QBManager

LAWS OF MOTION (ADDITIONAL)

372128 A uniform chain of length $L$ hangs partially from table and held in equilibrium by friction. If greatest length of chain that hangs without slipping is $l$ then the coefficient of friction between chain and table is

1

\frac{l}{2}

2

\frac{l}{L + l}

3

\frac{l}{L - l}

4

\frac{l}{L + 1}

Explanation:

C
Let, mass of chain $= m kg$
Per unit mass $= λ kg / m$
So, mass of hanging part $= λ l kg$
mass of upper part of chain $= λ (L - l) kg$
For hanging part,
$T = λ l g$
For part on table,
$N = (L - l) g λ$
$T = f \Rightarrow T = μ (L - l) g λ$
$\lg λ = μ (L - l) g λ$
$μ = \frac{l}{L - l}$

EAMCET-1995

LAWS OF MOTION (ADDITIONAL)

372129 A body of weight $50 N$ is placed on a smooth surface. If the force required to move the body on the rough surface is $30 N$ the coefficient of friction is

1 0.6

2 1.2

3 0.3

4 1.67

Explanation:

A Given,
Weight of body $(w) = 50 N$
Force $(F) = 30 N$
Coefficient of friction $(μ) = \frac{F}{N} [∵ N = w]$
$μ = \frac{30}{50}$
$μ = 0.6$

EAMCET-1996

LAWS OF MOTION (ADDITIONAL)

372130 A body of mass $10 kg$ lies on a rough horizontal surface. When a horizontal force $F$ Newton acts on it, it gets an acceleration of $5 {ms}^{- 2}$ and when the horizontal force is doubled, it gets an acceleration of $18 {ms}^{- 2}$ . Then, the coefficient of friction between the body and the horizontal surface is (assume $g = 10 {m s}^{- 2}$ )

1 0.2

2 0.8

3 0.4

4 0.6

Explanation:

B $Mass (m) = 10 kg$
Friction force $(f) = μ . mg$
When force is F Newton is acts,
acceleration $(a) = \frac{F - f}{m} = 5 m / s^{2}$
$F - f = 50 N$
When force is $2 F$ is act,
Then, acceleration $(a) = \frac{2 F - f}{m} = 18 m / s^{2}$
$2 F - f = 180 N$
From equation (i) and (ii), we get
$F = 130 N$
Put this value in equation (i), we get
$f = 80 N$
$μ N = 80 \Rightarrow μ = \frac{80}{N} = \frac{80}{mg} = \frac{80}{10 \times 10} = 0.8$

EAMCET-1997

LAWS OF MOTION (ADDITIONAL)

372131 An iron block of sides $50 cm \times 8 cm \times 15 cm$ has to be pushed along the floor. The force required will be minimum when the surface in contact with ground is

1

8 cm \times 15 cm

surface

2

5 cm \times 15 cm

surface

3

8 cm \times 5 cm

surface

4 Force is same for all surfaces

Explanation:

D Friction force is independent of the area of contact between the bodies surfaces. So the friction will be same for different areas in contact, its depend on roughness of surface of contact.
The force $(F)$ required to push the iron block should be more than static friction $f_{s} = μ_{s} N$
$F > f_{s}$

EAMCET-1998

LAWS OF MOTION (ADDITIONAL)

372132 To determine the coefficient of friction between a rough surface and a block, the surface is kept inclined at $45^{\circ}$ and the block is released from rest. The block takes a time $t$ in moving a distance $d$ . The rough surface is then replaced by a smooth surface and the same experiment is repeated. The block now takes a time $t / 2$ in moving down the same distance $d$ . The coefficient of friction is

1

3 / 4

2

5 / 4

3

1 / 2

4

1 / \sqrt{2}

Explanation:

A Given, $θ = 45^{\circ}$
Time takes moving distance ' $d$ ' on rough surface $= t$
Time takes moving distance ' $d$ ' on smooth surface $= t / 2$

