ICSE 2025 Physics Class 10 PDF Past Paper - Nodia

 ICSE 2025 PHYSICS Including Case Based Questions CLASS 10 Chapter-wise Question Bank Based on Previous 20 Year 54 Papers NODIA AND COMPANY ICSE Physics Question Bank Class 10 Edition August 2024 Copyright © By Nodia and Company Information contained in this book has been obta...

 ICSE 2025 PHYSICS Including Case Based Questions CLASS 10 Chapter-wise Question Bank Based on Previous 20 Year 54 Papers NODIA AND COMPANY ICSE Physics Question Bank Class 10 Edition August 2024 Copyright © By Nodia and Company Information contained in this book has been obtained by author, from sources believes to be reliable. However, neither Nodia and Company nor its author guarantee the accuracy or completeness of any information herein, and Nodia and Company nor its author shall be responsible for any error, omissions, or damages arising out of use of this information. This book is published with the understanding that Nodia and Company and its author are supplying information but are not attempting to render engineering or other professional services. ISBN : 978-9384843809 NODIA AND COMPANY MRP Rs 560.00 This book is available on amazon and flipkart only and not available in market. Published by : NODIA AND COMPANY 125, Sector 6, Vidyadhar Nagar, Jaipur 302039 Phone :+91 9024037387  CONTENTS CHAP 1. Force : Turning Forces and Uniform Circular Motion 5-46 CHAP 2. Work, Power and Energy 47-95 CHAP 3. Mechanics 96-135 CHAP 4. Light 136-219 CHAP 5. Sound 220-262 CHAP 6. Electricity and Magnetism 263-366 CHAP 7. Heat 367-405 CHAP 8. Modern Physics 406-438 ******** DON’T WASTE MONEY PRINTING NODIA BOOKS Purchase paperback books directly from Amazon. Click any subject link to purchase a book. NODIA ICSE NODIA ICSE NODIA ICSE NODIA ICSE CHAPTERWISE CHAPTERWISE CHAPTERWISE CHAPTERWISE 20 Years PYQ Bank 20 Years PYQ Bank 20 Years PYQ Bank 20 Years PYQ Bank 2024-2005 (54 Papers) 2024-2005 (54 Papers) 2024-2005 (54 Papers) 2024-2005 (54 Papers) Main, SQP and Comp Main, SQP and Comp Main, SQP and Comp Main, SQP and Comp MATHEMATICS PHSYSICS CHEMISTRY BIOLOGY Class 10 Class 10 Class 10 Class 10 NODIA ICSE NODIA ICSE NODIA ICSE NODIA ICSE CHAPTERWISE CHAPTERWISE CHAPTERWISE CHAPTERWISE 20 Years PYQ Bank 20 Years PYQ Bank 20 Years PYQ Bank 20 Years PYQ Bank 2024-2005 (54 Papers) 2024-2005 (54 Papers) 2024-2005 (54 Papers) 2024-2005 (54 Papers) Main, SQP and Comp Main, SQP and Comp Main, SQP and Comp Main, SQP and Comp HISTORY COMPUTER GEOGRAPHY HINDI & APPLICATION Class 10 CIVICS Class 10 Class 10 Class 10 NODIA ICSE NODIA ICSE NODIA ICSE NODIA ICSE CHAPTERWISE CHAPTERWISE CHAPTERWISE CHAPTERWISE 20 Years PYQ Bank 20 Years PYQ Bank 20 Years PYQ Bank 20 Years PYQ Bank 2024-2005 (54 Papers) 2024-2005 (54 Papers) 2024-2005 (54 Papers) 2024-2005 (54 Papers) Main, SQP and Comp Main, SQP and Comp Main, SQP and Comp Main, SQP and Comp ENGLISH LITERATURE COMMERICAL LANGUAGE IN ECONOMICS STUDIES Class 10 ENGLISH Class 10 Class 10 Class 10 For more details whatsapp at 94142 43489  NODIA APP Free PDF For All Study Material Search Play Store by NODIA ICSE CHAPTERWISE PYQ CLASS 10 PHYSICS PAGE 5 CHAPTER 1 FORCE TURNING FORCES AND UNIFORM CIRCULAR MOTION SUMMARY w = mg where, m is the mass of the object and g , is the acceleration due to gravity. 1. FORCE As g varies from place to place on the earth, the weight of an object of mass m also varies. Its Force is an external effort in the form of push SI unit is the unit of force, i.e., Newton. or pull which changes/tends to change the state of rest or of uniform motion of an object along 3. RIGID BODY a straight line or changes/tends to change the Every material object is made up of a large shape of the object. number of particles. A rigid body is one whose The SI unit of force is Newton (N). 1 Newton size and shape remain the same whatever force is that force which produces an acceleration of 1 be applied to different parts of it. In an actual ms-2 in an object of mass 1 kg. The CGS unit of practice, no material object is truly rigid. force is dyne. 1 dyne is that force which produces an acceleration of 1 cm2 in an object of mass 1 g. 4. MOTION OF RIGID BODY 1 N = 105 dyne A rigid object has two types of motion, viz. The gravitational units of force are kilogram translatory motion and rotatory motion. force (or kilogram weight) and gram force (or gram weight) which are symbolically represented 4.1 Translational Motion as 1 kg - f (or kg - wt) and g - f or (g - wt ) In a translational motion, on application of a respectively. force, an object starts moving bodily with all 1 kg - f (or kg - wt) = 9.8 N its particles moving along parallel lines in the 1 g - f ^or g wt h = 980 dyne direction of force with the same speed. In the mathematical form its is defined as the rate 4.2 Rotational Motion of change of linear momentum Change in momentum In a rotatory motion, when a force is applied on an Force, F \ time object at a fixed point, an object starts rotating p1 - p2 about that point with its particles moving along \ t concentric circles with different speeds. m (v - u ) \ t & F \ ma Rotational effect of a force depends on two factors, viz. F = kma (i) the magnitude of the force ]F g , and Where, k =1 (ii) perpendicular distance of its line of action F =ma from the axis or point about which rotation is taking place. 2. MASS AND WEIGHT 5. SOME DEFINITIONS RELATED TO FORCE The mass of an object is defined as the quantity Some definition regarding to the force is given of matter contained in the object. It is a scalar below : quantity. Its SI unit is kilogram (kg). (i) Point of action of force is that point on a rigid The weight of an object on the earth is the object where a force acts. gravitational pull exerted on it by the earth. (ii) Line of action of force is an imaginary line Weight is a vector quantity. It is expressed as CH 1 : FORCE TURNING FORCES AND UNIFORM CIRCULAR MOTION passing through point of action of force drawn Moment of a couple = either force # in the direction of force. perpendicular distance between the forces (iii) According to the principle of transmissibility, the effect of a force on a rigid object depends = force # couple arm upon its magnitude and on the line along = F#d which it acts and remains unaltered if the The SI unit of moment of couple is newton-metre point of application is shifted to some other denoted by N - m. Its CGS unit is dyne-cm. point on the same line of action. 