SATHEE: Kinematics Question 766

Kinematics Question 766

Question: A cyclist goes round a circular path of circumference 34.3 m in $\sqrt{22}$ sec. the angle made by him, with the vertical, will be [MH CET 2000]

Options:

A) $45^{o}$

B) $40^{o}$

C) $42^{o}$

D) $48^{o}$

Show Answer

Answer:

Correct Answer: A

Solution:

$2 π r = 34.3$

therefore $r = \frac{34.3}{2 π}$ and $v = \frac{2 π r}{T} = \frac{2 π r}{\sqrt{22}}$ Angle of binding $θ = \tan^{- 1} (\frac{v^{2}}{r g}) = 45^{\circ}$

Kinematics Question 766

Question: A cyclist goes round a circular path of circumference 34.3 m in $\sqrt{22}$ sec. the angle made by him, with the vertical, will be [MH CET 2000]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Kinematics Question 766

Question: A cyclist goes round a circular path of circumference 34.3 m in 22 sec. the angle made by him, with the vertical, will be [MH CET 2000]

Options:

Answer:

Solution:

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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Question: A cyclist goes round a circular path of circumference 34.3 m in $\sqrt{22}$ sec. the angle made by him, with the vertical, will be [MH CET 2000]