| ⇦ | 
| ⇨ |
What will be the ratio of the distance moved by a freely falling body from rest in 4th and 5th seconds of journey?
Options
(a) 4 : 5
(b) 7 : 9
(c) 16 : 25
(d) 1 : 1
Correct Answer:
7 : 9
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - A uniform rod of length l is placed with one end in contact with the horizontal
- A force F=ai+3j+6k is acting at a point r=2i-6j-12k. The value of α for
- Two bodies A(of mass 1kg) and B(of mass 3 kg) are dropped from heights of 16m and 25m
- A hemispherical bowl of radius R is rotating about its axis of symmetry
- A magnet of length 10 cm and magnetic moment 1 A-m² is placed
Question Type: Apply
(15)
Difficulty Level: Easy
(1008)
Topics: Motion in Plane
(93)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A uniform rod of length l is placed with one end in contact with the horizontal
- A force F=ai+3j+6k is acting at a point r=2i-6j-12k. The value of α for
- Two bodies A(of mass 1kg) and B(of mass 3 kg) are dropped from heights of 16m and 25m
- A hemispherical bowl of radius R is rotating about its axis of symmetry
- A magnet of length 10 cm and magnetic moment 1 A-m² is placed
Question Type: Apply (15)
Difficulty Level: Easy (1008)
Topics: Motion in Plane (93)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
By using,
Snth=u+a/2 (2n-1)=0+a/2 (2×4-1)=7a/2
Again,s=u+a/2 (2n-1)=0+a/2 (2×5-1)=9a/2
Now,S1:S2=7a/2:9a/2=7:9