46. The spring constant of a helical compression spring DOES NOT depend on

1. 1. coil diameter

   
   
   

   2. material strength

   
   
   

   3. number of active turns

   
   
   

   4. wire diameter

   
   
   

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1. View Hint View Answer Discuss in Forum

   Stiffness of helical spring
   

   k =	Gd4
   	64R3 n

   
   where G is Shear modulus
   d is Wire diameter
   R is Wire radius
   n is No. of active turns.

   Correct Option: B

   Stiffness of helical spring
   

   k =	Gd4
   	64R3 n

   
   where G is Shear modulus
   d is Wire diameter
   R is Wire radius
   n is No. of active turns.

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47. A weighing machine consists of a 2 kg pan resting on a spring. In this condition, with the pan resting on the spring, the length of the spring is 200 mm. When a mass of 20 kg is placed on the pan, the length of the spring becomes 100 mm. For the spring, the undeformed length L and the spring constant k (stiffness) are

1. 1. L = 220 mm, k= 1862 N/m

   
   
   

   2. L = 210 mm, k= 1960 N/m

   
   
   

   3. L = 200 mm, k = 1960 N/m

   
   
   

   4. L = 200 mm, k= 2156 N/m

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let initial length is δ without load, and stiffness is k
   ∴ 2g = k(δ – 0.2) ...(i)
   22g = k(δ – 0.1) ...(ii)
   Solving equations (i) and (ii), we get
   δ = 210 mm and k = 1960 N/m

   Correct Option: B

   Let initial length is δ without load, and stiffness is k
   ∴ 2g = k(δ – 0.2) ...(i)
   22g = k(δ – 0.1) ...(ii)
   Solving equations (i) and (ii), we get
   δ = 210 mm and k = 1960 N/m

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48. The deflection of a spring with 20 active turns under a load of 1000 N is 10 mm. The spring is made into two pieces each of 10 active coils and placed in parallel under the same load. The deflection of this system is

1. 1. 20 mm

   
   
   

   2. 10 mm

   
   
   

   3. 5 mm

   
   
   

   4. 2.5 mm

   
   
   

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1. View Hint View Answer Discuss in Forum

   Where; n = 20 P = 1000 N
   ∆x = 10 mm
   F = K∆x
   

   K =	P	= 100 N /mm
   	∆x

   
   When a spring is cut into two pieces, no of coils get halved
   ∴ Stiffness of each half coiled get doubled
   The stiffness when connected in parallel = 2k + 2k = 4k
   P = K∆x
   

   ∆x =	P	=	1000	= 2.5 mm
   	k		4 × 100

   Correct Option: D

   Where; n = 20 P = 1000 N
   ∆x = 10 mm
   F = K∆x
   

   K =	P	= 100 N /mm
   	∆x

   
   When a spring is cut into two pieces, no of coils get halved
   ∴ Stiffness of each half coiled get doubled
   The stiffness when connected in parallel = 2k + 2k = 4k
   P = K∆x
   

   ∆x =	P	=	1000	= 2.5 mm
   	k		4 × 100

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49. The figure shows arrangements of springs. They have stiffness k₁, and k₂ as marked.

    Which of the following arrangements offers a stiffness =	2k1k2	?
    	k1 + 2k2

1. 1. 

   
   
   

   2. 

   
   
   

   3. 

   
   
   

   4. 

   
   
   

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1. View Hint View Answer Discuss in Forum

   k =	2k1k2
   	k1 + 2k2

   
   

   	1	=	k1 + 2k2	=	1	+	1
   	k		(2k1)k2		2k2		k1

   
   

   	1	=	1	+	1
   	k		k1		2k2

   
   

   ∴ k =	2k1k2
   	k1 + 2k2

   Correct Option: D

   k =	2k1k2
   	k1 + 2k2

   
   

   	1	=	k1 + 2k2	=	1	+	1
   	k		(2k1)k2		2k2		k1

   
   

   	1	=	1	+	1
   	k		k1		2k2

   
   

   ∴ k =	2k1k2
   	k1 + 2k2

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50. A motor driving a soiid circular steel shaft transmits 40 kW of power at 500 rpm. If the diameter of the shaft is 40 mm, the maximum shear stress in the shaft is ______ MPa.

1. 1. 60

   
   
   

   2. 60.79

   
   
   

   3. 50.79

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   We know that, Power is given by
   

   P =	2πNT
   	60

   
   

   4 × 103 =	2π × 500 × T
   	60

   
   T = 763.94 Nm
   

   T =	π	d3τ
   	16

   
   

   763.94 × 10³ =	π	× 403 × τ
   	16

   
   τ = 60.79 × 10⁶ Pascal
   τ = 60.79 MPa

   Correct Option: B

   We know that, Power is given by
   

   P =	2πNT
   	60

   
   

   4 × 103 =	2π × 500 × T
   	60

   
   T = 763.94 Nm
   

   T =	π	d3τ
   	16

   
   

   763.94 × 10³ =	π	× 403 × τ
   	16

   
   τ = 60.79 × 10⁶ Pascal
   τ = 60.79 MPa

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