MCQ
The tension in the string connected between blocks is ......... $N$
- A$\frac{80}{3}$
- ✓$\frac{40}{3}$
- C$\frac{50}{7}$
- D$26$
✓
Answer
Correct option: B.
$\frac{40}{3}$
b
(b)
(b)
We have,
$f _1=20\,N , f _2=10\,N$
$m_1=2\,kg , m_2=4\,kg$
Let tension between the blocks is $T$
Here, $F_1 > F_2$
So, the net force acts in the right direction.
for block of mass $2 kg$,
$T - F _2=2 a$
$T-10=2 a \ldots \ldots \ldots(1)$
For the block of mass $4\,kg$,
$F_1-T=4 a$
$20-T=4 a .$
BY solving the above two equation the we get,
$10=6 a$
$\Rightarrow a=\frac{5}{3} m / s ^2$
Thus,
From equation $(1)$ we get,
$T=10+\frac{10}{3}$
$=\frac{40}{3} \,N$
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