Chapter 10 Physics Answers

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  • [FREE] Chapter 10 Physics Answers | HOT

    Which of the two figures in incorrect? Answer: Figure a is incorrect. As a result, the water should not rise higher in the tube where there is a kink i. The cylindrical tube of a spare pump has a cross-section of 8. If the liquid flow inside the...
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    The length of the slider is 30 cm. What is the surface tension of the film? Answer: In present case force of surface tension is balancing the weight of 1. What is the weight supported by a film of the same liquid at the same temperature in Fig. In...
  • Physics Answers

    Ignore the small change in the volume of the gas. Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill up to a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases? If so, why do the vessels filled with water to that same height give different readings on a weighing scale?
  • CBSE 10th Science Board Exam 2021: Important MCQs From Chapter 10 Light - Reflection & Refraction

    Answer: Pressure and therefore force on the two equal base areas are identical. But force is exerted by water on the sides of the vessels also, which has a non-zero vertical component when sides of the vessel are not perfectly normal to the base. This net vertical component of force by water on the sides of the vessel is greater for the first vessel than the second. Hence, the vessels weigh different even when the force on the base is the same in the two cases. During blood transfusion, the needle is inserted in a vein where the gauge pressure is Pa. At what height must the blood container be placed so that blood may just enter the vein? Discuss qualitatively. Answer: a If dissipative forces are present, then some forces in liquid flow due to pressure difference is spent against dissipative forces, due to which the pressure drop becomes large. Take viscosity of blood to be 2. Density of blood is 1.
  • NCERT Solutions For Class 12 Physics Chapter 10

    A plane is in level flight at constant speed and each of its wings has an area of 25 m2. Take the viscosity of air at the temperature of the experiment to be 1. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air. Neglecting upward thrust due to air, we find that terminal speed is Question A narrow tube of radius 1. By what amount does the mercury dip down in the tube relative to the liquid surface outside?
  • NCERT Solutions For Class 10 Science Chapter 10 Light Reflection And Refraction

    Surface tension of mercury at the temperature of the experiment is 0. Two narrow bores of diameters 3. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7. Take the angle of contact to be zero and density of water to be 1. Answer: Let rx be the radius of one bore and r2 be the radius of second bore of the U-tube. The, if h1 and h2 are the heights of water on two sides, then Question This density variation is called the law of atmospheres.
  • Exams 2021, Tests & Answers

    Obtain this law assuming that the temperature of atmosphere remains a constant isothermal conditions. Also, assume that the value of g remains constant. Assume that the balloon maintains a constant radius as it rises. How high does it rise? Answer: a We know that rate of decrease of density p of air is directly proportional to the height y.
  • Chapter 10 - Fluids

    A wave has a frequency of 60 Hz and a wave length of 2. What are the period and the velocity of propagation? A traveling wave with frequency of 40 Hz moves to the right. How long does it take for one complete wave length to pass a given point? As you stand on a pier, you notice that a crest of a wave passes every 1. If the distance between crests is 4. Standing on a cliff, you watch the ocean waves and notice that a new crest strikes the cliff every 3 seconds. What is the speed of the surface waves on the water? A body oscillates with simple harmonic motion along the x axis. Determine the amplitude, frequency and period. Find the frequency and period of the motion b. Find the amplitude of motion c. Find the phase angle d. The displacement of the particle b. The velocity of the particle c. The acceleration of the particle d. The period and amplitude of the motion.
  • Physics Class 9 Chapter 1 Worksheet

    List four properties of the image formed by a plane mirror. Properties of images formed by a plane mirrors: i The image formed by a plane mirror is virtual and erect. List the four properties of the image formed by a convex mirror. Properties of the image formed by a convex mirror: i This image is always virtual and erect. State the laws of refraction of light. Laws of refraction of light: The refraction of light obeys the following two laws: First law: The incident ray, the refracted ray and normal to the surface of separation at the point of incidence, all lie in the same plane.
  • NCERT Solutions For Class 10 Science Chapter 10

    Second law: The ratio of the sine of the angle of incidence and the sine of the angle of refraction is constant for a given pair of media. The ratio n2 is called refractive index of the second medium with respect to the first medium. What is meant by refraction of light? Refraction of light: The phenomenon of bending of light from its straight line path as it passes obliquely from one transparent medium to another is called refraction of light. The path of the ray of light in the first medium is called incident ray. The path of the ray of light in the second medium is called refracted ray. The angle between the incident ray and the normal at the surface of separation is called angle of incidence i. The angle between the refracted ray and the normal at the surface of separation is called angle of refraction r. Questions for Practice Give three points of difference between real and virtual images. Answer 1: i Here the rays actually meet at the image point.
  • OpenStax College Physics Solution, Chapter 10, Problem 16 (Problems & Exercises) (5:24)

    Question 2: Distinguish between a concave and convex lens. Answer 2: a It is thicker at the centre than at the edges. Concave lens: a It is thinner at the centre than at the edges. Questions from Board Papers Question 1: Define the principal focus of a concave mirror. Calculate the focal length of the mirror of radius of curvature of 20 cm. Answer 1: The principal focus of concave mirror is a point on its principal axis at which a beam of light parallel to the principal axis actually converges after reflection from the mirror. Question 2: State the type of mirror preferred as i rear view mirror in vehicles, ii shaving mirror. Justify your answer giving two reason in each case. Answer 2: i A convex mirror is preferred as a rear-view mirror because a It always forms an erect, virtual and diminished image of an object placed anywhere in front of it.
  • Chapter 10 Class 10 - Light - Reflection And Refraction

    Light rays that are parallel to the principal axis of a concave mirror converge at a specific point on its principal axis after reflecting from the mirror. This point is known as the principal focus of the concave mirror. The radius of curvature of a spherical mirror is 20 cm. What is its focal length? Name the mirror that can give an erect and enlarged image of an object. When an object is placed between the pole and the principal focus of a concave mirror, the image formed is virtual, erect, and enlarged. Why do we prefer a convex mirror as a rear-view mirror in vehicles? Convex mirrors give a virtual, erect, and diminished image of the objects placed in front of them. They are preferred as a rear-view mirror in vehicles because they give a wider field of view, which allows the driver to see most of the traffic behind him. A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? The light ray bends towards the normal.
  • Physics Chapter 10 Answers

    When a ray of light travels from an optically rarer medium to an optically denser medium, it gets bent towards the normal. Since water is optically denser than air, a ray of light travelling from air into the water will bend towards the normal. The refractive index of diamond is 2. What is the meaning of this statement? This suggests that the speed of light in diamond will reduce by a factor 2. Define 1 dioptre of power of a lens. Power of lens is defined as the reciprocal of its focal length. It is denoted by D.
  • Kerala Syllabus SSLC 10th Standard Physics Solutions Guide

    Question Since the flow of water in a river is rapid, way cannot be treated as streamlined motion, the theorem cannot be used. Glycerine flows steadily through a horizontal tube of length 1. If the amount of glycerine collected per second at one end is 4. Answer: Question Figures a and b refer to the steady flow of a non-viscous liquid. Which of the two figures in incorrect? Answer: Figure a is incorrect. As a result, the water should not rise higher in the tube where there is a kink i.
  • NCERT Solutions For Class 9 Science Chapter 10

    The cylindrical tube of a spare pump has a cross-section of 8. If the liquid flow inside the tube is 1. A U-shaped wire is dipped in a soap solution, and removed. A thin soap film formed between the wire and a light slider supports a weight of 1. The length of the slider is 30 cm. What is the surface tension of the film?
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    Answer: In present case force of surface tension is balancing the weight of 1. What is the weight supported by a film of the same liquid at the same temperature in Fig. In Fig. What is the pressure inside a drop of mercury of radius 3. The atmospheric pressure is 1. Also give the excess pressure inside the drop. Answer: Since data is correct up to three significant figures, we should write total pressure inside the drop as 1.
  • Physics Answers - Assignment Expert

    What is the excess pressure inside a bubble of soap solution of radius 5. If an air bubble of the same dimension were formed at depth of A tank with a square base of area 1. The bottom of the partition has a small-hinged door of area 20 cm2. The tank is filled with water in one compartment, and an acid of relative density 1. Compute the force necessary to keep the door close. A manometer reads the pressure of a gas in an enclosure as shown in Fig. The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury. Ignore the small change in the volume of the gas. Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill up to a particular common height.
  • NCERT Solutions For Class 12 Physics Chapter 10 Wave Optics In PDF

    Is the force exerted by the water on the base of the vessel the same in the two cases? If so, why do the vessels filled with water to that same height give different readings on a weighing scale? Answer: Pressure and therefore force on the two equal base areas are identical. But force is exerted by water on the sides of the vessels also, which has a non-zero vertical component when sides of the vessel are not perfectly normal to the base. This net vertical component of force by water on the sides of the vessel is greater for the first vessel than the second. Hence, the vessels weigh different even when the force on the base is the same in the two cases. During blood transfusion, the needle is inserted in a vein where the gauge pressure is Pa.
  • MP Board Class 12th Physics Important Questions Chapter 10 Wave Optics

    At what height must the blood container be placed so that blood may just enter the vein? Discuss qualitatively. Answer: a If dissipative forces are present, then some forces in liquid flow due to pressure difference is spent against dissipative forces, due to which the pressure drop becomes large. Take viscosity of blood to be 2. Density of blood is 1. A plane is in level flight at constant speed and each of its wings has an area of 25 m2. Take the viscosity of air at the temperature of the experiment to be 1. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air. Neglecting upward thrust due to air, we find that terminal speed is Question A narrow tube of radius 1. By what amount does the mercury dip down in the tube relative to the liquid surface outside?
  • Choose A Chapter From College Physics | OpenStax College Physics Answers

    Surface tension of mercury at the temperature of the experiment is 0. Two narrow bores of diameters 3. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7. Take the angle of contact to be zero and density of water to be 1. Answer: Let rx be the radius of one bore and r2 be the radius of second bore of the U-tube. The, if h1 and h2 are the heights of water on two sides, then Question This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant isothermal conditions. Also, assume that the value of g remains constant. Assume that the balloon maintains a constant radius as it rises. How high does it rise? Answer: a We know that rate of decrease of density p of air is directly proportional to the height y.
  • MP Board Class 12th Physics Important Questions Chapter 10 Wave Optics – MP Board Solutions

    Important Questions for practice 1. Find angle of deviation. Calculate the polarizing angle and the angle of refraction in glass corresponding to this. An object of length 2. The length of object is perpendicular to principal axis. Find the size of image. Is the image erect or inverted? A radio can tune into any station of frequency band 7. Find the corresponding wave length range. Describe an astronomical telescope and derive an expression for its magnifying power using a labelled ray diagram.
  • Choose A Chapter From OpenStax College Physics

    Questions from Board Papers 1. Show that maximum intensity in interference pattern is four times the intensity due to each slit if amplitude of light emerging from slits is same. A person looking at a mesh of crossed wire is able to see the vertical wire more distinctly than the horizontal wire. Which defect he is suffering from? How can this defect be corrected? How is a wave front different from a ray?
  • Physics MCQs For Class 12 With Answers Chapter 10 Wave Optics

    Draw the geometrical shape of the wave fronts when. Define diffraction. Deduce an expression for fringe width of the central maxima of the diffraction pattern, produced by single slit illuminated with monochromatic light source 5. State the condition under which the phenomenon of diffraction of light takes place. Derive an expression for the width of the central maximum due to diffraction of light at a single slit.
  • NCERT Solutions For Class 10 Science Chapter 10 Light Reflection And Refraction - Learn CBSE

    Also draw the intensity pattern with angular position. Important Questions on 12th Physics Chapter 10 Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism? The speed of light in glass is not independent of the colour of light. The refractive index of a violet component of white light is greater than the refractive index of a red component. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism. What is the Brewster angle for air to glass transition? Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment? Hence, the normal component of velocity increases while the component along the surface remains unchanged.
  • PHYSICS CHAPTER 10 SECTION 2 Worksheet 1

    The wave picture of light is consistent with the experimental results. For sound waves, the Doppler formula for frequency shift differs slightly between the two situations: i source at rest; observer moving, and ii source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium? No Sound waves can propagate only through a medium. The two given situations are not scientifically identical because the motion of an observer relative to a medium is different in the two situations. Hence, the Doppler formulas for the two situations cannot be the same. In case of light waves, sound can travel in a vacuum. In a vacuum, the above two cases are identical because the speed of light is independent of the motion of the observer and the motion of the source.
  • NCERT Solutions For Class 11 Physics Chapter 10 Mechanical Properties Of Fluids - Free PDF

    When light travels in a medium, the above two cases are not identical because the speed of light depends on the wavelength of the medium. In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band? In a single slit diffraction experiment, if the width of the slit is made double the original width, then the size of the central diffraction band reduces to half and the intensity of the central diffraction band increases up to four times. In what way is diffraction from each slit related to the interference pattern in a double-slit experiment? The interference pattern in a double-slit experiment is modulated by diffraction from each slit. The pattern is the result of the interference of the diffracted wave from each slit.
  • NCERT Solutions For Class 6 Science Chapter 10: Motion And Measurement Of Distances

    When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why? This is because light waves are diffracted from the edge of the circular obstacle, which interferes constructively at the centre of the shadow. This constructive interference produces a bright spot. Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily. Bending of waves by obstacles by a large angle is possible when the size of the obstacle is comparable to the wavelength of the waves. On the one hand, the wavelength of the light waves is too small in comparison to the size of the obstacle. Thus, the diffraction angle will be very small.
  • Class X Physics Notes

    Hence, the students are unable to see each other. On the other hand, the size of the wall is comparable to the wavelength of the sound waves. Thus, the bending of the waves takes place at a large angle. Hence, the students are able to hear each other. Ray optics is based on the assumption that light travels in a straight line. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification? The justification is that in ordinary optical instruments, the size of the aperture involved is much larger than the wavelength of the light used. When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen.
  • RBSE Solutions For Class 12 Physics Chapter 10 Alternating Current

    Suggest a possible explanation. Weak radar signals sent by a low flying aircraft can interfere with the TV signals received by the antenna. As a result, the TV signals may get distorted. Hence, when a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns.
  • Chapter 1 - Measurement - Problems - Page 8: 1a

    What is the justification of this principle? The principle of linear superposition of wave displacement is essential to our understanding of intensity distributions and interference patterns. This is because superposition follows from the linear character of a differential equation that governs wave motion. If y1 and y2 are the solutions of the second order wave equation, then any linear combination of y1 and y2 will also be the solution of the wave equation. Related Links.
  • RBSE Solutions For Class 12 Physics Chapter 10 Alternating Current

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