One end of a string of length 1.0 m is tied to a body of mass 0.5 kg. It is whirled in a vertical circle with angular frequency 4 rad/s. The tension in the string, when the body is at the lower most point of its motion, will be equal to (Take g=10 m/s2 )

Question

One end of a string of length 1.0 m is tied to a body of mass 0.5 kg. It is whirled in a vertical circle with angular frequency 4 rad/s. The tension in the string, when the body is at the lower most point of its motion, will be equal to (Take g=10 m/s2 ) (A) 3 N (B) 5 N (C) 8 N (D) 13 N

Tardigrade Admin · Accepted Answer

Tension T=m ω2 r+ m g =0.5 ×(4)2 × 1+0.5 × 10 =13 N

Q. One end of a string of length $1.0 m$ is tied to a body of mass $0.5 k g$ . It is whirled in a vertical circle with angular frequency $4 r a d / s$ . The tension in the string, when the body is at the lower most point of its motion, will be equal to (Take $g = 10 m / s^{2}$ )

3 N

5 N

8 N

13 N

Tension $T = m ω^{2} r + m g$
$= 0.5 \times (4)^{2} \times 1 + 0.5 \times 10$ $= 13 N$

Q. One end of a string of length 1.0m is tied to a body of mass 0.5kg. It is whirled in a vertical circle with angular frequency 4rad/s. The tension in the string, when the body is at the lower most point of its motion, will be equal to (Take g=10m/s2 )

3 N

5 N

8 N

13 N

Tension T=mω2r+mg =0.5×(4)2×1+0.5×10 =13N

Q. One end of a string of length $1.0 m$ is tied to a body of mass $0.5 k g$ . It is whirled in a vertical circle with angular frequency $4 r a d / s$ . The tension in the string, when the body is at the lower most point of its motion, will be equal to (Take $g = 10 m / s^{2}$ )

Tension $T = m ω^{2} r + m g$
$= 0.5 \times (4)^{2} \times 1 + 0.5 \times 10$ $= 13 N$