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4-1.Newton's Laws of Motion
medium
Consider the situation shown in figure. All the surfaces are smooth. The tension in the string connected to $2\,m$ is
A$\frac{{mg}}{3}$
B$\frac{{4mg}}{3}$
C$\frac{{2mg}}{3}$
D$mg$
Solution
${a_{A}=\frac{a_{B}+0}{2}}$
$2a_A=a_B$
$\mathrm{ma}_{\mathrm{B}}=\mathrm{m}\left(2 \mathrm{a}_{\mathrm{A}}\right)$
$\mathrm{mg}-\mathrm{T}=\mathrm{ma}_{\mathrm{B}}$ $…(1)$
$2 \mathrm{T}=2 \mathrm{m} \cdot \mathrm{a}_{\mathrm{A}}$ $…(2)$
From $2(\mathrm{i})+(\text { ii }),$ we get
$2 \mathrm{mg}=6 \mathrm{ma}_{\mathrm{A}}$
$\mathrm{T}^{\prime}=2 \mathrm{ma}_{\mathrm{A}}=\frac{2 \mathrm{mg}}{3}$
$2a_A=a_B$
$\mathrm{ma}_{\mathrm{B}}=\mathrm{m}\left(2 \mathrm{a}_{\mathrm{A}}\right)$
$\mathrm{mg}-\mathrm{T}=\mathrm{ma}_{\mathrm{B}}$ $…(1)$
$2 \mathrm{T}=2 \mathrm{m} \cdot \mathrm{a}_{\mathrm{A}}$ $…(2)$
From $2(\mathrm{i})+(\text { ii }),$ we get
$2 \mathrm{mg}=6 \mathrm{ma}_{\mathrm{A}}$
$\mathrm{T}^{\prime}=2 \mathrm{ma}_{\mathrm{A}}=\frac{2 \mathrm{mg}}{3}$
Standard 11
Physics
