QBManager

Laws of Motion

146228 A block of mass $10 kg$ is placed on a rough horizontal surface whose coefficient of friction is 0.5 . If a horizontal force of $100 N$ is applied on it, then acceleration of block will be:

1

10 m / s^{2}

2

5 m / s^{2}

3

15 m / s^{2}

4

0.5 m / s^{2}

Explanation:

B Given, $m = 10 kg$
$g = 10 {ms}^{- 2}$
$μ = 0.5$
$F = 100 N$

$∵ N = mg = 10 \times 10$
In horizontal direction,
$100 - f = ma$
$100 - μ N = ma$
$100 - 0.5 \times 10 \times 10 = 10 a$
$10 a = 100 - 50 \Rightarrow 50 = 10 a$
$\Rightarrow a = 5 m / s^{2}$

JCECE-2006

Laws of Motion

146229 A block of mass $2 kg$ rests on a plane inclined at an angle of $30^{\circ}$ with the horizontal. The coefficient of friction between the block and surface is 0.7 . The frictional force acting on the block is :

1

11.9 N

2

25 N

3

50 N

4

22.9 N

Explanation:

A Given,
$m = 2 kg$
$θ = 30^{\circ}$
$μ = 0.7$

According to figure,
$f = mg \sin θ$
$f = mg \sin 30^{\circ}$
$f = 2 \times 9.8 \times \frac{1}{2}$
$f = 9.8$
Limiting friction force between the block and the inclined plane is given as -
$f_{lim} = μ_{s} \cdot N$
$= μ_{s} \cdot mg \cos 30^{\circ}$
$= 0.7 \times 2 \times 9.8 \times \frac{\sqrt{3}}{2}$
$f_{lim} = 11.9 N$
The frictional force acting on the block is $11.9 N$ .

JCECE-2005

Laws of Motion

146230 A block of mass $0.1 kg$ is held against a wall by applying a horizontal force of $5 N$ on it. If $μ_{s}$ is between the wall and the block is 0.5 , the magnitude of the frictional force acting on the block is

1

0.98 N

2

0.49 N

3

4.9 N

4

2.5 N

Explanation:

A Given,
$F = 5 N$
$μ_{s} = 0.5$
$m = 0.1 kg$

Hence, friction force acting on the block is -
$f = m . g$
$f = 0.1 \times 9.8$
$f = 0.98 N$

COMEDK 2012

Laws of Motion

146231 A uniform chain of length $L$ is lying partly on a table, the remaining part hanging down from the edge of the table. If the coefficient of friction between the chain and the table is 0.5 , what is the minimum length of the chain that should lie on the table, to prevent the chain from slipping down to the ground?

1

L / 3

2

L / 2

3

2 L / 3

4

3 L / 4

Explanation:

C
Let, $x =$ length of the chain lies on the
$M =$ mass of chain
mass per unit length of chain $= \frac{m}{L}$
mass of the length $(L - x)$ of hanging chain $= \frac{m}{L} (L - x)$
mass of length $x$ of the chain $= \frac{m}{L} x$
Friction force on the upper part $=$ weight of hanging part.
Weight of length $x$ of the chain $=$ weight of length $(L -$ $x)$ of hanging chain
$μ . \frac{m}{L} x \cdot g = \frac{m}{L} (L - x) g$
$μ x = L - x$
$x (μ + 1) = L$
$x = \frac{L}{μ + 1}$
$x = \frac{L}{0.5 + 1}$
$x = \frac{2 L}{3}$

COMEDK 2015

Laws of Motion

146234 Assertion: Angle of repose is equal to the angle of limiting friction.
Reason: When the body is just at the point of motion, the force of friction in this stage is called limiting friction.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

Explanation:

B The maximum value of static friction which is generated between two surface is known as the limiting friction.

Angle of repose is defined as the minimum angle of inclination of a plane with horizontal such that a body on the plane just begin to slide down.
In limiting condition,
$F = mg \sin α$ and $R = mg \cos α$
Where, $α$ - angle of repose
So, $\frac{F}{R} = \tan α$
$∴ \frac{F}{R} = μ_{s} = \tan θ = \tan α (∵ \tan θ = μ_{s})$
or $θ = α$
i.e., angle of friction $=$ angle of repose

AIIMS-2008

Laws of Motion

146228 A block of mass $10 kg$ is placed on a rough horizontal surface whose coefficient of friction is 0.5 . If a horizontal force of $100 N$ is applied on it, then acceleration of block will be:

1

10 m / s^{2}

2

5 m / s^{2}

3

15 m / s^{2}

4

0.5 m / s^{2}

Explanation:

B Given, $m = 10 kg$
$g = 10 {ms}^{- 2}$
$μ = 0.5$
$F = 100 N$

$∵ N = mg = 10 \times 10$
In horizontal direction,
$100 - f = ma$
$100 - μ N = ma$
$100 - 0.5 \times 10 \times 10 = 10 a$
$10 a = 100 - 50 \Rightarrow 50 = 10 a$
$\Rightarrow a = 5 m / s^{2}$

JCECE-2006

Laws of Motion

146229 A block of mass $2 kg$ rests on a plane inclined at an angle of $30^{\circ}$ with the horizontal. The coefficient of friction between the block and surface is 0.7 . The frictional force acting on the block is :

1

11.9 N

2

25 N

3

50 N

4

22.9 N

Explanation:

A Given,
$m = 2 kg$
$θ = 30^{\circ}$
$μ = 0.7$

According to figure,
$f = mg \sin θ$
$f = mg \sin 30^{\circ}$
$f = 2 \times 9.8 \times \frac{1}{2}$
$f = 9.8$
Limiting friction force between the block and the inclined plane is given as -
$f_{lim} = μ_{s} \cdot N$
$= μ_{s} \cdot mg \cos 30^{\circ}$
$= 0.7 \times 2 \times 9.8 \times \frac{\sqrt{3}}{2}$
$f_{lim} = 11.9 N$
The frictional force acting on the block is $11.9 N$ .

JCECE-2005

Laws of Motion

146230 A block of mass $0.1 kg$ is held against a wall by applying a horizontal force of $5 N$ on it. If $μ_{s}$ is between the wall and the block is 0.5 , the magnitude of the frictional force acting on the block is

1

0.98 N

2

0.49 N

3

4.9 N

4

2.5 N

Explanation:

A Given,
$F = 5 N$
$μ_{s} = 0.5$
$m = 0.1 kg$

Hence, friction force acting on the block is -
$f = m . g$
$f = 0.1 \times 9.8$
$f = 0.98 N$

COMEDK 2012

Laws of Motion

146231 A uniform chain of length $L$ is lying partly on a table, the remaining part hanging down from the edge of the table. If the coefficient of friction between the chain and the table is 0.5 , what is the minimum length of the chain that should lie on the table, to prevent the chain from slipping down to the ground?

1

L / 3

2

L / 2

3

2 L / 3

4

3 L / 4

Explanation:

C
Let, $x =$ length of the chain lies on the
$M =$ mass of chain
mass per unit length of chain $= \frac{m}{L}$
mass of the length $(L - x)$ of hanging chain $= \frac{m}{L} (L - x)$
mass of length $x$ of the chain $= \frac{m}{L} x$
Friction force on the upper part $=$ weight of hanging part.
Weight of length $x$ of the chain $=$ weight of length $(L -$ $x)$ of hanging chain
$μ . \frac{m}{L} x \cdot g = \frac{m}{L} (L - x) g$
$μ x = L - x$
$x (μ + 1) = L$
$x = \frac{L}{μ + 1}$
$x = \frac{L}{0.5 + 1}$
$x = \frac{2 L}{3}$

COMEDK 2015

Laws of Motion

146234 Assertion: Angle of repose is equal to the angle of limiting friction.
Reason: When the body is just at the point of motion, the force of friction in this stage is called limiting friction.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

Explanation:

B The maximum value of static friction which is generated between two surface is known as the limiting friction.

Angle of repose is defined as the minimum angle of inclination of a plane with horizontal such that a body on the plane just begin to slide down.
In limiting condition,
$F = mg \sin α$ and $R = mg \cos α$
Where, $α$ - angle of repose
So, $\frac{F}{R} = \tan α$
$∴ \frac{F}{R} = μ_{s} = \tan θ = \tan α (∵ \tan θ = μ_{s})$
or $θ = α$
i.e., angle of friction $=$ angle of repose

AIIMS-2008

Laws of Motion

146228 A block of mass $10 kg$ is placed on a rough horizontal surface whose coefficient of friction is 0.5 . If a horizontal force of $100 N$ is applied on it, then acceleration of block will be:

1

10 m / s^{2}

2

5 m / s^{2}

3

15 m / s^{2}

4

0.5 m / s^{2}

Explanation:

B Given, $m = 10 kg$
$g = 10 {ms}^{- 2}$
$μ = 0.5$
$F = 100 N$

$∵ N = mg = 10 \times 10$
In horizontal direction,
$100 - f = ma$
$100 - μ N = ma$
$100 - 0.5 \times 10 \times 10 = 10 a$
$10 a = 100 - 50 \Rightarrow 50 = 10 a$
$\Rightarrow a = 5 m / s^{2}$

JCECE-2006

Laws of Motion

146229 A block of mass $2 kg$ rests on a plane inclined at an angle of $30^{\circ}$ with the horizontal. The coefficient of friction between the block and surface is 0.7 . The frictional force acting on the block is :

1

11.9 N

2

25 N

3

50 N

4

22.9 N

Explanation:

A Given,
$m = 2 kg$
$θ = 30^{\circ}$
$μ = 0.7$

According to figure,
$f = mg \sin θ$
$f = mg \sin 30^{\circ}$
$f = 2 \times 9.8 \times \frac{1}{2}$
$f = 9.8$
Limiting friction force between the block and the inclined plane is given as -
$f_{lim} = μ_{s} \cdot N$
$= μ_{s} \cdot mg \cos 30^{\circ}$
$= 0.7 \times 2 \times 9.8 \times \frac{\sqrt{3}}{2}$
$f_{lim} = 11.9 N$
The frictional force acting on the block is $11.9 N$ .

JCECE-2005

Laws of Motion

146230 A block of mass $0.1 kg$ is held against a wall by applying a horizontal force of $5 N$ on it. If $μ_{s}$ is between the wall and the block is 0.5 , the magnitude of the frictional force acting on the block is

1

0.98 N

2

0.49 N

3

4.9 N

4

2.5 N

Explanation:

A Given,
$F = 5 N$
$μ_{s} = 0.5$
$m = 0.1 kg$

Hence, friction force acting on the block is -
$f = m . g$
$f = 0.1 \times 9.8$
$f = 0.98 N$

COMEDK 2012

Laws of Motion

146231 A uniform chain of length $L$ is lying partly on a table, the remaining part hanging down from the edge of the table. If the coefficient of friction between the chain and the table is 0.5 , what is the minimum length of the chain that should lie on the table, to prevent the chain from slipping down to the ground?

1

L / 3

2

L / 2

3

2 L / 3

4

3 L / 4

Explanation:

C
Let, $x =$ length of the chain lies on the
$M =$ mass of chain
mass per unit length of chain $= \frac{m}{L}$
mass of the length $(L - x)$ of hanging chain $= \frac{m}{L} (L - x)$
mass of length $x$ of the chain $= \frac{m}{L} x$
Friction force on the upper part $=$ weight of hanging part.
Weight of length $x$ of the chain $=$ weight of length $(L -$ $x)$ of hanging chain
$μ . \frac{m}{L} x \cdot g = \frac{m}{L} (L - x) g$
$μ x = L - x$
$x (μ + 1) = L$
$x = \frac{L}{μ + 1}$
$x = \frac{L}{0.5 + 1}$
$x = \frac{2 L}{3}$

COMEDK 2015

Laws of Motion

146234 Assertion: Angle of repose is equal to the angle of limiting friction.
Reason: When the body is just at the point of motion, the force of friction in this stage is called limiting friction.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

Explanation:

B The maximum value of static friction which is generated between two surface is known as the limiting friction.

