A string of length l fixed at one end carries a mass m at the other end. The string makes (2/π )rev/s around the horizontal axis through the fixed end as shown in the figure, the tension in the string is

Question

A string of length l fixed at one end carries a mass m at the other end. The string makes (2/π )rev/s around the horizontal axis through the fixed end as shown in the figure, the tension in the string is (A)  16 ml  (B)  6 ml  (C)  5 ml  (D)  3 ml

Tardigrade Admin · Accepted Answer

T sin θ =(mv2/r) T cos θ =mg v=rω and sin θ =(r/l) Putting these values in Eq (i), we get T× (r/l)=mω 2r T=mω 2l ω =2π n T=m(2π n)2l T=m( 2π × (2/π ) )2l T=16 ml

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

$16 m l$

$6 m l$

$5 m l$

$3 m l$

$T sin θ = \frac{m v ^{2}}{r}$
$T cos θ = m g$
$v = r ω$
and $sin θ = \frac{r}{l}$
Putting these values in Eq (i), we get
$T \times \frac{r}{l} = m ω^{2} r$
$T = m ω^{2} l$
$ω = 2 πn$
$T = m (2 πn)^{2} l$
$T = m (2 π \times \frac{2}{π})^{2} l$
$T = 16 m l$

Q. A string of length l fixed at one end carries a mass m at the other end. The string makes π2​rev/s around the horizontal axis through the fixed end as shown in the figure, the tension in the string is

16ml

6ml

5ml

3ml