Time: 2 Hours <\/span> General Instructions to candidates:<\/span><\/p>\n Answer any 4 questions from question numbers 1 to 5. Each carry 1 score. (4 \u00d7 1 = 4)<\/span><\/p>\n Question 1. Question 2. Question 3. Question 4. Question 5. Answer any 4 questions from question numbers 6 to 10. Each carries 2 scores. (4 \u00d7 2 = 8)<\/span><\/p>\n Question 6. Question 7. Question 8. Question 9. Question 10. Answer any 4 questions from question numbers 11 to 15. Each carries 3 scores. (4 \u00d7 3 = 12)<\/span><\/p>\n Question 11. Question 12. Question 13. Question 14. Question 15. Answer any 4 questions from question numbers 16 to 20. Each carries 4 scores. (4 \u00d7 4 = 16)<\/span><\/p>\n Question 16. Question 17. c) Consider a vehicle along a curved road with angle of banking q. Then the normal reaction on the ground will be inclined at an angle q with the vertical.<\/p>\n The vertical component can be divided into N \u2018 Cosq (vertical component) and N sinq (horizontal component). Suppose the vehicle has a tendency to slip outward. Then the frictional force will be developed along the plane of road as shown in the figure. The frictional force can be divided into two components. Fcosq (horizontal component) and F sinq (vertical component). Question 18. Question 19. Question 20. Answer any 4 questions from question numbers 21 to 25. Each carries 5 scores. (4 \u00d7 5 = 20)<\/span><\/p>\n Question 21. Question 22. Question 23. b) Consider a tank containing a liquid of density r with a small hole in its side. Let Y1<\/sub> be the height of the hole, and y2<\/sub> be height of water in the tank. Applying Bernoulli’s equation at points (1) and (2) we get c) Torricelli’s law may be stated as the velocity of afflux through a hole at a depth \u2018h\u2019 will be equal to the velocity gained by a freely falling body when it travels a distance \u2018h\u2019.
\nCool off time: 15 Minutes<\/span>
\nMaximum: 60 Scores<\/span><\/p>\n\n
\nState TRUE or FALSE
\n\u201cSome conservation laws are true for one fundamental force, but not for the others\u201d.
\nAnswer:
\nTrue<\/p>\n
\nThe angle between A = i + j and B = i – j is
\na) 45\u00b0
\nb) 90\u00b0
\nc) 60\u00b0
\nd) 180\u00b0
\nAnswer:
\nb) 90\u00b0<\/p>\n
\nThree objects with a mass of 40kg each are placed in a straight line 50cm apart. What is the net gravitational force at the centre object due to the other two?
\nAnswer:
\nZero<\/p>\n
\nWhich one of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?
\na) a = 5x
\nb) a = -200x2<\/sup>
\nc) a = -5x
\nd) a = 100x3<\/sup>
\nAnswer:
\nc) a = – 5x<\/p>\n
\nThe stress required to double the length of a wire of Young\u2019s modulus Y is ………..
\na) Y\/2
\nb) 2Y
\nc) y
\nd) 4Y
\nAnswer:
\nc) y<\/p>\n
\nA car travelling at a speed 54 km\/hr is brought to rest in 90s. Find the distance travelled by the car before coming to rest.
\nAnswer:
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-1.png\")
<\/p>\n
\nThe parallelogram law is used to find the resultant of two vectors. Find the magnitude of the resultant of two vectors in terms of their magnitudes and angle between them.
\nAnswer:
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-2.png\")
\nConsider two vectors \\(\\overrightarrow{\\mathrm{A}}\\) (\\(=\\overrightarrow{\\mathrm{OP}}\\)) and B (\\(=\\overrightarrow{\\mathrm{OQ}}\\)) making an angle \u03b8. Using the parallelogram method of vectors, the resultant vector R can be written as,
\n\\(\\overrightarrow{\\mathrm{R}}\\) = \\(\\overrightarrow{\\mathrm{A}}\\) + \\(\\overrightarrow{\\mathrm{B}}\\)
\nSN is normal to OP and PM is normal to OS. From the geometry of the figure
\nOS2<\/sup> = ON2<\/sup> + SN2<\/sup>
\nbut ON = OP + PN
\nie. OS2<\/sup> = (OP + PN)2<\/sup> + SN2<\/sup> ……..(1)
\nFrom the triangle SPN, we get
\nPN = Bcos\u03b8 and SN = Bsin\u03b8
\nSubstituting these values in eq.(1), we get
\nOS2\u00a0<\/sup>= (OP + Bcos\u03b8)2<\/sup> + (Bsin\u03b8)2<\/sup>
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-3.png\")
<\/p>\n
\nSelect the CORRECT alternative.
\na) When a conservative force does positive work on a body, the potential energy of the body……….