Angle of repose is defined as the minimum angle of inclination of a plane with horizontal such that a body on the plane just begin to slide down.
In limiting condition,
$F = mg \sin α$ and $R = mg \cos α$
Where, $α$ - angle of repose
So, $\frac{F}{R} = \tan α$
$∴ \frac{F}{R} = μ_{s} = \tan θ = \tan α (∵ \tan θ = μ_{s})$
or $θ = α$
i.e., angle of friction $=$ angle of repose

AIIMS-2008

Laws of Motion

146228 A block of mass $10 kg$ is placed on a rough horizontal surface whose coefficient of friction is 0.5 . If a horizontal force of $100 N$ is applied on it, then acceleration of block will be:

1

10 m / s^{2}

2

5 m / s^{2}

3

15 m / s^{2}

4

0.5 m / s^{2}

Explanation:

B Given, $m = 10 kg$
$g = 10 {ms}^{- 2}$
$μ = 0.5$
$F = 100 N$

$∵ N = mg = 10 \times 10$
In horizontal direction,
$100 - f = ma$
$100 - μ N = ma$
$100 - 0.5 \times 10 \times 10 = 10 a$
$10 a = 100 - 50 \Rightarrow 50 = 10 a$
$\Rightarrow a = 5 m / s^{2}$

JCECE-2006

Laws of Motion

146229 A block of mass $2 kg$ rests on a plane inclined at an angle of $30^{\circ}$ with the horizontal. The coefficient of friction between the block and surface is 0.7 . The frictional force acting on the block is :

1

11.9 N

2

25 N

3

50 N

4

22.9 N

Explanation:

A Given,
$m = 2 kg$
$θ = 30^{\circ}$
$μ = 0.7$

According to figure,
$f = mg \sin θ$
$f = mg \sin 30^{\circ}$
$f = 2 \times 9.8 \times \frac{1}{2}$
$f = 9.8$
Limiting friction force between the block and the inclined plane is given as -
$f_{lim} = μ_{s} \cdot N$
$= μ_{s} \cdot mg \cos 30^{\circ}$
$= 0.7 \times 2 \times 9.8 \times \frac{\sqrt{3}}{2}$
$f_{lim} = 11.9 N$
The frictional force acting on the block is $11.9 N$ .

JCECE-2005

Laws of Motion

146230 A block of mass $0.1 kg$ is held against a wall by applying a horizontal force of $5 N$ on it. If $μ_{s}$ is between the wall and the block is 0.5 , the magnitude of the frictional force acting on the block is

1

0.98 N

2

0.49 N

3

4.9 N

4

2.5 N

Explanation:

A Given,
$F = 5 N$
$μ_{s} = 0.5$
$m = 0.1 kg$

Hence, friction force acting on the block is -
$f = m . g$
$f = 0.1 \times 9.8$
$f = 0.98 N$

COMEDK 2012

Laws of Motion

146231 A uniform chain of length $L$ is lying partly on a table, the remaining part hanging down from the edge of the table. If the coefficient of friction between the chain and the table is 0.5 , what is the minimum length of the chain that should lie on the table, to prevent the chain from slipping down to the ground?

1

L / 3

2

L / 2

3

2 L / 3

4

3 L / 4

Explanation:

C
Let, $x =$ length of the chain lies on the
$M =$ mass of chain
mass per unit length of chain $= \frac{m}{L}$
mass of the length $(L - x)$ of hanging chain $= \frac{m}{L} (L - x)$
mass of length $x$ of the chain $= \frac{m}{L} x$
Friction force on the upper part $=$ weight of hanging part.
Weight of length $x$ of the chain $=$ weight of length $(L -$ $x)$ of hanging chain
$μ . \frac{m}{L} x \cdot g = \frac{m}{L} (L - x) g$
$μ x = L - x$
$x (μ + 1) = L$
$x = \frac{L}{μ + 1}$
$x = \frac{L}{0.5 + 1}$
$x = \frac{2 L}{3}$

COMEDK 2015

Laws of Motion

146234 Assertion: Angle of repose is equal to the angle of limiting friction.
Reason: When the body is just at the point of motion, the force of friction in this stage is called limiting friction.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

Explanation:

B The maximum value of static friction which is generated between two surface is known as the limiting friction.

