# Kinematics (NEET Physics): Questions 131 - 135 of 229

Get 1 year subscription: Access detailed explanations (illustrated with images and videos) to **2099** questions. Access all new questions we will add tracking exam-pattern and syllabus changes. View Sample Explanation or View Features.

Rs. 550.00 or

## Question number: 131

» Kinematics » Average Speed and Instantaneous Velocity

### Question

A particle moves in a circle of radius 10cm with a linear speed of 20m/s. The angular velocity will be –

### Choices

Choice (4) | Response | |
---|---|---|

a. | 25 rad/s | |

b. | 75 rad/s | |

c. | 50 rad/s | |

d. | 200 rad/s |

## Question number: 132

» Kinematics » Motion in a Straight Line

### Question

If t _{1} be the time taken by a body to clear the top of a building and be the time spent in air, then t _{2}: t _{1} will be -

### Choices

Choice (4) | Response | |
---|---|---|

a. | 2: 1 | |

b. | 1: 1 | |

c. | 1: 2 | |

d. | 1: 4 |

## Question number: 133

» Kinematics » Motion in a Straight Line

### Assertion (Ꭺ)

A cyclist is cycling on rough horizontal circular track with increasing speed. Then the frictional force on cycle is always directed towards center of the circular track.

### Reason (Ꭱ)

For a particle moving in a circle, radial component of net force should be directed towards center. You have to select the correct choice.

### Choices

Choice (4) | Response | |
---|---|---|

a. | Both Ꭺ and Ꭱ are true and Ꭱ is the correct explanation of Ꭺ | |

b. | Ꭺ is true but Ꭱ is false | |

c. | Both Ꭺ and Ꭱ are true but Ꭱ is NOT the correct explanation of Ꭺ | |

d. | Both Ꭺ and Ꭱ are false |

## Question number: 134

» Kinematics » Average Speed and Instantaneous Velocity

### Question

The angular speed with which the earth would have to rotate on its axis so that a person on the equator would weigh (3/5) ^{th} as much as present, will be: (Take the equatorial radius as 6400 km)

### Choices

Choice (4) | Response | |
---|---|---|

a. | 7.8 x 10 | |

b. | 7.8 x 10 | |

c. | 8.7 x 10 | |

d. | 8.7 x 10 |

## Passage

Velocity at a general point P (x, y) for a horizontal projectile motion is given by

V = =

α is angle made by v with horizontal in clockwise direction Trajectory equation for a horizontal projectile motion is given by x = v _{x} t = ut

y =- ( ) gt ^{2 }

Eliminating t, We get y =-

## Question number: 135 (1 of 3 Based on Passage) Show Passage

» Kinematics » Average Speed and Instantaneous Velocity

### Question

An aeroplane is flying horizontally with a velocity of 720 km/h at an altitude of 490 m. When it is just vertically above the target a bomb is dropped from it. How far horizontally it missed the target?

### Choices

Choice (4) | Response | |
---|---|---|

a. | 2000 m | |

b. | 200 m | |

c. | 1000 m | |

d. | 100 m |