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iitjamjesthometask11

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ALL INDIA PHYSICS FORUM QUISTIONS AND ANSWERS OF Time : 1 Hours 30 minutes Subject : MECHANICS Full Marks: 50 GROUP—(A) M.C.Q TYPE QUESTIONSQUESTION NO.(1):-A piece of wire is bent in the shape of a parabola ������ = ������������2 (y-axis vertical) with abead of mass m on it. The bead can slide on the wire without friction. It stays atthe lowest point of the parabola when the wire is at rest. The wire is nowaccelerated parallel to the x-axis with a constant acceleration a. The distance ofthe new equilibrium position of the bead, where the bead can stays at rest withrespect to the wire, from the y axis is–––(������) ������ (������) ������ (������) 2������ (������) ������ ������������ 2������������ ������������ 4������������QUESTION NO.(2):- Stay connected with www.allindiaphysicsforum.com @ALL RIGHTS RESERVED FOR ALL INDIA PHYSICS FORUM

Three objects A, B and C are kept in a straight line on a frictionless horizontalsurface. These have masses m, 2m and m, respectively. The object A movestowards B with a speed 3 m/s and makes an elastic collision with it. Thereafter, Bmakes completely inelastic collision with C. All motions occur on the same straightline. Find the final speed (in m/s) of the object C ?(������) 4 ������ (������)6 ������ (������) 0 ������ (������) 10 ������ ������ ������ ������ ������QUESTION NO.(3):-This question has statement I and Statement II. Of the four choice given after thestatements, choose the one that best describes the two statements.Statement – I : A Point particle of mass m moving with speed v collides withstationary point particle of mass M. If the maximum energy loss possible is givenas������ (1 ������������2) ������ℎ������������ ������ = ( ������ ) 2 ������+������Statement – II : Maximum energy loss occurs when the particles get stuck togetheras a result of the collision.(A) Statement - I is true, Statement - II is true, statement - II is a correctexplanation of Statement - I(B) Statement - I is true, Statement - II is true, statement - II is not a correctexplanation of Statement – I(C) Statement - I is true, Statement - II is false(D) Statement – I is false, Statement – II is trueQUESTION NO.(4):-What is the minimum energy required to launch a satellite of mass m from thesurface of a planet of mass M and radius R in a circular orbit at an altitude of 2R ?(������) 5������������������ (������) 2������������������ (������) ������������������ (������) ������������������ 6������ 3������ 2������ 3������QUESTION NO.(5):- Stay connected with www.allindiaphysicsforum.com @ALL RIGHTS RESERVED FOR ALL INDIA PHYSICS FORUM

A hoop of radius r and mass m rotating with an angular velocity ������0 is placed on arough horizontal surface. The initial velocity of the centre of the hoop is zero.What will be the velocity of the centre of the hoop when it ceases to slip ?(������) ������������0 (������) ������������0 (������) ������������0 (������) ������������0 4 3 2QUESTION NO.(6):-Given in the figure are two blocks A and B of weight 20 N and 100 N, respectively.These are being pressed against a wall by a force F as shown. If the coefficient offriction between the blocks is 0.1 and between block B and the wall is 0.15, thefrictional force applied by the wall on block B is :(A) 100 N (B) 80 N (C) 120 N (D) 150 NQUESTION NO.(7):-A particle of mass m moving in the x direction with speed 2v is hit by anotherparticle of mass 2m moving in the y direction with speed v. If the collision isperfectly inelastic, the percentage loss in the energy during the collision is close to:(A) 44 % (B) 50 % (C) 56 % (D) 62 %QUESTION NO.(8):-Distance of the centre of mass of a solid uniform cone from its vertex is ������0 . If theradius of its base is R and its height is h the ������0 is equal to :(������) ℎ2 (������) 3ℎ2 (������) 3ℎ (������) 5ℎ 4������ 8������ 4 8 Stay connected with www.allindiaphysicsforum.com @ALL RIGHTS RESERVED FOR ALL INDIA PHYSICS FORUM

QUESTION NO.(9):-From a solid sphere of mass M and radius R a cube of maximum possible volume iscut. Moment of inertia of cube about an axis passing through its centre andperpendicular to one of its faces is :(������) ������������2 (������) ������������2 (������) 4������������2 (������) 4������������2 32√2 ������ 16√2 ������ 9√3 ������ 3√3 ������QUESTION NO.(10):-From a solid sphere of mass M and radius R, a spherical portion of radius ������ is 2removed, as shown in the figure. Taking gravitational potential V = 0 at r = ∞, thePotential at the centre of the cavity thus formed is :(G = gravitational constant)(������) −������������ (������) −������������ (������) −2������������ (������) −2������������ 2������ ������ 3������ ������Stay connected with www.allindiaphysicsforum.com @ALL RIGHTS RESERVED FOR ALL INDIA PHYSICS FORUM

