46. An annular disc has a mass m, inner radius R and outer radius 2R. The disc rolls on a flat surface without slipping. If the velocity of the center of mass is v, the kinetic energy of the disc is

1. 1. 9/16 mv²

   
   
   

   2. 11/16 mv²

   
   
   

   3. 13/16 mv²

   
   
   

   4. 15/16 mv²

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   I =	5mr2
   	2

   
   K.E. = (K.E.) _translation + (K.E.) _Rotation
   

   =		1	mv² +	1	Iω²
   		2		2

   
   Where V = 2 R ω
   

   ⇒ ω =	V
   	2R

   
   

   ∴ K.E. =		1	mv² +	5mV²
   		2		16

   
   

   K.F. =	13mV²
   	16

   
   

   Correct Option: C

   
   

   I =	5mr2
   	2

   
   K.E. = (K.E.) _translation + (K.E.) _Rotation
   

   =		1	mv² +	1	Iω²
   		2		2

   
   Where V = 2 R ω
   

   ⇒ ω =	V
   	2R

   
   

   ∴ K.E. =		1	mv² +	5mV²
   		2		16

   
   

   K.F. =	13mV²
   	16

   
   

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47. A stone with mass of 0.1 kg is catapulted as shown in the figure. The total force F_x (in N) exerted by the rubber band as a function of distance x (in m) is given by F_x = 300x². If the store is displaced by 0.1 m from the un-stretched position (x = 0) of the rubber band, the energy stored in the rubber band is
    

1. 1. 0.01 J

   
   
   

   2. 0.1 J

   
   
   

   3. 1 J

   
   
   

   4. 10 J

   
   
   

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1. View Hint View Answer Discuss in Forum

   Energy stored in the bar
   = Work done by the stone
   
   

   = 300 ×	x3
   	3

   
   = 100 × 0.1³
   = 0.1J

   Correct Option: B

   Energy stored in the bar
   = Work done by the stone
   
   

   = 300 ×	x3
   	3

   
   = 100 × 0.1³
   = 0.1J

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48. A ball is dropped from rest from a height of 1 min a frictionless tube as shown in the figure. If the tube profile is approximated by two straight lines (ignoring the curved portion), the total distance travelled (in m) by the ball is (correct to two decimal places).
    

1. 1. 2.414 m

   
   
   

   2. 1.414 m

   
   
   

   3. 3.414 m

   
   
   

   4. 5.414 m

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   	BC	= sin 45°
   	AB

   
   

   AB =		BC	=	1	= 1.414m
   		sin 45°		sin 45°

   
   Total travel, OA + AB = 1+1.414 = 2.414 m

   Correct Option: A

   
   

   	BC	= sin 45°
   	AB

   
   

   AB =		BC	=	1	= 1.414m
   		sin 45°		sin 45°

   
   Total travel, OA + AB = 1+1.414 = 2.414 m

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49. An ejector mechanism consists of a helical compression spring having a spring constant of K = 981 × 10³ N/m. It is pre-compressed by 100 mm from its free state. If it is used to eject a mass of 100 kg held on it, the mass will move up through a distance of
    

1. 1. 100 mm

   
   
   

   2. 5000 mm

   
   
   

   3. 581 mm

   
   
   

   4. 1000 mm

   
   
   

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1. View Hint View Answer Discuss in Forum

   From the conservation of energy after precompression,
   Initial energy = Final energy
   

   or		1	k(0.1)² =	1	+ kx² + mg(x + 0.1)
   		2		2

   
   or k(0.1)²= kx² + 2mg(x + 0.1)
   or 981 × 10³ × 10^{- 2} = 981 × 10³ × x² + 2 × 100 × 9.81(x + 0.1)
   or 100x² + 2x – 0.8 = 0
   

   ∴ x =	-2 ± √(2)² - 4 × 100 × 0.8
   	200

   
   Taking + ve sign, x = 100 mm

   Correct Option: A

   From the conservation of energy after precompression,
   Initial energy = Final energy
   

   or		1	k(0.1)² =	1	+ kx² + mg(x + 0.1)
   		2		2

   
   or k(0.1)²= kx² + 2mg(x + 0.1)
   or 981 × 10³ × 10^{- 2} = 981 × 10³ × x² + 2 × 100 × 9.81(x + 0.1)
   or 100x² + 2x – 0.8 = 0
   

   ∴ x =	-2 ± √(2)² - 4 × 100 × 0.8
   	200

   
   Taking + ve sign, x = 100 mm

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50. Block P of mass 2 kg slides down the surface and has a speed 20 m/s at the lowest point, Q where the local radius of curvature is 2 m as shown in the figure. Assuming g = 10 m/s², the normal force (in N) at Q is _______ (correct to two decimal places).
    
    

1. 1. 420 N

   
   
   

   2. 421 N

   
   
   

   3. 422 N

   
   
   

   4. 419 N

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   m = 2 kg, g = 10 m/s²
   FBD of mass P at point Q
   
   

   N = mg +		mv²	= 2 × 10 +	2 × 20 × 20	= 420 N
   		R		2

   
   

   Correct Option: A

   
   m = 2 kg, g = 10 m/s²
   FBD of mass P at point Q
   
   

   N = mg +		mv²	= 2 × 10 +	2 × 20 × 20	= 420 N
   		R		2

   
   

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