SATHEE: Laws of Motion - Result Question 62

Laws of Motion - Result Question 62

62. A body of mass $0.4 k g$ is whirled in a vertical circle making $2 r e v / s e c$ . If the radius of the circle is $1.2 m$ , then tension in the string when the body is at the top of the circle, is [1999]

(a) $41.56 N$

(b) $89.86 N$

(d) $115.86 N$

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Answer:

Correct Answer: 62. (a)

Solution:

(a) Given : Mass $(m) = 0.4 k g$ Its frequency $(n) = 2 r e v / s e c$

Radius $(r) = 1.2 m$ . We know that linear velocity of the body $(v) = ω r = (2 π n) r$ $= 2 \times 3.14 \times 2 \times 1.2 = 15.08 m / s$ .

Therefore, tension in the string when the body is at the top of the circle (T)

$\begin{aligned} = \frac{m v^{2}}{r} - m g = \frac{0.4 \times (15.08)^{2}}{2} - (0.4 \times 9.8) \\ = 45.78 - 3.92 = 41.56 N \end{aligned}$

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Laws of Motion - Result Question 62

62. A body of mass 0.4kg is whirled in a vertical circle making 2rev/sec. If the radius of the circle is 1.2m, then tension in the string when the body is at the top of the circle, is [1999]

Answer:

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62. A body of mass $0.4 k g$ is whirled in a vertical circle making $2 r e v / s e c$ . If the radius of the circle is $1.2 m$ , then tension in the string when the body is at the top of the circle, is [1999]