QUESTION NO.(11):-Mount Fuji has approximately the shape of a cone. The half-angle at the apex ofthis cone is 65 degrees, and the height of the apex is 3800 m. At what height is thecenter of mass? (Assume that the material in Mount Fuji has uniform density.)(������) 150 ������ ������������������������������ ������ℎ������ ������������������������ (������) 560������ ������������������������������ ������ℎ������ ������������������������(������) 950 ������ ������������������������������ ������ℎ������ ������������������������ (������) 847������ ������������������������������ ������ℎ������ ������������������������QUESTION NO.(12):-A bead as shown in the above figure , is free to slide down a smooth wire tightly stretchedbetween points ������1 ������������������ ������2 on a vertical circle of radius R. If the bead starts from rest at ������1 , thehighest point on the circle . What is its velocity on arriving at ������2 ?(������)2√������������ (������) 2(√������������) sin ������ (������)2 (√������������) cos ������ (������)2(√������������) cos ������Stay connected with www.allindiaphysicsforum.com @ALL RIGHTS RESERVED FOR ALL INDIA PHYSICS FORUM

QUESTION NO.(13):-A bullet of mass “m” travels at a very high velocity v (as shown in thefigure) and gets embedded inside the block of mass “M” initially at rest on a roughhorizontal floor. The block with the bullet is seen to move a distance “s ” along thefloor. Assuming ������ to be the coefficient of kinetic friction between the block andthe floor and “g ” the acceleration due to gravity .what is the velocity v of thebullet ?(������) ������+������ √2������������������ (������) ������−������ √2������������������ (������) ������(������+������) √2������������������ (������) ������ √2������������������ ������ ������ ������ ������QUESTION NO.(14):-A block of mass M is released from point P on a rough inclined plane withinclination angle ������, shown in the figure below. The co-efficient of friction is .������������ ������ < tan ������ , then the time taken by the block to reach another point Q on theinclined plane, where PQ = s , is(������) √������ cos ������ 2������ ������−������) (������)√������ cos ������ 2������ ������+������) (tan (tanStay connected with www.allindiaphysicsforum.com @ALL RIGHTS RESERVED FOR ALL INDIA PHYSICS FORUM

(������) √������ sin ������ 2������ ������−������) (������)√������ sin ������ 2������ ������+������) (tan (tanQUESTION NO.(15):-A particle P is projected from the earth surface at latitude 45° with escape velocityv = 11.19 km/s. The velocity direction makes an angle ������ with the local vertical. Theparticle will escape the earth’s gravitational field(������) ������������������������ ������ℎ������������ ������ = 0 (������) ������������������������ ������ℎ������������ ������ = 45°(������) ������������������������ ������ℎ������������ ������ = 90° (������) ������������������������������������������������������������������������ ������������ ������ℎ������ ������������������������������ ������������ ������ GROUP—(B) M.S.Q TYPE QUESTIONSQUESTION NO.(16):-A sphere is rolling without slipping on a fixed horizontal plane surface. In thefigure, A is the point of contact, B is the centre of the sphere and C is its topmostpoint. Then,Stay connected with www.allindiaphysicsforum.com @ALL RIGHTS RESERVED FOR ALL INDIA PHYSICS FORUM

(������) ���⃗��������� − ���⃗��������� = 2(���⃗��������� − ���⃗���������) (������) ���⃗��������� − ���⃗��������� = ���⃗��������� − ���⃗���������(������) |���⃗��������� − ���⃗���������| = 2|���⃗��������� − ���⃗���������| (������) |���⃗��������� − ���⃗���������| = 4|���⃗���������|QUESTION NO.(17):-Phase space diagrams are useful tools in analyzing all kinds of dynamical problems.They are especially useful in studying the changes in motion as initial position andmomentum are changed. Here we consider some simple dynamical systems in onedimension. For such systems, phase space is a plane in which position is plotted alonghorizontal axis and momentum is plotted along vertical axis. The phase space diagramis x(t) vs. p(t) curve in this plane. The arrow on the curve indicates the time flow. Forexample, the phase space diagram for a particle moving with constant velocity is astraight line as shown in the figure. We used the sign convention in which position ormomentum upwards (or to right) is positive and downwards (or to left) is negative.The phase space diagram for a ball thrown vertically up from ground is/are ––Stay connected with www.allindiaphysicsforum.com @ALL RIGHTS RESERVED FOR ALL INDIA PHYSICS FORUM

QUESTION NO.(18):-Phase space diagrams are useful tools in analyzing all kinds of dynamical problems.They are especially useful in studying the changes in motion as initial position andmomentum are changed. Here we consider some simple dynamical systems in onedimension. For such systems, phase space is a plane in which position is plottedalong horizontal axis and momentum is plotted along vertical axis. The phasespace diagram is x(t) vs. p(t) curve in this plane. The arrow on the curve indicatesthe time flow. For example, the phase space diagram for a particle moving withconstant velocity is a straight line as shown in the figure. We used the signconvention in which position or momentum upwards (or to right) is positive anddownwards (or to left) is negative.The phase space diagram for simple harmonic motion is a circle centered at theorigin. In the figure, the two circles represent the same oscillator but for differentinitial conditions, and ������1 ������������������ ������2 are the total mechanical energies respectively.Then–– Stay connected with www.allindiaphysicsforum.com @ALL RIGHTS RESERVED FOR ALL INDIA PHYSICS FORUM

(������) ������1 = √2 ������2 (������)������1 = 2������2 (������)������1 = 4������2 (������) ������1 = 16������2QUESTION NO.(19):-Phase space diagrams are useful tools in analyzing all kinds of dynamical problems.They are especially useful in studying the changes in motion as initial position andmomentum are changed. Here we consider some simple dynamical systems in onedimension. For such systems, phase space is a plane in which position is plotted alonghorizontal axis and momentum is plotted along vertical axis. The phase space diagramis x(t) vs. p(t) curve in this plane. The arrow on the curve indicates the time flow. Forexample, the phase space diagram for a particle moving with constant velocity is astraight line as shown in the figure. We used the sign convention in which position ormomentum upwards (or to right) is positive and downwards (or to left) is negative.Consider the spring mass system, with the mass submerged in water, as shown in thefigure. The phase space diagram for one cycle of this system is: Stay connected with www.allindiaphysicsforum.com @ALL RIGHTS RESERVED FOR ALL INDIA PHYSICS FORUM

GROUP—(C)NAT (NUMERICAL APTITUDE TEST) TYPE QUESTIONSQUESTION NO.(20):- Stay connected with www.allindiaphysicsforum.com @ALL RIGHTS RESERVED FOR ALL INDIA PHYSICS FORUM

Two books of mass 1 kg each are kept on a table, one over the other. Thecoefficient of friction on every pair of contacting surfaces is 0.3. The lower book ispulled with a horizontal force F . The minimum value of F for which slip occursbetween the two books is __________________ ?QUESTION NO.(21):-___________������������. ������2 is the moment of inertia of the centre of mass ������������������ of apropeller with three blades (treated as rods) of mass ������ = 12 ������������, length ������ =1.25 ������, at 120° relative to each other ?QUESTION NO.(22):-From previous problem , If a torque ������ = 3000 ������ acts on this, after______________second it will take to reach an angular velocity ������ =2000 ������������������/������ ?QUESTION NO.(23):-From previous problem , ________number of revolutions it will have madebefore reaching this ?QUESTION NO.(24):-The radius of the aorta is ~10 mm and the blood flowing through it has a speed~300 ������������. ������−1 . A capillary has a radius ~4×10−3 ������������ but there are literallybillions of them. The average speed of blood through the capillaries is ~5×10−4 ������. ������−1 ._________������2 is the effective cross sectional area of the capillaries ?QUESTION NO.(25):-The radius of the aorta is ~10 mm and the blood flowing through it has a speed~300 ������������. ������−1 . A capillary has a radius ~4×10−3 ������������ but there are literallybillions of them. The average speed of blood through the capillaries is ~5×10−4 ������. ������−1 ._________×109 is the number of capillaries ? Stay connected with www.allindiaphysicsforum.com @ALL RIGHTS RESERVED FOR ALL INDIA PHYSICS FORUM

GROUP—(A) M.C.Q TYPE QUESTIONSQUESTION NO.(1):- The correct option is (B).SOLUTION:-tan ������ = ������ ⋯ ⋯ ⋯ ⋯ ⋯ (1) ������������������������������������ , tan ������ = ������������ = 2������������ ⋯ ⋯ ⋯ ⋯ (2) ������������Equating the above two equations we have ,2������������ = ������ ������������������, ������ = ������ (Answer) 2������������QUESTION NO.(2):- The correct option is (A).SOLUTION:- After 1st collision ,������������������ = ���������������′��� + 2���������������′��� ⋯ ⋯ ⋯ ⋯ ⋯ (1)1 = ���������′���−���������′��� ������������, ���������′��� = 6 ������ ⋯⋯ ⋯ ⋯ (2) 0−������������ ������After the 2nd collision ,2���������������′��� = (2������ + ������)������������ ������������, ������������ = 2 ���������′��� = 4 ������/������ (Answer) 3QUESTION NO.(3):- The correct option is (D).SOLUTION:- Maximum energy loss when inelastic collision takes place, Stay connected with www.allindiaphysicsforum.com @ALL RIGHTS RESERVED FOR ALL INDIA PHYSICS FORUM

