the ray first meets the plane mirror. [ 72 cm ] Hints : Object distance (u) = 60 + 30 = 90 cm and focal length (f) = 40 cm. Obtain image distance from the mirror formula.
280. An object is situated 40 cm from a convex mirror. When a plane mirror is inserted between object and convex mirror at a distance 32 cm from the object, the image in the two images coincide. What is the focal length of the convex mirror?
[ – 60 cm ]
Hints: Object distance (u) = 40 cm and image distance (v) = – (32-8) = – 24 cm. Then apply mirror formula.
281. A small object placed in front of a convex mirror gives a real image four times the size of the object. When the object is moved 10 cm towards the mirror, a virtual image of the same magnification is formed. Find the focal length of the mirror. [20 cm] Hints: From first condition, u = x and m = 4. So from relation m = v/u, v = 4x . Use mirror formula and obtain ……….(i) From second condition, u = x – 10 , m = 5. So, v = – 4 (x – 10 ). Use mirror formula and obtain ……………(ii). Equate both equations to obtain focal length.
282. It is desired to cast the image of a candle magnified 4 times upon a wall 40 cm distance from the lamp. What type of spherical mirror is required and what is its position? [ 10.67 cm ]
Hints: Concave mirror is required to form magnified image. Let, u = x then v = x + 40. Obtain x from relation m = v/u and use mirror formula to obtain the object and image distance.
282. It is desired to cast the image of a candle magnified 4 times upon a wall 40 cm distance from the lamp. What type of spherical mirror is required and what is its position? [ 10.67 cm ]
Hints: Concave mirror is required to form magnified image. Let, u = x then v = x + 40. Obtain x from relation m = v/u and use mirror formula to obtain the object and image distance.
Refraction of Light
283. A ray of light is incident at 600 in air on an air – glass plane surface. Find the angle of refraction in the glass ( for glass= 1.5 ) Hints: Use Snell’s law to obtain angle of refraction.
284. A ray of light is incident in water at an angle of 300 on water air plane surface. Find the angle of refraction in the air ( for water = 4 / 3 ) Hints: Use Snell’s law , where is the refractive index of air with respect of water which is the reciprocal of refractive index of water with respect to air, .
285. A ray of light is incident in glass at an angle of (i) 300 , (i) 700 on a glass – water plane surface. Determine the angle of refraction in the water in each case ( g = 1.5, w = 1.33) Hints: Determine refractive index of glass with respect of water and apply snell’s law. In second case determine critical angle for water-glass interface and check whether there is total internal reflection or not. If TIR takes place, the ray reflects on the same denser medium.
286. What is the apparent position of an object below a rectangular block of glass 6cm thick if a layer of water 4 cm thick is on the top of the glass? (Refractive index of glass and water are 3/2 and 4/3 respectively)? Hints: Determine separately the apparent position of the object due to glass and water by using relation and finally add the apparent depths.
287. A tank whose bottom is a mirror is filled with water to a depth of 20 cm. A small object hangs motionless 8 cm under the surface of water. What is the apparent depth of the image when viewed at normal incidence? ( refractive index of water = 4/3) [ 24 cm ] Hints: Total real depth of the image = 20 + 12 = 32 cm. Using relation , apparent depth of the image can be determined.
288. Calculate the critical angel for (i) an air – glass surface, (ii) an air water surface, (iii) a water – glass surface. Draw diagrams in each case illustrating the total internal reflection of a ray incident on the surface (g = 1.5, w = 1.33) Hints: For critical angle, the angle of refraction in rarer medium will be 900. So, the snell’s law will be .So,
289. A microscope is focused on a scratch on the bottom of a beaker. Turpentine is poured in to the beaker to a depth of 4 cm, and it is found necessary to raise the microscope through a vertical distance of 1.28 cm to bring the scratch again into focus. Find the refractive index of the turpentine. Hints: Real depth = 4 cm. Apparent depth = 4 – 1.18 = 2.82 Now , refractive index
290. A concave mirror of small aperture and focal length 8 cm lies on a bench and a pin is moved vertically above it. At what point will image and object coincide if the mirror is filled with water of refractive index 4/3? [ 12 cm ]
291. Sky subtends angle 1800 to a man’s eye. What is the angle subtending by the sky with the fish’s eye? [ 840 ] Hints: Light coming from horizon HH’ refract into the water making angle of refraction equal to the critical angle. So, angle subtend by sky is twice of the critical angle.
292. Light from a luminous point on the lower face of a rectangular glass slab, 2.0 cm thick, strikes the upper face and the totally reflected rays outline a circle of 3.2 cm radius on the lower face. What is the refractive index of the glass?
284. A ray of light is incident in water at an angle of 300 on water air plane surface. Find the angle of refraction in the air ( for water = 4 / 3 ) Hints: Use Snell’s law , where is the refractive index of air with respect of water which is the reciprocal of refractive index of water with respect to air, .
285. A ray of light is incident in glass at an angle of (i) 300 , (i) 700 on a glass – water plane surface. Determine the angle of refraction in the water in each case ( g = 1.5, w = 1.33) Hints: Determine refractive index of glass with respect of water and apply snell’s law. In second case determine critical angle for water-glass interface and check whether there is total internal reflection or not. If TIR takes place, the ray reflects on the same denser medium.
286. What is the apparent position of an object below a rectangular block of glass 6cm thick if a layer of water 4 cm thick is on the top of the glass? (Refractive index of glass and water are 3/2 and 4/3 respectively)? Hints: Determine separately the apparent position of the object due to glass and water by using relation and finally add the apparent depths.
287. A tank whose bottom is a mirror is filled with water to a depth of 20 cm. A small object hangs motionless 8 cm under the surface of water. What is the apparent depth of the image when viewed at normal incidence? ( refractive index of water = 4/3) [ 24 cm ] Hints: Total real depth of the image = 20 + 12 = 32 cm. Using relation , apparent depth of the image can be determined.
288. Calculate the critical angel for (i) an air – glass surface, (ii) an air water surface, (iii) a water – glass surface. Draw diagrams in each case illustrating the total internal reflection of a ray incident on the surface (g = 1.5, w = 1.33) Hints: For critical angle, the angle of refraction in rarer medium will be 900. So, the snell’s law will be .So,
289. A microscope is focused on a scratch on the bottom of a beaker. Turpentine is poured in to the beaker to a depth of 4 cm, and it is found necessary to raise the microscope through a vertical distance of 1.28 cm to bring the scratch again into focus. Find the refractive index of the turpentine. Hints: Real depth = 4 cm. Apparent depth = 4 – 1.18 = 2.82 Now , refractive index
290. A concave mirror of small aperture and focal length 8 cm lies on a bench and a pin is moved vertically above it. At what point will image and object coincide if the mirror is filled with water of refractive index 4/3? [ 12 cm ]
291. Sky subtends angle 1800 to a man’s eye. What is the angle subtending by the sky with the fish’s eye? [ 840 ] Hints: Light coming from horizon HH’ refract into the water making angle of refraction equal to the critical angle. So, angle subtend by sky is twice of the critical angle.
292. Light from a luminous point on the lower face of a rectangular glass slab, 2.0 cm thick, strikes the upper face and the totally reflected rays outline a circle of 3.2 cm radius on the lower face. What is the refractive index of the glass?
293. ABCD is the plane of a glass cube. A horizontal beam of light enters the face AB at grazing incidence. Show that the angel with any ray emerging from BC would make with the normal to BC is given by Sin = Cot c , where c is the critical angel. What the greatest value that the refractive index of the glass may have if any of the light is to emerge from BC?
Hints: In first face AB, applying Snell’s law, ………….(i) In second face ,angle of incidence = 900 – c and angle of refraction = . Applying Snell’s law , .
Or, ………………(ii)
Equating equation (i) and (ii), the required expression can be obtained.
Hints: In first face AB, applying Snell’s law, ………….(i) In second face ,angle of incidence = 900 – c and angle of refraction = . Applying Snell’s law , .
Or, ………………(ii)
Equating equation (i) and (ii), the required expression can be obtained.
Prism
294. A ray of light is refracted through a prism of angle 700. If the angle of refraction in the glass at the first face is 280, what is the angle of incidence in the glass at the second face?
[ 420 ] Hints: The relation between angle of prism and angles of refraction is A = r1 + r2
295. (i) The angle of a glass prism is 600, and the minimum deviation of light through the prism is 390. Calculate (i) the refractive index of the glass. (ii) The refractive index of a glass prism is 1.66, and the angle of the prism is 600. Find the minimum deviation. [ 1.52, 52.20]
Hints: The relation between refractive index and minimum deviation is
296. A narrow beam of light is incident normally on one face of an equilateral prism (refractive index 1.45) and finally emerges from the prism. The prism is new surrounded by water (refractive index 1.33). What is the angle between the directions of the emergent beam in the two cases?
[ 49.2] Hints: In first case, angle of incidence on the second face will be 600 ( can be obtained geometrically) and the light undergoes into total internal reflection. When the prism is surrounded by water, the Snell’s law in the second face of prism will be , where is the refractive index of water with respect to glass. It is obtained by relation It gives angle of emergence on water r = 70.70 From geometry, angle between emergent rays in two conditions E1OE2 = 300 + (900 – 70.760) = 49.20
297. A certain prism is found to produce a minimum deviation of 510. While is produces a deviation of 620 48’ for two values of the angle of incidence, namely 400 6’ and 820 42’ respectively. Determine the refracting angle of the prism, the angle of incidence at minimum deviation and the refractive index of the material of the prism.
[600, 300, 1.65] Hints: The prism formula is A + = i1 + i2 , here, angle of emergence e is replaced by angle of incidence. It is due to the principle of reversibility of light. Also, in the case of minimum deviation r1 = r2. So, r = A/2.
Refractive index of the material of prism
298. PQR represents a right – angle in a prism of glass of refractive index 1.50. A ray of light enters into the prism through the hypotenuse QR at an angle of incidence I, and is reflected a the critical angle from PQ to PR calculate and draw a diagram showing the path of the ray through the prism . (Only rays in the plane of PQR need be considered.)
294. A ray of light is refracted through a prism of angle 700. If the angle of refraction in the glass at the first face is 280, what is the angle of incidence in the glass at the second face?
[ 420 ] Hints: The relation between angle of prism and angles of refraction is A = r1 + r2
295. (i) The angle of a glass prism is 600, and the minimum deviation of light through the prism is 390. Calculate (i) the refractive index of the glass. (ii) The refractive index of a glass prism is 1.66, and the angle of the prism is 600. Find the minimum deviation. [ 1.52, 52.20]
Hints: The relation between refractive index and minimum deviation is
296. A narrow beam of light is incident normally on one face of an equilateral prism (refractive index 1.45) and finally emerges from the prism. The prism is new surrounded by water (refractive index 1.33). What is the angle between the directions of the emergent beam in the two cases?
[ 49.2] Hints: In first case, angle of incidence on the second face will be 600 ( can be obtained geometrically) and the light undergoes into total internal reflection. When the prism is surrounded by water, the Snell’s law in the second face of prism will be , where is the refractive index of water with respect to glass. It is obtained by relation It gives angle of emergence on water r = 70.70 From geometry, angle between emergent rays in two conditions E1OE2 = 300 + (900 – 70.760) = 49.20
297. A certain prism is found to produce a minimum deviation of 510. While is produces a deviation of 620 48’ for two values of the angle of incidence, namely 400 6’ and 820 42’ respectively. Determine the refracting angle of the prism, the angle of incidence at minimum deviation and the refractive index of the material of the prism.
[600, 300, 1.65] Hints: The prism formula is A + = i1 + i2 , here, angle of emergence e is replaced by angle of incidence. It is due to the principle of reversibility of light. Also, in the case of minimum deviation r1 = r2. So, r = A/2.
Refractive index of the material of prism
298. PQR represents a right – angle in a prism of glass of refractive index 1.50. A ray of light enters into the prism through the hypotenuse QR at an angle of incidence I, and is reflected a the critical angle from PQ to PR calculate and draw a diagram showing the path of the ray through the prism . (Only rays in the plane of PQR need be considered.)
299. A is the vertex of a triangular glass prism, the angle at a being 300. A ray of light OP is incident at P on one of the face enclosing the angle A, in a direction such that the angle OPA = 400. Show that, if the refractive index of the glass is 1.50 the ray cannot emerge from the second face. [ 65.20]
Hints: For face AC, angle of incidence (I1 ) = 600 . Snell’s law gives the angle of refraction (r1) = 35.20. From geometry, angle of incidence in AB face (i2) = 65.20 The angle is more than the critical angle. So the light cannot escape from the second face.
300. The refracting angle of a prism is 62.00 and the refractive index of the glass for yellow light is 1.65. What is the smallest possible angle of incidence of a ray of this yellow light which is transmitted without total internal reflection? Explain what happens if white light is used instead and the angle of incidence is varied? [ 39.540 ] Hints: In second face, since grazing emergence is taking place, the angle of incidence on the face should be ‘c’. The relation between angle of prism and angles of incidence is A = r1 + r2 , which gives r1 = 22.70. Applying Snell’s law in first face, angle of incidence can be determined.
Hints: For face AC, angle of incidence (I1 ) = 600 . Snell’s law gives the angle of refraction (r1) = 35.20. From geometry, angle of incidence in AB face (i2) = 65.20 The angle is more than the critical angle. So the light cannot escape from the second face.
300. The refracting angle of a prism is 62.00 and the refractive index of the glass for yellow light is 1.65. What is the smallest possible angle of incidence of a ray of this yellow light which is transmitted without total internal reflection? Explain what happens if white light is used instead and the angle of incidence is varied? [ 39.540 ] Hints: In second face, since grazing emergence is taking place, the angle of incidence on the face should be ‘c’. The relation between angle of prism and angles of incidence is A = r1 + r2 , which gives r1 = 22.70. Applying Snell’s law in first face, angle of incidence can be determined.
301. Determine minimum deviation of an equatorial glass prism of refractive index 1.5 when it is immersed in a liquid of refractive index 1.33.
Hints: The refractive index , where is the refractive index of glass with respect to water. It is calculated by relation = x
Hints: The refractive index , where is the refractive index of glass with respect to water. It is calculated by relation = x
Lens
302. An object is placed (i) 12 cm (ii) 4 cm from a converging lens of focal length 6 cm. Calculate image position and magnification in each case. [ 12cm, 1 ; 12cm, 3 ] Hints: (a) Use lens formula (b) In second case, take focal length of concave lens negative.
303. The image of a real object in a diverging lens of focal length 10 cm is formed 4 cm from the lens. Find the object distance and the magnification. [ 6.67 cm, 0.6 ]
Hints: Use lens formula , take focal length of concave lens negative.
302. An object is placed (i) 12 cm (ii) 4 cm from a converging lens of focal length 6 cm. Calculate image position and magnification in each case. [ 12cm, 1 ; 12cm, 3 ] Hints: (a) Use lens formula (b) In second case, take focal length of concave lens negative.
303. The image of a real object in a diverging lens of focal length 10 cm is formed 4 cm from the lens. Find the object distance and the magnification. [ 6.67 cm, 0.6 ]
Hints: Use lens formula , take focal length of concave lens negative.
304. The image obtained with the converging lens is erected and three times magnified. The focal length of the lens is 20cm. Calculate object distance and image distance. [ 13.33, 40 cm ]
Hints: Magnification m = . Since m = 3, v = 3 u. Image is erected, so it must be virtual. Therefore image distance is negative. Now the lens formula will be , . Substituting value of f, object distance and hence image distance can be determined.
305. A beam of light converges to a point 9 cm behind (i) a converging lens of focal length 12 cm (ii) diverging lens of focal length 15 cm. Find the image position in each case. [ 5.1 cm, 22.5 cm]
Hints: In the absence of lens, the rays converge 9 cm behind it and image forms there. But when lens is introduced, the point where image forms, behaves as virtual object and image is formed at new position. So, take virtual object distance, u = – 9 cm.
306. If the moon subtends the angle 9.1 x 10-3 radian at the center of the lens of focal length 20 cm. Calculate the diameter of this image. With the screen removed, a second converging lens of focal length 5 cm is placed coaxial with the first and 24 cm from it opposite to the moon. Find position, nature and size of final image. [ 1.82mm, 20cm, 9.1 mm ] Hints: Since moon is very far, image forms at principal focus. If ‘d’ be the diameter of the image, then So, d = 9.1 x 10-3 x 20 = 1.82mm For Concave lens, Object distance (u) = 4 cm Focal length (f) = 5 cm Image distance (v) = ? Lens formula gives the virtual image distance 20 cm. Magnification m = v/u = 5 So, size of final image = 1.82mm x 5 = 9.1 mm
307. A converging lens of focal length 20 cm is arranged coaxially with the diverging lens of focal length 8 cm. A point object lies on the side of first lens and very far from it. What is the smallest possible distance between the lenses if the combination is to form a real image of the object? [ 12 cm ]
Hints: For converging lens, Object distance (u) = Focal length ( f ) = 20 cm Lens formula gives the image distance (v) = 20 cm. Let distance between two lenses be x. For diverging lens, Object distance (u) = – (20 – x ), negative sign for virtual image. Image distance (v) = , final rays are parallel to each other. Focal length (f) = – 12 cm
Lens formula gives the value of x
308. A converging lens of 6 cm focal length is mounted at 10 cm from screen placed right angles to the axis. A diverging lens of 12 cm focal length is then placed coaxially between the lens and the screen so that an image of an object 24 cm from the converging lens is focused on the screen. What is the distance between the two lenses? [ 4 cm ] Hints: For converging lens, Object distance (u) = 24 cm Focal length ( f ) = 6 cm Lens formula gives the image distance (v) = 8 cm. Let distance between two lenses be x. For diverging lens, Object distance (u) = – (8 – x ), negative sign for virtual image. Image distance (v) = 10 – x Focal length (f) = – 12 cm
Lens formula gives the value of x
309. Light from an object passes through a thin converging lens of focal length 20 cm, placed 24 cm from the object and then thin diverging lens of focal length 50 cm forming a real image 62.5 cm from the diverging lens. Find (a) position of the image due to first lens, (b) distance between the lenses, (c) magnification of the final image. [ 120 cm, 92.2 cm, 3.4 ] Hints: For converging lens, Object distance (u) = 24 cm Focal length ( f ) = 20 cm Lens formula gives the image distance (v) = 120 cm. Let distance between two lenses be x. For diverging lens, Object distance (u) = – (120 – x ), negative sign for virtual image. Image distance (v) = 62.5 cm Focal length (f) = – 50 cm
Lens formula gives the value of x Magnification of the final image m = m1 x m2, where m1 and m2 are magnification of converging and diverging lenses respectively.
310. An object is paced right angles to the axis of a converging lens of focal length 15 cm, and 22.5 cm from it. On the other side of the lens, a diverging lens of focal length 30 cm is placed coaxially. Find the position of the final image when (a) lenses are 15 cm apart and plane mirror is placed perpendicular to the axis 40 cm beyond the diverging lens, (b) when the mirror is removed and the lenses are 35 cm apart. [beside O,72.5 cm from O] Hints: For converging lens, Object distance (u) = 22.5 cm Focal length ( f ) = 15 cm Lens formula gives the image distance (v) = 45 cm. Given, distance between two lenses = 15 cm For diverging lens, Object distance (u) = – (45 – 15 ) = – 30 cm, negative sign for virtual image. Focal length (f) = – 30 cm Image distance (v) = ?
Lens formula gives the value of image distance = . It means final rays are parallel. When they fall on a plane mirror perpendicularly, they return along the same path. It is due to the reversibility of light. The returning light follows the same path and final image coincides with the object.
311. A thin equiconvex lens of refractive index 1.5 whose surface have radius of curvature 24 cm is placed on a horizontal plane mirror. When the space between the lenses and the mirror is filled with a liquid, a pin held 40 cm vertically above the lens is found to coincide with its own image. Calculate the refractive index of the liquid. [ 1.4 ] Hints: From Lens maker formula the focal length of convex lens = 1/24 cm Similarly, focal length of liquid plano-concave lens = , since 1/R2 = 0 Now, the combined focal length F = 40 cm, since object and image coincide at the same point. So,
312. A converging equiconvex lens of glass of refractive index 1.5 is laid on a horizontal plane mirror. A pin coincide with it’s inverted image when it is 1.0m above the lens. When some liquid is placed between the lens and the mirror the, pin has to be raised by 0.55m for the coincidence to occur again. What is the refractive index of the liquid? [ 1.35 ] Hints: Focal length of equiconvex lens = 1 m = 100 cm Focal length of combined lens = 0.55 m = 55 cm
313. The curved face of a planoconvex lens (ยต=1.5) is placed in contact with a plane mirror. An object at 20 cm distance coincides with the image produce by the reflection by the mirror. If the gap between the lens and the mirror is filled by liquid, the coincident object and image will be at distance 100 cm. What is the index of the liquid? [ 1.4 ] Hints: From Lens maker formula, the focal length of glass plano-convex lens …………………………..(i)
Similarly, focal length of liquid plano-concave lens = ………………………… (ii) Now, dividing (i) by (ii) .
, which gives the value of l
Hints: Magnification m = . Since m = 3, v = 3 u. Image is erected, so it must be virtual. Therefore image distance is negative. Now the lens formula will be , . Substituting value of f, object distance and hence image distance can be determined.
305. A beam of light converges to a point 9 cm behind (i) a converging lens of focal length 12 cm (ii) diverging lens of focal length 15 cm. Find the image position in each case. [ 5.1 cm, 22.5 cm]
Hints: In the absence of lens, the rays converge 9 cm behind it and image forms there. But when lens is introduced, the point where image forms, behaves as virtual object and image is formed at new position. So, take virtual object distance, u = – 9 cm.
306. If the moon subtends the angle 9.1 x 10-3 radian at the center of the lens of focal length 20 cm. Calculate the diameter of this image. With the screen removed, a second converging lens of focal length 5 cm is placed coaxial with the first and 24 cm from it opposite to the moon. Find position, nature and size of final image. [ 1.82mm, 20cm, 9.1 mm ] Hints: Since moon is very far, image forms at principal focus. If ‘d’ be the diameter of the image, then So, d = 9.1 x 10-3 x 20 = 1.82mm For Concave lens, Object distance (u) = 4 cm Focal length (f) = 5 cm Image distance (v) = ? Lens formula gives the virtual image distance 20 cm. Magnification m = v/u = 5 So, size of final image = 1.82mm x 5 = 9.1 mm
307. A converging lens of focal length 20 cm is arranged coaxially with the diverging lens of focal length 8 cm. A point object lies on the side of first lens and very far from it. What is the smallest possible distance between the lenses if the combination is to form a real image of the object? [ 12 cm ]
Hints: For converging lens, Object distance (u) = Focal length ( f ) = 20 cm Lens formula gives the image distance (v) = 20 cm. Let distance between two lenses be x. For diverging lens, Object distance (u) = – (20 – x ), negative sign for virtual image. Image distance (v) = , final rays are parallel to each other. Focal length (f) = – 12 cm
Lens formula gives the value of x
308. A converging lens of 6 cm focal length is mounted at 10 cm from screen placed right angles to the axis. A diverging lens of 12 cm focal length is then placed coaxially between the lens and the screen so that an image of an object 24 cm from the converging lens is focused on the screen. What is the distance between the two lenses? [ 4 cm ] Hints: For converging lens, Object distance (u) = 24 cm Focal length ( f ) = 6 cm Lens formula gives the image distance (v) = 8 cm. Let distance between two lenses be x. For diverging lens, Object distance (u) = – (8 – x ), negative sign for virtual image. Image distance (v) = 10 – x Focal length (f) = – 12 cm
Lens formula gives the value of x
309. Light from an object passes through a thin converging lens of focal length 20 cm, placed 24 cm from the object and then thin diverging lens of focal length 50 cm forming a real image 62.5 cm from the diverging lens. Find (a) position of the image due to first lens, (b) distance between the lenses, (c) magnification of the final image. [ 120 cm, 92.2 cm, 3.4 ] Hints: For converging lens, Object distance (u) = 24 cm Focal length ( f ) = 20 cm Lens formula gives the image distance (v) = 120 cm. Let distance between two lenses be x. For diverging lens, Object distance (u) = – (120 – x ), negative sign for virtual image. Image distance (v) = 62.5 cm Focal length (f) = – 50 cm
Lens formula gives the value of x Magnification of the final image m = m1 x m2, where m1 and m2 are magnification of converging and diverging lenses respectively.
310. An object is paced right angles to the axis of a converging lens of focal length 15 cm, and 22.5 cm from it. On the other side of the lens, a diverging lens of focal length 30 cm is placed coaxially. Find the position of the final image when (a) lenses are 15 cm apart and plane mirror is placed perpendicular to the axis 40 cm beyond the diverging lens, (b) when the mirror is removed and the lenses are 35 cm apart. [beside O,72.5 cm from O] Hints: For converging lens, Object distance (u) = 22.5 cm Focal length ( f ) = 15 cm Lens formula gives the image distance (v) = 45 cm. Given, distance between two lenses = 15 cm For diverging lens, Object distance (u) = – (45 – 15 ) = – 30 cm, negative sign for virtual image. Focal length (f) = – 30 cm Image distance (v) = ?
Lens formula gives the value of image distance = . It means final rays are parallel. When they fall on a plane mirror perpendicularly, they return along the same path. It is due to the reversibility of light. The returning light follows the same path and final image coincides with the object.
311. A thin equiconvex lens of refractive index 1.5 whose surface have radius of curvature 24 cm is placed on a horizontal plane mirror. When the space between the lenses and the mirror is filled with a liquid, a pin held 40 cm vertically above the lens is found to coincide with its own image. Calculate the refractive index of the liquid. [ 1.4 ] Hints: From Lens maker formula the focal length of convex lens = 1/24 cm Similarly, focal length of liquid plano-concave lens = , since 1/R2 = 0 Now, the combined focal length F = 40 cm, since object and image coincide at the same point. So,
312. A converging equiconvex lens of glass of refractive index 1.5 is laid on a horizontal plane mirror. A pin coincide with it’s inverted image when it is 1.0m above the lens. When some liquid is placed between the lens and the mirror the, pin has to be raised by 0.55m for the coincidence to occur again. What is the refractive index of the liquid? [ 1.35 ] Hints: Focal length of equiconvex lens = 1 m = 100 cm Focal length of combined lens = 0.55 m = 55 cm
313. The curved face of a planoconvex lens (ยต=1.5) is placed in contact with a plane mirror. An object at 20 cm distance coincides with the image produce by the reflection by the mirror. If the gap between the lens and the mirror is filled by liquid, the coincident object and image will be at distance 100 cm. What is the index of the liquid? [ 1.4 ] Hints: From Lens maker formula, the focal length of glass plano-convex lens …………………………..(i)
Similarly, focal length of liquid plano-concave lens = ………………………… (ii) Now, dividing (i) by (ii) .
, which gives the value of l
314. An achromatic lens having a focal length 20 cm is to be constructed by combining a crown lens and a flint lens. What must be the focal length of the component lenses if the dispersive power of the crown and flint glasses are 0.0158 and 0.0324 respectively? [ 10.2 cm, -21 cm] Hints: Apply the condition of achromatism and obtain f1 = – 0.488 f2. Also, . Substituting value of f1 and F = 20, f1 and f2 can be determined.
Defect of vision
Defect of vision
315. A person can focus objects only when they lie between 50 cm and 300 cm from his eye. What spectacles should he use (i) to see up to infinity, (ii) to reduce his least distance of distinct vision to 25 cm. Find the range of vision using each pair. Hints: (i) To extend far point from 300 cm to infinity, concave lens should be used. For concave lens, object distance (u) = image distance (v) = – 300 cm , virtual. Lens formula gives the focal length of the lens, equal to – 300 cm. (i) To reduce near point from 50 cm to 25 cm, convex mirror should be used. For concave lens, object distance (u) = 25 image distance (v) = – 50 cm , virtual. Lens formula gives the focal length of the lens, equal to 50 cm.
316. A certain person can see clearly objects lying between 20 cm and 200 cm from his eye. What spectacle is required to enable him to see distance object clearly? Also calculate the least distance of distinct vision after wearing them?
[-0.5D, 22.2cm]
Hints: (i) To extend far point from 200 cm to infinity, concave lens should be used. For concave lens, object distance (u) = image distance (v) = – 200 cm , virtual. Lens formula gives the focal length of the lens, equal to – 300 cm and the power of the lens P = 1/f = – 0.5 D
316. A certain person can see clearly objects lying between 20 cm and 200 cm from his eye. What spectacle is required to enable him to see distance object clearly? Also calculate the least distance of distinct vision after wearing them?
[-0.5D, 22.2cm]
Hints: (i) To extend far point from 200 cm to infinity, concave lens should be used. For concave lens, object distance (u) = image distance (v) = – 200 cm , virtual. Lens formula gives the focal length of the lens, equal to – 300 cm and the power of the lens P = 1/f = – 0.5 D
317. In order to correct his near point to 25 cm , a man given spectacles with converging lens of 50 cm focal length an to correct his far point to infinity he is given diverging lens of focal length 200 cm. Determine his near and far point not wearing the glass. [50 cm, -200 cm]
318. A person can not see objects in nearer than 500 cm from the eye. Determine the focal length and the power of the glass which enable him to read a book 25 cm from his eye. [ 3.8 D ]
319. An elderly person can not see clearly without the use of spectacles, objects nearer than 200 cm. What spectacles will he need to reduce this distance to 25 cm? If his eyes can focus rays which are converging to points not less than 150 cm behind them, calculate his range of distinct vision when using spectacles. [ 28.6 cm, 25-35.3cm ]
Optical instruments
318. A person can not see objects in nearer than 500 cm from the eye. Determine the focal length and the power of the glass which enable him to read a book 25 cm from his eye. [ 3.8 D ]
319. An elderly person can not see clearly without the use of spectacles, objects nearer than 200 cm. What spectacles will he need to reduce this distance to 25 cm? If his eyes can focus rays which are converging to points not less than 150 cm behind them, calculate his range of distinct vision when using spectacles. [ 28.6 cm, 25-35.3cm ]
Optical instruments
320. A simple astronomical telescope in normal adjustment has an objective of focal length 100 cm and eyepiece of focal length 5 cm. (a) where is the final image formed ? (b) calculate the angular magnification. (c) how would you increase the resolving power of the telescope? [ infinity, 20] Hints: (i) In normal adjustment, final image forms at infinity. (ii) Angular magnification in normal adjustment m =
321. Calculate the distance of the eye ring from the eye piece of a simple astronomical telescope in normal adjustment whose objective and the eyepiece have focal lengths of 80 cm and 10 cm respectively. [ 11.25 cm ] Hints: Eye ring is that point where eye lens forms the image of objective lens. So, for eye lens, object distance (u) = 80 + 10 = 90 cm image distance (v) = ? focal length (f) = 10 cm
322. A refracting telescope has its objective of focal length 1.0 m and eye piece of focal length 2 cm. A real image of sun, 10 cm in diameter is formed in the screen 24 cm from the eye piece. What angle does the sun subtend the objective? [0.0091 rad] Hints: For eyepiece Object distance (u) = ? Image distance (v) = 24 cm Focal length (f) = 2 cm Lens formula gives u = 2.18 cm Now So, object size O=0.91cm Angle made by objective with lens Tan = 0.91/100 = 0.0091 radian.
323. A telescope is made of an object glass of focal length 20 cm and eye piece of 5 cm, both conversing lenses. Find the magnifying power in the following cases (a) when the eye is focused to receive parallel rays. (b) when the eye sees the image situated at the nearest distance of distinct vision which may be taken 25 cm. [4, 4.8 ] Hints: (a) magnification of telescope when image is formed at is m = (b) magnification when image is formed at near point is m =
324. A telescope consist two thin converging lenses of focal lengths 0.3 m and 0.03 m separated by 0.33 m. It is focused on the moon which subtends an angle 0.5 0 at the objective. Find the angle subtends to the observer’s eye by the image of the moon formed by the instrument. [ 50 ] Hints: Magnification of telescope when image is formed at is m = So,
325. A telescope objective has focal length 96 cm and diameter 12 cm. Calculate the focal length and minimum diameter of a eyepiece lens for use with the telescope, if the magnifying power required is x24, and all the light transmitted by the objective from a distance point on the telescope axis is to fall on the eyepiece.
326. An astronomical telescope has objective of focal length 60 cm and an eyepiece of focal length 3 cm is focused on the moon so that the final image is formed at the minimum distance of distinct image from the eyepiece. Assuming that the diameter of the moon subtends an angle of 0.5 0 at the objective, calculate the (a) angular magnification, (b) actual size of the moon on seen, (c) How, with the same lenses, could an image of the moon 10 cm in diameter be formed on a photographic plate? [ 22.4, 4.95 cm] Hints: (a) Magnification when image is formed at near point is m = = 22.4 (b) Magnification m = . So, = m x = 11.20 If ‘d’ be the actual size of the image seen, then Tan 11.2 = h / D
321. Calculate the distance of the eye ring from the eye piece of a simple astronomical telescope in normal adjustment whose objective and the eyepiece have focal lengths of 80 cm and 10 cm respectively. [ 11.25 cm ] Hints: Eye ring is that point where eye lens forms the image of objective lens. So, for eye lens, object distance (u) = 80 + 10 = 90 cm image distance (v) = ? focal length (f) = 10 cm
322. A refracting telescope has its objective of focal length 1.0 m and eye piece of focal length 2 cm. A real image of sun, 10 cm in diameter is formed in the screen 24 cm from the eye piece. What angle does the sun subtend the objective? [0.0091 rad] Hints: For eyepiece Object distance (u) = ? Image distance (v) = 24 cm Focal length (f) = 2 cm Lens formula gives u = 2.18 cm Now So, object size O=0.91cm Angle made by objective with lens Tan = 0.91/100 = 0.0091 radian.
323. A telescope is made of an object glass of focal length 20 cm and eye piece of 5 cm, both conversing lenses. Find the magnifying power in the following cases (a) when the eye is focused to receive parallel rays. (b) when the eye sees the image situated at the nearest distance of distinct vision which may be taken 25 cm. [4, 4.8 ] Hints: (a) magnification of telescope when image is formed at is m = (b) magnification when image is formed at near point is m =
324. A telescope consist two thin converging lenses of focal lengths 0.3 m and 0.03 m separated by 0.33 m. It is focused on the moon which subtends an angle 0.5 0 at the objective. Find the angle subtends to the observer’s eye by the image of the moon formed by the instrument. [ 50 ] Hints: Magnification of telescope when image is formed at is m = So,
325. A telescope objective has focal length 96 cm and diameter 12 cm. Calculate the focal length and minimum diameter of a eyepiece lens for use with the telescope, if the magnifying power required is x24, and all the light transmitted by the objective from a distance point on the telescope axis is to fall on the eyepiece.
326. An astronomical telescope has objective of focal length 60 cm and an eyepiece of focal length 3 cm is focused on the moon so that the final image is formed at the minimum distance of distinct image from the eyepiece. Assuming that the diameter of the moon subtends an angle of 0.5 0 at the objective, calculate the (a) angular magnification, (b) actual size of the moon on seen, (c) How, with the same lenses, could an image of the moon 10 cm in diameter be formed on a photographic plate? [ 22.4, 4.95 cm] Hints: (a) Magnification when image is formed at near point is m = = 22.4 (b) Magnification m = . So, = m x = 11.20 If ‘d’ be the actual size of the image seen, then Tan 11.2 = h / D
327. A telescope constructed from two converging lenses, one of focal length 250 cm and other of 2 cm, is used to observe a planet which subtends an angle of 5 x 10-5 radian. Explain how these two lenses would be placed for normal adjustment and calculate the angle subtended at the eye of the observer by the image. Hints: Magnification of telescope when image is formed at is m = So,
In normal adjustment, final image forms at infinity. So, the distance between two lenses will be f0 + fe
328. A converging lens of focal length 5 cm is used as a magnifying glass. If the near point of the observer is 25 cm from the eye and the lens is held close to the eye, calculate (a) distance of the object from the lens, (b) angular magnification.
329. State and explain how you would arrange simple converging lenses, of focal lengths 2 cm and 5 cm, to act as a compound microscope with magnifying power x 42, the final image being 25 cm from the eye lens.
330. A point object is placed on the axis of, and 3.6 cm from, a thin converging lens of focal length 3 cm. A second thin converging lens of focal length 16.0 cm is placed coaxial with the first and 26.0 cm from it in the side of objective. Find the final image produced by the two lenses.
331. If the final image formed coincides with the object, and is at the least distance of distinct vision when the object is 4 cm from the objective, calculate the focal length of the objective and the eye lenses, assuming magnifying power of the microscope 14.
332. A compound microscope has lenses of focal length 1 cm and 3 cm. An objective is placed 1.2 cm from the object lens. If virtual image is formed 25 cm from the eye, calculate the separation of the lenses.
In normal adjustment, final image forms at infinity. So, the distance between two lenses will be f0 + fe
328. A converging lens of focal length 5 cm is used as a magnifying glass. If the near point of the observer is 25 cm from the eye and the lens is held close to the eye, calculate (a) distance of the object from the lens, (b) angular magnification.
329. State and explain how you would arrange simple converging lenses, of focal lengths 2 cm and 5 cm, to act as a compound microscope with magnifying power x 42, the final image being 25 cm from the eye lens.
330. A point object is placed on the axis of, and 3.6 cm from, a thin converging lens of focal length 3 cm. A second thin converging lens of focal length 16.0 cm is placed coaxial with the first and 26.0 cm from it in the side of objective. Find the final image produced by the two lenses.
331. If the final image formed coincides with the object, and is at the least distance of distinct vision when the object is 4 cm from the objective, calculate the focal length of the objective and the eye lenses, assuming magnifying power of the microscope 14.
332. A compound microscope has lenses of focal length 1 cm and 3 cm. An objective is placed 1.2 cm from the object lens. If virtual image is formed 25 cm from the eye, calculate the separation of the lenses.
Photometry
333. A lamp whose luminous intensity is 100 candela is placed one meter above the horizontal round table of 1 m radius. Calculate the illuminance at a point directly under the table and also the edge of the table. [ 100, 35.36 lm/m2]
334. Two lamps 25 and 50 candela are 100 cm apart. At which point between them will a screen be equally illuminated on both side ? [ 41.4 cm]
335. Two identical lamps are placed at 60 cm apart. Where should a screen be placed between them so that the intensity on one side of the screen is four times that of the other side? [ 20 cm]
336.
Revision
HSEB Exam Questions
337. How long will the light take in traveling a distance of 500 m in water? Given refractive index of water is 1.33. [ 2.2 x 10-6 sec]
338. A glass prism of angle 720 and index of refraction 1.66 is immersed in a liquid
339. Calculate the focal length of a concave mirror when an object placed at a distance of 40 cm makes image equal to the size of the object. [ 20 cm ]
340. If two identical lens are 1 m apart, where should be the screen be placed between them so that the intensity on one side of the serene be four times the intensity of the other side. [ 0.29 m ]
341. An object 10 cm high is placed in front of a convex mirror of focal length 20 cm and the object is 30 cm from the mirror. Find the height of the image. [ 0.04 m]
342. What is the apparent position of an object below a rectangular block of glass 6 cm thick if a layer of water 4 cm thick is on the top of the glass ? Refractive index of glass 3/2 and that of water is 4/3. [ 0.07 m from top]
343. A microscope is focused on a scratch on the bottom of a beaker. Turpentine is poured into the beaker to a depth of 4 cm and it is found necessary to raise the microscope through a vertical distance of 1.25 cm to bring the scratch again into focus. Find the refractive index of the turpentine. [ 1.45 ]
344. An erect image, three times the size of the object is obtained with a concave mirror of radius of curvature 36 cm. What is the position of the object? [ 12 cm]
345. The radii of curvature of the faces of a thin converging meniscus lens of glass of refractive index 1.5 are 15 cm and 30 cm. What is the focal length of the lens when it is completely immersed in water of refractive index 4/3 ? [ 2.35 m] Hints: Apply Lens Makers formula where ยต be the refractive index of glass with respect to water.
346. A glass prism of angle 720 and index of refraction 1.66 is immersed in a liquid of refractive index 1.33. Find the angle of minimum deviation for a parallel beam of light passing through the prism. [ 22.40 ]
347. Light from a luminous point on the lower face of a rectangular glass slab, 2 cm thick, strikes the upper face and the totally refractive rays outline a circle of 3.2 cm radius on the lower face. What is the refractive index of the glass? [ 1.6 ]
348. A narrow beam of light incident normally on one face of an equilateral prism of refractive index 1.45, being surrounded by water of refractive index 1.33. At what angle the ray of light emerges out? [ 710 ]
333. A lamp whose luminous intensity is 100 candela is placed one meter above the horizontal round table of 1 m radius. Calculate the illuminance at a point directly under the table and also the edge of the table. [ 100, 35.36 lm/m2]
334. Two lamps 25 and 50 candela are 100 cm apart. At which point between them will a screen be equally illuminated on both side ? [ 41.4 cm]
335. Two identical lamps are placed at 60 cm apart. Where should a screen be placed between them so that the intensity on one side of the screen is four times that of the other side? [ 20 cm]
336.
Revision
HSEB Exam Questions
337. How long will the light take in traveling a distance of 500 m in water? Given refractive index of water is 1.33. [ 2.2 x 10-6 sec]
338. A glass prism of angle 720 and index of refraction 1.66 is immersed in a liquid
339. Calculate the focal length of a concave mirror when an object placed at a distance of 40 cm makes image equal to the size of the object. [ 20 cm ]
340. If two identical lens are 1 m apart, where should be the screen be placed between them so that the intensity on one side of the serene be four times the intensity of the other side. [ 0.29 m ]
341. An object 10 cm high is placed in front of a convex mirror of focal length 20 cm and the object is 30 cm from the mirror. Find the height of the image. [ 0.04 m]
342. What is the apparent position of an object below a rectangular block of glass 6 cm thick if a layer of water 4 cm thick is on the top of the glass ? Refractive index of glass 3/2 and that of water is 4/3. [ 0.07 m from top]
343. A microscope is focused on a scratch on the bottom of a beaker. Turpentine is poured into the beaker to a depth of 4 cm and it is found necessary to raise the microscope through a vertical distance of 1.25 cm to bring the scratch again into focus. Find the refractive index of the turpentine. [ 1.45 ]
344. An erect image, three times the size of the object is obtained with a concave mirror of radius of curvature 36 cm. What is the position of the object? [ 12 cm]
345. The radii of curvature of the faces of a thin converging meniscus lens of glass of refractive index 1.5 are 15 cm and 30 cm. What is the focal length of the lens when it is completely immersed in water of refractive index 4/3 ? [ 2.35 m] Hints: Apply Lens Makers formula where ยต be the refractive index of glass with respect to water.
346. A glass prism of angle 720 and index of refraction 1.66 is immersed in a liquid of refractive index 1.33. Find the angle of minimum deviation for a parallel beam of light passing through the prism. [ 22.40 ]
347. Light from a luminous point on the lower face of a rectangular glass slab, 2 cm thick, strikes the upper face and the totally refractive rays outline a circle of 3.2 cm radius on the lower face. What is the refractive index of the glass? [ 1.6 ]
348. A narrow beam of light incident normally on one face of an equilateral prism of refractive index 1.45, being surrounded by water of refractive index 1.33. At what angle the ray of light emerges out? [ 710 ]
Electrostatics
349. A body is charged to 6 C. Calculate the number of electrons that are taken out from the body.
[ 3.75 x 1013] Hints: Use the mathematical expression of quantization of charge is q = ne, where ‘n’ be the number of basic charge and ‘e’ is the charge of an electron equal to 1.6 x 10-19 C.
350. A body of mass 5mg is charged to 1200C. Does its mass change after charging? If so, calculate the magnitude of the mass after charging.
