Question

Q] A string of length (L) fixed at one end carries a mass m at the other. The string makes 2/π revolutions/sec around the vertical axis through the fixed end. The tension in the
string is ....​

Answer 1

${\boxed{\underline{\tt{\orange{Required \: \: answer \: \: is \: \: as \: \: follows:-}}}}}$

❀ 16ml

━━━━━━━━━━━━━━━━━━━━━━

By the diagram, It is given that

$\leadsto \: \displaystyle \sf{T sin\theta = \frac{m {v}^{2} }{r}-----(1)}$

$\leadsto\displaystyle\sf{T cos\theta=mg-----(2)}$

Where linear velocity

$\leadsto\displaystyle\sf{v = r\omega}$

and,

$\leadsto\displaystyle\sf{sin\theta = \frac{r}{l} }$

Put these values in Eq (1), we get,

$: \longrightarrow\displaystyle\sf{T × \frac{r}{l} = m { \omega}^{2} r}$

$:\longrightarrow\displaystyle\sf{T = m { \omega}^{2} l }$

We know that ω = 2πn, we have

$\therefore\displaystyle\sf{T = m(2\pi n)^2l}$

$:\longrightarrow\displaystyle\sf{T = m {(2 \pi \times \frac{2}{ \pi}) }^{2}l }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\boxed{\underline{\purple{ \huge{\rm{T = 16ml}}}}}}$

Answer 2

${ \underline{ \underline{ \huge{ \tt{ \purple{SOLUTION}}}}}}$

${ \leadsto{ \green{ \sf{angular \: frequency \: (f) = (\frac{2}{\pi}) rev \: {sec}^{ - 1}}} }}$

${ \boxed{ \boxed{ \red { \sf{angular \: velocity \: (w) = 2\pi \times f}}}}}$

${ \implies{ \blue{ \sf{ \: w = 2 \times \pi \times \frac{2}{\pi}} \: rad \: {sec}^{ - 1} }}}$

${ \implies{ \sf{ \blue{ \: w = 4 \: rad \: {sec}^{ - 1}}}}}$

${ \rightarrow{ \sf{ \: \: from \: the \: given \: figure}}} \\ \\ \: \: \: \: \: { \green{ \tt{ \: \: r = l \sin(</p><p>θ) }}}$

${ \rightarrow{ \tt{ \green { \: T \sin(θ )= \: m \: {w}^{2} r}}}}$

${ \rightarrow{ \tt{ \green{ \: </p><p>T \sin(θ) = m \: {</p><p>w}^{2} l \sin(θ) }}}}$

${ \rightarrow{ \tt{ \green{ \: </p><p>T \: = \: m {(4)}^{2} \: l}}}}$

${ \rightarrow{ \boxed{ \boxed{ \tt{ \red{ \: T \: = \: 16 \: m \: l}}}}}}$