\ni) increases
\nii) decreases
\niii) remains unaltered
\nb) Work done by a body against friction always results in a loss of its……..
\ni) kinetic energy
\nii) potential energy
\nc) The rate of change of total momentum of a system of many particle system is proportional to the ……… on the system.
\ni) external force
\nii) sum of the internal forces
\nd) The quantity which is conserved in an inelastic collision of two bodies is …….
\ni) total kinetic energy
\nii) total linear momentum
\nAnswer:
\na) decreases
\nb) kinetic energy
\nc) external force
\nd) total linear momentum<\/p>\n
\nA steel wire of length 1.5 m and diameter 0.25 cm is loaded with a force of 98 N. The increase in length of the wire 1.5 \u00d7 10-4<\/sup> m. Calculate the tensile stress and the fractional change in length of the wire.
\nAnswer:
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-4.png\")
<\/p>\n
\nAccording to the kinetic theory of gases, gas molecules are always in random motion.
\na) State the law of equipartition of energy.
\nb) Write the average value of energy of a molecule for each vibrational mode.
\nAnswer:
\na) The total kinetic energy of a molecule is equally divided among the different degrees freedom.
\nb) E = \\(\\frac{1}{2}\\)KB<\/sub>T \u00d7 2 = KB<\/sub>T<\/p>\n
\nTwo parallel rail tracks run north-south. Train A moves north with a speed of 15 m\/s and train B moves south with a speed of 25 m\/s.
\na) What is the velocity of B with respect to A?
\nb) What is the velocity of ground with respect to B?
\nc) What is the velocity of a monkey running on the roof of the train A against its motion (with a velocity of 5m\/s with respect to the train A) as observed by a man standing on the ground?
\nAnswer:
\na) VBA<\/sub> = VB<\/sub> – -VA<\/sub>\u00a0= 25 + 15 = 40 m\/s
\nb) VgB<\/sub>\u00a0= Vg<\/sub> – VB<\/sub> = 0 – 25 = -25 m\/s
\nResultant velocity = 15 – 5
\n(opposite to direction) = 10 m\/s<\/p>\n
\nAn insect trapped in a circular groove of radius 12cm moves along the groove steadily and completes 7 revolutions in 100s.
\na) What is the linear speed of the motion?
\nb) Is the acceleration vector a constant vector? What is its magnitude?
\nAnswer:
\na) R = 12 \u00d7 10-2<\/sup>m
\n7 revolutions = 100 sec
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-5.png\")
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-6.png\")
<\/p>\n
\nA car and a truck have the same kinetic energies at a certain instant while they are moving along two parallel roads.
\na) Which one will have greater momentum?
\nb) If the mass of truck is 100 times greater than that of the car, find the ratio of velocity of the truck to that of the car.
\nAnswer:
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-7.png\")
<\/p>\n
\nA girl rotates on a swivel chair as shown below.
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-8.png\")
\na) What happens to her angular speed when she stretches her arms?
\nb) Name and state the conservation law applied for your justification.
\nAnswer:
\na) The angular speed decreases
\nb) Law of conservation of angular momentum. According to this law, the angular momentum of a body is conserved when external torque is zero.<\/p>\n
\nA solid sphere of mass m and radius R starts from rest and rolls down along an inclined plane of height h without slipping as shown below:
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-9.png\")
\na) Calculate the kinetic energy of the sphere when it reaches the ground
\nb) Find the velocity when it reaches the base.
\nAnswer:
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-10.png\")
<\/p>\n
\nA man jumping out of a slow moving bus falls forward.
\na) This is due to ……..
\nb) Which Newton\u2019s law gives the above concept? State the law.
\nc) What is the net force acting on a book at rest on the table?
\nAnswer:
\na) Inertia\/inertia of motion
\nb) First law of Newton. The law states that every body continues its state of rest or uniform motion along straight line unless it is compelled by an. external unbalanced force to change that state.
\nc) Zero<\/p>\n
\nTo reduce friction and accident by skidding, the roads are banked at curves.
\na) What is meant by banking of roads?
\nb) Sketch the schematic diagram of a vehicle on a banked road with friction and mark the various forces.
\nc) Derive an expression for maximum safe speed of a vehicle on a banked road with friction.
\nAnswer:
\na) Theputeredgeofthe road is raised slightly above the inner edge. This is called banking of road.
\nb)
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-11.png\")
<\/p>\n
\nFrom the figure are get
\nN cos\u03b8 = F sin\u03b8 + mg
\nN cos\u03b8 – F sin\u03b8 = mg ……..(1)
\nThe component Nsin\u03b8 and Fsin\u03b8 provide centripetal force. Hence
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-12.png\")
\nThis is the maximum speed at which vehicle can move over a banked curved road.<\/p>\n
\nFor a liquid-gas interface, the convex side has a high pressure than the concave side.
\na) Derive an expression for excess pressure inside a drop.
\nb) Which is better, washing of cloth in cool soap water or warm soap water? Why?
\nAnswer:
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-13.png\")
\nConsider a drop of liquid of radius r. Let Pi<\/sub> and Po<\/sub> be the values of pressure inside and outside the drop.
\nLet the radius of liquid of drop increases by a small amount Dr under the pressure difference.
\nThe outward force acting on the surface of the drop, f = pressure difference \u00d7 surface area,
\nie. f = (pi<\/sub>\u00a0– po<\/sub>)4pr2<\/sup>
\nIf the radius of liquid drop is increased by Dr due to the above force
\nTheworkdone(DW) = f.Dr
\n= (Pi<\/sub> – Po<\/sub>)4pr2<\/sup>Dr ……….(1)
\nThe increase in surface area of the drop
\n= 4p(r + Dr)2<\/sup> – 4pr2<\/sup>
\n= 4p(r2<\/sup> + 2rDr + Dr2<\/sup>) – 4pr2<\/sup> = 8prDr
\n[ Neglecting Dr2<\/sup>]
\nIf S is the surface tension of the liquid, the workdone to increase the surface area,
\nDW = Increase in surface area \u00d7 surface tension
\nDW = 8prDr.S ………(2)
\nFrom equation (1) and (2), we get
\nPi<\/sub> – Po<\/sub> = \\(\\frac{2 \\mathrm{~S}}{\\mathrm{r}}\\)
\nb) Warm soap water, because temperature reduces surface tension.<\/p>\n
\nLinear expansion is change in length of an object with temperature.
\na) Write the equation for coefficient of linear expansion.
\nb) Show that the coefficient of volume expansion is thrice its coefficient of linear expansion.
\nc) The absolute zero is……..
\n-273.15\u00b0C, -273.15K, -273.15\u00b0F, 0\u00b0C,
\nAnswer:
\na) \u03b1 = \\(\\frac{\\Delta \\ell}{\\ell \\Delta T}\\)
\nb) Consider a cube of length ‘l’. Due to the increase in temperature \u2018\u0394T\u2019, length of cube increases by \u0394l in all directions.
\nCoefficient of linear expansion, \u03b1l = \\(\\frac{\\Delta \\ell}{\\ell \\Delta \\mathrm{T}}\\)
\nIncrease in area of cube \u0394A
\n= Final area – initial area
\n= (l + \u0394l)2<\/sup> – l2<\/sup> = 2 \u00d7 l \u00d7 \u0394l
\n[Neglecting \u0394l2<\/sup>]
\nArea expansivity
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-14.png\")
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-15.png\")
\nc) -273.15\u00b0C<\/p>\n
\nThe simplest example of simple harmonic motion is the oscillations of a simple pendulum.
\na) Derive an expression for the period of oscillation of a simple pendulum.
\nb) In a simple pendulum made of a metallic wire, what will happen to the period when temperature increases? Give a reason.
\nAnswer:
\na)
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-16.png\")
\nConsider a mass m suspended from one end of a string of length L fixed at the other end as shown in figure. Suppose P is the instantaneous position of the pendulum. At this instant its string makes an angle \u03b8 with the vertical.
\nThe forces acting on the bob are (1) weight of bob Fg<\/sub> (mg) acting vertically downward. (2) Tension T in the string.
\nThe gravitational force Fg<\/sub> can be divided into a radial component Fg<\/sub>Cos\u03b8 and tangential component Fg<\/sub>Sin\u03b8. The radial component is cancelled by the tension T. But the tangential component Fg<\/sub>Sin\u03b8 produces a restoring torque.
\nRestoring torque \u03c4 = – Fg\u00a0<\/sub>sin\u03b8.L
\n\u03c4 = -mgsin\u03b8.L ………(1)
\n-ve sign shown that the torque and angular displacement \u03b8 are oppositely directed. For rotational motion of bob,
\n\u03c4 = I\u03b1 ……….(2)
\nWhere I is moment of inertia about the point of suspension and \u03b1 is angular acceleration. From eq (1) and eq (2).
\nI\u03b1 = – mgsin\u03b8.L
\nIf we assume that the displacement \u03b8 is small, sin\u03b8 \u2248 \u03b8.
\n\u2234 I\u03b1 = – mg\u03b8.L
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-17.png\")
\nb) Period increases because as temperature increases length of metal wire also increases.<\/p>\n
\na) The centripetal force depends on mass of the body, velocity and radius of circular path. Find the expression for the centripetal force acting on the body using the principle of dimensional analysis. (Take constant k = 1)
\nb) When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72″ of arc. Calculate the diameter of the Jupiter.
