The coefficient of restitution of a perfectly plastic impact is (A) 0 (B) 1 (C) 2 (D) 3

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1 CHAPTER ENGINEERING MECHANICS YEAR 0 TWO MARKS Common Data For Q. and Two steel truss members, AC and BC, each having cross sectional area of 00 mm, are subjected to a horizontal force F as shown in figure. All the joints are hinged. MCQ. If F kn, the magnitude of the vertical reaction force developed at the point B in kn is (A) 0.63 (B) 0.3 (C).6 (D).46 MCQ. The maximum force F is kn that can be applied at C such that the axial stress in any of the truss members DOES NOT exceed 00 MPa is (A) 8.7 (B).5 (C) 4.4 (D).30 YEAR 0 ONE MARK MCQ.3 MCQ.4 The coefficient of restitution of a perfectly plastic impact is (A) 0 (B) (C) (D) 3 A stone with mass of 0. kg is catapulted as shown in the figure. The total force F x (in N) exerted by the rubber band as a function of distance x Published by: NODIA and COMPANY ISBN:

2 PAGE 7 ENGINEERING MECHANICS CHAP (in m) is given by Fx 300x. If the stone is displaced by 0. m from the un-stretched position ( x 0) of the rubber band, the energy stored in the rubber band is (A) 0.0 J (C) J (B) 0. J (D) 0 J YEAR 0 TWO MARKS MCQ.5 A kg block is resting on a surface with coefficient of friction μ 0.. A force of 0.8 N is applied to the block as shown in the figure. The friction force is (A) 0 (C) 0.98 N (B) 0.8 N (D). N YEAR 009 ONE MARK MCQ.6 A block weighing 98 N is resting on a horizontal surface. The coefficient of friction between the block and the horizontal surface is μ 0.. A vertical cable attached to the block provides partial support as shown. A man can pull horizontally with a force of 00 N. What will be the tension, T (in N) in the cable if the man is just able to move the block to the right? (A) 76. (B) 96.0 Published by: NODIA and COMPANY ISBN:

3 CHAP ENGINEERING MECHANICS PAGE 73 (C) 48.0 (D) 98.0 YEAR 009 TWO MARKS MCQ.7 A uniform rigid rod of mass M and length L is hinged at one end as shown in the adjacent figure. A force P is applied at a distance of L/ 3 from the hinge so that the rod swings to the right. The reaction at the hinge is (A) P (B) 0 (C) P/ 3 (D) P/ 3 YEAR 008 ONE MARK MCQ.8 A straight rod length Lt, () hinged at one end freely extensible at the other end, rotates through an angle θ () t about the hinge. At time t, Lt () m, Lt o () m/s, θ () t π rad and θ () t 4 o rad/s. The magnitude of the velocity at the other end of the rod is (A) m/s (B) m/s (C) 3 m/s (D) m/s YEAR 008 TWO MARKS MCQ.9 A circular disk of radius R rolls without slipping at a velocity V. The magnitude of the velocity at point P (see figure) is Published by: NODIA and COMPANY ISBN:

4 PAGE 74 ENGINEERING MECHANICS CHAP (A) 3 V (B) 3 V/ (C) V / (D) V/ 3 MCQ.0 Consider a truss PQR loaded at P with a force F as shown in the figure - The tension in the member QR is (A) 0.5 F (C) 0.73 F (B) 0.63 F (D) 0.87 F YEAR 007 ONE MARK MCQ. During inelastic collision of two particles, which one of the following is conserved? (A) Total linear momentum only (B) Total kinetic energy only (C) Both linear momentum and kinetic energy (D) Neither linear momentum nor kinetic energy YEAR 007 TWO MARKS MCQ. A block of mass M is released from point P on a rough inclined plane with inclination angle θ, shown in the figure below. The co-efficient of friction is μ. If μ < tan θ, then the time taken by the block to reach another point Q on the inclined plane, where PQ s, is (A) s g cos θ( tan θ μ) (B) s g cos θ( tan θ+ μ) Published by: NODIA and COMPANY ISBN:

5 CHAP ENGINEERING MECHANICS PAGE 75 (C) s g sin θ( tan θ μ) (D) s g sin θ( tan θ+ μ) YEAR 006 TWO MARKS MCQ.3 MCQ.4 If a system is in equilibrium and the position of the system depends upon many independent variables, the principles of virtual work states that the partial derivatives of its total potential energy with respect to each of the independent variable must be (A).0 (B) 0 (C).0 (D) 3 If point A is in equilibrium under the action of the applied forces, the values of tensions T AB and T AC are respectively (A) 50 N and 300 N (C) 450 N and 50 N (B) 300 N and 50 N (D) 50 N and 450 N YEAR 005 ONE MARK MCQ.5 MCQ.6 The time variation of the position of a particle in rectilinear motion is given 3 by x t + t + t. If v is the velocity and a is the acceleration of the particle in consistent units, the motion started with (A) v 0, a 0 (B) v 0, a (C) v, a 0 (D) v, a A simple pendulum of length of 5 m, with a bob of mass kg, is in simple harmonic motion. As it passes through its mean position, the bob has a speed of 5 m/s. The net force on the bob at the mean position is (A) zero (B).5 N (C) 5 N (D) 5 N YEAR 005 TWO MARKS MCQ.7 Two books of mass kg each are kept on a table, one over the other. The Published by: NODIA and COMPANY ISBN:

6 PAGE 76 ENGINEERING MECHANICS CHAP coefficient of friction on every pair of contacting surfaces is 0.3. The lower book is pulled with a horizontal force F. The minimum value of F for which slip occurs between the two books is (A) zero (B).06 N (C) 5.74 N (D) 8.83 N MCQ.8 A shell is fired from a cannon. At the instant the shell is just about to leave the barrel, its velocity relative to the barrel is 3 m/s, while the barrel is swinging upwards with a constant angular velocity of rad/s. The magnitude of the absolute velocity of the shell is (A) 3 m/s (C) 5 m/s (B) 4 m/s (D) 7 m/s MCQ.9 An elevator (lift) consists of the elevator cage and a counter weight, of mass m each. The cage and the counterweight are connected by chain that passes over a pulley. The pulley is coupled to a motor. It is desired that the elevator should have a maximum stopping time of t seconds from a peak speed v. If the inertias of the pulley and the chain are neglected, the minimum power that the motor must have is MCQ.0 (A) (C) mv mv t (B) (D) mv t mv t A kg mass of clay, moving with a velocity of 0 m/s, strikes a stationary wheel and sticks to it. The solid wheel has a mass of 0 kg and a radius of Published by: NODIA and COMPANY ISBN:

7 CHAP ENGINEERING MECHANICS PAGE 77 m. Assuming that the wheel is set into pure rolling motion, the angular velocity of the wheel immediately after the impact is approximately (A) zero (B) 3 rad/s (C) 0 rad/s (D) 0 rad/s 3 3 YEAR 004 ONE MARK MCQ. The figure shows a pin-jointed plane truss loaded at the point M by hanging a mass of 00 kg. The member LN of the truss is subjected to a load of (A) 0 Newton (C) 98 Newtons in compression (B) 490 Newtons in compression (D) 98 Newtons in tension YEAR 004 TWO MARKS MCQ. An ejector mechanism consists of a helical compression spring having a 3 spring constant of k 98 # 0 N/m. It is pre-compressed by 00 mm from its free state. If it is used to eject a mass of 00 kg held on it, the mass will move up through a distance of Published by: NODIA and COMPANY ISBN:

8 PAGE 78 ENGINEERING MECHANICS CHAP (A) 00 mm (C) 58 mm (B) 500 mm (D) 000 mm MCQ.3 A rigid body shown in the figure (a) has a mass of 0 kg. It rotates with a uniform angular velocity ω. A balancing mass of 0 kg is attached as shown in figure (b). The percentage increase in mass moment of inertia as a result of this addition is (A) 5% (B) 50% (C) 00% (D) 00% MCQ.4 The figure shows a pair of pin-jointed gripper-tongs holding an object weighting 000 N. The coefficient of friction (μ) at the gripping surface is 0. XX is the line of action of the input force and YY is the line of application of gripping force. If the pin-joint is assumed to be frictionless, the magnitude of force F required to hold the weight is (A) 000 N (C) 500 N (B) 000 N (D) 5000 N Published by: NODIA and COMPANY ISBN:

9 CHAP ENGINEERING MECHANICS PAGE 79 YEAR 003 ONE MARK MCQ.5 A truss consists of horizontal members (AC,CD, DB and EF) and vertical members (CE and DF) having length l each. The members AE, DE and BF are inclined at 45c to the horizontal. For the uniformly distributed load p per unit length on the member EF of the truss shown in figure given below, the force in the member CD is MCQ.6 (A) pl (C) 0 (B) pl (D) pl 3 A bullet of mass m travels at a very high velocity v (as shown in the figure) and gets embedded inside the block of mass M initially at rest on a rough horizontal floor. The block with the bullet is seen to move a distance s along the floor. Assuming μ to be the coefficient of kinetic friction between the block and the floor and g the acceleration due to gravity what is the velocity v of the bullet? (A) (C) M+ m μgs m μ( M+ m) μgs m (B) M m μgs m (D) M μgs m YEAR 003 TWO MARKS Common Data For Q. Data for Q. 7 & 8 are given below. Solve the problems and choose correct answers. A reel of mass m and radius of gyration k is rolling down smoothly from rest with one end of the thread wound on it held in the ceiling as depicated in the figure. Consider the thickness of thread and its mass negligible in comparison with the radius r of the hub and the reel mass m. Symbol g represents the acceleration due to gravity. Published by: NODIA and COMPANY ISBN:

10 PAGE 80 ENGINEERING MECHANICS CHAP MCQ.7 MCQ.8 The linear acceleration of the reel is gr (A) ( r k + ) grk (C) ( r k + ) The tension in the thread is mgr (A) ( r k + ) mgk (C) ( r k + ) gk (B) ( r k + ) mgr (D) ( r k + ) mgrk (B) ( r k + ) mg (D) ( r k + ) YEAR 00 ONE MARK MCQ.9 A particle P is projected from the earth surface at latitude 45c with escape velocity v.9 km/ s. The velocity direction makes an angle α with the local vertical. The particle will escape the earth s gravitational field (A) only when α 0 (B) only when α 45c (C) only when α 90c (D) irrespective of the value of α MCQ.30 The area moment of inertia of a square of size unit about its diagonal is (A) (B) 3 4 (C) (D) 6 Published by: NODIA and COMPANY ISBN:

11 CHAP ENGINEERING MECHANICS PAGE 8 YEAR 00 TWO MARKS MCQ.3 For the loading on truss shown in the figure, the force in member CD is MCQ.3 (A) zero (B) kn (C) kn (D) kn Bodies and shown in the figure have equal mass m. All surfaces are smooth. The value of force P required to prevent sliding of body on body is (A) P mg (B) P mg (C) P mg (D) P mg MCQ.33 Mass M slides in a frictionless slot in the horizontal direction and the bob of mass m is hinged to mass M at C, through a rigid massless rod. This system is released from rest with θ 30c. At the instant when θ 0c, the velocities of m and M can be determined using the fact that, for the system (i.e., m and M together) Published by: NODIA and COMPANY ISBN:

12 PAGE 8 ENGINEERING MECHANICS CHAP (A) the linear momentum in x and y directions are conserved but the energy is not conserved. (B) the linear momentum in x and y directions are conserved and the energy is also conserved. (C) the linear momentum in x direction is conserved and the energy is also conserved. (D) the linear momentum in y direction is conserved and the energy is also conserved. ********* Published by: NODIA and COMPANY ISBN:

13 CHAP ENGINEERING MECHANICS PAGE 83 SOLUTION SOL. Option (A) is correct. From above figure. Three forces are acting on a common point. Hence by Lami s Theorem. F T T sin( 05c) sin 0c sin 35c & T F sin 35c sin 05c sin 05c T kn Hence vertical reaction at B, R NT T cos 30c # cos30c kn SOL. Option (B) is correct. From Previous question F T sin 05c sin 0c T sin 0c F F sin 35 # and T ( )F T > T σ 00 MPa (given) As we know F σ # A & F max σ max # A T 00 # F 00 # 00 F 00 # N.5 kn Published by: NODIA and COMPANY ISBN:

