MHT CET · Physics · Laws of Motion
Two blocks \(A\) and \(B\) each of \(20 \mathrm{~kg}\) lying on a frictionless table are connected by a light string. The system is pulled horizontally with an acceleration of \(2 \mathrm{~ms}^{-2}\) by a force \(F\) on \(B\). The tension in the string will be
- A \(10 \mathrm{~N}\)
- B \(40 \mathrm{~N}\)
- C \(100 \mathrm{~N}\)
- D \(120 \mathrm{~N}\)
Answer & Solution
Correct Answer
(B) \(40 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
The given situation can be shown as
\(\text { The tension } T=M_{A} \times a\)
here,
\(\begin{aligned}
M_{A} &=20 \mathrm{~kg} \text { and } a=2 \mathrm{~ms}^{-2} \\
T &=20 \times 2=40 \mathrm{~N}
\end{aligned}\)
\(\text { The tension } T=M_{A} \times a\)
here,
\(\begin{aligned}
M_{A} &=20 \mathrm{~kg} \text { and } a=2 \mathrm{~ms}^{-2} \\
T &=20 \times 2=40 \mathrm{~N}
\end{aligned}\)
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