Angle of repose is defined as the minimum angle of inclination of a plane with horizontal such that a body on the plane just begin to slide down.
In limiting condition,
$F = mg \sin α$ and $R = mg \cos α$
Where, $α$ - angle of repose
So, $\frac{F}{R} = \tan α$
$∴ \frac{F}{R} = μ_{s} = \tan θ = \tan α (∵ \tan θ = μ_{s})$
or $θ = α$
i.e., angle of friction $=$ angle of repose

AIIMS-2008

Laws of Motion

146228 A block of mass $10 kg$ is placed on a rough horizontal surface whose coefficient of friction is 0.5 . If a horizontal force of $100 N$ is applied on it, then acceleration of block will be:

1

10 m / s^{2}

2

5 m / s^{2}

3

15 m / s^{2}

4

0.5 m / s^{2}

Explanation:

B Given, $m = 10 kg$
$g = 10 {ms}^{- 2}$
$μ = 0.5$
$F = 100 N$

$∵ N = mg = 10 \times 10$
In horizontal direction,
$100 - f = ma$
$100 - μ N = ma$
$100 - 0.5 \times 10 \times 10 = 10 a$
$10 a = 100 - 50 \Rightarrow 50 = 10 a$
$\Rightarrow a = 5 m / s^{2}$

JCECE-2006

Laws of Motion

146229 A block of mass $2 kg$ rests on a plane inclined at an angle of $30^{\circ}$ with the horizontal. The coefficient of friction between the block and surface is 0.7 . The frictional force acting on the block is :

1

11.9 N

2

25 N

3

50 N

4

22.9 N

Explanation:

A Given,
$m = 2 kg$
$θ = 30^{\circ}$
$μ = 0.7$

According to figure,
$f = mg \sin θ$
$f = mg \sin 30^{\circ}$
$f = 2 \times 9.8 \times \frac{1}{2}$
$f = 9.8$
Limiting friction force between the block and the inclined plane is given as -
$f_{lim} = μ_{s} \cdot N$
$= μ_{s} \cdot mg \cos 30^{\circ}$
$= 0.7 \times 2 \times 9.8 \times \frac{\sqrt{3}}{2}$
$f_{lim} = 11.9 N$
The frictional force acting on the block is $11.9 N$ .

JCECE-2005

Laws of Motion

146230 A block of mass $0.1 kg$ is held against a wall by applying a horizontal force of $5 N$ on it. If $μ_{s}$ is between the wall and the block is 0.5 , the magnitude of the frictional force acting on the block is

1

0.98 N

2

0.49 N

3

4.9 N

4

2.5 N

Explanation:

A Given,
$F = 5 N$
$μ_{s} = 0.5$
$m = 0.1 kg$

Hence, friction force acting on the block is -
$f = m . g$
$f = 0.1 \times 9.8$
$f = 0.98 N$

COMEDK 2012

Laws of Motion

146231 A uniform chain of length $L$ is lying partly on a table, the remaining part hanging down from the edge of the table. If the coefficient of friction between the chain and the table is 0.5 , what is the minimum length of the chain that should lie on the table, to prevent the chain from slipping down to the ground?

1

L / 3

2

L / 2

3

2 L / 3

4

3 L / 4

Explanation:

C
Let, $x =$ length of the chain lies on the
$M =$ mass of chain
mass per unit length of chain $= \frac{m}{L}$
mass of the length $(L - x)$ of hanging chain $= \frac{m}{L} (L - x)$
mass of length $x$ of the chain $= \frac{m}{L} x$
Friction force on the upper part $=$ weight of hanging part.
Weight of length $x$ of the chain $=$ weight of length $(L -$ $x)$ of hanging chain
$μ . \frac{m}{L} x \cdot g = \frac{m}{L} (L - x) g$
$μ x = L - x$
$x (μ + 1) = L$
$x = \frac{L}{μ + 1}$
$x = \frac{L}{0.5 + 1}$
$x = \frac{2 L}{3}$

COMEDK 2015

Laws of Motion

146234 Assertion: Angle of repose is equal to the angle of limiting friction.
Reason: When the body is just at the point of motion, the force of friction in this stage is called limiting friction.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

Explanation:

B The maximum value of static friction which is generated between two surface is known as the limiting friction.

Angle of repose is defined as the minimum angle of inclination of a plane with horizontal such that a body on the plane just begin to slide down.
In limiting condition,
$F = mg \sin α$ and $R = mg \cos α$
Where, $α$ - angle of repose
So, $\frac{F}{R} = \tan α$
$∴ \frac{F}{R} = μ_{s} = \tan θ = \tan α (∵ \tan θ = μ_{s})$
or $θ = α$
i.e., angle of friction $=$ angle of repose

AIIMS-2008

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