21. A drum of radius R and mass M, rolls down without slipping along an inclined plane of angle θ. The frictional force

1. 1. dissipates energy as heat

   
   
   

   2. decreases the rotational motion

   
   
   

   3. decreases the rotational and translational translational motion

   
   
   

   4. converts translational energy to rotational energy

   
   
   

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1. View Hint View Answer Discuss in Forum

   Net work done by frictional force when drum rolls down without slipping is zero.
   W_net = 0
   
   W_trans. + W_rot. = 0; ∆K_trans. + ∆K_rot. = 0
   ∆K_trans = –∆K_rot.
   i.e., converts translation energy to rotational energy.

   Correct Option: D

   Net work done by frictional force when drum rolls down without slipping is zero.
   W_net = 0
   
   W_trans. + W_rot. = 0; ∆K_trans. + ∆K_rot. = 0
   ∆K_trans = –∆K_rot.
   i.e., converts translation energy to rotational energy.

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22. A small object of uniform density rolls up a curved surface with an initial velocity ‘ν’. It reaches upto a maximum height of 3v²/4g with respect to the initial position. The object is a

1. 1. solid sphere

   
   
   

   2. hollow sphere

   
   
   

   3. disc

   
   
   

   4. ring

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   From law of conservation of mechanical energy
   

   	1	Iω² + 0 +	1	mv² = mg ×	3v²
   	2		2		4g

   
   

   ⇒	1	Iω² =	3	mv² -	1	mv²
   	2		4		2

   
   

   =	mv²		3	- 1
   	2		2

   
   

   or,	1	I	V²	=	mv²	or, I =	1	mR²
   	2		R²		4		2

   
   Hence, object is a disc.

   Correct Option: C

   
   From law of conservation of mechanical energy
   

   	1	Iω² + 0 +	1	mv² = mg ×	3v²
   	2		2		4g

   
   

   ⇒	1	Iω² =	3	mv² -	1	mv²
   	2		4		2

   
   

   =	mv²		3	- 1
   	2		2

   
   

   or,	1	I	V²	=	mv²	or, I =	1	mR²
   	2		R²		4		2

   
   Hence, object is a disc.

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23. The ratio of the accelerations for a solid sphere (mass ‘m’ and radius ‘R’) rolling down an incline of angle ‘θ’ without slipping and slipping down the incline without rolling is :

1. 1. 5 : 7

   
   
   

   2. 2 : 3

   
   
   

   3. 2 : 5

   
   
   

   4. 7 : 5

   
   
   

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1. View Hint View Answer Discuss in Forum

   For solid sphere rolling without slipping
   slipping on inclined plane, acceleration
   

   a1 =	g sin θ
   	1 +	K²
   		R²

   
   For solid sphere slipping on inclined plane without rolling, acceleration
   a₂ = g sin θ
   

   Therefore required ratio =	a1
   	a2

   
   

   =	1			=	1					=	5
   		1 +	K²		1 +	2					7
   			R²			5

   Correct Option: A

   For solid sphere rolling without slipping
   slipping on inclined plane, acceleration
   

   a1 =	g sin θ
   	1 +	K²
   		R²

   
   For solid sphere slipping on inclined plane without rolling, acceleration
   a₂ = g sin θ
   

   Therefore required ratio =	a1
   	a2

   
   

   =	1			=	1					=	5
   		1 +	K²		1 +	2					7
   			R²			5

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24. A disk and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first ?

1. 1. Disk

   
   
   

   2. Sphere

   
   
   

   3. Both reach at the same time

   
   
   

   4. Depends on their masses

   
   
   

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1. View Hint View Answer Discuss in Forum

   Time of descent ∝	K²
   	R²

   
   

   Order of value of	K²
   	R²

   
   

   for disc;	K²	=	1	= 0.5
   	R²		2

   
   

   for sphere;	K²	=	2	= 0.4
   	R²		5

   
   (sphere) < (disc)
   ∴ Sphere reaches first

   Correct Option: B

   Time of descent ∝	K²
   	R²

   
   

   Order of value of	K²
   	R²

   
   

   for disc;	K²	=	1	= 0.5
   	R²		2

   
   

   for sphere;	K²	=	2	= 0.4
   	R²		5

   
   (sphere) < (disc)
   ∴ Sphere reaches first

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25. A ring of mass m and radius r rotates about an axis passing through its centre and perpendicular to its plane with angular velocity ω. Its kinetic energy is

1. 1. 1	mr²ω²
      2

   
   
   

   2. mrω²

   
   
   

   3. mr²ω²

   
   
   

   4. 1	mrω²
      2

   
   
   

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1. View Hint View Answer Discuss in Forum

   Kinetic energy =	1	Iω²
   	2

   
   and for ring I = mr²
   

   Hence, KE =	1	mr²ω²
   	2

   Correct Option: A

   Kinetic energy =	1	Iω²
   	2

   
   and for ring I = mr²
   

   Hence, KE =	1	mr²ω²
   	2

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