1. For the truss shown in the figure, the magnitude of the force (in kN) in the member SR is
   
   

1. 1. 10
      

   
   
   

   2. 14.14
      

   
   
   

   3. 20
      

   
   
   

   4. 28.28
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   R_p + R_x = 30000
   ∑M_P = 0
   R_x × 3 = 2 × 3000
   ⇒ R_x = 20 kN and R_p = 10 kN
   For balance at ‘x' → F_Rx = 20 kN
   at ‘RT’ → F_RT cos 45 = 20
   
   

   ⇒ FRT =	20	...(i)
   	cos 45

   
   Also F_SR = F_RT cos 45... (ii)
   From equations (i) and (ii),
   F_SR = 20 kN
   

   Correct Option: C

   
   R_p + R_x = 30000
   ∑M_P = 0
   R_x × 3 = 2 × 3000
   ⇒ R_x = 20 kN and R_p = 10 kN
   For balance at ‘x' → F_Rx = 20 kN
   at ‘RT’ → F_RT cos 45 = 20
   
   

   ⇒ FRT =	20	...(i)
   	cos 45

   
   Also F_SR = F_RT cos 45... (ii)
   From equations (i) and (ii),
   F_SR = 20 kN
   

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2. Consider a fly wheel whose mass M is distributed almost equally between a heavy, ring-like rim of radius P and a concentric disk-like feature of radius R/2. Other parts of the flywheel, such as spokes, etc, have negligible mass. The best approximation for α, if the moment Of inertia of the flywheel about its axis of rotation is expressed as αMR₂, is _______.

1. 1. 0.56

   
   
   

   2. 0.51

   
   
   

   3. 0.55

   
   
   

   4. 0.59

   
   
   

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1. View Hint View Answer Discuss in Forum

   Moment of Inertia of fly wheel
   I = m_rR²
   R = mean Radius of Rim
   

   = M		R + R/2		2
   		2

   
   = 0.56 MR²
   ⇒ α = 0.56

   Correct Option: A

   Moment of Inertia of fly wheel
   I = m_rR²
   R = mean Radius of Rim
   

   = M		R + R/2		2
   		2

   
   = 0.56 MR²
   ⇒ α = 0.56

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3. A circular solid disc of uniform thickness 20 mm, radius 200 mm and mass 20 kg, is used as a flywheel. If it rotates at 600 rpm, the kinetic energy of the flywheel, in Joules is

1. 1. 395

   
   
   

   2. 790

   
   
   

   3. 1580

   
   
   

   4. 3160

   
   
   

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1. View Hint View Answer Discuss in Forum

   K.E. of fly wheel =	1	ω2
   	2

   
   

   where, I =	mR2	=	20 × 0.22	= 0.4 kg- m2
   	2		2

   
   

   ω =	2πN	=	2 × π × 600	= 62.83 rad/s
   	60		60

   
   

   K.E.=	1	× 0.4 × 62.832 = 790 J
   	2

   Correct Option: B

   K.E. of fly wheel =	1	ω2
   	2

   
   

   where, I =	mR2	=	20 × 0.22	= 0.4 kg- m2
   	2		2

   
   

   ω =	2πN	=	2 × π × 600	= 62.83 rad/s
   	60		60

   
   

   K.E.=	1	× 0.4 × 62.832 = 790 J
   	2

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4. A two member truss PQR is supporting a load W. The axial forces in members PQ and QR are respectively
   
   

1. 1. 2W tensile and √3W compressive
      

   
   
   

   2. √3W tensile and 2W compressive
      

   
   
   

   3. √3W compressive and 2W tensile
      

   
   
   

   4. 2W compressive and √3W tensile

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Lami’s theorem is used to find the F
   

   	W	=	TPQ	=	TQR
   	sin 30°		sin 60°		sin 270°

   
   ⇒ T_PQ = √3W (T)
   ⇒ T_QR = – 2 W = 2W(C)

   Correct Option: C

   
   Lami’s theorem is used to find the F
   

   	W	=	TPQ	=	TQR
   	sin 30°		sin 60°		sin 270°

   
   ⇒ T_PQ = √3W (T)
   ⇒ T_QR = – 2 W = 2W(C)

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5. Two identical trusses support a load of 100 N as shown in the figure. The length of each truss is 1.0 m, cross-sectional area is 200 mm²; Young's modulus E = 200 GPa. The force in the truss AB (in N) is
   
   

1. 1. F = 150 N

   
   
   

   2. F = 10 N

   
   
   

   3. F = 100 N

   
   
   

   4. None of above

   
   
   

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1. View Hint View Answer Discuss in Forum

   ⇒ 2F sin 30 = 100
   ⇒ F = 100 N
   

   Correct Option: C

   ⇒ 2F sin 30 = 100
   ⇒ F = 100 N
   

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