9. PRINCIPLE OF MOMENTS 6. MOMENT OF FORCE OR TORQUE According to the principle of moments, the sum of The turning effect of a force acting on an object anticlockwise moments of forces about any point about an axis is called moment of force or torque. must be equal to the sum of clockwise moments of Mathematically, moment of force or the torque the forces about the same point. exerted by the force about the point is the product of the applied force ^F h and the perpendicular 10. EQUILIBRIUM OF AN OBJECT distance of its line of action from the point about An object is said to be in equilibrium if a number which the object is free to rotate. of forces acting on it produce no change in its Moment of force or torque produced by a state of rest or of uniform motion (translational force about an axis is equal to the product of the or rotational). magnitude of force and its lever arm about the 1. An object in equilibrium is not necessarily in axis. It is represented by t (tau). rest while an object at rest is necessarily in Torque (t) = Force ^F h # Lever arm ^d=h equilibrium. Moment of force or torque is a vector quantity. Its 2. An object is said to be in static equilibrium SI unit is newton-metre (N-m) and CGS unit is if it remains in the state of rest under the dyne-centimetre (dyne-cm). influence of applied forces. 1 N - m = 107 dyne - cm 3. An object is said to be in dynamic equilibrium Moment of force is assigned a positive sign if the if it remains in the state of uniform motion turning tendency of the force is anticlockwise and (translational or rotational). a negative sign if the turning tendency of the force Under the action of a set of parallel forces, the is clockwise. two conditions for an object to be in equilibrium are: 7. COUPLE (i) The resultant of all the forces acting on an Couple is a pair of equal and opposite forces whose object must be equal to zero. lines of action of forces are not the same. Couple (ii) The resultant moment of all the forces acting causes the object to execute rotational motion. on the object about a point should be zero Arm of the couple is the perpendicular i.e.the sum of the anticlockwise moments distance between two equal and unlike parallel about any point must be equal to the sum of forces. the clockwise moments about the same point. 8. MOMENT OF A COUPLE 11. CENTRE OF GRAVITY The moment of a couple is equal to the product of Centre of gravity of an object is a point at which either of the forces and the perpendicular distance its total weight may be supposed to act. (called the arm of the couple) between their lines of action, i.e., ICSE CHAPTERWISE PYQ CLASS 10 PHYSICS PAGE 7 The position of the centre of gravity of some 12. STABLE EQUILIBRIUM regular shaped objects having uniform density is An object is said to be in stable equilibrium if it given in the following table: has a tendency to return to its original equilibrium position after being slightly disturbed e.g., a cone Object Figure of the Centre of lying on its base. object gravity Conditions for an object in stable equilibrium are: Uniform rod Mid-point (i) The centre of gravity of the object must be as of rod low as possible. Circular disc Geometric (ii) The vertical imaginary line from the centre of centre gravity must pass through the base or area of the support of the object. 13. UNSTABLE EQUILIBRIUM Sphere (solid Geometric An object is said to be in unstable equilibrium if or hollow) centre of it does not regain its original equilibrium position the sphere after being slightly disturbed. Conditions for an object to be in unstable Cylinder Mid-point equilibrium are: on the (i) The centre of gravity must be at its highest axis of the position, and cylinder (ii) On tilting the object slightly, the centre of gravity is lowered but when left the centre of Circular ring Centre of gravity tends to lower itself further and hence the ring the object topples. 14. NEUTRAL EQUILIBRIUM Rectangle or The point of An object is said to be in neutral equilibrium, parallelogram intersection if after being slightly disturbed, the body does of the not have any tendency either to regain its original diagonals equilibrium position or to tilt further from its new Triangular The point of position. lamina intersection Conditions for an object to be in neutral of medians equilibrium are: (i) The centre of gravity should remain unaffected on tilting the body, i.e. neither lowered nor raised. Hollow cone At a height (ii) The vertical line joining the centre of gravity (h/3) from and centre of the earth must always fall the base on within the base of the object. its axis 15. UNIFORM CIRCULAR MOTION If an object moves on a circular path with a Solid cone At a height constant speed, its motion is said to be a uniform (h/4) from circular motion in a plane. In uniform circular the base on motion, the speed of the body remains constant its axis but direction changes continuously, due to which velocity changes continuously, i.e. there is an acceleration in the uniform circular motion which is always directed towards the centre of the circle. Cube/Cuboid At the It is called centripetal acceleration. point of intersection 15.1 Centripetal Force of diagonals If a body performing uniform circular motion, it CH 1 : FORCE TURNING FORCES AND UNIFORM CIRCULAR MOTION is acted upon by a force which is directed towards F= 5 N the centre of the circle is called centripetal force. Thus (a) is correct option. Centripetal force is not a new force. Frictional force, gravitational force, electrical and magnetic 4. The condition for equilibrium is : force may act as a centripetal force. (a) The resultant of moments of all the forces acting on the body about the turning point 15.2 Centrifugal Force (a pseudo force) should be zero. Centrifugal force (a pseudo force) is an apparent (b) The resultant of all the forces acting on the force in a certain situation that a object appears body be zero only. to be acted upon by a force while there is no force (c) Both (a) and (b) acting on the object. (d) None of the above Ans : COMP 2024 An object is in equilibrium in a reference MULTIPLE CHOICE QUESTION coordinate system when all external forces (including moments) acting on it are balanced. This means that the net result of all the external 1. A moment of couple has a tendency to rotate forces and moments acting on this object is zero. the body in an anticlockwise direction. Then the Thus (a) is correct option. moment of couple is taken as: 5. A moment of couple has a tendency to rotate (a) positive (b) negative the body in an anticlockwise direction. Then the (c) maximum (d) zero moment of couple is taken as: Ans : SQP 2025 (a) positive (b) negative (c) maximum (d) zero If couple has a tendency to rotate the body in anti-clockwise direction then its moment is Ans : SQP 2024 taken positive and if the tendency of rotation is When the body tends to rotate in the anticlockwise clockwise then the moment is negative. direction under the influence of a Couple, then its Thus (a) is correct option. moment is called positive moment. 