������������ = (������ + ������)������′ ������������, ������′ = ( ������ ) ������ ⋯ ⋯ ⋯ ⋯ (1) ������2������2 (������+������)2 ������+������������������ = 1 ������������2 ������������������ ������������ = 1 (������ + ������)������′2 = 1 (������ + ������) = 1 ������������2 ( ������ ) 2 2 2 2 ������+������Loss of energy , = ������������ − ������������ = 1 ������������2 (1 − ������ ) = ( ������ ) ×1 ������������2(Answer) 2 ������+������ ������+������ 2QUESTION NO.(4):- The correct option is (A).SOLUTION:- The kinetic energy at altitude 2R , = ������������������ ⋯ ⋯ ⋯ ⋯ ⋯ (1) 6������ ������������������The gravitational potential energy at altitude 2R= − 3������ ⋯ ⋯ ⋯ ⋯ ⋯ (2)∴ ������������������������������ ������������������������������������ = ������. ������ + ������. ������ = − ������������������ ⋯ ⋯ ⋯ ⋯ (3) 6������ ������������������Potential energy at the surface is= − ������ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (4)∴ ������������������������������������������������ ������������������������������������ = ������������������ − ������������������ = 5������������������ (Answer) ������ 6������ 6������QUESTION NO.(5):- The correct option is (C).SOLUTION:-From the law of conservation of angular momentum ,������������0 = ������������ + ������������������∴ ������������2������0 = ������������2������ + ������������������ = 2������������������∴ ������ = ������������0 (Answer) 2QUESTION NO.(6):- The correct option is (C).SOLUTION:- Stay connected with www.allindiaphysicsforum.com @ALL RIGHTS RESERVED FOR ALL INDIA PHYSICS FORUM

For Block A : ������1������ = ������1������������������, 20 = 0.1×������ ������������, ������ = 200 ������ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (1)Frictional force on block A in upward direction = ������1������ = 0.1×200 = 20 ������Block A exert a frictional force of 20 N on block B in down-ward direction.For block B :∴ ������2������ = ������2������ + ������1������ = 100 + 20 = 120 ������ (Answer)QUESTION NO.(7):- The correct option is (C).SOLUTION:-K.E of the system before collision ,������������ = 1 ������(2������)2 + 1 2������. ������2 = 3������������2 ⋯ ⋯ ⋯ ⋯ (1) 2 2Total momentum of the system before collision is = √(2������������)2 + (2������������)2 = √2 .2������������From the law of conservation of linear momentum we can write that ,3������������ = √2 .2������������ ������������, ������ = 2√2 ������ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (2) 3K.E of the system after collision ,������������ = 1 ×3������× (2√2 ������ 2 = 4 ������������2 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (3) 23 ) 3 Stay connected with www.allindiaphysicsforum.com @ALL RIGHTS RESERVED FOR ALL INDIA PHYSICS FORUM

Therefore , the loss of kinetic energy due to the collision is given by ,������������ − ������������ = 3������������2 − 4 ������������2 = 5 ������������2 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (4) 3 3∴ ������������������������������������������������������������ ������������������������ ������������ ������������������������������������������ ������������������������������������ ������������������ ������������ ������ℎ������ ������������������������������������������������������ = ������������−������������ ×100 ������������= 5 ×100 = 55.6 ≈ 56 (Answer) 9QUESTION NO.(8):- The correct option is (B).QUESTION NO.(9):- The correct option is (C).SOLUTION:-Figure alongside shows a solid sphere of mass M. The radius of the sphere is R.The volume of the sphere is������ = 4 ������������3 ⋯ ⋯ ⋯ ⋯ ⋯ (1) 3 ������ ������ 3������The density of the sphere is , ������ = ������ = 43������������3 = 4������������3 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (2)From this solid sphere a cube of maximum possible volume is cut.Therefore , 2������ = √3 ������ , where a is the length of the side of the cube ofmaximum volume.∴ ������ = 2������ ⋯ ⋯ ⋯ ⋯ ⋯ (3) √3 (2������)3Mass of the cube is , ������′ = ������������3 = 3������ × = 2������ ⋯ ⋯ ⋯ ⋯ ⋯ (4) 4������������3 √3 √3 ������The moment of inertia of the cube is,������ = ������′ ������2 = 2������ × 1 × (2������)2 = 8������������2 = 4������������2 (Answer) 6 √3 ������ 6 √3 18√3 ������ 9√3 ������QUESTION NO.(10):- The correct option is (B).SOLUTION:- Potential at internal point of solid sphere at a distance ‘r’ , Stay connected with www.allindiaphysicsforum.com @ALL RIGHTS RESERVED FOR ALL INDIA PHYSICS FORUM

