SATHEE: Rotational Motion Question 156

Rotational Motion Question 156

Question: ABC is a uniform wire. If centre of mass of wire lies vertically below point A, then $\frac{B C}{A B}$ Disclose to:

Options:

A)1.85

B) 1.5

C)1.37

D) 3

Show Answer

Answer:

Correct Answer: C

Solution:

[c] Centre of mass

$x_{c m} = \frac{x}{2} \frac{(ρ x) (\frac{x}{2}) \frac{1}{2} + ρ y (\frac{y}{2})}{ρ (x + y)} \Rightarrow \frac{1}{2} + \frac{y}{x} = \frac{y^{2}}{x^{2}}$

$∴ \frac{B C}{A B} = \frac{y}{2}$

$= \frac{1 + \sqrt{3}}{2} = 1.37$

Rotational Motion Question 156

Question: ABC is a uniform wire. If centre of mass of wire lies vertically below point A, then $\frac{B C}{A B}$ Disclose to:

Options:

Answer:

Solution:

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Rotational Motion Question 156

Question: ABC is a uniform wire. If centre of mass of wire lies vertically below point A, then BCAB Disclose to:

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: ABC is a uniform wire. If centre of mass of wire lies vertically below point A, then $\frac{B C}{A B}$ Disclose to: