Question

The ratio of tensions in the string connected to the block of mass $m_2$ in figure(i) and figure(ii) respectively is (friction is absent everywhere) : $[m_1=50kg,m_2=80kg$ and $F=1000N]$

• A. 7 :2
• B. 2 : 7
• C. 3 :4
• D. 4 :3

Answer 1

Answer: (C)

In figure (i): Let tension in the string is $T_1$
$F-T_1=m_{1}a$
$\Rightarrow 1000-T_1=50a\dots$(1)
$T_1-m_{2}g=m_{2}a$
$\Rightarrow T_1-80g=80a\dots$(2)
From (1) and $(2)T_1=\frac{12000}{13}N$
In figure (ii): Let tension in the string is $T_{2^{\ast }}$
$F+m_{1}g-T_2=m_{1}a$
$\Rightarrow 1000+50g-T_2=50a\dots$(3)
$T_2-80g=80a\dots$(4)
From (3) and (4)
$T_2=\frac{16000}{13}N$
$\Rightarrow \frac{T_1}{T_2}=\frac{12000}{16000}$
$=\frac{3}{4}$