Question

Two lead balls of masses m and 5m having radii R and 2R are separated by 12R. If they attract each other by gravitational force, the distance covered by small sphere before they touch each other is:

A
10 R
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B
7.5 R
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
9 R
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D
2.5 R
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

The correct option is B 7.5 REffective distance =9RDistance travelled by smaller mass=xx=(m2m1+m2)(9R)The gravitational force between masses m and 5m is given by:FG=G(m)(5m)(12R−x)2=5Gm2(12R−x)2The acceleration of mass m is given by: a(m)=FGm=5Gm(12R−x)2 The acceleration of mass 5m is given by:a(5m)=FG5m=Gm(12R−x)2Distance travelled by mass m is given by:x=12a(m)t2x=125Gm(12R−x)2t2 ................................(1)The distance covered by mass 5m is given by:9R−x=12a(5m)t29R−x=12Gm(12R−x)2t2 .................... (2)Dividing 1 by 2:x9R−x=125Gm(12R−x)2t212Gm(12R−x)2t2=5∴x=45R−5x6x=45Rx=7.5R

Suggest Corrections
0
Related Videos
Kepler's Law
PHYSICS
Watch in App
Explore more