96. In a certain slider-crank mechanism, lengths of crank and connecting rod are equal. If the crank rotates with a uniform angular speed of 14 rad/s and the crank length is 300 mm, the maximum acceleration of the slider (in m/s2) is _________.

1. 1. 112.6 m/s²

   
   
   

   2. 117 m/s²

   
   
   

   3. 117.6 m/s²

   
   
   

   4. 7.6 m/s²

   
   
   

---

1. View Hint View Answer Discuss in Forum

   a_max = 2rω²
   (where θ = 0) i.e at Inner dead centre
   = 2 × 0.3 × 14²
   a_max = 117.6 m/s²

   Correct Option: C

   a_max = 2rω²
   (where θ = 0) i.e at Inner dead centre
   = 2 × 0.3 × 14²
   a_max = 117.6 m/s²

---

97. A slider-crank mechanism with crank radius 60 mm and connecting rod length 240 mm is shown in figure. The crank is rotating with a uniform angular speed of 10 rad/s, counter clockwise. For the given configuration, the speed (in m/s) of the slider is _______.
    

1. 1. 0.6 m/s

   
   
   

   2. 0.3 m/s

   
   
   

   3. 0.1 m/s

   
   
   

   4. 0.1 m/s

   
   
   

---

1. View Hint View Answer Discuss in Forum

   
   L = 240 mm
   r = 60 mm
   

   n =	L	= 4
   	r

   
   ω = 10 rad/s
   

   cosθ =	sinθ
   	n

   
   θ = tan⁻¹(4) = 75.96°
   

   ∴  VS = ωr		sinθ +	sin2θ
   			2n

   
   

   ∴  VS = 10 × 0.06		sin75.96° +	sin151.92°
   			2 × 4

   
   = 0.6 × 1.0288
   = 0.6 m/s

   Correct Option: A

   
   L = 240 mm
   r = 60 mm
   

   n =	L	= 4
   	r

   
   ω = 10 rad/s
   

   cosθ =	sinθ
   	n

   
   θ = tan⁻¹(4) = 75.96°
   

   ∴  VS = ωr		sinθ +	sin2θ
   			2n

   
   

   ∴  VS = 10 × 0.06		sin75.96° +	sin151.92°
   			2 × 4

   
   = 0.6 × 1.0288
   = 0.6 m/s

---

---

98. An offset slider-crank mechanism is shown in the figure at an instant. Conventionally, the Quick Return Ratio (QRR) is considered to be greater than one. The value of QRR is ______.
    

1. 1. 255

   
   
   

   2. 1.255

   
   
   

   3. 5.255

   
   
   

   4. 1255

   
   
   

---

1. View Hint View Answer Discuss in Forum

   AB = stroke length
   AO = 40 – 20 = 20
   BO = 40 + 20 = 60
   ∠ AOB = ∠ BOC – ∠ AOC
   = 80.41 – 60 = 20.41
   

   QRR =	180 + φ	=	180 + 20.41	= 1.255.
   	180 − φ		180 − 20.41

   
   

   Correct Option: B

   AB = stroke length
   AO = 40 – 20 = 20
   BO = 40 + 20 = 60
   ∠ AOB = ∠ BOC – ∠ AOC
   = 80.41 – 60 = 20.41
   

   QRR =	180 + φ	=	180 + 20.41	= 1.255.
   	180 − φ		180 − 20.41

   
   

---

99. Consider the triangular formed by the connecting rod and the crank of an IC engine as the two sides of the triangle. If the maximum area of this triangle occurs when the crank is 75°, the ratio of connecting rod length to crank radius to

1. 1. 5

   
   
   

   2. 4

   
   
   

   3. 3.73

   
   
   

   4. 3

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Δ =	1	(PQ) (PR) sin A
   	2

   
   Area will be maximum when A = 90°
   i.e. PQR is a right angled triangle.
   ∴  Ratio of connecting rod length to crank radius,
   

   l	= tan 75° = 3.732;
   r

   
   l = 3.732 r
   

   Correct Option: C

   Δ =	1	(PQ) (PR) sin A
   	2

   
   Area will be maximum when A = 90°
   i.e. PQR is a right angled triangle.
   ∴  Ratio of connecting rod length to crank radius,
   

   l	= tan 75° = 3.732;
   r

   
   l = 3.732 r
   

---

---

100. The cross head velocity in the slider crank mechanism, for the position shown in figure
     

1. 1. V_ccos(90 − α + β)cosβ

   
   
   

   2. V_ccos(90 − α + β)secβ

   
   
   

   3. V_ccos(90 − α − β)cosβ

   
   
   

   4. V_ccos(90 − α − β)secβ

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

---