������ = − ������������ [3 − 2������������22] ⋯ ⋯ ⋯ ⋯⋯ ⋯ ⋯ (1) ������ 2 ������ = ������At 2 8������������22]������1 = − ������������ [23 − = − 11 × ������������ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (2) ������ 8 ������Because of sphere removed.������2 = 3× ���������8��� = 3 × ������������ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (3) 2 ������ 8 ������ 2 − 11 × ������������ + 3 × ������������ − ������������Net potential, ������ = ������1 + ������2 = 8 ������ 8 ������ = ������ (Answer)QUESTION NO.(11):- The correct option is (C ).∅ = 65° , ������ = ������ tan ∅ , ������������ = ������(������������2)������������ = ������(������������2������������������2∅)������������������ = ∫0ℎ ������������ ∫0ℎ ������×������(������������2������������������2∅)������������ ∫0ℎ ������ ������������ ∫0ℎ ������(������������2������������������2∅)������������ [43������������43]ℎ0������������������ = ∫0ℎ ������������ = = = 3ℎ (������������������������ ������ℎ������ ������������������) 4ℎ = 3800 ������ , ������������������ = 950 ������ ������������������������������ ������ℎ������ ������������������������ (answer)QUESTION NO.(12):- The correct option is (D ).QUESTION NO.(13):- The correct option is (B ).QUESTION NO.(14):- The correct option is (A ). Stay connected with www.allindiaphysicsforum.com @ALL RIGHTS RESERVED FOR ALL INDIA PHYSICS FORUM

QUESTION NO.(15):- The correct option is (D ). GROUP—(B) M.S.Q TYPE QUESTIONSQUESTION NO.(16):- The correct options are (B) & (C ).QUESTION NO.(17):- The correct option is (D).SOLUTION:- For ball thrown up, initially momentum has maximum positive valueand position is zero. Then as ball moves up, the momentum decreases & positionincreases. At highest position, momentum is zero, position is maximum. Then ballreturns during which position decreases and momentum increases in negative.QUESTION NO.(18):- The correct option is (C).SOLUTION:- ������1 = (2������)2 = 4 ������������, ������1 = 4������2 (Answer) ������2 ������QUESTION NO.(19):- (B).SOLUTION:- The mass has zero momentum at highest position. From here, it movesdown, during which position decreases but momentum increases in negative value.After one cycle, its position is slightly less than initial value (damped oscillation). GROUP—(C)NAT (NUMERICAL APTITUDE TEST) TYPE QUESTIONSQUESTION NO.(20):- KEY/ RANGE: 8.80 to 8.90QUESTION NO.(21):- KEY/ RANGE: 19 to 20QUESTION NO.(22):- KEY/ RANGE: 13 to 14QUESTION NO.(23):- KEY/ RANGE: 2000 to 2001QUESTION NO.(24):- KEY/ RANGE: 0.18 to 0.25Aorta , ������������ = 10×10−3������ = 10 ������������ , ������������ =cross sectional area of aortaCapillaries , ������������ = 4×10−6������ = 0.004 ������������ , ������������ =cross sectional area of capillaries������������ = 0.300 ������. ������−1 , ������������ = 5×10−4������. ������−1Assume steady flow of an ideal fluid and apply the equation of continuity,������ = ������������ = ������������������������������������������������ = (������������×������������) = (������������×������������) ������������, ������������ = (������������×������������) = ���������������2���(������������������������) = 0.20 ������2 ������������Stay connected with www.allindiaphysicsforum.com @ALL RIGHTS RESERVED FOR ALL INDIA PHYSICS FORUM

QUESTION NO.(25):- KEY/ RANGE: 3 to 5������������ = ������������������������������ ������������, ������ = ������������ = ������.������ = ������×������������������ ������������������������ {������(������×������������−������)������} Stay connected with www.allindiaphysicsforum.com @ALL RIGHTS RESERVED FOR ALL INDIA PHYSICS FORUM


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