CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If a body travels a distance of 97.52 m along a straight line in 2.54 s, then its speed in appropriate significant figures is

A
38.4 m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
38.3937 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
38.394 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
38.39 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 38.4 m/s
We know that, the speed of a body is given by,
v=dt, where d is distance travelled and t is time taken.
v=97.522.54=38.3937 m/s
We know that in multiplication or division, the final answer should have as many significant figures as in given data with minimum number of significant figures.
Here, 2.54 s have the minimum number of significant figures equal to three. Therefore, the final answer should have three significant figures.
On rounding off 38.3937 m/s to three significant figures, we get, 38.4 m/s.

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Significant Figures
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon