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Rotational Motion

269593 The centre of mass of a non uniform rod of length $L$ whose mass per unit length $λ = \frac{K x^{2}}{L}$ , Where $k$ is a constant and $x$ is the distance from one end is :

1

\frac{3 L}{4}

2

\frac{L}{8}

3

\frac{K}{L}

4

\frac{3 K}{L}

Explanation:

$d m = \frac{k x^{2}}{L} d x; x_{c m} = \frac{\int_{\frac{L}{L} x d m}^{L}}{\int_{0}^{L} d m} = \frac{\int_{0}^{L} \frac{k x^{3}}{L} d x}{\int_{0}^{L} \frac{k x^{2}}{L} d x} = \frac{G^{4} L^{4}}{◻^{3} ◻} = \frac{3 L}{4}$

Rotational Motion

269594 A rope of length $30 cm$ is on a horizontal table with maximum length hanging from edge $A$ of the table. The coefficient of friction between the rope and table is 0.5 . The distance of centre of mass of the rope from $A$ is

1

\frac{5 \sqrt{15}}{3} c

2

\frac{5 \sqrt{17}}{3} cm

3

\frac{5 \sqrt{19}}{3} cm

4

\frac{7 \sqrt{17}}{3} cm

Explanation:

Fractional length hanging,
$\frac{l}{L} = \frac{μ}{1 + μ} \Rightarrow \frac{l}{30} = \frac{0.5}{1 + 0.5} \Rightarrow l = 10 cm$
let ' $ρ$ ' be the mass per unit length. The coordinates of $20 ρ$ and $10 ρ$ are $(10, 0)$ and $(0, 5)$ respectively from ' $A$ '.
Distance of C.M from A, $r_{cm} = \sqrt{X_{c m}^{2} + y_{c m}^{2}}$

Rotational Motion

269595 As shown in figure from a uniform rectangular sheet a triangular sheet is removed from one edge. The shift of centre of mass is

1

4.2 cm

2

- 4.2 cm

3

6.67 cm

4

- 6.67 cm

Explanation:

shift $= \frac{- mass of removed part \times d}{Mass of remaining part}$ Here $d = 20 cm$

Rotational Motion

269596 A circular disc of radius $R$ is removed from a bigger circular disc of radius $2 R$ such that the circumference of the discs coincide . The centre of mass of the new disc is $α R$ from the centre of the bigger disc. The value of $α$ is

1

1 / 3

2

1 / 2

3

1 / 6

4

1 / 4

Explanation:

Shift of centre of mass $x = \frac{r^{2} a}{R^{2} - r^{2}}$
Where $r =$ radius of removed disc
$R =$ radius of original disc
$a =$ distance between the centres
Note:In this question shift must be $\propto R$ for exact approach to the solution

Rotational Motion

269593 The centre of mass of a non uniform rod of length $L$ whose mass per unit length $λ = \frac{K x^{2}}{L}$ , Where $k$ is a constant and $x$ is the distance from one end is :

1

\frac{3 L}{4}

2

\frac{L}{8}

3

\frac{K}{L}

4

\frac{3 K}{L}

Explanation:

$d m = \frac{k x^{2}}{L} d x; x_{c m} = \frac{\int_{\frac{L}{L} x d m}^{L}}{\int_{0}^{L} d m} = \frac{\int_{0}^{L} \frac{k x^{3}}{L} d x}{\int_{0}^{L} \frac{k x^{2}}{L} d x} = \frac{G^{4} L^{4}}{◻^{3} ◻} = \frac{3 L}{4}$

Rotational Motion

269594 A rope of length $30 cm$ is on a horizontal table with maximum length hanging from edge $A$ of the table. The coefficient of friction between the rope and table is 0.5 . The distance of centre of mass of the rope from $A$ is

1

\frac{5 \sqrt{15}}{3} c

2

\frac{5 \sqrt{17}}{3} cm

3

\frac{5 \sqrt{19}}{3} cm

4

\frac{7 \sqrt{17}}{3} cm

Explanation:

Fractional length hanging,
$\frac{l}{L} = \frac{μ}{1 + μ} \Rightarrow \frac{l}{30} = \frac{0.5}{1 + 0.5} \Rightarrow l = 10 cm$
let ' $ρ$ ' be the mass per unit length. The coordinates of $20 ρ$ and $10 ρ$ are $(10, 0)$ and $(0, 5)$ respectively from ' $A$ '.
Distance of C.M from A, $r_{cm} = \sqrt{X_{c m}^{2} + y_{c m}^{2}}$

Rotational Motion

269595 As shown in figure from a uniform rectangular sheet a triangular sheet is removed from one edge. The shift of centre of mass is

1

4.2 cm

2

- 4.2 cm

3

6.67 cm

4

- 6.67 cm

Explanation:

shift $= \frac{- mass of removed part \times d}{Mass of remaining part}$ Here $d = 20 cm$

Rotational Motion

269596 A circular disc of radius $R$ is removed from a bigger circular disc of radius $2 R$ such that the circumference of the discs coincide . The centre of mass of the new disc is $α R$ from the centre of the bigger disc. The value of $α$ is

1

1 / 3

2

1 / 2

3

1 / 6

4

1 / 4

Explanation:

Shift of centre of mass $x = \frac{r^{2} a}{R^{2} - r^{2}}$
Where $r =$ radius of removed disc
$R =$ radius of original disc
$a =$ distance between the centres
Note:In this question shift must be $\propto R$ for exact approach to the solution

Rotational Motion

269593 The centre of mass of a non uniform rod of length $L$ whose mass per unit length $λ = \frac{K x^{2}}{L}$ , Where $k$ is a constant and $x$ is the distance from one end is :

1

\frac{3 L}{4}

2

\frac{L}{8}

3

\frac{K}{L}

4

\frac{3 K}{L}

Explanation:

$d m = \frac{k x^{2}}{L} d x; x_{c m} = \frac{\int_{\frac{L}{L} x d m}^{L}}{\int_{0}^{L} d m} = \frac{\int_{0}^{L} \frac{k x^{3}}{L} d x}{\int_{0}^{L} \frac{k x^{2}}{L} d x} = \frac{G^{4} L^{4}}{◻^{3} ◻} = \frac{3 L}{4}$

Rotational Motion

269594 A rope of length $30 cm$ is on a horizontal table with maximum length hanging from edge $A$ of the table. The coefficient of friction between the rope and table is 0.5 . The distance of centre of mass of the rope from $A$ is

1

\frac{5 \sqrt{15}}{3} c

2

\frac{5 \sqrt{17}}{3} cm

3

\frac{5 \sqrt{19}}{3} cm

4

\frac{7 \sqrt{17}}{3} cm

Explanation:

Fractional length hanging,
$\frac{l}{L} = \frac{μ}{1 + μ} \Rightarrow \frac{l}{30} = \frac{0.5}{1 + 0.5} \Rightarrow l = 10 cm$
let ' $ρ$ ' be the mass per unit length. The coordinates of $20 ρ$ and $10 ρ$ are $(10, 0)$ and $(0, 5)$ respectively from ' $A$ '.
Distance of C.M from A, $r_{cm} = \sqrt{X_{c m}^{2} + y_{c m}^{2}}$

Rotational Motion

269595 As shown in figure from a uniform rectangular sheet a triangular sheet is removed from one edge. The shift of centre of mass is

1

4.2 cm

2

- 4.2 cm

3

6.67 cm

4

- 6.67 cm

Explanation:

shift $= \frac{- mass of removed part \times d}{Mass of remaining part}$ Here $d = 20 cm$

Rotational Motion

269596 A circular disc of radius $R$ is removed from a bigger circular disc of radius $2 R$ such that the circumference of the discs coincide . The centre of mass of the new disc is $α R$ from the centre of the bigger disc. The value of $α$ is

1

1 / 3

2

1 / 2

3

1 / 6

4

1 / 4

Explanation:

Shift of centre of mass $x = \frac{r^{2} a}{R^{2} - r^{2}}$
Where $r =$ radius of removed disc
$R =$ radius of original disc
$a =$ distance between the centres
Note:In this question shift must be $\propto R$ for exact approach to the solution

Rotational Motion

269593 The centre of mass of a non uniform rod of length $L$ whose mass per unit length $λ = \frac{K x^{2}}{L}$ , Where $k$ is a constant and $x$ is the distance from one end is :

1

\frac{3 L}{4}

2

\frac{L}{8}

3

\frac{K}{L}

4

\frac{3 K}{L}

Explanation:

$d m = \frac{k x^{2}}{L} d x; x_{c m} = \frac{\int_{\frac{L}{L} x d m}^{L}}{\int_{0}^{L} d m} = \frac{\int_{0}^{L} \frac{k x^{3}}{L} d x}{\int_{0}^{L} \frac{k x^{2}}{L} d x} = \frac{G^{4} L^{4}}{◻^{3} ◻} = \frac{3 L}{4}$

Rotational Motion

269594 A rope of length $30 cm$ is on a horizontal table with maximum length hanging from edge $A$ of the table. The coefficient of friction between the rope and table is 0.5 . The distance of centre of mass of the rope from $A$ is

1

\frac{5 \sqrt{15}}{3} c

2

\frac{5 \sqrt{17}}{3} cm

3

\frac{5 \sqrt{19}}{3} cm

4

\frac{7 \sqrt{17}}{3} cm

Explanation:

Fractional length hanging,
$\frac{l}{L} = \frac{μ}{1 + μ} \Rightarrow \frac{l}{30} = \frac{0.5}{1 + 0.5} \Rightarrow l = 10 cm$
let ' $ρ$ ' be the mass per unit length. The coordinates of $20 ρ$ and $10 ρ$ are $(10, 0)$ and $(0, 5)$ respectively from ' $A$ '.
Distance of C.M from A, $r_{cm} = \sqrt{X_{c m}^{2} + y_{c m}^{2}}$

Rotational Motion

269595 As shown in figure from a uniform rectangular sheet a triangular sheet is removed from one edge. The shift of centre of mass is

1

4.2 cm

2

- 4.2 cm

3

6.67 cm

4

- 6.67 cm

Explanation:

shift $= \frac{- mass of removed part \times d}{Mass of remaining part}$ Here $d = 20 cm$

Rotational Motion

269596 A circular disc of radius $R$ is removed from a bigger circular disc of radius $2 R$ such that the circumference of the discs coincide . The centre of mass of the new disc is $α R$ from the centre of the bigger disc. The value of $α$ is

1

1 / 3

2

1 / 2

3

1 / 6

4

1 / 4

Explanation:

Shift of centre of mass $x = \frac{r^{2} a}{R^{2} - r^{2}}$
Where $r =$ radius of removed disc
$R =$ radius of original disc
$a =$ distance between the centres
Note:In this question shift must be $\propto R$ for exact approach to the solution

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