A stone of mass 1 kg tied to the end of a string of length 1 m , is whirled in a horizontal circle with a uniform angular velocity 2 rad s- 1 . The tension of the string is (in newton)

Question

A stone of mass 1 kg tied to the end of a string of length 1 m , is whirled in a horizontal circle with a uniform angular velocity 2 rad s- 1 . The tension of the string is (in newton) (A) 2 (B) (1/3) (C) 4 (D) (1/4)

Tardigrade Admin · Accepted Answer

Tension in string will provide required centripetal force. The tension of the string, T=mrω 2 =1× 1× (. 2 .)2=4 N

Q. A stone of mass $1 k g$ tied to the end of a string of length $1 m$ , is whirled in a horizontal circle with a uniform angular velocity $2 r a d s^{- 1}$ . The tension of the string is (in newton)

$2$

$\frac{1}{3}$

$4$

$\frac{1}{4}$

Tension in string will provide required centripetal force.
The tension of the string,
$T = m r ω^{2}$
$= 1 \times 1 \times (2)^{2} = 4 N$

Q. A stone of mass 1kg tied to the end of a string of length 1m , is whirled in a horizontal circle with a uniform angular velocity 2rads−1 . The tension of the string is (in newton)

2

31​

4

41​

Tension in string will provide required centripetal force. The tension of the string, T=mrω2 =1×1×(2)2=4N

Q. A stone of mass $1 k g$ tied to the end of a string of length $1 m$ , is whirled in a horizontal circle with a uniform angular velocity $2 r a d s^{- 1}$ . The tension of the string is (in newton)

$2$

$\frac{1}{3}$

$4$

$\frac{1}{4}$

Tension in string will provide required centripetal force.
The tension of the string,
$T = m r ω^{2}$
$= 1 \times 1 \times (2)^{2} = 4 N$