2. Which of the following is a class III lever? Thus (a) is correct option. (a) Pair of scissors (b) Wheelbarrow 6. Calculate the force which will produce a moment (c) Crowbar (d) Human forearm of force of 1575 dyne-cm, when the perpendicular Ans : MAIN 2024 distance between point of application of force and turning point is 45 cm. For class III lever force is applied in between the (a) 75 dyne (b) 45 dyne fulcrum and the load. (c) 35 dyne (d) 25 dyne In human forearm, elbow is the pivot, bicep muscle applies the force and load is placed on the Ans : SQP 2024 palm. Hence, it is a class III lever. Thus (d) is correct option. F=? Moment of force = 1575 dyne-cm 3. A door lock is opened by turning the lever (handle) of length 0.2 m. If the moment of force produced Perpendicular distance = 45 cm is 1 Nm, then the minimum force required is F= moment of force (a) 5 N (b) 10 N Perpendicular dis tan ce (c) 20 N (d) 0.2 N F = 1575 45 Ans : MAIN 2024 = 35 dyne Moment of Force = 1 N-m Thus (c) is correct option. Length of handle = 0.2 m Moment of a force = Force × perpendicular distance from the line of action of the force 1 = F × 0.2 ICSE CHAPTERWISE PYQ CLASS 10 PHYSICS PAGE 9 7. The diagram below shows the balanced position at which the entire weight of a body may be of a meter scale. considered as concentrated. Thus (b) is correct option. 9. Moment of force is given by (a) force × parallel distance between axis of Which one of the following diagrams shows the rotation. correct position of the scale, when it is supported (b) force × perpendicular distance. at the centre? (c) Torque × perpendicular distance. (d) None of these Ans : SQP 2018 Moment of force = force × perpendicular distance (a) Thus (b) is correct option. 10. When a body moves in a circular path, outward force is called a............... (a) Centripetal force (b) Reaction force (b) (c) Centrifugal force (d) Pseudo force Ans : MAIN 2015 (c) Centripetal force is the component of force acting on an object in curvilinear motion which is (d) directed towards the axis of rotation or centre of curvature. Centrifugal force is a pseudo force in a circular motion which acts along the radius and is directed away from the centre of the circle. Ans : MAIN 2021 SEM - 1 Thus (c) is correct option. 11. Clockwise moments are considered to be............... (a) zero (b) positive (c) negative (d) all of these Ans : COMP 2012 Let us suppose left side have mass m and right side have mass m l. As a matter of convention, an anticlockwise Consider centre of mass at 20 cm and at 70 cm for moment is taken as positive and a clockwise left and right side respectively. moment is taken as negative. By moment of force, Thus (c) is correct option. mg × 20 = m l g # (30) 12. A body is acted upon by two unequal and opposite 3 forces along different lines of action of force. The m = 2 ml body will have Here, left hand side is more heavy, so it will bend (a) only rectilinear motion or rotate more towards left side as compared to (b) only rotatory motion right side when fulcrum is placed at 50 cm mark. (c) only translatory motion Thus (a) is correct option. (d) both (a) and (b) 8. The weight of an object lies at the............... Ans : SQP 2010 (a) Geometric centre always Since the forces are unequal, net force will be (b) Centre of gravity non-zero and will result in translation motion. (c) Centre of buoyancy Also, since they are acting on two sides of the (d) Centre of mass pivot point, they develop moments in the same direction and the net moment will be non-zero. Ans : COMP 2000 Thus, it results in rotational motion also. The centre of gravity is an imaginary point Thus (b) is correct option. CH 1 : FORCE TURNING FORCES AND UNIFORM CIRCULAR MOTION 13. As a body moves in a circular path, inward seeking The speed and the angular velocity remain force is called............... constant in case of uniform circular motion. (a) Centripetal force (b) Centrifugal force Thus (b) is correct option. (c) Tension force (d) Tangential force 18. The centre of gravity of a cricket ball is at : Ans : MAIN 2009 (a) at any point on its surface A centripetal force is a net force that acts on an (b) at its bottom touching the ground object to keep it moving along a circular path. (c) its geometric centre Thus (a) is correct option. (d) its top most point 14. Centre of gravity of a solid and hollow sphere are Ans : MAIN 2023............... For a uniform ball, the centre of gravity lies at the (a) almost the same point of intersection of any two of its diameters, (b) the same which is at the geometrical centre of the ball. (c) different Thus (c) is correct option. (d) Geometric centre 19. Lesser the force applied............... is the moment Ans : COMP 2008 of force. Due to their weights there would be a small (a) lesser (b) equal difference in relative position of center of gravity. (c) greater (d) unit Thus (a) is correct option. Ans : COMP 2000 15. The point of action of force on a rigid body is : Moment of force is directly proportional to force (a) Fixed point, but can be transferred anywhere Thus (a) is correct option. opposite to the direction of force. (b) Fixed point on rigid body 20. The moment of couple is mathematically the : (c) Fixed point but can be transferred any where (a) product of one force and the perpendicular along the line of action of force. distance between the point of application of (d) Fixed point but can be transferred anywhere force and turning point. along the direction of force. (b) product of both forces and the perpendicular distance between them. Ans : MAIN 2020 (c) product of one force and the perpendicular The point of action of force on a rigid body is a distance between two forces. fixed point but can be transferred anywhere along (d) None of the above. the line of action of force. Ans : SQP 2014 Thus (a) is correct option. Moment of couple = force × couple arm. 16. The turning effect produced in a rigid body Thus (b) is correct option. around a fixed point by the application of force is called 21. A force F acts on a rigid body capable of turning (a) movement of force around a fixed point. The moment of force (b) moment of couple depends upon (c) turning force (a) magnitude of perpendicular distance between (d) none of these the point of action of force and the turning point Ans : COMP 2013 (b) magnitude of force F The turning effect of force is called Torque (c) both (a) and (b) Thus (a) is correct option. (d) none of these 17. A body is describing a uniform circular motion. Ans : COMP 2012 Which of the following quantities is/are constant Moment of force about the point is the product of (a) velocity (b) speed force and the perpendicular distance between the (c) acceleration (d) both (a) and (b) axis of rotation and line of action of force. Hence, it depends on both of them. Ans : SQP 2020 Thus (c) is correct option. ICSE CHAPTERWISE PYQ CLASS 10 PHYSICS PAGE 11 22. In a uniform circular motion : unit is Newton-meter (Nm). (a) the motion of body is accelerated Thus (d) is correct option. (b) speed of body continuously changes because the direction of motion changes 26. The weight of a uniform half meter scale would (c) velocity of body continuously changes because act at............... mark. the direction of motion changes (a) 50 cm (b) 25 cm (d) both (b) and (c) (c) 100 cm (d) 0 cm Ans : COMP 2017 Ans : SQP 2004 In a uniform circular motion, the speed is constant Centre of gravity would be the midpoint of 50 cm. but the direction of motion is continuously Thus (b) is correct option. changing due to which velocity continuously 27. The centre of gravity of a regular object will changes. Therefore, uniform circular motion is depend on............... also called as accelerated motion. (a) Density of the body Thus (d) is correct option. (b) Volume of the body 23................ moments are considered to be positive. (c) distribution of mass (a) Total (b) Clockwise (d) All of these (c) Anticlockwise (d) None of these Ans : COMP 2017 Ans : MAIN 2010 Centre of gravity of given mass depend on its Since moments are vectors hence are associated shape and on the distribution of mass. with directions too. Thus (c) is correct option. Thus (c) is correct option. 28. Uniform linear motion has............... 24. Manoj of mass 55 kg and Ashok of mass 40 kg are (a) Variable speed, variable velocity sitting on a see saw at a distance of 2 m and 1.5 m (b) Constant speed, constant velocity respectively from the centre of the see saw in an (c) Constant speed, variable velocity amusement park as shown in the figure. Is the see (d) Variable speed, constant velocity saw in rotational equilibrium? Ans : SQP 2009 Same speed and same direction as it is linear motion. Thus (b) is correct option. 29. The centre of gravity of a regular object would lie............... (a) outside the body (b) inside the body (c) on the body (d) all of these (a) data incomplete (b) Yes Ans : MAIN 2006 (c) No (d) None of the above 2D objects on the body while 3D objects outside Ans : COMP 2022 as well as inside the body. According to the principle of moments the see saw Thus (d) is correct option. in rotational equilibrium Thus (c) is correct option. 30. Huge trailer trucks have their steering wheel of............... diameter. 25. The unit of moment of force in SI system is : (a) least (b) large (a) N-cm (b) dyne-m (c) small (d) all of these (c) dyne-cm (d) N-m Ans : SQP 2018 Ans : MAIN 2014 Larger couple arm large is moment of couple. Moment of force is defined as the product of force Thus (b) is correct option. and perpendicular distance from the axis; so its SI CH 1 : FORCE TURNING FORCES AND UNIFORM CIRCULAR MOTION 31. Centre of gravity is............... when it is a hollow Ans : SQP 2011 cone. Moment of force = F # OA (a) centre of vertical axis (b) raised from the base 2 = F # 0.5 (c) lower towards the base (d) none of these Ans : MAIN 2016 Lighter the object the more is the height of CG from the base. Thus (b) is correct option. 32. For equilibrium,............... sum of all moments F = 20 = 40 N 0.5 is equal to............... sum of all anticlockwise Thus (b) is correct option. moments. (a) Arithmetic, arithmetic 35. The relation between CGS and SI unit of moment (b) Algebraic, algebraic of force is (c) Total, none of (a) 1 Nm = 105 dyne cm (d) All of these (b) 1 Nm = 105 dyne Ans : SQP 2000 (c) 1 Nm = 107 dyne cm (d) 1 dyne-cm = 107 Nm Since moments are vectors hence are associated with directions too. Ans : SEM-I 2021 Thus (a) is correct option. 1 newton = 105 dyne 33. A force of 50 N produces a moment of force of 10 1 Nm = 105 × 102 dyne-cm Nm in a rigid body. Calculate the perpendicular distance between the point of application of force 1 Nm = 107 dyne-cm and the turning point is 45 cm. Thus (c) is correct option. (a) 1.5 m (b) 0.1 m 36. Find the resultant moment of couple if two equal (c) 0.2 m (d) 0.8 m forces each of 6 N are acting at a distance of 3 cm Ans : COMP 2000 each from the centre. (a) 3.6 Nm (b) 0.36 Nm Force F = 50 N (c) 360 Nm (d) 36 Nm Moment of force = 10 Nm Ans : COMP 2019 Moment of force = F # Perpendicular distance Distance given is radius but diameter is couple Hence, Perpendicular distance = 10 = 0.2 m arm so multiply by 2. 50 Couple arm = 3 × 2 = 6 cm = 0.06 m Thus (c) is correct option. Moment of couple is given by 34. The diagram alongside shows a force F acting = force × couple arm at point A, such that it produces a moment of = 6 × 0.06 = 0.36 Nm force of 20 Nm in clockwise direction. Calculate Thus (b) is correct option. the magnitude of force F. 37. A uniform metre scale is balanced at 20 cm mark, when a weight of 100 gf is suspended from one end. Where must the weight be suspended ? Calculate the weight of the metre scale. (a) 77.77 gf (b) 11.11 gf (c) 88.88 gf (d) 66.66 gf Ans : SQP 2008 (a) 30 N (b) 40 N (c) 10 N (d) 20 N Let W be the weight of metre scale. As the scale is ICSE CHAPTERWISE PYQ CLASS 10 PHYSICS PAGE 13 balanced at 20 cm and 100 gf is suspended on one end (0 mark). The weight of longer arm i.e. BC is balanced by 100 gf. (a) 10 cm (b) 60 cm (c) 30 cm (d) 40 cm Ans : SQP 2000 Clockwise moment = Anti clockwise moment Two forces F1 = F2 are parallel but in opposite W # 30 = 20 # 100 direction constitute a couple W = 2000 30 = 66.66 gf Weight of 100 gf should be suspended at zero mark. Thus (d) is correct option. 