Question

A string of length 1 fixed at one end carries a mass m at the other end. The string makes $\frac{2}{\pi }rev/s$around the horizontal axis through the fixed end as shown in the figure, the tension in the string is

• A. $16ml$
• B. $6ml$
• C. $5ml$
• D. $3ml$

Answer 1

Answer: (A)

$16ml$

$Tsin\theta =\frac{m{v}^2}{r}$
$Tcos\theta =mg$
$v=r\omega$ and $sin\theta =\frac{r}{r}$
Putting these values in Eq (i), we get
$T\times \frac{r}{r}=m{\omega }^{2}r$
$T=m{\omega }^{2}l$
$\omega =2\pi n$
$T=m(2\pi n{)}^{2}l$
$T=m{(2\pi \times \frac{2}{\pi })}^{2}l$
$T=16ml$