Two masses are connected by a string which passes over a pulley acceleration upward at a rate `A` shown. If `a_(1)` and `a_(2)` be the accelerations of bodies 1 and 2 respectively then,

Options

A. Relation (ii) and (iii) always follow.
B. Relations (ii) and (iv) always follow.
C. Only relation (i) always follows.
D. Only relation (iv) always follows.

Answer

D

Explanation

`a_(1)=((m_(2)-m_(1))g)/((m_(1)+m_(2)))`
and `a_(2)=((m_(2)-M_(1))g)/(m_1)`
Hence, `(a_1)/(a_2)=(m_1)/((m_(1)+m_(2))) = (1)/((1+(m_2)/(m_1))) implies (a_1)/(a_2) lt1`
As `T_(1) = (2m_(1)m_(2)g)/((m_(1)+m_(2)))` and `T_(2)=m_(2)g`
Hence, `(T_1)/(T_2) = (2m_(1))/((m_(1)+m_(2))) = (2)/((1+(m_2)/(m_1)))`
Therefore, `T_(1)//T_(2)` will depend upon the value of `m_(1)` and `m_(2)`
`N_(1)=2T_(1), N_(2)=2T_(2)`
So the relation of `N_(1)//N_(2)` will be same as `T_(1)//T_(2)`.