Plus One Physics Questions Bank 2023 PDF

Chapter 2 Units and Measurement 1.Name the fundamental(base) quantities and units according to SI system. 2.Define angle 3.Define solid angle 4.Write the dimensional formulae of following derived quantities. Area -L2 Work or energy - ML2 T...

Chapter 2 Units and Measurement 1.Name the fundamental(base) quantities and units according to SI system. 2.Define angle 3.Define solid angle 4.Write the dimensional formulae of following derived quantities. Area -L2 Work or energy - ML2 T −2 Volume -L3 Power - ML2 T −3 Density -ML−3 Pressure - ML−1 T −2 Velocity- LT −1 Stress - ML−1 T −2 Acceleration - LT −2 Modulus of elasticity- ML−1 T −2 Momentum - MLT −1 Force - MLT −2 5.Write two physical quantities having no unit and dimension Relative density, strain ………………………………………………………………………………………………………………………………….. Seema Elizabeth, MARM Govt HSS Santhipuram, Thrissur 1 Downloaded from www.hssreporter.com 6.Write two physical quantities that have unit but no dimension Plane angle, solid angle, angular displacement 7. 8. 9.Name and state the principle used to check the correctness of an equation. Principle of homogeneity of dimensions. For an equation to be correct the dimensions of each terms on both sides of the equation must be the same Or The magnitudes of physical quantities may be added or subtracted only if they have the same dimensions 10. Using the method of dimension check whether the equation is dimensionally correct or not Since the dimensions of all terms are not same the equation is not correct 2 Downloaded from www.hssreporter.com 11. Using the method of dimension check whether the equation is dimensionally correct or not Since the dimensions of all terms on both sides are same the equation is dimensionally correct correct 12. Using the method of dimension check whether the equation is dimensionally correct or not 13.Check the dimensional correctness of the equation E=m𝐜 𝟐 14.In the given equation v = x + at , find the dimensions of x. (where v= velocity , a=acceleration , t=time) 15. In the given equatio x= a + bt + c𝐭 𝟐 , find the dimensions of a,b and c. (where x is in meters and t in seconds) 3 Downloaded from www.hssreporter.com  𝒂 16.The Van der waals equation of 'n' moles of a real gas is (P+ 𝟐)(V−b)=nRT. Where P 𝑽 is the pressure, V is the volume, T is absolute temperature, R is molar gas constant and a, b, c are Van der waal constants. Find the dimensional formula for a and b. 𝐚 (P+ 𝟐)(V−b)=nRT. 𝐕 By principle of homegeneity, the quantities with same dimensions can be added or subtracted. a [P] =[ 2 ] V [a] =[PV 2 ] =ML−1 T −2 x L6 [a] = ML5 T −2 [b] = [V] [b] =L3 17.Derive the equation for kinetic energy E of a body of mass m moving with velocity v 18.Suppose that the period of oscillations of a simple pendulum depends on its mass of the bob(m),length(l) and acceleration due to gravity(g).Derive the expression for its time period using the method of dimensions. 𝑙 T=k√ 𝑔 4 Downloaded from www.hssreporter.com 19.Write any two limitations of dimensional analysis. 20.Find the number of significant figures in following numbers 0.02380 - 4 23.08 - 4 23.80 - 4 2380 -3 43.00 - 4 4300 -2 2 4.700 × 10 -4 −3 4.700 × 10 -4 21.If mass of an object is measured to be, 4.237 g (four significant figures) and its volume is measured to be 2.51cm3(3 significant figures), then find its density in appropriate significant figures. 𝑚𝑎𝑠𝑠 4.237 𝑔 Density = = = 1.688047 𝑣𝑜𝑙𝑢𝑚𝑒 2.51𝑐𝑚3 As per rule the final result should be rounded to 3 significant figures. So the answer is 1.69 g/ 𝑐𝑚3 22.Find the sum of the numbers 436.32 g, 227.2 g and 0.301 g to appropriate significant figures. 436.32 g + (2 decimal places) 227.2 g + (1 decimal place) 0.301 g (3 decimal places) _______________ 663.821 g As per rule ,the final result should be rounded to 1 decimal place. So the answer 663.8 g 5 Downloaded from www.hssreporter.com  Chapter 3 Motion in a Straight Line 1.Define instantaneous velocity? The velocity at an instant is called instantaneous velocity. Δx v = lim Δt→0 Δt 𝐝𝐱 v= 𝐝𝐭 2.When does the average velocity become equal to instantaneous velocity? When the time interval Δt → 0 ,the average velocity becomes equal to instantaneous velocity. Δx dx v = lim = Δt→0 Δt dt 3.The speedometer of a vehicle shows ………………….. Instantaneous speed. 4.Define uniform motion If an object moving along the straight line covers equal distances in equal intervals of time, it is said to be in uniform motion along a straight line. 5.. Draw the position -time graph of an object moving with 6.. The position -time graph of an object in uniform motion is------------- Ans: A straight line inclined to the time axis 7. The slope of position-time graph gives ------------- Ans: Velocity 8. The velocity -time graph of an object in uniform motion is------------- A straight line parallel to the time axis 6 Downloaded from www.hssreporter.com 9. The area under velocity -time graph gives -------------- Displacement 10. The slope of velocity-time graph gives --------------- Acceleration 11. Define average accelaration The average acceleration over a time interval is defined as the ratio of change in velocity to the time interval. 𝐯𝟐 −𝐯𝟏 𝚫𝐯 𝐚⃗ = = 𝐭 𝟐 −𝐭 𝟏 𝚫𝐭 12.Draw the position- time graph of an object moving with 13. Draw the velocity- time graph of an object moving with (a) uniform positive acceleration (b) uniform negative acceleration 14.Draw the velocity- time graph of a stone thown vertiaccly upwrds and comes back. 7 Downloaded from www.hssreporter.com 15.Draw the speed- time graph of a stone thown vertiaccly upwrds and comes back. 16. Draw the velocity-time graph of a freely falling body.( A stone vertically falling downwards) 17. Is it possible for a body to have zero velocity with a nonzero acceleration. Give an example. Yes. When a body is thrown upwards ,at the highest point of projection, its velocity is zero , but it has an acceleration. 18.The velocity -time graph of a ball thrown vertically upward with an initial velocity is shown in figure. a)What is themagnitude of initial velocity of ball? b)Calculate the distance travelled by the ball during 20seconds from the grah. c) Calculate the acceleration of the ball from the graph a)Initial velocity=100m/s b)Distance travelled = area of graph 1 1 = b1h1 + b2h2 2 2 1 1 = x10 x 100 + x 10 x100 2 2 =1000m c) Acceleration = slope v −v 0−100 = 2 1= =-10 m/s2 t2 −t1 10−0 8 Downloaded from www.hssreporter.com 19. (a)Draw the velocity-time graph of a body with uniform aceeleration. (b) Using the graph obtain (i) Velocity - time relation (ii) Displacement -tme relation (iii) Displacement velocity relation 9 Downloaded from www.hssreporter.com 20. An object is under freefall. Draw its (a) Acceleration -time graph (b) Velocity- time graph (c) Displacement-time graph 21.Velocity – time graph of a body is given below a) Which portion of the graph represents uniform retardation? (i) OA (ii)AB (iii) BC (iv) OC b) Find the displacement in time 2s to 7s. c) A stone is dropped from a height h. Arrive at an expression for the time taken to reach the ground. a)BC b) Displacement = area of rectangle. =6x5= 30m 10 Downloaded from www.hssreporter.com c) s = ut +½ at2 -h= 0 - ½ gt2 2h t2 = g 2h t =√ g 22. 11 Downloaded from www.hssreporter.com  Chapter 4 Motion in a Plane 1. Differentiate scalar and vector quantities A scalar quantity has only magnitude and no direction. Eg. distance , speed, mass , temperature, time ,work ,power, energy, pressure, frequency, angular frequency etc. A vector quantity has both magnitude and direction and obeys the triangle law of addition or the parallelogram law of addition. Eg. displacement, velocity, acceleration , momentum, force, angular velocity, torque, angular momentum etc. 2.When two vectors are said to be equal? Two vectors A and B are said to be equal if, and only if, they have the same magnitude and the same direction. 3.What do you mean by null vectors or zero vector? A Null vector or Zero vector is a vector having zero magnitude and is represented by O or Ō. The result of adding two equal and opposite vectors will be a Zero vector Eg: When a body returns to its initial position its displacement will be a zero vector. 4.What are unit vectors? A unit vector is a vector of unit magnitude and points in a particular direction. It has no dimension and unit. It is used to specify a direction only. ̅ ̂= 𝐀 𝐀 |𝐀| 5.The position vector of a particle P located in an x-y plane is shown in figure. a)Redraw the figure by showing the rectangular components. b)Write the position vector in terms of rectangular components. a) b) 𝑟̅ = 𝑟 cos 𝜃𝑖̂ + 𝑟 sin 𝜃𝑗̂ 12 Downloaded from www.hssreporter.com 6.State triangle law of vector addition. If two vectors are represented in magnitude and direction by the two sides of a triangle taken in order ,then their resultant is given by the third side of the triangle taken in reverse order. 7.State parallelogram law of vector addition If two vectors are represented in magnitude and direction by the adjacent sides of a parallelogram ,then their resultant is given by the diagonal of the parallelogram. 8.Two vectors A and B are given below. Redraw the figure and show the vector sum using parallelogram method. 9.Write the equation to find the magnitude of resultant of two vectors A and B R = √A2 + B 2 + 2ABcosθ 10.Derive the expression for magnitude of resultant of two vectors by analytical method. Write the expression for direction of resultant vector. SNP , cos θ = PN / PS sin θ = SN /PS cos θ = PN / B sin θ = SN / B PN = B cos θ SN = B sin θ From the geometry of the figure, OS 2 = ON 2 + SN 2 OS 2 = (OP + PN) 2 + SN 2 OS 2 = (A + B cos θ ) 2 + (B sin θ ) 2 R 2 = A 2 +2AB cos θ + B 2 cos 2θ +B 2sin2 θ R 2 = A 2 + B 2 + 2AB cos θ 𝐑 = √𝐀𝟐 + 𝐁 𝟐 + 𝟐𝐀𝐁𝐜𝐨𝐬𝛉 𝑺𝑵 Direction , tan𝜶 = 𝑶𝑵 𝐁 𝐬𝐢𝐧 𝛉 tan𝛂 = 𝐀+𝐁 𝐜𝐨𝐬 𝛉 13 Downloaded from www.hssreporter.com 11.What is the trajectory(path) followed by a projectile? Parabola 12. Draw the trajectory of a projectile 13.A stone is thrown up with a velocity u , which makes an angle 𝛉 with the horizontal. a)What are the magnitudes of horizontal and vertical components of velocity? b)How do these components vary with time? a) Horizontal component - u cos θ and vertical component - u sin θ b) Horizontal component- u cos θ remains constant with time. vertical component first deceases, becomes zero at the highest point of projection and then increases in reverse direction. 14.What are the values of these components at the highest point of projection? At the highest point, Horizontal component= u cos θ Vertical component = zero 15. A projectile has an acceleration of ……… in vertical direction and……… acceleration in horizontal direction 9.8m 𝑠 −2 , zero 16. Show that the path of the projectile is a parabola. Displacement of the projectile after a time t x= ucosθ t x t= ucosθ 1 y= u sinθ t − g t 2 2 x 1 x 2 y= u sinθ ( ) − 2 g (ucosθ) ucosθ g y= tanθ x − x2 2 u2 cos2 θ 14 Downloaded from www.hssreporter.com 17. Derive the equation for Time of flight, Horizontal range and Maximum height of a projectile. Time of Flight of a projectile (T) Consider the motion in vertical direction, s = ut +½ at2 s=0, u = u sin θ , a =-g , t = T 0 = u sin θ T - ½ gT2 ½ gT2 =u sin θ T 2 u sin θ T= g Horizontal range of a projectile (R) Horizontal range = Horizontal component of velocity x Time of flight 2 u sin θ R = u cos θ x g u2 x 2 sinθ cos θ R= g u2 sin 2θ R= g Maximum height of a projectile (H) Consider the motion in vertical direction to the highest point v2 – u2 = 2as u = u sin θ, v = 0 , a = -g , s = H 0 - u sin θ = -2 g H 2 2 u2 sin2 θ H= 2g 18.What is the angle of projection for maximum horizontal range 𝟒𝟓𝟎 19. What is the maximum value of horizontal range Range is maximum when θ=450 u2 sin 90 R= g u2 Rmax = g 15 Downloaded from www.hssreporter.com 20. Find the angle of projection for which the range will be same as that in case of θ=𝟑𝟎𝟎 for a given velocity of projection. For a given velocity of projection range will be same for angles 𝜽 and ( 90-𝜽 ) Here θ=300 90-𝜽 =90-30 =600 The range will be same for 300 and 600 ,for a given velocity of projection. 21.A cricket ball is thrown at a speed of 28 m s –1 in a direction 30° above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point where the ball returns to the same level. (a) H = u2 sin 2θ 2g H = 282 sin2 30 2 x 9.8 H = 10 m (b) T = 2 u sin θ g T = 2x 28 sin30 9.8 T = 2.9 s (c) R = u2 sin 2θ g R = 282 sin60 9.8 R = 69 m 22.What do you mean by uniform circular motion? When an object follows a circular path at a constant speed, the motion of the object is called uniform circular motion. Note:- ▪ The speed is constant in uniform circular motion. ▪ The velocity is not constant in uniform circular motion. ▪ The acceleration is not constant in uniform circular motion. 23.Give an example for a motion in which speed is constant, but velocity varying. Uniform circular motion 16 Downloaded from www.hssreporter.com 24.Give an example for a motion in which speed is constant, still accelerating. Uniform circular motion 25.What is period in uniform circular motion The time taken by an object to make one revolution is known as its time period T. Unit-second 26.What is Frequency in uniform circular motion The number of revolutions made in one second is called its frequency. 1 υ= 𝑇 unit - hertz (Hz) 27.Define Angular velocity Angular velocity is the time rate of change of angular displacement Δθ dθ 2π ω= or ω = or ω = or ω = 2π υ Δ𝑡 d𝑡 𝑇 Unit is rad/s 28.Derive the relation connecting angular velocity and linear velocity arc angle = radius Δr Δθ= r Δr=r Δθ Δr Linear velocity v = Δ𝑡 rΔθ v= Δ𝑡 Δθ But ω = Δ𝑡 v=rω 29.Define angular acceleration The rate of change of angular velocity is called angular acceleration. dω α = dt dθ ω= d𝑡 d dθ α= ( ) dt 𝑑𝑡 𝑑2𝜃 α= 𝑑𝑡 2 17 Downloaded from www.hssreporter.com 30.Define Centripetal acceleration A body in uniform circular motion experiences an acceleration , which is directed towards the centre along its radius.This is s called centripetal acceleration. 31.Derive the expression for centripetal acceleration. Δv Δr = v r vΔr Δv= r Δv vΔr = Δt r Δt v a= xr r 𝐯𝟐 a= 𝐫 If R is the radius of circular path, then centripetal acceleration. v2 ac = R ac = ω 2R ac = v ω 32.An insect trapped in a circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100 s. (a) What is the angular speed, and the linear speed of the motion? (b) Is the acceleration vector a constant vector ? What is its magnitude ? 100 Period, T= s 7 (a) The angular speed ω is given by 2π 2π 2π x7 ω= = 100 = =0.44 rad/s 𝑇 100 7 The linear speed v is : v =ω R = 0.44 × 0.12 = 5.3 x 10 -2 m s -1 (b) The direction of velocity v is along the tangent to the circle at every point. The acceleration is directed towards the centre of the circle. Since this direction changes continuously, acceleration here is not a constant vector. a = ω 2 R = (0.44 ) 2 x0.12 = 2.3x10 -2 m s -2 18 Downloaded from www.hssreporter.com  Chapter 5 Laws of Motion 1.State Galileo’s Law Inertia If the net external force is zero, a body at rest continues to remain at rest and a body in motion continues to move with a uniform velocity. 2.What do you mean by inertia of a body? If the net external force is zero, a body at rest continues to remain at rest and a body in motion continues to move with a uniform velocity. This property of the body is called inertia. Inertia means ‘resistance to change’. ▪ Suppose a person is standing in a stationary bus and the driver starts the bus suddenly. He gets thrown backward with a jerk. This is due to his inertia of rest. ▪ If a person is standing in a moving bus and if the bus suddenly stops he is thrown forward. This is due to his inertia of motion. 3.State Newton’s first law of motion (Law of inertia) Every body continues to be in its state of rest or of uniform motion in a straight line unless compelled by some external force to change that state. 4.Newton’s first law of motion describes ………………. Inertia 5.Define momentum Momentum, P of a body is defined to be the product of its mass m and velocity v, and is denoted by p. p=mv 6.State Newton’s Second Law f Motion. Write its mathematical expression. ∆p F∝ ∆t ∆p dp F= or F= ∆t dt 7.Why a seasoned cricketer draws his hands backwards during a catch? By Newton's second law of motion , ∆p F= ∆t When he draws his hands backwards, the time interval (∆t) to stop the ball increases.Then force decreases and it does not hurt his hands. 19 Downloaded from www.hssreporter.com 8. Derive of Equation of force from Newton's second law of motion By Newton's second law of motion , 9.Define newton 10. A bullet of mass 0.04 kg moving with a speed of 90 m/s enters a heavy wooden block and is stopped after a distance of 60 cm. What is the average resistive force exerted by the block on the bullet? 11.Show that Newton’s second Law is consistent with the first law. ( or starting from Newton’s second Law arrive at Newton’s first law) Zero acceleration implies the state of rest or uniform linear motion. i.e, when there is no external force , the body will remain in its state of rest or of uniform motion in a straight line. This is Newtons first law of motion. 20 Downloaded from www.hssreporter.com 12.Define Impulse Impulse is the the product of force and time duration, which is the change in momentum of the body. Impulse = Force × time duration I=Fxt Unit = kg m s −1 13. Define Impulsive force. A large force acting for a short time to produce a finite change in momentum is called an impulsive force. Eg: A cricket ball hitting a bat 14. Using Newtons second law of motion arrive at Impulse momentum Principle Impulse is equal to the change in momentum of the body. By Newton's second law of motion, dp F= dt F x dt = dp I = dp Impulse = change in momentum 15.A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 m 𝐬 −𝟏. If the mass of the ball is 0.15 kg, determine the impulse imparted to the ball. Impulse = change of momentum Change in momentum = final momentum – initial momentum Change in momentum = 0.15 × 12–(0.15×-12) Impulse = 3.6 N s 16.Eventhough action and reaction are equal and opposite they do not cancel each other.Why? Action and reaction forces act on different bodies, not on the same body. So they do not cancel each other , eventhough they are equal and opposite. 17. A man of mass 70 kg stands on a weighing scale in a lift which is moving, (a) upwards with a uniform speed of 10 m 𝐬 −𝟏 (b) downwards with a uniform acceleration of 5 m 𝐬 −𝟐 (c) upwards with a uniform acceleration of 5 m 𝐬 −𝟐 What would be the readings on the scale in each case? (d) What would be the reading if the lift mechanism failed and it falls down freely under gravity? Take g=10 m 𝐬 −𝟐 (a)When lift moves with uniform speed , a=0 R= mg = 70 x 10=700 N Reading= 700/ 10= 70 kg (b)Acceleration a = 5m s-2 downwards R= m(g-a) =70 ( 10-5)= 70x5= 350N Reading = 350/ 10 = 35 kg 21 Downloaded from www.hssreporter.com  (c) Acceleration a = 5m s-2 upwards R= m(g+a) = 70(10+5) = 70x 15= 1050N Reading = 1050/ 10 = 105 kg (d) when lift falls freely a=g R= m(g-g) = 0 Reading = 0 18.State the Law of Conservation of Momentum The total momentum of an isolated system of interacting particles is conserved. Or When there is no external force acting on a system of particles ,their total momentum remains constant. 19.Prove law of conservation of momentum Using Newton’s second law of motion dp By Newton's second law of motion , F= dt When F = 0 dp =0 dt dp = 0 , p=constant Thus when there is no external force acting on a system of particles, their total momentum remains constant. 20. Prove law of conservation of momentum Using Newton’s third law of motion Total Final momentum = Total initial momentum i.e. , the total final momentum of the isolated system equals its total initial momentum. 21.Explain the recoil of gun using law of conservation of linear momentum By the law of conservation of momentum,as the system is isolated, the momentum remains constant 22 Downloaded from www.hssreporter.com  If pb and pg are the momenta of the bullet and gun after firing pb + pg = 0 pb = - pg The negative sign shows that gun recoils to conserve momentum. 22.Obtain the expression for Recoil velocity and muzzle velocity Momentum of bullet after firing , pb = mv Recoil momentum of the gun after firing , pg = MV pb = - pg mv = −MV −mv Recoil velocity of gun , V= M −MV Muzzle velocity of bullet , v= m M= mass of gun, V= recoil velocity of bullet m= mass of bullet, v=muzzle velocity of bullet 23.A shell of mass 0.020kg is fired by a gun of mass 100kg. If the muzzle speed of the shell is 80 m/s, what is the recoil speed of the gun? 𝑚𝑣 0.020 𝑥 80 V= = =0.016m/s 𝑀 100 24.Write the condition for equilibrium when two forces F1 and F2 act on a particle 25.Write the condition for equilibrium when three forces F1 , F2 and F3 act on a particle. 26. The maximun value of limiting friction is called ………………. Limiting friction or Limiting static friction. 27.State the law of static friction The law of static friction may thus be written as , fs ≤ μs N 0r ( fs )max = μs N where μs the coefficient of static friction, 23 Downloaded from www.hssreporter.com 28.State the Law of Kinetic Friction fk = μk N where μk the coefficient of kinetic friction, 29.Write the characteristics of static friction ▪ The maximum value of static friction is ( fs )max ▪ The limiting value of static friction ( fs )max , is independent of the area of contact. ▪ The limiting value of static friction ( fs )max , varies with the normal force(N) ( fs )max αN ( 𝐟𝐬 )𝐦𝐚𝐱 = 𝛍𝐬 𝐍 30.Write the characteristics of kinetic friction ▪ Kinetic friction is independent of the area of contact. ▪ Kinetic friction is independent of the velocity of sliding. ▪ Kinetic friction , fk varies with the normal force(N) fk αN fk = μk N 31.Show that 𝛍𝐬 = 𝐭𝐚𝐧 𝛉 (the coefficient of static friction is equal to the tangent of angle of friction) when a body just begins to slide on an inclined surface The forces acting on a block of mass m When it just begins to slide are (i) the weight, mg (ii) the normal force, N (iii) the maximum static frictional force ( fs )max In equilibrium, the resultant of these forces must be zero. m g sin θ = ( fs )max But ( fs )max = μs N mg sin θ= μs N------------(1) mg cos θ = N-------------(2) (1) mg sin θ μs N Eqn(2) -------- = m g cos θ N μs = tan θ 32.Disadvantages of friction In a machine with different moving parts, friction opposes relative motion and thereby dissipates power in the form of heat, etc. Friction produces wear and tear. 24 Downloaded from www.hssreporter.com 33.Advantages of friction Kinetic friction is made use of by brakes in machines and automobiles. We are able to walk because of static friction. The friction between the tyres and the road provides the necessary external force to accelerate the car. 34.Methods to reduce friction (1)Lubricants are a way of reducing kinetic friction in a machine. (2)Another way is to use ball bearings between two moving parts of a machine. (3) A thin cushion of air maintained between solid surfaces in relative motion is another effective way of reducing friction. 35.A car moving on a curved level road. What are the various forces acting on the car? Three forces act on the car. (i) The weight of the car, mg (ii) Normal reaction, N (iii) Frictional force, fs 36.Derive the expression for maximum safe speed on a curved level road N= mg The static friction provides the centripetal acceleration mv2 fs = R But , fs ≤ μs N mv2 ≤ μs mg (N=mg) R 2 v ≤ μs Rg 𝐯𝐦𝐚𝐱 = √𝛍𝐬 𝐑𝐠 37. a) What do you mean by banking of curved roads? b) Obtain the expression for maximum permissible speed of a vehicle on a banked road. c)Write the expression for optimum speed (without considering frictional force) a)Raising the outer edge of a curved road above the inner edge is called banking of curved roads. 25 Downloaded from www.hssreporter.com b) N cos θ = mg +f sin θ N cos θ - f sin θ =mg --------------(1) The centripetal force is provided by the horizontal components of N andfs. mv2 N sin θ + f cos θ = -------------(2) R Eqn(1) N cos θ − f sin θ mg ----- = mv2 Eqn(2) N sin θ + f cos θ R Dividing throughout by N cos θ f 1 − tan θ Rg N f = tan θ + v2 N 1 −μs tan θ Rg = tan θ + μs v2 2 Rg(μs +tan θ ) v = 1 −μs tan θ Rg(μs +tan θ ) vmax = √ 1 −μs tan θ c)If friction is absent, μs = 0 Then Optimum speed, voptimum = √Rg tan θ 38.A circular racetrack of ra dius 300 m is banked at an angle of 15°. If the coefficient of friction between the wheels of a race-car and the road is 0.2, what is the (a) optimum speed of the race car to avoid wear and tear on its tyres, and (b) maximum permissible speed to avoid slipping ? (a) (b) 26 Downloaded from www.hssreporter.com  Chapter 6 Work ,Energy and Power ⃗ and 𝐁 1.Define scalar product or Dot Product two vectors 𝐀 ⃗⃗. ⃗A ⋅ B ⃗ = AB cosθ 2.Some properties of dot product. 𝑖̂ ⋅ 𝑖̂ = 𝑗̂ ⋅ 𝑗̂ = 𝑘̂ ⋅ 𝑘̂ = 1 𝑖̂ ⋅ 𝑗̂ = 𝑗̂ ⋅ 𝑘̂ = 𝑘̂ ⋅ 𝑖̂ = 0 ⃗ ⋅A A ⃗ = A2 ⃗ 𝐚𝐧𝐝 𝐁 3.If two vectors 𝐀 ⃗⃗ are perpendicular, then their dot product will be………….. zero ( ⃗A ⋅ B ⃗ = A B cos 90 = 0) 4.Define work. The work done by the force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement. W=F ⃗ ⋅ ⃗d 5.Write the situations in which work done by a body is zero (i) when the displacement is zero. (ii )when the force is zero. (iii) the force and displacement are mutually perpendicular W=Fd cos 90 = 0. 6.Give an example for zero Work. When you push hard against a rigid brick wall, the force you exert on the wall does no work. A weightlifter holding a 150 kg mass steadily on his shoulder for 30 s does no work on the load during this time. 7.Give an example for Positive Work Workdone by Gravitational force on a freely falling body is positive 8.Give an example of Negative work The the work done by friction is negative (cos 180 o = –1). 9.Find out the sign of the workdone in following cases. a)Workdone by a man in lifting a bucket out of a well. Positive b)Workdone by friction on a body sliding down an inclined plane Negative c)Workdone by an applied force on a body moving on a rough horizontal surface Positive 27 Downloaded from www.hssreporter.com d)Workdone by the resistive force on air on a vibrating pendulum Negative e)Workdone gravitational force during the motion of an object on a horizontal surface. Zero 10. What is the work done by centripetal force on a body moving in circular path Zero. Here θ = 90 o , W =Fd cos 90 = 0. 11. 1 horse power,1HP= -----------Watt. 746W 12. 1 kilowatt-hour, 1kWh = ------------------ J 3.6 × 106 J 13. Kilowatt-hour is the unit of ------------- Energy 14.Find the workdone by a force 𝐅 = (𝟑𝐢̂+ 𝟒𝐣̂ - 𝟓𝐤 ̂ ) N , if the displacement Produced is ̂ ) m. 𝐝 = (𝟓𝐢̂+ 𝟒𝐣̂ + 3𝐤 W =F ⃗ ⃗.d 𝑊 = 𝐹𝑥 𝑑𝑥 + 𝐹𝑦 𝑑𝑦 + 𝐹𝑧 𝑑𝑧 = (3 x 5 )+ (4x4 ) + (-5 x3) W= 16 J 15. The energy possessed by a body by virtue of its motion is called----------- Kinetic energy 16. The energy stored by virtue of the position or configuration of a body(state of strain) is called------------- Potential Energy. 17. Calculate the work done in lifting a body of mass 10kg to a height of 10m above the ground W= F x d = mg x h =10 x 9.8 x10 =980J 18. Two bodies of masses 𝐦𝟏 𝐚𝐧𝐝 𝐦𝟐 have same momenta. What is the ratio of their kinetic energies? 1 p2 KE, K = mv 2 = 2 2m p2 p2 K1 = K2 = 2 m1 2 m2 K1 / K 2 = m2 /m1 K1 : K 2 = m2 : m1 19. A light body and heavy body have same momenta, Which one has greater kinetic energy? p2 KE = 2m 1 KE∝ m Lighter body will have more Kinetic energy. 20.Power is the scalar product of force and -------------- Velocity ( P= F.v) 28 Downloaded from www.hssreporter.com 21.Write the characteristics of conservative forces. ▪ A conservative force can be derived from a scalar quantity. − 𝑑V F= where V is a scalar dx ▪ The work done by a conservative force depends only upon initial and final positions of the body ▪ The work done by a conservative force in a cyclic process is zero 22.Give two examples for conservative forces Eg: Gravitational force, Spring force. 23.Give two examples for non conservative forces. Frictional force , air resistance are non conservative forces. 24. State and prove the law of conservation of mechanical energy for a freely falling body. The total mechanical energy of a system is conserved if the forces, doing work on it, are conservative. Consider a body of mass m falling freely from a height h At Point A PE = mgh KE = 0 (since v=0) TE= mgh + 0 TE= mgh-----------(1) At Point B PE = mg (h-x) KE = ½ mv 2 v 2 = u2 + 2as u=0, a=g , s= x v 2 = 2gx KE = ½ m x 2gx =mgx TE = mg (h-x) + mgx TE = mgh--------------(2) At Point C PE = 0 (Since h=0 KE = ½ mv 2 v 2 = u2 + 2as u=0, a=g , s= h 2 v = 2gh KE = ½ m x 2gh=mgh TE = 0 + mgh TE = mgh--------------(3) From eqns (1), (2) and (3), it is clear that the total mechanical energy is conserved during the free fall. 29 Downloaded from www.hssreporter.com 25.Show that the gravitational potential energy of the object at height h, is completely converted to kinetic energy on reaching the ground. PE at a height h, V = mgh When the object is released from a height it gains KE K = ½ mv 2 v 2 = u2 + 2as u=0, a=g , s=h 2 v = 2gh K = ½ m x 2gh K= mgh 26.The energy possessed by a body by virtue of its motion is called …………….. Kinetic energy 27.The eneegy possessed by a body by virtue of the position or configuration of a body is called ………………….. Potential energy. 28.A body at a height h above the surface of earth possesses ………………… due to its position. Potential energy. 29.A Stretched or compressed spring possesses ………………… due to its state of strain. potential energy 30.State and prove work-energy theorem The work-energy theorem can be stated as :The change in kinetic energy of a particle is equal to the work done on it by the net force. Proof For uniformly accelerated motion v 2 − u 2 = 2 as 1 Multiplying both sides by 𝑚, we have 2 1 1 mv − mu2 = mas = Fs 2 2 2 Kf -Ki = W Change in KE = Work 30 Downloaded from www.hssreporter.com 31.In a ballistics demonstration a police officer fires a bullet of mass 50.0 g with speed 200 m s-1 on soft plywood of thickness 2.00 cm. The bullet emerges with only 10% of its initial kinetic energy. What is the emergent speed of the bullet ? 32.State Hooke’s law for a spring Hooke’s law states that ,for an ideal spring, the spring force F is proportional displacement x of the block from the equilibrium position. F = − kx The constant k is called the spring constant. Its unit is N𝑚−1 33.Derive the expression for potential energy of a spring Then the spring force F = − kx The work done by the spring force is x W = ∫0 F dx x W = − ∫0 kx dx 1 W = − kx 2 2 This work is stored as potential energy of spring 1 PE = kx 2 2 34.Draw the graphical variation of kinetic Energy and potential of a spring 31 Downloaded from www.hssreporter.com 35.Write Einstein’s mass energy relation. E = m 𝐜𝟐 36.Find the amount of energy is associated with 1 kilogram of matter E = m c2 E = 1×(3 × 108 )2 J E= 9 × 1016 J. 37.Write the expression for instantaneous power in dot product form dW P= dt The work done, dW = F. dr. dr P =F. dt P= F. v Instantaneous power is the dot product of force and velocity. 38.An elevator can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 m s–1. The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power. The downward force on the elevator is F = m g + Frictional Force = (1800 × 10) + 4000 = 22000 N Power, P = F. v = 22000 × 2 = 44000 W In horse power, power = 44000/746 =59 hp 39.Differentiate Elastic and inelastic collisions.Give examples for each. The collisions in which both linear momentum and kinetic energy are conserved are called elastic collisions. Eg: Collision between sub atomic particles The collisions in which linear momentum is conserved, but kinetic energy is not conserved are called inelastic collisions.. Part of the initial kinetic energy is transformed into other forms of energy such as heat,sound etc.. Eg: Collision between macroscopic objects 40.For elastic collisions in one dimension show that the relative velocity before collision is numerically equal to relative velocity after collision. 32 Downloaded from www.hssreporter.com By the conservation of momentum m1 u1 + m2 u2 = m1 v1 + m2 v2 --------------(1) m1 u1 − m1 v1 = m2 v2 − m2 u2 m1 (u1 − v1 ) = m2 (v2 − u2 )----------------(2) By the conservation of kinetic energy 1 1 1 1 m1 u12 + m2 u22 = m1 v12 + m2 v22 -----------(3) 2 2 2 2 1 1 1 1 m u − m1 v1 = m2 v2 − m2 u22 2 2 2 2 1 1 2 2 2 1 1 m (u1 − v1 ) = m2 (v2 − u22 ) 2 2 2 2 1 2 m1 (u12 − v12 ) = m2 (v22 − u22 ) -------------(4) (4) m1 (u21 −v21 ) m2 (v22 −u22 ) Eqn ------------ = (2) m1 (u1 −v1 ) m2 (v2 −u2 ) (u1 +v1 ) (u1 −v1 ) (v2 +u2 )(v2 −u2 ) = (u1 −v1 ) (v2 −u2 ) u1 + v1 = v2 + u2 -------------(5) u1 − u2 = −(v1 − v2 )--------(6) i.e., relative velocity before collision is numerically equal to relative velocity after collision. 