96. Match the rolling element bearings with the most appropriate loading condition
    

    Bearing-type	Loading condition
    (a) Ball bearing	(p) Tangential load
    (b) Roller bearing	(q) Radial load
    (c) Needle bearing	(r) Heavy radial load with impact
    (d) Taper roller bearing	(s) Light radial load with space limitation
    	(t) Heavy radial and axial load
    	(u) Fatigue load

1. 1. a – p, b – r, c –s, d – q
      

   
   
   

   2. a – r, b –p, c –s, d – t
      

   
   
   

   3. a – q, b – r, c –s, d – t
      

   
   
   

   4. a – p, b – r, c –t, d – s
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   a – q, b – r, c –s, d – t
   

   Correct Option: C

   a – q, b – r, c –s, d – t
   

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97. If the load on a ball bearing is reduced to half, the life of the ball bearing will
    

1. 1. increase 8 times
      

   
   
   

   2. increase 4 times
      

   
   
   

   3. increase 2 times
      

   
   
   

   4. not change

   
   
   

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1. View Hint View Answer Discuss in Forum

   LP³ = C
   

   ⇒ L1 P13 = L2		P1		3
   		2

   
   ⇒ L₂ = 8L₁

   Correct Option: A

   LP³ = C
   

   ⇒ L1 P13 = L2		P1		3
   		2

   
   ⇒ L₂ = 8L₁

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98. The expected life of a ball bearing subjected to a load of 9800 N and working at 1000 rpm is 3000 hours. What is the expected life of the same bearing for a similar load of 4900 N and speed of 2000 rpm?
    

1. 1. Unchanged
      

   
   
   

   2. 12000 hours
      

   
   
   

   3. 1500 hours
      

   
   
   

   4. 6000 hours

   
   
   

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1. View Hint View Answer Discuss in Forum

   L (P)³ = constant
   

   	L1	=		P2		3
   	L2			P1

   
   

   ⇒	1000 × 60 × 3000	=		4900		3
   	2000 × 60 × t2			9800

   
   t₂ = 12000 hours
   

   Correct Option: B

   L (P)³ = constant
   

   	L1	=		P2		3
   	L2			P1

   
   

   ⇒	1000 × 60 × 3000	=		4900		3
   	2000 × 60 × t2			9800

   
   t₂ = 12000 hours
   

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99. A bolted Joint is shown below. The maximum shear stress, in MPa, in the bolts at A and B, respectively are
    
    

1. 1. 242.6 , 42.5
      

   
   
   

   2. 42.5 , 242.6
      

   
   
   

   3. 42.5 , 42.5
      

   
   
   

   4. 60.15 , 343.64

   
   
   

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1. View Hint View Answer Discuss in Forum

   FA = FB = FC =	10	kN
   	3

   
   

   FA =	P.e.r	=	10 × 103 × 150 × 40
   	rA² + rB² + rC²		40² + 0² + 40²

   
   F'_A = 18.75 kN = F'_C
   Shear load at A and C
   F_R = √(F_A)² + (F'_A)² = √(3.33)² + (18.75)²
   F_R = 19.04 KN
   

   τmax =	19.04 × 103
   	(π / 4) × 102

   
   ( τ_max )A,C = 242.5 MPa
   ( τ_max )B = 42.44 MPa
   

   Correct Option: A

   FA = FB = FC =	10	kN
   	3

   
   

   FA =	P.e.r	=	10 × 103 × 150 × 40
   	rA² + rB² + rC²		40² + 0² + 40²

   
   F'_A = 18.75 kN = F'_C
   Shear load at A and C
   F_R = √(F_A)² + (F'_A)² = √(3.33)² + (18.75)²
   F_R = 19.04 KN
   

   τmax =	19.04 × 103
   	(π / 4) × 102

   
   ( τ_max )A,C = 242.5 MPa
   ( τ_max )B = 42.44 MPa
   

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100. In a bolted Joint two smooth members are connected with an axial tightening force of 2200 N. If the bolt used has metric threads of 4 mm pitch, the torque required for achieving the tightening force is
     
     

1. 1. 0.7 Nm
      

   
   
   

   2. 1.0 Nm
      

   
   
   

   3. 1.4 Nm
      

   
   
   

   4. 2.8 Nm
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   Tightening force = 2200 N
   Pitch = 4 mm = 0.004 m
   

   Torque = Tightening force ×	0.004	= 1.4 N-m
   	2π

   Correct Option: C

   Tightening force = 2200 N
   Pitch = 4 mm = 0.004 m
   

   Torque = Tightening force ×	0.004	= 1.4 N-m
   	2π

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