[ 4.99 mg] Hints: The decrease in mass due to charging is m = n × me , where n is number of electrons that are taken out from the body and me is the mass of an electron. From quantization of charge, n = q/e. So, the decrease in mass be m = q/e x me
351. Calculate the total flux generated from a charge of magnitude 88.5 nC. If the charge is surrounded by a sphere, calculate the flux passing through a section of the sphere which subtends an angle 2/5 steradian to the charge.
[10000Wb,1000Wb]
Hints: Total flux generated from a charged body is given by Gauss’s law i.e. = . The emitted flux scattered around 4 steradian angle. Applying unitary method flux through the given section can be determined.
352. If 250 weber flux passes through a cross-section of a cylinder having radius 20 cm, calculate the electric flux density at that point
[ 1990 wb/m2]
Hints: Flux density is the flux per unit area. So, E = /A
353. Two charges 6C and 9C are placed 10 cm apart in air. Calculate the electrostatic repulsive force between them. What will be the new force when they placed in water in same distance away? ( of water is 80 )
[48.6N, 0.6N]
Hints: Electric force between two charges, according to Coulomb’s law is F = When charges are placed in dielectric, the electric force decreases by dielectric constant times. So, the new force F’ = F/
354. Two electrons are placed 1m away in air. Compare the electric and gravitational force between them.
[ – 4.17 x 1042 :1 ]
Hints: Electric force between two electrons Fe = ……..(i), where e = 1.6 x 10-19 C Gravitational force between two electrons Fg = ………..(ii), where m = 9.1×10-31Kg Divide (i) by (ii) to obtain the result. Negative sign in ratio is required to show that electric force in this case is repulsive while gravitational force is attractive i.e. they are opposite in nature.
355. Two point charges 6 x 10-8 C and -9 x 10-8 C are 50 cm apart in vacuum. At what point on the line joining them are the (a) electric field intensity and (b) potential zero?
[222 cm, 20cm from first]
[ 3.75 x 1013] Hints: Use the mathematical expression of quantization of charge is q = ne, where ‘n’ be the number of basic charge and ‘e’ is the charge of an electron equal to 1.6 x 10-19 C.
350. A body of mass 5mg is charged to 1200C. Does its mass change after charging? If so, calculate the magnitude of the mass after charging.
[ 4.99 mg] Hints: The decrease in mass due to charging is m = n × me , where n is number of electrons that are taken out from the body and me is the mass of an electron. From quantization of charge, n = q/e. So, the decrease in mass be m = q/e x me
351. Calculate the total flux generated from a charge of magnitude 88.5 nC. If the charge is surrounded by a sphere, calculate the flux passing through a section of the sphere which subtends an angle 2/5 steradian to the charge.
[10000Wb,1000Wb]
Hints: Total flux generated from a charged body is given by Gauss’s law i.e. = . The emitted flux scattered around 4 steradian angle. Applying unitary method flux through the given section can be determined.
352. If 250 weber flux passes through a cross-section of a cylinder having radius 20 cm, calculate the electric flux density at that point
[ 1990 wb/m2]
Hints: Flux density is the flux per unit area. So, E = /A
353. Two charges 6C and 9C are placed 10 cm apart in air. Calculate the electrostatic repulsive force between them. What will be the new force when they placed in water in same distance away? ( of water is 80 )
[48.6N, 0.6N]
Hints: Electric force between two charges, according to Coulomb’s law is F = When charges are placed in dielectric, the electric force decreases by dielectric constant times. So, the new force F’ = F/
354. Two electrons are placed 1m away in air. Compare the electric and gravitational force between them.
[ – 4.17 x 1042 :1 ]
Hints: Electric force between two electrons Fe = ……..(i), where e = 1.6 x 10-19 C Gravitational force between two electrons Fg = ………..(ii), where m = 9.1×10-31Kg Divide (i) by (ii) to obtain the result. Negative sign in ratio is required to show that electric force in this case is repulsive while gravitational force is attractive i.e. they are opposite in nature.
355. Two point charges 6 x 10-8 C and -9 x 10-8 C are 50 cm apart in vacuum. At what point on the line joining them are the (a) electric field intensity and (b) potential zero?
[222 cm, 20cm from first]
Hints: Since two charges are opposite in nature, potential will be zero at any point between them (at point N) and intensity will be zero outside them towards the smaller charge ( at point M). (a) Let at point M, ‘x’ distance from q1, resultant field intensity is zero. Intensity at M due to charge q1, E1 = =
Intensity at M due to charge q2, E2 = = According to question,
(b) Let at point N, ‘x’ distance from q1, resultant potential is zero. Potential at N due to charge q1, V1 = =
Potential at N due to charge q2,V2 = = According to question,
Intensity at M due to charge q2, E2 = = According to question,
(b) Let at point N, ‘x’ distance from q1, resultant potential is zero. Potential at N due to charge q1, V1 = =
Potential at N due to charge q2,V2 = = According to question,
356. Two charged metal balls, each of mass 0.05 g are suspended from the same point by means of two silk threads each 5 cm long. Calculate the charge on each if they repel each other to a distance of 6 cm. [ 1.22×10-8 C] Hints: In equilibrium condition, T Cos = mg …………..(i) and T Sin = F ………….(ii). Dividing, the electrostatic force F = mg Tan. Or, = mg Tan, where Tan = p/b = 3/4 .
357. Two charges of magnitude 9C and -9C are placed two vertices of an equilateral triangle of side 1m. What is the electric field intensity at third vertex? If 6C charge is placed at the third vertex, calculate the force experienced by the charge. [ 81000N/C, 0.48N] Hints: Determine electric field intensity due to first and second charge separately. The angle between them is 1200. So, apply parallelogram law of vector addition to get resultant intensity. Intensity at C due to charge A, E1 = = = 81000 N/C Intensity at C due to charge B, E2 = = = 81000 N/C The angle between E1 and E2 is 1200. So, resultant intensity E = If 6ยตC charge is placed in third vertex, force experienced by the charge F = Eq.
358. Find the potential and field due to a charged hollow sphere of radius 20 cm at the following distances (a) 10 cm from the centre, (b) 25 cm from the sphere if charge on the sphere is 6C.
[2.7 x 105V, 2.16 x 105V]
Hints: (a) the region inside the hollow sphere is the equi-potential region and the potential is equal to that of the surface. So, take distance equal to the radius of the sphere. Since there is no charge inside the hollow sphere, due to skin effect, intensity is zero. (b) Outside the sphere, potential V = and field intensity E = , where r = 25 cm = 0.25m.
359. A charged oil drop remains stationary when situated between two parallel metal plates 25 mm apart and a pd of 1000V is applied to the plates. Find the charge on the oil drop if it has a mass of 5 x 10-15 kg . Also calculate the number of basic charge present.
[ 1.25×10-18C, 8 ]
Hints: Downward gravitational force = mg. Upward electrostatic force = Eq. Since the body is stationary, downward gravitational force ‘mg’ must be equal to the upward electric force ‘Eq’ So, Eq = mg Or, x q = mg Or, q = The number of basic charge, according to quantization of charge is, n = q/e.
360. A charged oil drop of radius 1.3 x 10-6 m is prevent from falling under gravity by the vertical field between two parallel plates charged to a pd of 8340 V. If the plate separation is 16 mm and density of oil is 920 kg/m3, calculate the magnitude of charge present. [ e ] Hints: Determine mass of the oil drop by mass = density x volume and follow the above process. i.e. m = x v = x 4/3 r3
361. An electron starts from a point of a conductor and reaches a second conductor with a velocity 107 m/s , calculate the pd in volts. [281V] Hints: Use principle of conservation of energy i.e. KE ½ mv2 = PE Ve So, pd V =
362. A high voltage terminal of a generator consist of a spherical conducting shell of radius 0.5 m. Estimate the maximum potential to which it can be raised in air for which electric breakdown occurs when the electric intensity exceeds 3 x 106 V/m? [1.5 x 106 V] Hints: Maximum electric field intensity ( dielectric strength of air) E = …….(i) Maximum PD due to the charge V = ……………..(ii) Dividing (i) by (ii), we get V = E r
363. Two positive point charges 10 and 14 micro coulomb are 20 cm apart. Calculate the work done if the distance between them was reduced to half the original distance? [ 6.3 J ] Hints: Make any one of the charges stationary and bring another charge to half of the distance. Use formula, Work done W = V q Where V is potential difference between two points due to stationary charge and q is the moving charge.
364. What is the potential gradient between two parallel plane conductors when their separation is 20 mm and pd 400 V is applied to them? Calculate the force on an oil drop between the plates if the drop carries charge 8 x 10-19 C?
[ 1.2 x 104 V/m, 1.6 x 10-14 N]
Hints: Potential gradient E = V / d, where V is the potential difference and d is the plate separation. Force experienced by charge particle in an electric field is F = E q
357. Two charges of magnitude 9C and -9C are placed two vertices of an equilateral triangle of side 1m. What is the electric field intensity at third vertex? If 6C charge is placed at the third vertex, calculate the force experienced by the charge. [ 81000N/C, 0.48N] Hints: Determine electric field intensity due to first and second charge separately. The angle between them is 1200. So, apply parallelogram law of vector addition to get resultant intensity. Intensity at C due to charge A, E1 = = = 81000 N/C Intensity at C due to charge B, E2 = = = 81000 N/C The angle between E1 and E2 is 1200. So, resultant intensity E = If 6ยตC charge is placed in third vertex, force experienced by the charge F = Eq.
358. Find the potential and field due to a charged hollow sphere of radius 20 cm at the following distances (a) 10 cm from the centre, (b) 25 cm from the sphere if charge on the sphere is 6C.
[2.7 x 105V, 2.16 x 105V]
Hints: (a) the region inside the hollow sphere is the equi-potential region and the potential is equal to that of the surface. So, take distance equal to the radius of the sphere. Since there is no charge inside the hollow sphere, due to skin effect, intensity is zero. (b) Outside the sphere, potential V = and field intensity E = , where r = 25 cm = 0.25m.
359. A charged oil drop remains stationary when situated between two parallel metal plates 25 mm apart and a pd of 1000V is applied to the plates. Find the charge on the oil drop if it has a mass of 5 x 10-15 kg . Also calculate the number of basic charge present.
[ 1.25×10-18C, 8 ]
Hints: Downward gravitational force = mg. Upward electrostatic force = Eq. Since the body is stationary, downward gravitational force ‘mg’ must be equal to the upward electric force ‘Eq’ So, Eq = mg Or, x q = mg Or, q = The number of basic charge, according to quantization of charge is, n = q/e.
360. A charged oil drop of radius 1.3 x 10-6 m is prevent from falling under gravity by the vertical field between two parallel plates charged to a pd of 8340 V. If the plate separation is 16 mm and density of oil is 920 kg/m3, calculate the magnitude of charge present. [ e ] Hints: Determine mass of the oil drop by mass = density x volume and follow the above process. i.e. m = x v = x 4/3 r3
361. An electron starts from a point of a conductor and reaches a second conductor with a velocity 107 m/s , calculate the pd in volts. [281V] Hints: Use principle of conservation of energy i.e. KE ½ mv2 = PE Ve So, pd V =
362. A high voltage terminal of a generator consist of a spherical conducting shell of radius 0.5 m. Estimate the maximum potential to which it can be raised in air for which electric breakdown occurs when the electric intensity exceeds 3 x 106 V/m? [1.5 x 106 V] Hints: Maximum electric field intensity ( dielectric strength of air) E = …….(i) Maximum PD due to the charge V = ……………..(ii) Dividing (i) by (ii), we get V = E r
363. Two positive point charges 10 and 14 micro coulomb are 20 cm apart. Calculate the work done if the distance between them was reduced to half the original distance? [ 6.3 J ] Hints: Make any one of the charges stationary and bring another charge to half of the distance. Use formula, Work done W = V q Where V is potential difference between two points due to stationary charge and q is the moving charge.
364. What is the potential gradient between two parallel plane conductors when their separation is 20 mm and pd 400 V is applied to them? Calculate the force on an oil drop between the plates if the drop carries charge 8 x 10-19 C?
[ 1.2 x 104 V/m, 1.6 x 10-14 N]
Hints: Potential gradient E = V / d, where V is the potential difference and d is the plate separation. Force experienced by charge particle in an electric field is F = E q
Capacitor
365. A capacitor has a charge of 50 C when a p.d. of 50 V is applied to it. Calculate the capacitance of the capacitor.
[ 1 F ] Hints: Charge stored by capacitor is given by q = CV. So, C = q / V
366. Three capacitors of capacitance 2F, 4F and 6F are connected in (a) series and (b) parallel. Calculate resultant capacitance. If the combination is connected to a source of 12 V, calculate total energy stored in each case.
[ 12/11 and 12F,0.00086J] Hints: At first, determine equivalent capacitance using relation for series and C = C1 +C2 + C3 for parallel combination. The energy stored by the capacitor is given by U = ½ CV2, where C be the equivalent capacitance in each case.
367. Three capacitors of capacitance 3F, 10F and 15F are connected in series with a 15 V battery. What is (a) the charge in 3 F capacitor? (b) pd across 10F capacitor? (c) energy stored in 15 F capacitor?
[3V, 30C, 3×10-5 J] Hints: (a) At first, determine equivalent capacitance by relation and apply q = CV to determine charge stored by the capacitor. In series combination, each capacitor stores equal amount of charge. (b) The relation q = C2 V2 gives the pd across the second capacitor and (c) U3 = q2/2C3 gives the energy stored by the third capacitor.
368. 8 drops of water are equally charged. They combine together to form a bigger drop. Compare the capacity and the potential of the bigger drop with that of smaller drop.
[ 2:1, 4:1 ]
365. A capacitor has a charge of 50 C when a p.d. of 50 V is applied to it. Calculate the capacitance of the capacitor.
[ 1 F ] Hints: Charge stored by capacitor is given by q = CV. So, C = q / V
366. Three capacitors of capacitance 2F, 4F and 6F are connected in (a) series and (b) parallel. Calculate resultant capacitance. If the combination is connected to a source of 12 V, calculate total energy stored in each case.
[ 12/11 and 12F,0.00086J] Hints: At first, determine equivalent capacitance using relation for series and C = C1 +C2 + C3 for parallel combination. The energy stored by the capacitor is given by U = ½ CV2, where C be the equivalent capacitance in each case.
367. Three capacitors of capacitance 3F, 10F and 15F are connected in series with a 15 V battery. What is (a) the charge in 3 F capacitor? (b) pd across 10F capacitor? (c) energy stored in 15 F capacitor?
[3V, 30C, 3×10-5 J] Hints: (a) At first, determine equivalent capacitance by relation and apply q = CV to determine charge stored by the capacitor. In series combination, each capacitor stores equal amount of charge. (b) The relation q = C2 V2 gives the pd across the second capacitor and (c) U3 = q2/2C3 gives the energy stored by the third capacitor.
368. 8 drops of water are equally charged. They combine together to form a bigger drop. Compare the capacity and the potential of the bigger drop with that of smaller drop.
[ 2:1, 4:1 ]
Hints: Let, v be the volume of a smaller drop. Then the volume of the larger drop V = 8 x v
Or,
Or, R = 2r, i.e. radius of the larger drop is twice of the smaller drop.
Or,
Or, R = 2r, i.e. radius of the larger drop is twice of the smaller drop.
Capacitance of smaller spherical drop C1 = 40r, where ‘r’ is radius of smaller drop. Capacitance of larger spherical drop C2 = 40 R, where ‘R’ is radius of larger drop. Dividing, Potential of a smaller drop ……….(i)
Potential of larger ……….(ii)
Divide to obtain result.
369. 2F and 5F capacitors are charged to 500V and 100 V battery respectively. The capacitors are then joined positive to positive and negative to negative plate. Calculate common potential and loss of energy.
[ 0.114 J] Hints: Common pd across the combination is and the loss of energy due to combination is
370. Two capacitors of 4F and 12F are connected in series and the combination is joined across a 200V battery. They are now isolated and connected in parallel. Calculate the common potential and energy loss. [ 75 V, 0.015 J ] Hints: At first determine equivalent capacitance using relation . In series combination, each capacitor stores equal magnitude of charge. So, charge stored by each capacitor q = CV. Now, Common pd across the combination is . Initial energy stored by capacitor U1 = , where C is the equivalent capacitance. Final energy stored by capacitor U2 = , where Vc is common PD. Loss of energy is the difference between initial and final energy, i.e. U =U1 – U2 Alternative method: Energy loss U = , where V1 and V2 are pd across C1 and C2. Since V1 and V2 are not known, they are replaced by charge as U =
371. A 21V battery is connected across 3F and 4F capacitors in series. Calculate the pd across each capacitor and their total energy. The capacitors are disconnected and then joined plates of like charge together. Calculate the charge on 3F and energy stored in 4F capacitor. [ 30.8C,2.1 x 10-4 J] Hints: Determine equivalent capacitance and the charge stored by each capacitor. Using relation q = CV, determine pd across each capacitor. Determine common pd across the combination, . Relation q = CV gives charge and U = ½ CV2 gives energy stored by capacitor, where V is the common pd.
372. A capacitor of capacitance 60F is connected to 100 V dc supply. (i) Calculate energy stored by the capacitor. (ii) If the capacitor is disconnected to source and connected to a chargeless capacitor of capacitance 30F, calculate the new energy on each capacitor and loss of energy. [ 0.3J, 0.13J, 0.067J, 0.0103J ] Hints: (ii) Use formula to calculate common pd. Since second capacitor is chargeless, pd across it is zero. Q = CV gives the new charge stored by each capacitor.
373. A sheet of paper 40 mm wide and 1.5 x 10-2 mm thick is used as a dielectric of capacitor of capacitance 2 F which has foil of same wide. If dielectric constant of paper is 2.5, calculate the length of paper required. [ 34 m ] Hints: Capacitance of the parallel plate capacitor C = and the A = l × b gives the length of the sheet.
