In Class 11 Final Exams there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Downloads of CBSE Class 11 Physics Chapter 4 Case Study and Passage-Based Questions with Answers were Prepared Based on the Latest Exam Pattern. Students can solve Class 11 Physics Case Study Questions Motion In A Plane to know their preparation level.
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In CBSE Class 11 Physics Paper, There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.
Motion In A Plane Case Study Questions With Answers
Here, we have provided case-based/passage-based questions for Class 11 Physics Chapter 4 Motion In A Plane
Case Study/Passage-Based Questions
Case Study 1: When an object follows a circular path at a constant speed, the motion of the object is called uniform circular motion. The word “uniform” refers to the speed, which is uniform (constant) throughout the motion. Suppose an object is moving with uniform speed v in a circle of radius R Since the velocity of the object is changing continuously in direction, the object undergoes acceleration. Let us find the magnitude and the direction of this acceleration. Thus, the acceleration of an object moving with speed v in a circle of radius R has a magnitude V2/R and is always directed towards the center. This is why this acceleration is called centripetal acceleration (a term proposed by Newton). A thorough analysis of centripetal acceleration was first published in 1673 by the Dutch scientist Christiaan Huygens (1629-1695) but it was probably known to Newton also some years earlier. “Centripetal” comes from a Greek term that means ‘center-seeking’. Since v and R are constant, the magnitude of the centripetal acceleration is also constant. However, the direction changes pointing always toward the center. Therefore, centripetal acceleration is not a constant vector. We can express centripetal acceleration ac in terms of angular speed as
ac = ω2R
The time taken by an object to make one revolution is known as its time period T and the number of revolutions made in one second is called its frequency v (=1/T). However, during this time the distance moved by the object is s = 2πR. Therefore, v = 2πR/T =2πRv In terms of frequency n, we have
ω = 2πv
v = 2πRv
ac = 4π2v2R
1) SI unit of angular velocity is
a) Rev/sec
b) m/s
c) m/s2
d) None of these
Answer: a) Rev/sec
2) A centripetal acceleration is not a constant vector. True or false?
a) True
b) False
Answer: a) True
3) What does the term “uniform” refer to in uniform circular motion?
a) Constant acceleration
b) Constant radius
c) Constant velocity
d) Constant speed
Answer: d) Constant speed
4) How is the direction of centripetal acceleration characterized?
a) Always pointing outward
b) Always pointing toward the center
c) Constant in one direction
d) Changing randomly
Answer: b) Always pointing toward the center
5) Who first published a thorough analysis of centripetal acceleration?
a) Isaac Newton
b) Albert Einstein
c) Christiaan Huygens
d) Galileo Galilei
Answer: c) Christiaan Huygens
Case Study/Passage-Based Questions
Case Study 2: We consider the motion of a projectile. An object that is in flight after being thrown or projected is called a projectile. Such a projectile might be a football, a cricket ball, a baseball or any other object. The motion of a projectile may be thought of as the result of two separate, simultaneously occurring components of motion. One component is along a horizontal direction without any acceleration and the other is along the vertical direction with constant acceleration due to the force of gravity. It was Galileo who first stated this independency of the horizontal and the vertical components of projectile motion in his Dialogue on the great world systems.
Horizontal range of a projectile: The horizontal distance traveled by a projectile from its initial position (x = y = 0) to the position where it passes y = 0 during its fall is called the horizontal range, R. It is the distance traveled during the time of flight Tf. Therefore, the range
R is R = (v0 cos θ0) (Tf)
R = (v0 cos θ0) (2 v0 sin θ0)/g
R = (v02 sin 2 θ0)/g
This shows that for a given projection velocity, R is maximum when sin 2θ0 is maximum, i.e., when θ0 = 450. The maximum horizontal range is, therefore
R= v02/g
Maximum height of a projectile: Maximum height that can be achieved during projectile and it is given by
Hm = (v0 sin θ0)2/2g
Who first stated the independency of horizontal and vertical components of projectile motion?
a) Isaac Newton
b) Albert Einstein
c) Galileo Galilei
d) Johannes Kepler
Answer: c) Galileo Galilei
The horizontal range is directly dependent on:
a) The initial velocity and angle of projection
b) The maximum height and time of flight
c) Gravity and maximum height
d) Only the initial velocity
Answer: a) The initial velocity and angle of projection
At what angle θ is the horizontal range of the projectile maximum?
a) 30°
b) 45°
c) 60°
d) 90°
Answer: b) 45°
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