38. A uniform metre scale balances horizontally on a Moment of couple = F # = distance AB knife edge placed at 55 cm mark, when a mass of 25 g is supported from one end. Draw the diagram 9 Nm = 15 N # 2r (radius) of the arrangement. Calculate mass of the scale. r = 9 = 0.30 m (a) 400 g (b) 275 g 30 (c) 100 g (d) 225 g = 30 cm Ans : MAIN 2010 Thus (c) is correct option. 40. A force of 50 dyne acts on a rigid body, such that the perpendicular distance between the fulcrum and the point of application of force is 75 cm. Calculate the moment of force. (a) 2500 dyne cm (b) 3750 dyne cm (c) 3975 dyne cm (d) 1580 dyne cm Ans : MAIN 2015 As the metre scale is balanced at 55 cm mark Force F = 50 dyne i.e. large arm is balanced by arm BC and 25 g is Perpendicular distance = 75 cm suspended at one end i.e. at 100 cm mark Moment of force = F # Perpendicular distance Hence, Anticlockwise moment = clockwise moment = 50 # 75 W # (55 - 50) = 25 # (100 - 55) = 3750 dyne cm 5W = 25 # 45 Thus (b) is correct option. W = 25 # 45 = 225 g 41. Find the maximum force required to open a nut 5 by a spanner of length 50 cm producing a torque Thus (d) is correct option. of 50 Nm? 39. Two forces F1 and F2 are applied on a circular body (a) 0.1 N (b) 10 N such that moment of couple is 9 Nm in clockwise (c) 100 N (d) None of these direction. Calculate the radius of circular body. Ans : COMP 2000 CH 1 : FORCE TURNING FORCES AND UNIFORM CIRCULAR MOTION Moment of force = force × perpendicular distance Moment of couple = F # diameter AB 50 30 = F # (2 # 1.5) 50 = Force× 100 50×100 = F = 30 = 10 N Force = 100 N 3 50 Thus (a) is correct option. Thus (c) is correct option. 44. A see-saw 8 m long is balanced in the middle. 42. Study the diagram given below and calculate the Two children of mass 30 kgf and 40 kgf are sitting moment of couple. on the same side of the fulcrum at a distance of 1.5 m and 3.5 m from the fulcrum respectively. Where must a lady weighing 60 kgf sit from the fulcrum, so as to balance the see-saw ? (a) 6.15 m (b) 4.15 m (c) 3.08 m (d) 2.17 m Ans : MAIN 2011 (a) 2 Nm (b) 1 Nm (c) 6 Nm (d) 4 Nm Ans : COMP 2023 F1 = F2 = F = 5 N forces being equal opposite and parallel Couple arm = 1.2 m Hence, the moment of couple = F # Perpendicular distance = 5 # 1.2 = 6 Nm Thus (c) is correct option. Let lady of 60 kgf sits at a distance from 4 metre 43. Two forces F1 = F2 are applied on a wheel of mark 1.5 m radius, such that moment of couple is 30 Hence, Anti-clock wise moment = clockwise Nm. Calculate the magnitude of each of the force. moment 60 # x = (30 # 1.5) + (40 # 3.5) 60x = 45 + 140 = 185 x = 185 60 = 18.5 = 3.08 m 6 (a) 10 N (b) 60 N Hence, Lady must sit on opposite side of children (c) 20 N (d) 40 N at 3.08 m from fulcrum F. Thus (c) is correct option. Ans : SQP 2017 Anticlockwise direction 45. From the given figure, calculate moment of force about (i) P and (ii) Q. F1 = F2 = F in opp. direction and are parallel Hence constitute a couple. ICSE CHAPTERWISE PYQ CLASS 10 PHYSICS PAGE 15 (a) 10 N-m clockwise, 5 Nm anti-clockwise 47. When a car of mass M passes through a convex (b) 5 N-m clockwise, 10 Nm anti-clockwise bridge of radius r with velocity v , then it exerts a (c) 10 N-m clockwise, 10 Nm anti-clockwise force on it. What is the magnitude of the force ? 2 2 (d) 5 N-m clockwise, 5 Nm clockwise (a) Mv (b) Mg - Mv r r Ans : COMP 2015 2 From the given figure, (c) Mg + Mv (d) Mg r (i) At point P two torques t 1 and t 2 acting from distances 0.05 m and (0.2 - 0.05 = 0.15) Ans : COMP 2017 0.15 m. When a car of mass M passes through a convex Hence, bridge of radius r with velocity v , it exerts a force Torque, t 1 = force # distance on it, in following ways (i) The weight mg of the car acting vertically = 25 # 0.05 downwards. = 1.25 N clockwise (ii) Normal reaction R of the road on the car, Torque, t 2 = 25 # (0.2 - 0.05) acting vertically upwards. As the radius of the bridge is r , hence the centripetal force = 25 # 0.15 is along the surface of the road, towards the = 3.75 N clockwise centre of turn. 2 Hence, Total torque = t 1 + t 2 Hence, the magnitude of force = Mg - Mv. r = 1.25 + 3.75 Thus (b) is correct option. = 5 Nm, clockwise 48. Two forces each of magnitude 2 N act vertically (ii) Now, at point Q two torques t 1 and t 2 will be upward and downward respectively on two ends t 1 = F1 # 0 = 0 of a uniform rod of length 1m, freely pivoted at its centre. Determine the resultant moment of forces t 2 = 25 # 0.2 about the mid-point of the rod. = 25 # 2 (a) 6 Nm (b) 2 Nm 10 (c) 4 Nm (d) 1 Nm = 5 N, clockwise Ans : SQP 2006 Hence, net torque = t 1 + t 2 As the two equal forces 2 N are acting at the ends = 0 + 5 = 5 Nm, clockwise of pivoted rod AB. These constitute a couple in Thus (d) is correct option. anti-clockwise direction. 46. The length of seconds hand of a clock is 10 cm. The angular speed of the tip of the hand is (a) p rad s-1 (b) p rad s-1 300 40 (c) p rad s-1 (d) p rad s-1 3000 30 Ans : MAIN 2000 Moment of force at A about Length of seconds hand of a clock = 10 cm. O = 2 # 1 = 1 Nm Angular speed, w = q 2 t Moment of force at B about But q = 2p , t = 60 s O = 2 # 1 = 1 Nm 2 w = 2p 60 Hence, Resultant moment of force 1 + 1 = 2 Nm Thus (b) is correct option. = p rad s-1 30 Thus (d) is correct option. CH 1 : FORCE TURNING FORCES AND UNIFORM CIRCULAR MOTION 49. A tap is used to fill bucket of water under a tap as Ans : SQP 2018 shown in the figure. Name the force acting on the tap which produces rotational motion? Let w be the mass of metre scale acting a mid point 50 cm. Clock wise moment = 40 # (80 - 60) = 40 # 20 = 800 gf cm...(1) (a) Centripetal force (b) Tension force Anticlockwise moments (c) Centrifugal force (d) Couple = 5 # (60 - 10) + w # (60 - 50) Ans : MAIN 2007 = (250 + 10w) gf cm...(2) While opening a tap the system of forces acting Anticlockwise moment = clock wise moment are two equal opposite and parallel forces. Thus (d) is correct option. 250 + 10w = 800 10w = 800 - 250 = 550 50. A mechanic can open a nut by applying 120 N force while using a lever of 50 cm length. How w = 550 = 55 gf 10 long should the handle be, if he wishes to open, it Thus (d) is correct option. by applying a force of only 40 N ? (a) 1.5 m (b) 1 m 52. A couple of 15 N force acts on a rigid body, such (c) 2.5 m (d) 2 m that the arm of couple is 85 cm. Calculate the Ans : COMP 2010 moment of couple in S.I. system. (a) 25.00 dyne (b) 15.80 dyne A nut cracker is an example of second class of (c) 37.50 dyne (d) 12.75 dyne lever, where fulcrum is one side and load (nut) is between fulcrum and effort. Ans : COMP 2023 According to the question, Force F = 15 N Moment of force = F # d Arm of the couple = 85 cm = 85 m = 120 # 50 = 60 Nm 100 100 Hence, Moment of the couple = F # force arm As the same moment of force acts on the nutcracker by applying a force of 40 N, then = 15 # 85 100 applying formula. Moment of force = F # d = 12.75 Nm Thus (d) is correct option. 60 = 40 # d d = 60 = 3 = 1.5 m 53. What type of motions are exhibited by a vehicle 40 2 and its wheels? Thus (a) is correct option. (a) Translatory and rotatory (b) Rotatory 51. A uniform metre scale is balanced at 60 cm mark, (c) Translatory when weights of 5 gf and 40 gf are suspended at 10 (d) None of these cm mark and 80 cm mark respectively. Calculate the weight of the metre scale. Ans : SQP 2022 (a) 25 gf (b) 15 gf As the vehicle is moving in forward direction, (c) 37 gf (d) 55 gf but wheels are revolving i.e., executing circular DON’T WASTE MONEY PRINTING NODIA BOOKS Purchase paperback books directly from Amazon. Click any subject link to purchase a book. NODIA ICSE NODIA ICSE NODIA ICSE NODIA ICSE CHAPTERWISE CHAPTERWISE CHAPTERWISE CHAPTERWISE 20 Years PYQ Bank 20 Years PYQ Bank 20 Years PYQ Bank 20 Years PYQ Bank 2024-2005 (54 Papers) 2024-2005 (54 Papers) 2024-2005 (54 Papers) 2024-2005 (54 Papers) Main, SQP and Comp Main, SQP and Comp Main, SQP and Comp Main, SQP and Comp MATHEMATICS PHSYSICS CHEMISTRY BIOLOGY Class 10 Class 10 Class 10 Class 10 NODIA ICSE NODIA ICSE NODIA ICSE NODIA ICSE CHAPTERWISE CHAPTERWISE CHAPTERWISE CHAPTERWISE 20 Years PYQ Bank 20 Years PYQ Bank 20 Years PYQ Bank 20 Years PYQ Bank 2024-2005 (54 Papers) 2024-2005 (54 Papers) 2024-2005 (54 Papers) 2024-2005 (54 Papers) Main, SQP and Comp Main, SQP and Comp Main, SQP and Comp Main, SQP and Comp HISTORY COMPUTER GEOGRAPHY HINDI & APPLICATION Class 10 CIVICS Class 10 Class 10 Class 10 NODIA ICSE NODIA ICSE NODIA ICSE NODIA ICSE CHAPTERWISE CHAPTERWISE CHAPTERWISE CHAPTERWISE 20 Years PYQ Bank 20 Years PYQ Bank 20 Years PYQ Bank 20 Years PYQ Bank 2024-2005 (54 Papers) 2024-2005 (54 Papers) 2024-2005 (54 Papers) 2024-2005 (54 Papers) Main, SQP and Comp Main, SQP and Comp Main, SQP and Comp Main, SQP and Comp ENGLISH LITERATURE COMMERICAL LANGUAGE IN ECONOMICS STUDIES Class 10 ENGLISH Class 10 Class 10 Class 10 For more details whatsapp at 94142 43489 ICSE CHAPTERWISE PYQ CLASS 10 PHYSICS PAGE 17 motion. Ans : SQP 2017 Hence, option (a), a vehicle and its wheels execute Mass of the particle = m translatory and rotatory motion. As, the particle is executing a uniform circular Thus (a) is correct option. 2 motion, it will be accelerated with vr directed 54. A rectangular thin plate of dimension 3m × 4m is towards the centre as it is moving with a circle of balanced on the fulcrum as shown below Find the radius r and moving with uniform speed. resultant moment of force? According to Newton’s second law of motion, 2 Centripetal force, Fc = mv r Rearranging, it can be written as, 2 Fc = mv $ m r m (mv) 2 p2 = = mr mr (a) 6 N-m clockwise Thus (d) is correct option. (b) 5 N-m anticlockwise (c) 9 N-m anticlockwise 57. 10-10 Nm is equal to............... dyne cm. (d) 9 N-m clockwise (a) 0.001 (b) 10-4 (c) 10 (d) None of these Ans : MAIN 2010 The longer length is the length of rectangle Ans : COMP 2014 at which 6 N force acts producing a clockwise 1 Nm = 107 dyne cm moment while 5 N force acts along the breadth of the rectangle producing anti clockwise moments. 10 -10 Nm = 107 ×10 -10 Use the sign convention and calculate the resultant = 10 -3 = 0.001 dyne-cm force. Thus (a) is correct option. Thus (d) is correct option. 58. The diagram along side shows a force F = 5 N 55. The gate of a building is 3 m broad. It can be acting at point A produces a moment of force of opened by applying 100 N force at the middle 6 Nm about point O. What is the diameter of the of the gate. Calculate the least force required to wheel ? open this gate, (at a point 6 m) (a) 20 N (b) 30 N (c) 50 N (d) 10 N Ans : COMP 2011 Torque required to open the door will remain constant. r1 # F1 = r2 # F2 (a) 3.4 m (b) 2.4 m 3 # 100 = 6 # F2 (c) 4.6 m (d) 1.0 m F2 = 3 # 100 = 50 N 6 Ans : SQP 2004 Thus (c) is correct option. Force, F = 5 N 56. A practice of mass m is executing a uniform Moment of force = 6 Nm circular motion on a path of radius r. If p is the magnitude of its linear momentum, then what is the radial force acting on the particle? p (a) (b) rm2 m p p2 (c) r (d) p mr CH 1 : FORCE TURNING FORCES AND UNIFORM CIRCULAR MOTION r = OA = ? Perpendicular distance = 1.75 m Moment of force = Force × OA Force F = 80 N 6 = OA # 5 Moment of force = F # perpendicular distance OA = 6 = 1.2 m = 80 # 1.75 5 = 140.00 = 140 Nm Hence, diameter = 2r Thus (d) is correct option. = 2 # 1.2 = 2.4 m Thus (b) is correct option. 61. Calculate the length of the arm of couple, if a force of 13 N produces a moment of couple of 59. Two particles of equal masses are revolving in 14.3 Nm. circular paths of radii r1 and r2 respectively, with (a) 2.75 m (b) 3.75 m the same speed. The ratio of their centripetal (c) 1.1 m (d) 1.5 m forces is Ans : SQP 2018 (a) a r1 k r2 2 (b) r2 r1 Force, F = 13 N (c) a r2 k (d) r2 2 r1 r1 Moment of couple = 14.3 Nm Moment of couple = F # arm of couple Ans : MAIN 2006 14.3 = 13 # arm of couple Centripetal force is given by, FC = mv 2 Hence, Arm of couple = 14.3 = 1.1 m r 13 Thus (c) is correct option. Where, m = mass of the particle v = velocity of the particle 62. A uniform metre scale of weight 50 gf is balanced at the 40 cm mark, when a weight of 100 gf is r = radius of circular path suspended at the 5 cm mark. Where must a weight Now according to the question, of 80 gf be suspended to balance the metre scale ? 2 For mass m1 , F1 = m1 v2 1...(i) (a) 66.00 cm (b) 70.00 cm r1 (c) 77.5 cm (d) 10.00 cm 2 For mass m2 , F2 = m2 v2 2...(ii) Ans : COMP 2014 r2 Hence, the ratio of their centripetal force is obtained dividing equation (i) and (ii). Hence, m1 v 12 F1 = r1 F2 m2 v 22 r2 But as the masses of both the particles are same i.e., m1 = m2 = m and speeds are same i.e., v1 = v2 = v Then, F1 = mv2 r2 = r2 F2 r1 # mv2 r1 Let a wt. of 80 gf be placed at a distance x from Thus (d) is correct option. 