wiz-icon
MyQuestionIcon
MyQuestionIcon
4
You visited us 4 times! Enjoying our articles? Unlock Full Access!
Question

A metallic rod of length L and mass M is moving under the action of two unequal forces F1 and F2 (directed opposite to each other) acting at its ends along its length. Ignore gravity and any external magnetic field. If specific charge of electrons is (e/m), then the potential difference between the ends of the rod is steady state must be:

A
|F1F2| mL/eM
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(F1F2) mL/eM
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
[mL/eM] ln[F1/F2]
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D None

For the entire rod, assuming an acceleration a

F1F2=M×a(1)

For the same rod (therefore same acceleration)

Consider only length x

F1T=(M×xL)a.

Plugging in a=(F1F2M) from equation 1,

F1T=(M×xL)×(F1F2M)

T=F1(F1F2)×xL


Hence, The Option D is correct None


flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Electromagnetic Induction and Electric Generators
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon