A stone ties to a rope is rotated in a vertical circle with uniform speed. If the difference between the maximum and minimum tension in the rope is $20 N$ , mass of the stone in $k g$ is:
( $g = 10 {m s}^{- 2}$ )

Question

A stone ties to a rope is rotated in a vertical circle with uniform speed. If the difference between the maximum and minimum tension in the rope is

20 N

, mass of the stone in

k g

is:
(

g = 10 {m s}^{- 2}

)

Toppr Admin · Accepted Answer

Let the mass of the stone be m- and the angular speed with which it rotates be -x3C9-Thus at the bottom most point-xA0-Tb-x2212-mg-m-x3C9-2lAnd at the top most point-Tt-mg-m-x3C9-2l-x27F9-Tb-x2212-Tt-2mg-20-x27F9-mg-10-x27F9-m-1-xA0-kg

A stone ties to a rope is rotated in a vertical circle with uniform speed. If the difference between the maximum and minimum tension in the rope is $20 N$ , mass of the stone in $k g$ is:
( $g = 10 {m s}^{- 2}$ )
$0.75$
$1.0$
$1.5$
$0.5$

Let the mass of the stone be $m$ , and the angular speed with which it rotates be $ω$ .
Thus at the bottom most point,
$T_{b} - m g = m ω^{2} l$
And at the top most point,
$T_{t} + m g = m ω^{2} l$
$⟹ T_{b} - T_{t} = 2 m g = 20$
$⟹ m g = 10$
$⟹ m = 1 k g$

A stone ties to a rope is rotated in a vertical circle with uniform speed. If the difference between the maximum and minimum tension in the rope is 20 N, mass of the stone in kg is: (g=10ms−2)0.751.01.50.5

Let the mass of the stone be m, and the angular speed with which it rotates be ω.Thus at the bottom most point, Tb−mg=mω2lAnd at the top most point,Tt+mg=mω2l⟹Tb−Tt=2mg=20⟹mg=10⟹m=1 kg

A stone ties to a rope is rotated in a vertical circle with uniform speed. If the difference between the maximum and minimum tension in the rope is $20 N$ , mass of the stone in $k g$ is:
( $g = 10 {m s}^{- 2}$ )
$0.75$
$1.0$
$1.5$
$0.5$

Let the mass of the stone be $m$ , and the angular speed with which it rotates be $ω$ .
Thus at the bottom most point,
$T_{b} - m g = m ω^{2} l$
And at the top most point,
$T_{t} + m g = m ω^{2} l$
$⟹ T_{b} - T_{t} = 2 m g = 20$
$⟹ m g = 10$
$⟹ m = 1 k g$