21. What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop ?

1. 1. √gR

   
   
   

   2. √2gR

   
   
   

   3. √3gR

   
   
   

   4. √5gR
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   To complete the loop a body must enter a vertical loop of radius R with the minimum velocity v = √5gR.
   

   Correct Option: D

   To complete the loop a body must enter a vertical loop of radius R with the minimum velocity v = √5gR.
   

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22. Two stones of masses m and 2m are whirled in horizontal circles, the heavier one in radius 1/2 and the lighter one in radius r. The tangential speed of lighter stone is n times that of the value of heavier stone when they experience same centripetal forces. The value of n is :

1. 1. 3

   
   
   

   2. 4

   
   
   

   3. 1

   
   
   

   4. 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   According to question, two stones experience same centripetal force
   i.e. F_C1 = F_C2
   

   or,	mv1²	=	2mv2²
   	r		r/2

   
   or, v₁² = 4v₂²
   So, V₁ = 2V₂ i.e., n = 2

   Correct Option: D

   According to question, two stones experience same centripetal force
   i.e. F_C1 = F_C2
   

   or,	mv1²	=	2mv2²
   	r		r/2

   
   or, v₁² = 4v₂²
   So, V₁ = 2V₂ i.e., n = 2

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23. A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m/s. A bob is suspended from the roof of the car by a light wire of length 1.0 m. The angle made by the wire with the vertical is

1. 1. 0°

   
   
   

   2. π

      3

   
   
   

   3. π

      6

   
   
   

   4. π

      4

      
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given; speed = 10 m/s; radius r = 10 m
   Angle made by the wire with the vertical
   

   tanθ =	v²	=	10²	= 1
   	rg		10 × 10

   
   

   ⇒ θ = 45° =	π
   	4

   
   
   

   Correct Option: D

   Given; speed = 10 m/s; radius r = 10 m
   Angle made by the wire with the vertical
   

   tanθ =	v²	=	10²	= 1
   	rg		10 × 10

   
   

   ⇒ θ = 45° =	π
   	4

   
   
   

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24. A car of mass 1000 kg negotiates a banked curve of radius 90 m on a frictionless road. If the banking angle is 45°, the speed of the car is :

1. 1. 20 ms^–1

   
   
   

   2. 30 ms^–1

   
   
   

   3. 5 ms^–1

   
   
   

   4. 10 ms^–1

   
   
   

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1. View Hint View Answer Discuss in Forum

   For banking tan θ =	V²	= 1
   	90 × 10

   
   V = 30 m/s
   

   Correct Option: C

   For banking tan θ =	V²	= 1
   	90 × 10

   
   V = 30 m/s
   

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25. A car of mass m is moving on a level circular track of radius R. If μ_s represents the static friction between the road and tyres of the car, the maximum speed of the car in circular motion is given by :

1. 1. √μ_smRg

   
   
   

   2. √Rg/μ_s

   
   
   

   3. √mRg/μ_s

   
   
   

   4. √μ_sRg
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   For smooth driving maximum speed of car v then
   

   	mv²	= μmg
   	R

   
   v = √μ_sRg
   

   Correct Option: D

   For smooth driving maximum speed of car v then
   

   	mv²	= μmg
   	R

   
   v = √μ_sRg
   

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