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) : $\left[ m _{1}=50 kg , m _{2}=80 kg \right.$ and $F =1000 N ]$

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