\nAnswer:
\na) The centripetal force depends on mass (m), radius (r) and velocity (v)
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-18.png\")
\nEquating we get
\na = 1
\nb + c = 1
\nc = 2
\n\u2234 b = -1
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-19.png\")
<\/p>\n
\nSatellites are objects which revolve around the earth.
\na) The direction of revolution of geosynchronous satellite is from ……….
\ni) east to west
\nii) west to east
\niii) north to south
\niv) south to north
\nb) Derive an expression for total energy of an orbiting satellite.
\nc) What is the magnitude of the angular velocity for a geosynchronous satellite?
\nAnswer:
\na) West to east
\nb) The kinetic energy of the satellite in a circular orbit at height h from surface of earth with speed
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-20.png\")
<\/p>\n
\nThe flow of an ideal fluid in a pipe of varying cross section is shown below.
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-21.png\")
\na) Write Bernoulli\u2019s equation.
\nb) Find the speed of efflux using Bernoulli\u2019s principle.
\nc) StateTorricelli\u2019s law.
\nAnswer:
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-22.png\")
\nConsider an incompressible liquid flowing through a tube of non uniform cross section from region 1 to region 2. Let P1<\/sub> be the pressure, A1<\/sub> the area of cross section and V1<\/sub> the speed of flow at the region 1. The corresponding values at region 2 are P2<\/sub>, A2<\/sub> and V2<\/sub> respectively. Region 1 is at a height h, and region 2 is at a height h2.
\nThe workdone on the liquid in a time \u0394t at the region 1 is given by
\nW1<\/sub> = force \u00d7 distance
\n= P A1<\/sub>\u0394x1<\/sub>
\n= P1<\/sub>\u0394V1<\/sub>
\n(\u2235A1<\/sub>\u0394x1<\/sub> = \u0394V)
\nWhere \u0394x1 is the displacement produced at region 1, during the time interval \u0394t Similarly the workdone in a time \u0394t at the region 2 is given by,
\nW2<\/sub> = -P2<\/sub> A2<\/sub> \u0394x2<\/sub>
\nW2<\/sub> = -P2<\/sub>\u0394V2<\/sub>
\n[Here -ve sign appears as the direction of \\(\\overrightarrow{\\mathrm{p}}\\) and \u0394x are in opposite directions.]
\nNet workdone
\nDW = P1<\/sub>\u0394V1<\/sub> – P2<\/sub>\u0394V2<\/sub>
\nAccording the equation of continuity
\n\u0394V1<\/sub> = \u0394V2<\/sub> = \u0394V
\n\u2234 \u0394W= P1<\/sub>\u0394V – P2<\/sub>\u0394V
\n\u0394W = (P1<\/sub> – P2<\/sub>)\u0394V ……….(1)
\nThis work done changes the kinetic energy, pressure energy and potential energy of the fluid.
\nIf \u0394m is the mass of liquid passing through the pipe in a time \u0394t. the change in Kinetic energy is given by \u0394k.E = \\(\\frac{1}{2} \\Delta \\mathrm{mV}_{2}^{2}-\\frac{1}{2} \\Delta \\mathrm{mV}_{1}^{2}\\)
\n\u0394k.E = \\(\\frac{1}{2} \\Delta \\mathrm{m}\\left(\\mathrm{V}_{2}^{2}-\\mathrm{V}_{1}^{2}\\right)\\) ……….(2)
\nChange in gravitational potential energy is given by
\n\u0394p.E = \u0394mgh2<\/sub> – \u0394mgh1<\/sub>.
\n\u0394p.E= \u0394mg(h2<\/sub> – h1<\/sub>) ……….(3)
\nAccording to work-energy theorem work done is equal to the change in kinetic energy plus the change in potential energy.
\nie; \u0394w = \u0394kE + \u0394PE ……….(4)
\nSubstituting eq. 1, 2 and 3 in eq. 4, we get
\n\\(\\left(P_{1}-P_{2}\\right) \\Delta V=\\frac{1}{2} \\Delta m\\left(V_{2}^{2}-V_{1}^{2}\\right)+\\Delta m g\\left(h_{2}-h_{1}\\right)\\)
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-23.png\")
<\/p>\n
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-24.png\")
\n(\u2235P1<\/sub> = Pa, the atmospheric pressure).
\nIf the cross sectional area of the tank A2 is much larger than that of the hole (ie; A2<\/sub>>>A1<\/sub>), we take
\nv2<\/sub> \u2248 0
\n\u2234 eq(1) can be written as.
\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-25.png\")
<\/p>\n
\nie; v = \\(\\sqrt{2 \\mathrm{gh}}\\)<\/p>\n
![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-26.png\")
![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-27.png\")
<\/p>\n![\"Plus](\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Plus-One-Physics-Improvement-Question-Paper-Say-2018-28.png\")
<\/p>\n