SATHEE: Rotational Motion Question 207

Rotational Motion Question 207

Question: The moment of inertia of a uniform semicircular wire of mass m and radius r, about an axis passing through its centre of mass and perpendicular to its plane is $m r^{2} (1 - \frac{k}{π^{2}})$ then find the value of k.

Options:

A) 2

B) 3

C) 4

D) 5

Show Answer

Answer:

Correct Answer: C

Solution:

[c] Moment of inertia about z-axis,

$I_{z} = m r^{2}$

(about centre of mass) Applying parallel axes theorem,

$I_{z} = I_{c m} + m k^{2}$

$I_{c m} = I_{z} - m {(\frac{2}{π} r)}^{2}$

$= m r^{2} - \frac{m 4 r^{2}}{π^{2}} = m r^{2} (1 - \frac{4}{π^{2}})$

i.e., k=4

Rotational Motion Question 207

Question: The moment of inertia of a uniform semicircular wire of mass m and radius r, about an axis passing through its centre of mass and perpendicular to its plane is $m r^{2} (1 - \frac{k}{π^{2}})$ then find the value of k.

Options:

Answer:

Solution:

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Rotational Motion Question 207

Question: The moment of inertia of a uniform semicircular wire of mass m and radius r, about an axis passing through its centre of mass and perpendicular to its plane is mr2(1−kπ2) then find the value of k.

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: The moment of inertia of a uniform semicircular wire of mass m and radius r, about an axis passing through its centre of mass and perpendicular to its plane is $m r^{2} (1 - \frac{k}{π^{2}})$ then find the value of k.