41.For elastic collisions of a moving mass 𝐦𝟏 with the stationary mass 𝐦𝟐 write the expression for momentum conservation and kinetic energy conservation Equation for conservation of momentum in x direction m1 u1 = m1 v1 cosθ1 + m2 v2 cosθ2 Equation for conservation of momentum in y direction 0 = m1 v1 sinθ1 − m2 v2 sin θ2 Equation for conservation of kinetic energy,(KE is a scalar quantity) 1 1 1 m1 u12 = m1 v12 + m2 v22 2 2 2 33 Downloaded from www.hssreporter.com  𝐂𝐡𝐚𝐩𝐭𝐞𝐫 𝟕 𝐒𝐲𝐬𝐭𝐞𝐦𝐬 𝐨𝐟 𝐏𝐚𝐫𝐭𝐢𝐜𝐥𝐞𝐬 𝐚𝐧𝐝 𝐑𝐨𝐭𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐌𝐨𝐭𝐢𝐨𝐧 1.Define centre of mass The centre of is a hypothetical point where the entire mass of an object may be assumed to be concentrated to visualise its motion. 2.Write the expression for position vector of centre of mass of an n particle system ⃗R = m1r⃗1+m2⃗r2+⋯……..+mn⃗rn M where M =m1 + m2 +…….+mn 3.Write the expression for velocity of centre of mass of an n particle system ⃗V = m1v⃗1+m2v⃗2+⋯……..+mnv⃗n --------------(2) M where M =m1 + m2 +…….+mn 4.Write the expression for acceleration of centre of mass of an n particle system ⃗⃗⃗A = m1 a⃗1+m2 a⃗2+⋯……..+mna⃗n ………………(3) M where M =m1 + m2 +…….+mn 5.When the total external force on the system is zero the velocity of the centre of mass remains constant or the CM of the system is in…………………… uniform motion. 6.Define vector product or cross product of two vectors 𝐀 ⃗ and 𝐁 ⃗⃗ Vector product of two vectors ⃗A and B ⃗ is defined as ⃗A x B ⃗ = AB 𝐬𝐢𝐧 𝜽 n̂ where A and B are magnitudes of ⃗A and B⃗ 𝜽 is the angle between A ⃗ and B ⃗ 𝑛̂ is the unit vector perpendicular to the plane containing ⃗A and B⃗ 7.Some properties of vector product ▪ ⃗A x ⃗A = ⃗0 ▪ 𝑖̂ × 𝑖̂ = 0 , 𝑗̂ × 𝑗̂ = 0 , 𝑘̂ × 𝑘̂ = 0 ▪ 𝑖̂ × 𝑗̂ = 𝑘̂ , 𝑗̂ × 𝑘̂ = 𝑖̂, 𝑘̂ × 𝑖̂ = 𝑗̂ ▪ 𝑗̂ × 𝑖̂ = −𝑘̂ , 𝑘̂ × 𝑗̂ = −𝑖̂, 𝑖̂ × 𝑘̂ = −𝑗̂ 𝟖.. 𝐖𝐫𝐢𝐭𝐞 𝐭𝐡𝐞 𝐫𝐞𝐥𝐚𝐭𝐢𝐨𝐧 𝐜𝐨𝐧𝐧𝐞𝐜𝐭𝐢𝐧𝐠 𝐚𝐧𝐠𝐮𝐥𝐚𝐫 𝐯𝐞𝐥𝐨𝐜𝐢𝐭𝐲 𝐚𝐧𝐝 𝐢𝐭𝐬 𝐥𝐢𝐧𝐞𝐚𝐫 𝐯𝐞𝐥𝐨𝐜𝐢𝐭𝐲. ⃗⃗v = ω ⃗⃗ × r 𝟗. 𝐃𝐞𝐟𝐢𝐧𝐞 𝐀𝐧𝐠𝐮𝐥𝐚𝐫 𝐚𝐜𝐜𝐞𝐥𝐞𝐫𝐚𝐭𝐢𝐨𝐧 Angular acceleration α ⃗ is defined as the time rate of change of angular velocity. ⃗⃗⃗ 𝑑𝜔 𝛼= unit rad/𝑠 2 or rad 𝑠 −2 𝑑𝑡 10.The rotational analogue of force is ---------------- Torque or Moment of force 34 Downloaded from www.hssreporter.com 11.Write the equation for torque or moment of force 𝜏 = 𝑟 x ⃗⃗F 12.The handle of the door is always fixed at the edge of the door which is located at a maximum possible distance away from hinges. Give reason. 𝜏=𝑟xF ⃗⃗ =rF sin𝜃 When r is maximum ,torque will be maximum. The handle is fixed at the edge to increase r and hence to make torque maximum. 13.Find the torque of a force 𝟕𝐢̂ + 𝟑𝐉̂ − 𝟓𝐤 ̂ about the origin. The force acts on a particle whose position vector is 𝐢̂ − 𝐉̂ + 𝐤 ̂. 𝜏=𝑟xF ⃗⃗ 𝜏 = (î − Ĵ + k̂) x(7î + 3Ĵ − 5k̂) + - + î 𝐽̂ 𝑘̂ 𝜏 =|1 −1 1 | 7 3 −5 τ⃗ = î [(−1 x − 5) −(3 x 1) ] - Ĵ [(1x-5)- (7x1)] + k̂ [(1x3) -(7x-1)] τ⃗ = î [5 - 3] - Ĵ [-5 – 7] + k̂ [3 – -7] τ⃗ = 2î +12 Ĵ + 10k̂ 14.Angular momentum is the rotational analogue of --------------- linear momentum. 15. Write the relation connecting angular momentum and linear momentum. 𝑙 =r×p ⃗ 16.The moment of linear momentum is called --------------- Angular momentum 17.Write the relation connecting torque and angular momentum 𝑑𝑙 𝜏= 𝑑𝑡 18. Deduce the relation connecting torque and angular momentum (or) Show that the time rate of change of the angular momentum of a particle is equal to the torque acting on it. 𝑙 =𝑟×𝑝 𝑑𝑙 d Differentiating = (r×p ⃗ ) 𝑑𝑡 dt 𝑑𝑙 dr⃗ ⃗ dp = ×p ⃗ + rx 𝑑𝑡 dt dt dr⃗ ⃗ dp ⃗ = mv p ⃗ , =v ⃗ , ⃗⃗ =F dt dt 𝑑𝑙 =v ⃗ × mv ⃗⃗ ⃗ + rxF 𝑑𝑡 ⃗ = 0 , (r x ⃗⃗F =τ⃗ ) ⃗ ×v v 𝑑𝑙 =0 + 𝜏 𝑑𝑡 𝒅𝒍 ⃗ = 𝝉 𝒅𝒕 35 Downloaded from www.hssreporter.com 19.The time rate of change of the angular momentum of a particle is equal to the --------- ------ acting on it. Torque 20.State and prove the law of conservation of angular momentum If the total external torque on a system of particles is zero, then the total angular momentum of the system is conserved i.e, remains constant. ⃗ dL τ⃗ext = dt If external torque, τ⃗ext = 0 , ⃗ dL =0 dt ⃗L = constant 21.Write an example of a motion in which angular momentum remains constant Motion of planets around sun. 22.What do you mean by equilibrium of a rigid body? A rigid body is said to be in mechanical equilibrium, if it is in both translational equilibrium and rotational equilibrium. 23.What is translational equilibrium? When the total external force on the rigid body is zero, then the total linear momentum of the body does not change with time and the body will be in translational equilibrium. 24.What is rotational equilibrium? When the total external torque on the rigid body is zero, the total angular momentum of the body does not change with time and the body will be in rotational equilibrium. 25.What is partial equilibrium? A body may be in partial equilibrium, i.e., it may be in translational equilibrium and not in rotational equilibrium, or it may be in rotational equilibrium and not in translational equilibrium. 26.Define a couple. A pair of equal and opposite forces with different lines of action is known as a couple. A couple produces rotation without translation. 