Question 14 of 210

Three blocks $A,B$ and $C$ are pulled on a horizontal smooth surface by a force of $80N$ as shown in figure. The tensions ${T}_{1}\text{and}{T}_{2}$ in the string are respectively:

A$40N,64N$B$60N,80N$C$88N,96N$D$80N,100N$

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Answer: (A) $40N,64N$

Since, the length of the string will remain same. Hence, all blocks will have the same acceleration.
${a}_{A}={a}_{B}={a}_{C}=\dfrac{F}{5+3+2}=\dfrac{80}{10}=8m{s}^{-2}$

Therefore, ${T}_{1}=5\times 8=40N$.
Now,

${T}_{2}-{T}_{1}=3\times 8 \Rightarrow {T}_{2}=64N$