374. Two plane parallel conducting 15 mm apart are held horizontal one above the other. The upper plate is maintained +1500 V while the lower is earthen. Calculate the number of electrons which must be attached to a small oil drop of mass 4.9 x 10-15 Kg, if it remains stationary in the air between the plates. If the potential of the upper plate is suddenly changed to -1500V, what is the initial acceleration of the oil drop? [ 2g ] Hints: Since the oil drop is stationary, gravitational downward force must be equal to the upward electric force. That is mg = Eq. In second case, both forces are along the downward direction, total force is equal to F = ( mg + Eq ) and the total force divided by mass gives acceleration.
375. Calculate the equivalent capacitance of the shown figures, each capacitor has capacitance C.
Potential of larger ……….(ii)
Divide to obtain result.
369. 2F and 5F capacitors are charged to 500V and 100 V battery respectively. The capacitors are then joined positive to positive and negative to negative plate. Calculate common potential and loss of energy.
[ 0.114 J] Hints: Common pd across the combination is and the loss of energy due to combination is
370. Two capacitors of 4F and 12F are connected in series and the combination is joined across a 200V battery. They are now isolated and connected in parallel. Calculate the common potential and energy loss. [ 75 V, 0.015 J ] Hints: At first determine equivalent capacitance using relation . In series combination, each capacitor stores equal magnitude of charge. So, charge stored by each capacitor q = CV. Now, Common pd across the combination is . Initial energy stored by capacitor U1 = , where C is the equivalent capacitance. Final energy stored by capacitor U2 = , where Vc is common PD. Loss of energy is the difference between initial and final energy, i.e. U =U1 – U2 Alternative method: Energy loss U = , where V1 and V2 are pd across C1 and C2. Since V1 and V2 are not known, they are replaced by charge as U =
371. A 21V battery is connected across 3F and 4F capacitors in series. Calculate the pd across each capacitor and their total energy. The capacitors are disconnected and then joined plates of like charge together. Calculate the charge on 3F and energy stored in 4F capacitor. [ 30.8C,2.1 x 10-4 J] Hints: Determine equivalent capacitance and the charge stored by each capacitor. Using relation q = CV, determine pd across each capacitor. Determine common pd across the combination, . Relation q = CV gives charge and U = ½ CV2 gives energy stored by capacitor, where V is the common pd.
372. A capacitor of capacitance 60F is connected to 100 V dc supply. (i) Calculate energy stored by the capacitor. (ii) If the capacitor is disconnected to source and connected to a chargeless capacitor of capacitance 30F, calculate the new energy on each capacitor and loss of energy. [ 0.3J, 0.13J, 0.067J, 0.0103J ] Hints: (ii) Use formula to calculate common pd. Since second capacitor is chargeless, pd across it is zero. Q = CV gives the new charge stored by each capacitor.
373. A sheet of paper 40 mm wide and 1.5 x 10-2 mm thick is used as a dielectric of capacitor of capacitance 2 F which has foil of same wide. If dielectric constant of paper is 2.5, calculate the length of paper required. [ 34 m ] Hints: Capacitance of the parallel plate capacitor C = and the A = l × b gives the length of the sheet.
374. Two plane parallel conducting 15 mm apart are held horizontal one above the other. The upper plate is maintained +1500 V while the lower is earthen. Calculate the number of electrons which must be attached to a small oil drop of mass 4.9 x 10-15 Kg, if it remains stationary in the air between the plates. If the potential of the upper plate is suddenly changed to -1500V, what is the initial acceleration of the oil drop? [ 2g ] Hints: Since the oil drop is stationary, gravitational downward force must be equal to the upward electric force. That is mg = Eq. In second case, both forces are along the downward direction, total force is equal to F = ( mg + Eq ) and the total force divided by mass gives acceleration.
375. Calculate the equivalent capacitance of the shown figures, each capacitor has capacitance C.
376. In figure below PD is 100V, capacitors are 30 F and 60 F. If S2 is left open and S1 is closed, calculate the quantity of the charge on each capacitor in the given figure. If S1 is now opened and S2 is closed how many charge will flow through the 10 ohm resistor. [ 0.002C ] Hints: Two capacitors are is series, so determine equivalent capacitance. q = CV gives the charge stored by each capacitor. The equal amount of charge stored by one capacitor flows through the resistor. In series combination, charge or current doesn’t increase which increase in parallel combination.
377. A capacitor of capacitance 9F is charged from a source of emf 200V. It is now disconnected from the source and connected in parallel with 3F uncharged capacitor. The second capacitor is now removed and discharged. What charge remains in the first capacitor? How many times would the process have to be performed to reduce the charge on 9F capacitor below 50% of its initial value? [ 1012 C ] Hints: Determine common pd after the combination of second capacitor. q =CV gives the charge remaining in the first capacitor. Repeat the process.
377. A capacitor of capacitance 9F is charged from a source of emf 200V. It is now disconnected from the source and connected in parallel with 3F uncharged capacitor. The second capacitor is now removed and discharged. What charge remains in the first capacitor? How many times would the process have to be performed to reduce the charge on 9F capacitor below 50% of its initial value? [ 1012 C ] Hints: Determine common pd after the combination of second capacitor. q =CV gives the charge remaining in the first capacitor. Repeat the process.
Revision for ambitious students
378. Two charges 10 C and 20 C are placed at a separation of 2 m. Find the electric potential due to the pair at the middle of the line joining the two charges. [2.7 x105 V ]
378. Two charges 10 C and 20 C are placed at a separation of 2 m. Find the electric potential due to the pair at the middle of the line joining the two charges. [2.7 x105 V ]
379. A charge Q is to be divided on two parts. What should be the value of charge so that the force between them maximum. [ Q / 2 each ]
380. A charged particle of mass 1.0 g is suspended through a silk thread of length 40 cm in a horizontal electric field of 4 x 104 N/C. If the particle stays at a distance 24 cm from the wall in equilibrium, find the charge on the particle. [ 1.8 x 10-7 ]
381. Charges 2 x 10-6 and 1 x 10-6 are placed at A and B points of a square of side 5 cm. How much work will be done against the electric field in a moving of a charge 10-6 from C to D? [ 0.053 J ]
382. Two charged particles are placed at a distance 1.0 cm away from each other . What is minimum possible value of force between them? [ 2.3 x 10-24 N ]
383. A particle having the charge 2 C and the mass 100 g is placed at the bottom of the smooth inclined plane of inclination 300 . Where should the another particle having same charge and mass be placed on the incline so that it may remain in equilibrium? [ 26.8 cm ]
384. Two particles A and B, each having the charge Q are placed d cm apart. Where a particle of charge q should be placed on the perpendicular bisector of AB so that it experience maximum force? Also calculate the force. [ d/22, 3.08F ]
385. A charge of magnitude 1.0 C is placed at the corner of a cube. How much flux passes through the cube? [ 1.4 x 103 wb]
386. An inclined plane is making an angle of 300 with the horizontal electric field E of 100 V / m. A particle of mass 1kg and charge 0.01C allowed to slide down from a height of 1m . If the coefficient of friction is 0.2, find the time it will take the particle to reach the bottom. [2.13m/s2 and 1.3 s]
387. A parallel plate capacitor has plates of area 200 cm2 and separation between the plates 1.00 mm. What PD will be developed if a charge of 1.00 nC is given to the capacitor? If the plate separation is now increased to 2.00 mm, what will be the new PD ? [ 5.6V, 11.3V ]
388. A 50 V battery is connected to a 2 F capacitor. Calculate energy in the capacitor. If the terminals of the battery are removed and mica sheet of dielectric constant 5 is placed between the plates, find the loss of energy. [ 2.5 x 10-3, 2 x 10-3 J]
389. Find the capacitance between A and B. Find the charges on the three capacitors. Taking the potential of B is zero, find the potential of D. [ 6 F, 48 & 96 C, 16 V ]
390. Calculate the equivalent capacitance of the given circuit if each has capacitance ‘C’. [ Ans : 2C ]
391. A pendulum bob of mass 50g is suspended by a thread of length 1m in a region having vertical upward electric field 100 N / C. If the bob has charge 6 micro coulomb, what is the time period of oscillation ? What is the new time period if the field is horizontal ?
392. In adjacent figure, if value of each capacitor is C, calculate the equivalent capacitance.
393. Two capacitor of capacitance 4 and 12 micro coulomb are connected in series and the combination is further connected to a source of PD 200V. The charged capacitors are now isolated and connected in parallel. Calculate the common potential. [HSEB –2002]
Answer of Objective questions
Q.N Ans Q.N Ans Q.N Ans Q.N Ans Q.N Ans
15 b 20 b 25 a
16 a 21 d 26 c
17 d 22 b 27 c
18 c 23 a 28
19 b 24 b 29
15 b 20 b 25 a
16 a 21 d 26 c
17 d 22 b 27 c
18 c 23 a 28
19 b 24 b 29
