| ⇦ | 
| ⇨ |
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: - The half-life period of a radio-active element X is same as the mean life time
- Two charged particles have charges and masses in the ratio 2:3 and 1:4 respectively.
- When you make ice cubes, the entropy of water
- In Moseley’s law √ν=a(z-b), the values of the screening constant for K-series
- For a thermocouple, the neutral temperature is 270⁰C when its cold junction is at 20⁰C.
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
- The half-life period of a radio-active element X is same as the mean life time
- Two charged particles have charges and masses in the ratio 2:3 and 1:4 respectively.
- When you make ice cubes, the entropy of water
- In Moseley’s law √ν=a(z-b), the values of the screening constant for K-series
- For a thermocouple, the neutral temperature is 270⁰C when its cold junction is at 20⁰C.
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