SATHEE: Rotational Motion Question 174

Rotational Motion Question 174

Question: If the linear density (mass per unit length) of a rod of length 3m is proportional to x, where x is the distance from one end of the rod, the distance of the centre of gravity of the rod from this end is

Options:

A) $2.5$ m

B) $1$ m

C) $1.5$ m

D) $2$ m

Show Answer

Answer:

Correct Answer: D

Solution:

[d] Consider an element of length $d x$ at a distance $x$ from end $A$ .

Here, mass per unit length $λ$ of rod

$λ \propto x \Rightarrow λ = k x$

$∴ d m = λ d x = k x d x$

Position of centre of gravity of rod from end A.

$x_{C G} = \frac{\int_{0}^{L} x d m}{\int_{0}^{L} d m}$

$∴ x_{C G} = \frac{\int_{0}^{3} x (k x d x)}{\int_{0}^{3} k x d x} = \frac{{\frac{x^{3}}{3}}^{3}}{{\frac{x^{2}}{2}}_{0}^{3}} = \frac{\frac{(3)^{3}}{3}}{\frac{(3)^{3}}{2}} = 2 m$

Rotational Motion Question 174

Question: If the linear density (mass per unit length) of a rod of length 3m is proportional to x, where x is the distance from one end of the rod, the distance of the centre of gravity of the rod from this end is

Options:

Answer:

Solution:

Engineering Preparation

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NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Rotational Motion Question 174

Question: If the linear density (mass per unit length) of a rod of length 3m is proportional to x, where x is the distance from one end of the rod, the distance of the centre of gravity of the rod from this end is

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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