27.State the principle of moments. The lever is a system in mechanical equilibrium. For rotational equilibrium the sum of moments must be zero, d1 F1 − d2 F2 = 0 𝐝𝟏 𝐅𝟏 = 𝐝𝟐 𝐅𝟐 36 Downloaded from www.hssreporter.com 28.What is centre of gravity of a body? The Centre of gravity of a body is the point where the total gravitational torque on the body is zero. 29. Moment of Inertia is the rotational analogue of ------------ Mass. 30.The rotational analogue of mass is called--------------------- Moment of Inertia 31. Mass is a measure of ------------------ and moment of inertia is a measure of ------------ ------------- Inertia , Rotational inertia 32. Writ the expression for moment of inertia of a particle of mass m rotating about an axis I =mr 2 33. Write the equation for rotational kinetic energy. 1 Rotational kE = Iω2 2 34. What do you mean by radius of gyration ? The radius of gyration can be defined as the distance of a mass point from the axis of roatation whose mass is equal to the whole mass of the body and whose moment of inertia is equal to moment of inertia of the whole body about the axis. I =Mk 2 I k=√ M 35.The moment of inertia of a disc of mass ‘M’ and radius R about an axis passing 𝐌𝐑𝟐 through its centre and perpendicular to its plane is. 𝟐 What is the radius of gyration of this case. I k=√ M MR2 R2 𝑅 k=√ 2 =√ = M 2 √2 36.What is a flywheel? A disc with a large moment of inertia is called a flywheel. It is used in machines, that produce rotational motion. 37 Downloaded from www.hssreporter.com 37.Write the rotational analogue of the following equations in translational motion 38.A cord of negligible mass is wound round the rim of a flywheel mounted on a horizontal axle as shown in figure. Calculate the angular acceleration of the wheel if steady pull of 25N is applied on the cord. Moment of inertia of flywheel about its 𝑴𝑹𝟐 axis= 𝟐 MR2 20 x 0.22 I= = =0.4 kgm2 2 2 Torque , τ = force xperpendicular distance τ = 25 x0.2 = 5Nm τ=I∝ τ 5 ∝= = =12.5rad/s2 I 0.4 39.A girl rotates on a swivel chairas shown below. a)what happens to her angular speed when she stretches her arms b) what happens to her angular speed when she folds her arms c)Name and state the conservation law applied for your justification a)When she stretches her arms, the moment of inertia increases and hence the angular speed decreases. b)When she folds her arms, the moment of inertia decreases and hence the angular speed increases. c) Law of conservation of angular momentum. If the total external torque on a system of particles is zero, then the total angular momentum of the system is conserved i.e, remains constant. ( If τ⃗ext = 0 , ⃗L = 𝐼𝜔⃗ = constant.When I increases 𝜔 ⃗ decreases and vice versa) 38 Downloaded from www.hssreporter.com  Chapter 8 Gravitation 1.State Kepler’s first law of planetary motion(Law of orbits ) All planets move in elliptical orbits with the Sun situated at one of the foci of the ellipse. 2.State Kepler’s second law of planetary motion(Law of areas) The line that joins any planet to the sun sweeps equal areas in equal intervals of time. ⃗⃗ ΔA i.e, areal velocity is constant Δt 3.Kepler’s second law (law of areas) is a consequence of conservation of ………………………. angular momentum. 4.Prove Kepler’s second law of planetary mtion. The area swept out by the planet of mass m in time interval Δt is ΔA⃗ = 1 (r × v⃗ Δt) 2 ⃗ = mv p ⃗ , ⃗ p ⃗ = v m ⃗⃗ ΔA 1 ⃗ p = (r × ) Δt 2 m ⃗ =r×p L ⃗ ⃗⃗ ΔA ⃗L = Δt 2m Here angular momentum,⃗⃗⃗L is a constant. ⃗⃗ ΔA = constant Δt This is the law of areas. 5.State Kepler’s third law of planetary motion(Law of periods) The square of the time period of revolution of a planet is proportional to the cube of the semi- major axis of the ellipse traced out by the planet. T 2 ∝ a3 6.Prove Kepler’s third law of planetary mtion. (R+h)3 Period of a satellite, T= 2π√ GM 3 2 (R+h) T 2 = 4π GM 4π2 T2= (R + h)3 GM T 2 ∝ (R + h)3 T 2 ∝ a3 Which is Kepler’s Law of Periods. 39 Downloaded from www.hssreporter.com 7.State Universal Law of Gravitation Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. 𝐦𝟏 𝐦𝟐 𝐅=𝐆 𝟐 𝐫 8. The value of Gravitational Constant. G = 6.67×10−11 N m2 /kg 2 9.Define acceleration due to gravity of the Earth The acceleration gained by a body due to the gravitational force of earth is called acceleration due to gravity. 10. Obtain the expression for acceleration due to gravity on the surface of the earth (or) Obtain the relation connecting g and G. Consider a body of mass m on the surface of earth of mass M and radius R. The gravitational force between body and earth is given by GMm F = 2 -----------(1) R By Newton’s second law F=mg where g is acceleration due to gravity F g= m GM From Eq (1) g= R2 11.The average value of g on the surface of earth is -------. 9.8ms −2. 12. Acceleration due to gravity is independent of------------( mass of the body/mass of earth). mass of the body 13.A man can lift a mass of 15kg on earth.What will be the maximum mass that can be lifted by him by applying the same force on moon. 6x15 =90kg 1 (Acceleration due to gravity on the surface of moon is times that on earth. So he can 6 lift 6 times massive objects on the surface of moon) 14.A mass of 30kg is taken from earth to moon. What will be its mass and weight on the surface of moon Mass on the moon=30kg (mass remains the same) 30 𝑥9.8 Weight on the moon = =49N 6 40 Downloaded from www.hssreporter.com 15.Obtain the expression for Acceleration due to gravity at a height h above the surface of the earth. Acceleration due to gravity on the surface of earth GM g = 2 ------------(1) R Acceleration due to gravity at a height above the surface of earth GM gh = 2 ----------(2) (R+h) For , h