14 PAGE 84 ENGINEERING MECHANICS CHAP SOL.3 Option (A) is correct. From the Newton s Law of collision of Elastic bodies. Velocity of separation e # Velocity of approach ( V V) eu ( U) Where e is a constant of proportionality & it is called the coefficient of restitution and its value lies between 0 to. The coefficient of restitution of a perfectly plastic impact is zero, because all the K.E. will be absorbed during perfectly plastic impact. SOL.4 SOL.5 Option (B) is correct. Given : F x 300x, Position of x is, x 0 to x 0. The energy stored in the rubber band is equal to work done by the stone. Hence de Fdx x Integrating both the sides & put the value of F & limits E 0. de 300xdx 0# # E 300 x : 3 D 300 (.) 0 3 ; 0. Joule 3 E Option (B) is correct. Given : m kg, μ 0. ; From FBD : R N mg 0 Now static friction force, f S μrn μmg 0. # # N Applied force F 0.8 N is less then, the static friction f S 0.98 N F < f S So, we can say that the friction developed will equal to the applied force F 0.8 N SOL.6 Option (C) is correct. Given : W 98 N, μ 0. First of all we have to make a FBD of the block Here, R N Normal reaction force Published by: NODIA and COMPANY ISBN:

15 CHAP ENGINEERING MECHANICS PAGE 85 T Tension in string Using the balancing of forces, we have Σ F x 0 : μ R N 00 N R N N μ 0. and Σ F y 0 or downward forces upward forces W T+ R N & T W R N N SOL.7 Option (B) is correct. When rod swings to the right, linear acceleration a and angular acceleration α comes in action. Centre of gravity (G ) acting at the mid-point of the rod. Let R be the reaction at the hinge. Linear acceleration a r. α L # α L a...(i) and about point G, for rotational motion / Μ G I G # α R L P L b + ML a l b6l b L l From equation (i) R + P 3 Ma 3 a 3 M R + M P...(ii) Published by: NODIA and COMPANY ISBN:

16 PAGE 86 ENGINEERING MECHANICS CHAP By equilibrium of forces in normal direction to the rod / F m 0 : P R Ma M 3R P b + M M l From equation (ii) P R 3R+ P & R 0 So, reaction at the hinge is zero. SOL.8 Option (D) is correct. Let : V t Tangential Velocity V r Relative Velocity V Resultant Velocity Let rod of length Lt () increases by an amount T Lt (). : Given Lt () m, Lt () m/sec, θ() t π rad, θ : () t rad/sec 4 Time taken by the rod to turn 4 π rad is, t dis tan ce θ() t velocity : θ() t π/ 4 π sec 4 So, increase in length of the rod during this time will be Δ Lt () Lt ()# t π π 4 # meter 4 Rod turn π radian. So, increased length after π sec, (New length) 4 4 π a k m Now, tangential velocity, V t R. ω. 785 #. 785 m/sec ω θ o () t Radial velocity, V r Lt : () m/sec Published by: NODIA and COMPANY ISBN:

17 CHAP ENGINEERING MECHANICS PAGE 87 Therefore, the resultant velocity will be V R V + V (. 785) + ().04- m/sec t r SOL.9 Option (A) is correct. When disc rolling along a straight path, without slipping. The centre of the wheel O moves with some linear velocity and each particle on the wheel rotates with some angular velocity. Thus, the motion of any particular on the periphery of the wheel is a combination of linear and angular velocity. Let wheel rotates with angular velocityω rad/sec. So, ω V...(i) R Velocity at point P is, V P ω # PQ...(ii) From triangle OPQ PQ ( OQ) + ( OP) OQ # OP # cos( + POQ) ( R) + ( R) RRcos 0c ( R) + ( R) + ( R) 3 R...(iii) From equation (i), (ii) and (iii) V P V R # 3 R 3 V SOL.0 Option (B) is correct. The forces which are acting on the truss PQR is shown in figure. We draw a perpendicular from the point P, that intersects QR at point S. Published by: NODIA and COMPANY ISBN:

18 PAGE 88 ENGINEERING MECHANICS CHAP Let PS QS a R Q & R R are the reactions acting at point Q & R respectively. Now from the triangle PRS tan 30c PS & SR PS a 3 a.73a SR tan 30c Taking the moment about point R, RQ # ( a+ 73. a) F# 73. a R Q 73. a F 73. F F 73. a 73. From equilibrium of the forces, we have R + R F R Q R R F RQ F F F To find tension in QR we have to use the method of joint at point Q, and Σ Fy 0 and, Σ F x 0 FQP sin 45c R Q F QP F F FQP cos 45c F QR & F QR F # F F 3 SOL. SOL. Option (A) is correct. In both elastic & in inelastic collision total linear momentum remains conserved. In the inelastic collision loss in kinetic energy occurs because the coefficient of restitution is less than one and loss in kinetic energy is given by the relation, T KE.. mm ( m m ) ( u u ) ( e + ) Option (A) is correct. First of all we resolve all the force which are acting on the block. Published by: NODIA and COMPANY ISBN:

19 CHAP ENGINEERING MECHANICS PAGE 89 Given : PQ s where N Normal fraction force μ < tan θ Now from Newton s second law, F ma mg sin θ μn ma a Acceleration of block mg sin θ μmg cos θ ma N mgcos θ gsin θ μgcos θ a a g cos θ sin θ : μ cos θ D g cos θ( tan θ μ)...(i) From the Newton;s second law of Motion, s ut + at 0 + gcos θ( tan θ μ) t u 0 t s g cos θ( tan θ μ) SOL.3 Option (B) is correct. If a system of forces acting on a body or system of bodies be in equilibrium and the system has to undergo a small displacement consistent with the geometrical conditions, then the algebraic sum of the virtual works done by all the forces of the system is zero and total potential energy with respect to each of the independent variable must be equal to zero. SOL.4 Option (A) is correct. First we solve this problem from Lami s theorem. Here three forces are given. Now we have to find the angle between these forces Applying Lami s theorem, we have F TAB TAC sin 90c sin 0c sin 50c 600 TAB TAC 3/ / T AB 600 # N 3 3. T AC N Published by: NODIA and COMPANY ISBN:

20 PAGE 90 ENGINEERING MECHANICS CHAP Alternative : Now we using the Resolution of forces. Resolve the T AB & T AC in x & y direction (horizontal & vertical components) We use the Resolution of forces in x & y direction Σ F x 0, TAB cos 60c TAC cos 30c T T AB AC Σ 0, T sin 60c+ T sin 30c 600 N F y Now, AB AC T T 3 AB + AC 600 N 3 # 3...(i) 3 T AB + T AC 00 N T 3 T AB TAB + 00 N 3 4T AB 00 3 T AB AC TAB From equation (i) 3 50 N and T TAB AC N 3 3 SOL.5 Option (D) is correct. Given ; x t 3 + t + t We know that, v dx d 3 ( t + t + t) 6t + t+...(i) dt dt We have to find the velocity & acceleration of particle, when motion stared, So at t 0, v Again differentiate equation (i) w.r.t. t At t 0, a a dv dx t dt + dt Published by: NODIA and COMPANY ISBN:

21 CHAP ENGINEERING MECHANICS PAGE 9 SOL.6 Option (A) is correct. We have to make the diagram of simple pendulum Here, We can see easily from the figure that tension in the string is balanced by the weight of the bob and net force at the mean position is always zero. SOL.7 Option (D) is correct. Given : m m kg, μ 03. The FBD of the system is shown below : For Book () Σ F y 0 R N mg...(i) Then, Friction Force F N μrn μmg From FBD of book second, Σ F x 0, F μrn+ μrn Σ F y 0, R N RN + mg mg+ mg mg...(ii) For slip occurs between the books when F $ μrn+ μrn$ μmg + μ# mg F $ μ (3 mg) $ 03. ( 3# # 98. ) $ 88. It means the value of F is always greater or equal to the 8.8, for which slip Published by: NODIA and COMPANY ISBN:

22 PAGE 9 ENGINEERING MECHANICS CHAP occurs between two books. So, F 8.83 N SOL.8 Option (C) is correct. Given : ω rad/sec, r m Tangential velocity of barrel, V t rω # 4 m/sec V V i+ Vj 3i+ 4j Resultant velocity of shell, V () 3 + () m/sec r t SOL.9 SOL.0 Option (C) is correct. Given : Mass of cage & counter weight m kg each Peak speed V Initial velocity of both the cage and counter weight. V V m/sec Final velocity of both objects V 0 Initial kinetic Energy, E mv + mv mv Final kinetic Energy E m() 0 + m() 0 0 Now, Power Rate of change of K.E. E E mv t t Option (B) is correct. Given : m kg, V 0 m/ sec, m 0 kg, V Velocity after striking the wheel r meter Applying the principal of linear momentum on the system dp 0 & P constant dt Initial Momentum Final Momentum m# V ( m+ m) V V mv # 0 0 ( m+ m) + 0 Published by: NODIA and COMPANY ISBN:

23 CHAP ENGINEERING MECHANICS PAGE 93 Now after the collision the wheel rolling with angular velocity ω. So, V rω & ω V r # It is nearly equal to 3. / SOL. Option (A) is correct. First of all we consider all the forces, which are acting at point L. Now sum all the forces which are acting along x direction, F LK F LM Both are acting in opposite direction Also summation of all the forces, which are acting along y-direction. F LN 0 Only one forces acting in y-direction So the member LN is subjected to zero load. SOL. Option (A) is correct. 3 Given : k 98 # 0 N/m, xi 00mm0. m, m 00 kg Let, when mass m 00 kg is put on the spring then spring compressed by x mm. From the conservation of energy : Energy stored in free state Energy stored after the mass is attach. ( KE..) i ( KE..) f + ( PE..) f kxi kx mg( x 0.) + + kx i kx + mg( x + 0.) Substitute the values, we get 98 # 0 3 #(.) 0 3 (98 # 0 # x ) + [ # 00 # 9.8 #( x+ 0.)] 3 0 # x + ( x+ 0. ) 0 000x + x x + x Solving above equation, we get! () 4 000( 9.) 8 x # # 000!! On taking -ve sign, we get Published by: NODIA and COMPANY ISBN:

24 PAGE 94 ENGINEERING MECHANICS CHAP x , m 00 mm (-ve sign shows the compression of the spring) SOL.3 SOL.4 Option (B) is correct. Given : First Mass, m 0 kg Balancing Mass, m 0 kg We know the mass moment of inertia, I mk Where, k Radius of gyration Case (I) : When mass of 0 kg is rotates with uniform angular velocity ω I mk 0 # (0.) 0 # kg m k 0. m Case (II) : When balancing mass of 0 kg is attached then moment of inertia I 0 #( 0. ) + 0 #( 0. ) Here k 0. m and k 0. m Percent increase in mass moment of inertia, I I I I # 04. # 00 50% # Option (D) is correct. Given : Weight of object W 000 N Coefficient of Friction μ 0. First of all we have to make the FBD of the system. Here, R N Normal reaction force acting by the pin joint. F μr N Friction force In equilibrium condition of all the forces which are acting in y direction. μr + μr 000 N N N μ R N 000 N R N N μ Taking the moment about the pin, we get 0000 # 50 F # 300 F 5000 N Published by: NODIA and COMPANY ISBN:

25 CHAP ENGINEERING MECHANICS PAGE 95 SOL.5 Option (A) is correct. Given : AC CD DB EF CE DF l At the member EF uniform distributed load is acting, the U.D.L. is given as p per unit length. So, the total load acting on the element EF of length l Lord per unit length # Total length of element p# l pl This force acting at the mid point of EF. From the FBD we get that at A and B reactions are acting because of the roller supports, in the upward direction. In equilibrium condition, Upward force Downward forces Ra+ Rb pl...(i) And take the moment about point A, pl l l # b + l Rb( l+ l+ l) pl # l 3 pl Rb # 3l & R b Substitute the value of R b in equation (i), we get pl Ra + pl pl pl pl R a pl Rb At point A we use the principal of resolution of forces in the y-direction, pl / Fy 0 : FAE sin 45c Ra pl pl pl F AE # sin 45c # pl pl And F AC F cos 45 AE c # At C, No external force is acting. So, pl F F AC CD Published by: NODIA and COMPANY ISBN:

26 PAGE 96 ENGINEERING MECHANICS CHAP SOL.6 Option (A) is correct. Given : Mass of bullet m Mass of block M Velocity of bullet v Coefficient of Kinematic friction μ Let, Velocity of system (Block + bullet) after striking the bullet We have to make the FBD of the box after the bullet strikes, u Friction Force (Retardation) F r Applying principal of conservation of linear momentum, dp 0 or P mv cons tan t. dt So, mv ( M+ m) u u mv...(i) M + m And, from the FBD the vertical force (reaction force), R N ( M+ m) g F r μrn μ( M+ m) g Frictional retardation a Fr μ( M m) g + μg...(ii) ( m+ M) M+ m Negative sign show the retardation of the system (acceleration in opposite direction). From the Newton s third law of motion, V f u + as V f Final velocity of system (block + bullet) 0 u + as 0 u as #( μg) # s μgs From equation (ii) Substitute the value of u from equation (i), we get mv am+ mk μgs mv ( M+ m) μgs v μgs( M + m) m Published by: NODIA and COMPANY ISBN:

27 CHAP ENGINEERING MECHANICS PAGE 97 v μgs M+ m # b m l M m m + μgs SOL.7 Option (A) is correct. Given : Mass of real m Radius of gyration k We have to make FBD of the system, Where, T Tension in the thread mg Weight of the system Real is rolling down. So Angular acceleration ( α ) comes in the action From FBD, For vertical translation motion, mg T ma...(i) and for rotational motion, Σ M G I G α T# r mk # a IG mk, α ar / r T mk # a...(ii) r From equation (i) & (ii) Substitute the value of T in equation (i), we get mg mk # a ma r mg a mk ; m r + E a gr...(iii) k + r SOL.8 Option (C) is correct. From previous question, T mg ma Substitute the value of a from equation (iii), we get gr mg( k + r ) mgr mgk T mg m # ( k + r ) ( k + r ) k + r Published by: NODIA and COMPANY ISBN:

28 PAGE 98 ENGINEERING MECHANICS CHAP SOL.9 Option (D) is correct. We know that a particle requires the velocity of. km/ s for escape it from the earth s gravitational field. The angle α does not effect on it. SOL.30 Option (C) is correct. The BD is the diagonal of the square ABCD and CBD 45c. From the T BCE sin 45c CE & CE sin 45c BC # unit where CE is the height of the triangle T BCD. Now, the area moment of inertia of a triangle about its base BD is bh 3 where b base of triangle & h height of triangle So, the triangle T ABD are same and required moment of inertia of the square ABCD about its diagonal is, 3 I 3 # ( BD) ( CE) # # 6 # # c m unit SOL.3 Option (A) is correct. The reactions at the hinged support will be in only vertical direction as external loads are vertical. Now, consider the FBD of entire truss. In equilibrium of forces. Ra+ Rf + kn...(i) Taking moment about point A, we get Rf # 3L # L+ # L 3L kn R f From equation (i), R a kn Published by: NODIA and COMPANY ISBN:

29 CHAP ENGINEERING MECHANICS PAGE 99 First consider the FBD of joint A with the direction of forces assumed in the figure. Resolving force vertically, we get R a FAB sin 45c F AB kn (Compression) sin 45c Resolving forces horizontally F AC FAB cos 45c # kn (Tension) Consider the FBD of joint B with known value of force F AB in member AB Resolving forces vertically, F BC FAB cos 45c # kn (Tension) Resolving forces horizontally, F BD FAB sin 45c # kn (Compression) Consider the FBD of joint C with known value of force F BC and F AC Resolving forces vertically, F + F sin 45 BC CD c Published by: NODIA and COMPANY ISBN:

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34 PAGE 04 ENGINEERING MECHANICS CHAP + F sin 45c & F 0 CD CD SOL.3 Option (D) is correct. From the FBD of the system. mg cos 45c R N All surfaces are smooth, so there is no frictional force at the surfaces. The downward force mg sin 45c is balanced by P cos 45c. mg sin 45c P cos 45c mg # P # & P mg SOL.33 Option (C) is correct. In this case the motion of mass m is only in x -direction. So, the linear momentum is only in x -direction & it remains conserved. Also from Energy conservation law the energy remains constant i.e. energy is also conserved. ********** Published by: NODIA and COMPANY ISBN: