SATHEE: Work Power and Energy 4 Question 4

Work Power and Energy 4 Question 4

5. A stone tied to a string of length $L$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position, and has a speed $u$ . The magnitude of the change in its velocity as it reaches a position, where the string is horizontal, is

(a) $\sqrt{u^{2} - 2 g L}$

(b) $\sqrt{2 g L}$

(d) $\sqrt{2 (u^{2} - g L)}$

(1998, 2M)

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Answer:

Correct Answer: 5. (d)

Solution:

From energy conservation, $v^{2} = u^{2} - 2 g L \dots (i)$

Now, since the two velocity vectors shown in figure are mutually perpendicular, hence the magnitude of change of velocity will be given by $| Δ v | = \sqrt{u^{2} + v^{2}}$

Substituting value of $v^{2}$ from Eq.(i), we get

$| Δ v | = \sqrt{u^{2} + u^{2} - 2 g L} = \sqrt{2 (u^{2} - g L)}$

Work Power and Energy 4 Question 4

Answer:

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Work Power and Energy 4 Question 4

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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