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Two masses m(1) and m(2) are attached to...

Two masses `m_(1)` and `m_(2)` are attached to a spring balance S as shown in Figure. If `m_(1)gt m_(2)` then the reading of spring balance will be

A

`(m_(1)-m_(2))`

B

`(m_(1)+m_(2))`

C

`(2m_(1)m_(2))/(m_(1)+m_(2))`

D

`(m_(1)m_(2))/(m_(1)+m_(2))`

Text Solution

Verified by Experts

The correct Answer is:
C
`F_("net")=ma, " From FBD of "m_(1), m_(1)g-T_(1)=m_(1)a`
From FBD of `m_(2), T_(2)-m_(2)g=m_(2)a`
take `T_(2)=T_(2)`, solving the above eq.s, we get .a.
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