141. Consider the system of two wagons shown in figure. The natural frequencies of this system are
     
     

1. 1. 0 , √2k / m
      

   
   
   

   2. √k / m , √2k / m
      

   
   
   

   3. √k / m , √k / 2m
      

   
   
   

   4. 0 , √k / 2m

   
   
   

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1. View Hint View Answer Discuss in Forum

   Natural frequencies will be ( 0 , √2k / m )
   
   For small displacement
   

   θ =	a	=	b
   	300		150

   
   Taking moment about hinged point ‘O’
   k.a (0.3) = Wb
   ⇒ K(0.3)² = 300 × 0.150
   

   K =	300 × 0.150	= 500 N / m
   	0.32

   
   

   Correct Option: A

   Natural frequencies will be ( 0 , √2k / m )
   
   For small displacement
   

   θ =	a	=	b
   	300		150

   
   Taking moment about hinged point ‘O’
   k.a (0.3) = Wb
   ⇒ K(0.3)² = 300 × 0.150
   

   K =	300 × 0.150	= 500 N / m
   	0.32

   
   

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142. A mass of 1 kg is suspended by means of 3 springs as shown in figure. The spring constant K₁, K₂ and K₃ are respectively 1 kN/m, 3 kN/m and 2 kN/m. The natural frequency of the system is approximately
     
     

1. 1. 46.90 rad/s
      

   
   
   

   2. 52.44 rad/s
      

   
   
   

   3. 60.55 rad/s
      

   
   
   

   4. 77.46 rad/s

   
   
   

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1. View Hint View Answer Discuss in Forum

   K₁ and K₂ are in series,
   

   ∴	1	=	1	+	1	=	1	+	1	=	4
   	Keq1		K1		K2		1		3		3

   
   K_eq1 and K₃ are in parallel arrangement,
   

   Keq = Keq1 + K3 =	3	+ 2 =	11	kN / m
   	4		4

   
   Natural frequency, ω = √K_eq / m = √(11 × 10³) / (4 × 1) = 52.44 rad / s

   Correct Option: B

   K₁ and K₂ are in series,
   

   ∴	1	=	1	+	1	=	1	+	1	=	4
   	Keq1		K1		K2		1		3		3

   
   K_eq1 and K₃ are in parallel arrangement,
   

   Keq = Keq1 + K3 =	3	+ 2 =	11	kN / m
   	4		4

   
   Natural frequency, ω = √K_eq / m = √(11 × 10³) / (4 × 1) = 52.44 rad / s

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143. If the pressure angle of the rack is 20°, the force acting along the line of action between the rack and the gear teeth is
     

1. 1. 250 N
      

   
   
   

   2. 342 N
      

   
   
   

   3. 532 N
      

   
   
   

   4. 600 N

   
   
   

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1. View Hint View Answer Discuss in Forum

   P cos φ = F
   ∴ Force acting along the line of action,
   

   P =	F	=	500	= 532 N
   	cos φ		cos 20°

   Correct Option: C

   P cos φ = F
   ∴ Force acting along the line of action,
   

   P =	F	=	500	= 532 N
   	cos φ		cos 20°

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144. If the drive efficiency is 80%, the torque required on the input shaft to create 1000 N output thrust is
     

1. 1. 20 Nm
      

   
   
   

   2. 25 Nm
      

   
   
   

   3. 32 Nm
      

   
   
   

   4. 50 Nm

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given: Module m = 2,
   

   	D	= 2
   	T

   
   ∴ D = 80 × 2 = 160 mm
   2F = 1000, or F = 500 N
   
   Let T₁ be the torque applied by motor.
   T₂ be the torque applied by gear.
   ∴ Power transmission = 80%
   
   

   T1 ω1 =	2T2 × ω2
   	0.8

   
   

   or T1 =	2 × F × (D / 2)	×	ω2
   	0.8		ω1

   
   

   = 2 × 500 ×	0.16	×	1	×	1	= 25 N-m
   	2		0.8		4

   
   

   Correct Option: B

   Given: Module m = 2,
   

   	D	= 2
   	T

   
   ∴ D = 80 × 2 = 160 mm
   2F = 1000, or F = 500 N
   
   Let T₁ be the torque applied by motor.
   T₂ be the torque applied by gear.
   ∴ Power transmission = 80%
   
   

   T1 ω1 =	2T2 × ω2
   	0.8

   
   

   or T1 =	2 × F × (D / 2)	×	ω2
   	0.8		ω1

   
   

   = 2 × 500 ×	0.16	×	1	×	1	= 25 N-m
   	2		0.8		4

   
   

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145. A 1.5 kW motor is running at 1440 rev/min. It is to be connected to a stirrer running at 36 rev/ min. The gearing arrangement suitable for this application is
     

1. 1. differential gear
      

   
   
   

   2. helical gear
      

   
   
   

   3. spur gear
      

   
   
   

   4. worm gear

   
   
   

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1. View Hint View Answer Discuss in Forum

   Power of motor = 1.5 kW
   N = 1440 rev/min
   Speed of stirrer = 36 rev/min
   

   S.R.D =	1440	= 40
   	36

   
   ∴ Worm gear is used.

   Correct Option: D

   Power of motor = 1.5 kW
   N = 1440 rev/min
   Speed of stirrer = 36 rev/min
   

   S.R.D =	1440	= 40
   	36

   
   ∴ Worm gear is used.

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