40 cm mark. Clock wise moments = Anticlockwise moment 60. The perpendicular distance between the point of application of force and the turning point is (80 # x) + 50 # (50 - 40) = 100 # (40 - 5) 1.75 m, when a force of 80 N acts on a rigid body. 80x + 500 = 3500 Calculate the moment of force. (a) 370 Nm (b) 180 Nm 80x = 3500 - 500 (c) 220 Nm (d) 140 Nm = 3000 Ans : COMP 2019 x = 3000 = 37.5 cm 80 ICSE CHAPTERWISE PYQ CLASS 10 PHYSICS PAGE 19 Hence, A is at 40 + 37.5 = 77.5 cm mark M = 250 # 30 = 750 gf Thus (c) is correct option. 10 Thus (c) is correct option. 63. Essential characteristic of equilibrium is (a) Kinetic energy is equals to zero. 65. Force in linear motion has its analogue in (b) Velocity equals to zero. rotational motion ? (c) Acceleration equals to zero. (a) Torque (d) Momentum equals to zero. (b) Angular momentum (c) Weight Ans : MAIN 2012 (d) Moment of inertia Essential characteristics of equilibrium is the total Ans : MAIN 2007 force i.e., the vector sum of ail forces acting on the rigid body is zero. Force in linear motion has its analogue with torque n in rotational motion, because moment of force = i.e., / Fi = 0 = F1 + F2 +.... + Fn force #the perpendicular distance of the line of i=1 As mass cannot be zero, hence acceleration equals action of the force moving towards the point. to zero. Thus (a) is correct option. Thus (c) is correct option. 64. A uniform wooden beam AB , 80 cm long and weighing 250 gf, is supported on a wedge as shown ASSERTION AND REASON in the figure. Calculate the greatest weight which can be placed on end A without causing the beam to tilt. 66. Assertion : Uniform circular motion is an accelerated motion. Reason : In uniform circular motion, an object moves with constant speed but variable velocity. (a) Both assertion and reason are true. (b) Both assertion and reason are false. (c) Assertion is false but reason is true. (d) Assertion is true but reason is false. Ans : In uniform circular motion, the body moves with (a) 375 gf (b) 560 gf constant speed in circular path, but its direction of (c) 750 gf (d) 100 gf motion keeps on changing continuously. Thus, the velocity of the object in uniform circular motion Ans : SQP 2015 is variable and such motion is called accelerated motion. Thus (a) is correct option. 67. Assertion : The moment of force for force 10 N applied at a distance of 20 cm from the pivot will be 0.02 Nm. Reason : The moment of force is ratio of perpendicular distance of the line of action of the force for axis of rotation and magnitude of force. (a) Both Assertion and reason are true (b) Both Assertion and reason are false (c) Assertion is false but reason is true Let M gf be suspended at A as shown in the (d) Assertion is true but reason is false above figure. Hence, Ans : Anticlockwise wise moment = clockwise moment The moment of force is the product of M # 10 = 250 # (40 - 10) CH 1 : FORCE TURNING FORCES AND UNIFORM CIRCULAR MOTION perpendicular distance of the line of action of the motion only. force for axis of rotation and magnitude of force. (a) Both assertion and reason are true F = 10 N, r = 20 cm = 0.2 m (b) Both assertion and reason are false (c) Assertion is false but reason is true Moment of force = F × r = 10 × 0.2 = 2 Nm (d) Assertion is true but reason is false Thus (b) is correct option. Ans : 68. Assertion : The moment of force depends Linear motion is also known as translational upon magnitude of the force applied and the motion. When force acting on stationary object perpendicular distance of the line of action of makes it to move in a straight path then the body force from the axis of rotation. is said to have translational or linear motion. Reason : The opening and shutting of the door is And when a free body is acted upon by two possible due to the moment of force. unequal forces in opposite direction, but not in (a) Both assertion and reason are true one line, then the body possesses both rotational (b) Both assertion and reason are false as well as translational motion. (c) Assertion is true but reason is false Thus (d) is correct option. (d) Assertion is false but reason is true Ans : The turning effect on the body about its axis is due to the moment of force or torque applied on ONE MARK QUESTIONS the body. To open and shut the door the force is applied normally on the door at its handle which provided at the maximum distance from DIRECTION : Fill in the blanks : the hinges. Thus (a) is correct option. 71. When a stone tied to a string is rotated in a horizontal plane, the tension in the string provides 69. Assertion : A duster lying on the table is an............... force necessary for circular motion. example of static equilibrium. (Centripetal, centrifugal) Reason : When a body remains in the same state of motion under the influence of several forces Ans : MAIN 2024 then the body is in static equilibrium. Centripetal (a) Both assertion and reason are true (b) Both assertion and reason are false 72. In a beam balance when the beam is balanced in a (c) Assertion is false but reason is true horizontal position, it is in............... equilibrium. (d) Assertion is true but reason is false (stable/unstable) Ans : Ans : MAIN 2015 When a body remains in a state of rest under the Stable influence of several forces, then the body is in static equilibrium. Thus, the duster which lie on 73. A............... handle with a screw jack is used so the table in state of rest is the example of static that the required force decreases. (Long/short) equilibrium. Ans : SQP 2023 When a body remains in the same state of motion Long under the influence of several forces then the body is said to be in dynamic equilibrium. 74. The centre of gravity of a............... (Uniform/ Thus (d) is correct option. Non Uniform) ring is situated at its geometrical centre. 70. Assertion : When a force acts on a stationary rigid body which is free to move and if body starts Ans : MAIN 2015 moving in straight path in direction of applied Uniform force, then it is said to be linear motion. Reason : If free body is acted upon by two unequal 75................ is defined as a set of two equal and forces in opposite direction, but not in one line, opposite forces acting along different lines of the effect is that the body will possess rotational action. (Couple/Moment). ICSE CHAPTERWISE PYQ CLASS 10 PHYSICS PAGE 21 Ans : MAIN 2019 Distribution of mass. Couple 84. The position of the centre of gravity of a body 76. The centre of gravity of an object of given mass remains unchanged even when the body is depends on its............... (shape/Density) deformed.’ State whether the statement is true or false. Ans : MAIN 2016 Ans : COMP 2013 Shape False DIRECTION : Answer the following : 85. Where is the centre of gravity of a uniform ring situated? 77. Name the unit of physical quantity obtained by Ans : MAIN 2000 the formula 2 K/V2 , where K : Kinetic energy, V : Linear velocity. The centre of gravity of a uniform ring is situated at its geometrical centre. Ans : COMP 2000 Js m. –2 –2 86. What is the SI unit of the moment of force? Ans : COMP 2015 78. Where is the centre of gravity of a uniform ring situated ? N-m. Ans : COMP 2013 87. Define equilibrium. At its geometric centre. Ans : SQP 2015 79. What do you mean by kilogram force and write It is a state in which net force acting on a body its relation with newton? or system is zero. Ans : MAIN 2015 88. Why do we use a long handle with a screw jack? One kilogram force (kg f ) is the force required to Ans : COMP 2012 lift a body of mass 1 kilogram. In a screw jack, a long handle helps us to apply 1 kg f = 9.8 N less force F at a good distance from the axis of rotation making d= large so as to give a large 80. Which type of motions are exhibited by a vehicle turning effect. and its wheels? 89. Can the couple acting on a rigid object produce Ans : SQP 2010 translatory motion? The motion of vehicle is translatory and the Ans : MAIN 2014 motion of wheels is rotatory. No, the couple acting on a rigid object cannot 81. Why do we use a long handle with a screw jack ? produce translatory motion, it can cause only Ans : COMP 2007 rotatory motion. A long handle with a screw jack is used so that 90. A small coin is placed on a rotating record. The the required force decreases. coin does not slip on the record. Where does it get the centripetal force from ? 82. What do you mean by axis of rotation? Ans : SQP 2020 Ans : SQP 2005 The necessary centripetal force is provided by the Axis of rotation is the line or point about which force of friction between the coin and the record. the object rotates. It may be within the object or outside the object. 91. Can the centre of gravity of a body be outside it? If yes, give one example. 83. On what factor does the position of the centre of gravity of a body depend? Ans : MAIN 2021 Ans : SQP 2015 Yes, the CG of a body can be outside the body. CH 1 : FORCE TURNING FORCES AND UNIFORM CIRCULAR MOTION The CG of a uniform ring is at its centre, a point TWO MARKS QUESTIONS which is not on the body. 92. What do you mean by axis of rotation? 99. A hollow ice cream cone has height 6 cm. Ans : COMP 2017 1. Where is the position of its centre of gravity from the broad base? Axis of rotation is the line or point about which 2. Will its position change when it is filled the object rotates. It may be within the object or completely with honey? Write Yes or No. outside the object. Ans : MAIN 2024 93. Can the moment of a force be zero even if the force is not zero? If so, when? 1. The centre of gravity from the broad base is at a height h = 6 = 2 cm. Ans : SQP 2018 3 3 2. Yes. Yes, if the line of action passes through the point about which moment is to be calculated. 100. (i) Define moment of force. (ii) Write the relationship between SI and CGS 94. Why is it easier to open the cap of a pen with two unit of moment of force. fingers than with one finger? Ans : SQP 2020 Ans : MAIN 2000 (i) It is defined as the product of force and the This is so because when we use two fingers, a arm of the couple. couple is formed. This couple enables us to open (ii) 1 nm = 107 dyne-cm. the cap of the pen easily. 101. (i) Define couple. 95. Why it is difficult to open the door by pushing it (ii) State its SI unit. or pulling it at the hinge ? Explain. Ans : MAIN 2019 Ans : COMP 2015 (i) It is defined as a set of two equal and opposite This is so because near the hinge, the distance of forces acting along different lines of action. the point of application of force from the axis of (ii) SI unit is N m. rotation becomes negligible, so the force applied increases. 102. A brass ball is hanging from a stiff cotton thread. Draw a neat labelled diagram showing the forces 96. Can a simple machine act as a force multiplier acting on the brass ball and the cotton thread. and speed multiplier simultaneously in the same Ans : COMP 2017 machine? The diagram is as shown : Ans : SQP 2018 No, Machines which are force multipliers do not gain in speed and vice-versa. 97. What types of motions are executed by a vehicle and its wheels ? Ans : COMP 2011 The vehicle executes translatory but the wheels execute rotatory motion. 98. Mention the names of a few bodies which can be considered as rigid bodies for practical purposes. Ans : COMP 2017 In actual practice a rigid body is not possible 103. Why are railway wagons not loaded beyond a but for practical purposes stone, metal, glass and certain limit? wood are considered as rigid bodies. Ans : COMP 2005 ICSE CHAPTERWISE PYQ CLASS 10 PHYSICS PAGE 23 Railway wagons are not loaded beyond a certain Ans : SQP 2020 limit because it raises the centre of gravity of the False (Because position of centre of gravity of a wagon. Thus, while going around the sharp turns, body depends on distribution of mass in it so, if the line joining the centre of gravity and centre of body is deformed mass distribution varies and so the earth may falls outside the base of the wagon position of centre of gravity also varies). which can cause derailment. 108. On what factor does the centre of gravity of an 104. When a body is placed on a table top, it exerts a object depend? Give the example of objects whose force equal to its weight downwards on the table centre of gravity lie outside the object. top but does not move or fall. Ans : MAIN 2009 The centre of gravity of an object of given mass depends on its shape, e.g., the centre of gravity of a wire is at its mid-point but if this wire is bent into the form of a circle, its centre of gravity will be at the centre of the circle. Other examples in which centre of gravity lies outside the material are centre of gravity of ring, hollow sphere, L-shaped objects, etc. (i) Name the force exerted by the table top. (ii) What is the direction of the force ? 109. Where does t