The breaking strength of a string is 55 kg wt- The maximum permissible speed of a stone of mass 5 kg which is revolved in a vertical circle of radius 4 m with the of this string is -g - 10m-s-2-10 m-sec15 m-sec20 m-sec25 m-sec

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Toppr Admin · Accepted Answer

Tmax-55-xA0-kg-xA0-wt-xA0-xA0- -xA0- -xA0- -xA0- -55-xA0-N-T is maximum at bottom point-where-xA0-T-x2212-mg-mv2R

The breaking strength of a string is $55 k g$ wt. The maximum permissible speed of a stone of mass $5 k g$ which is revolved in a vertical circle of radius $4 m$ with the help of this string is $(g = 10 m / s^{2})$ :

$10 m / s e c$
$15 m / s e c$
$20 m / s e c$
$25 m / s e c$

$T_{m a x} = 55 k g w t .$
$= 55 N .$
T is maximum at bottom point.
where,
$T - m g = \frac{m v^{2}}{R}$

The breaking strength of a string is 55 kg wt. The maximum permissible speed of a stone of mass 5 kg which is revolved in a vertical circle of radius 4 m with the help of this string is (g=10m/s2):10 m/sec15 m/sec20 m/sec25 m/sec

Tmax=55 kg wt. =55 N.T is maximum at bottom point.where, T−mg=mv2R

The breaking strength of a string is $55 k g$ wt. The maximum permissible speed of a stone of mass $5 k g$ which is revolved in a vertical circle of radius $4 m$ with the help of this string is $(g = 10 m / s^{2})$ :

$10 m / s e c$
$15 m / s e c$
$20 m / s e c$
$25 m / s e c$

$T_{m a x} = 55 k g w t .$
$= 55 N .$
T is maximum at bottom point.
where,
$T - m g = \frac{m v^{2}}{R}$