Case (I) when plane is rough,
$N = mg \cos θ$
$f_{s} = μ N$
$f_{s} = μ mg \cos θ$
$m g \sin θ - f_{s} = m a$
$mg \sin θ - μ mg \cos θ = ma$
$a = g \sin θ - μ g \cos θ$
$∴ s = ut + \frac{1}{2} {at}^{2} ($ where, $u = 0)$
$d = 0 \times t_{1} + \frac{1}{2} (g \sin θ - μ g \cos θ) t_{1}^{2}$ $\sqrt{\frac{2 d}{g \sin θ - μ g \cos θ}} = t_{1}$
Case (II), When $t_{2} = t / 2$
$s = ut + \frac{1}{2} {at}^{2}$
where, $u = 0, s = - d$
$a = - g \sin θ$ (plane is smooth)
$∴ d = \frac{1}{2} g \sin θ t_{2}^{2}$
$\frac{2 d}{g \sin θ} = t_{2}^{2}$
$t_{2} = \sqrt{\frac{2 d}{g \sin θ}}$
Given
$\frac{t_{1}}{2} = t_{2}$
$\frac{1}{2} \sqrt{\frac{2 d}{g \sin θ - μ g \cos θ}} = \sqrt{\frac{2 d}{g \sin θ}}$
$\frac{1}{4} \times \frac{2 d}{g \sin θ - μ g \cos θ} = \frac{2 d}{g \sin θ}$
$\frac{1}{4 \sin θ - 4 μ \cos θ} = \frac{1}{\sin θ}$
$4 \sin θ - 4 μ \cos θ = \sin θ$
$3 \sin θ = 4 μ \cos θ$
$\frac{3}{4} \tan θ = μ$
$μ = \frac{3}{4} \tan 45^{\circ} (∵ θ = 45^{\circ})$
$∴ μ = 3 / 4$

EAMCET-2008

LAWS OF MOTION (ADDITIONAL)

372128 A uniform chain of length $L$ hangs partially from table and held in equilibrium by friction. If greatest length of chain that hangs without slipping is $l$ then the coefficient of friction between chain and table is

1

\frac{l}{2}

2

\frac{l}{L + l}

3

\frac{l}{L - l}

4

\frac{l}{L + 1}

Explanation:

C
Let, mass of chain $= m kg$
Per unit mass $= λ kg / m$
So, mass of hanging part $= λ l kg$
mass of upper part of chain $= λ (L - l) kg$
For hanging part,
$T = λ l g$
For part on table,
$N = (L - l) g λ$
$T = f \Rightarrow T = μ (L - l) g λ$
$\lg λ = μ (L - l) g λ$
$μ = \frac{l}{L - l}$

EAMCET-1995

LAWS OF MOTION (ADDITIONAL)

372129 A body of weight $50 N$ is placed on a smooth surface. If the force required to move the body on the rough surface is $30 N$ the coefficient of friction is

1 0.6

2 1.2

3 0.3

4 1.67

Explanation:

A Given,
Weight of body $(w) = 50 N$
Force $(F) = 30 N$
Coefficient of friction $(μ) = \frac{F}{N} [∵ N = w]$
$μ = \frac{30}{50}$
$μ = 0.6$

EAMCET-1996

LAWS OF MOTION (ADDITIONAL)

372130 A body of mass $10 kg$ lies on a rough horizontal surface. When a horizontal force $F$ Newton acts on it, it gets an acceleration of $5 {ms}^{- 2}$ and when the horizontal force is doubled, it gets an acceleration of $18 {ms}^{- 2}$ . Then, the coefficient of friction between the body and the horizontal surface is (assume $g = 10 {m s}^{- 2}$ )

1 0.2

2 0.8

3 0.4

4 0.6

Explanation:

B $Mass (m) = 10 kg$
Friction force $(f) = μ . mg$
When force is F Newton is acts,
acceleration $(a) = \frac{F - f}{m} = 5 m / s^{2}$
$F - f = 50 N$
When force is $2 F$ is act,
Then, acceleration $(a) = \frac{2 F - f}{m} = 18 m / s^{2}$
$2 F - f = 180 N$
From equation (i) and (ii), we get
$F = 130 N$
Put this value in equation (i), we get
$f = 80 N$
$μ N = 80 \Rightarrow μ = \frac{80}{N} = \frac{80}{mg} = \frac{80}{10 \times 10} = 0.8$

EAMCET-1997

LAWS OF MOTION (ADDITIONAL)

372131 An iron block of sides $50 cm \times 8 cm \times 15 cm$ has to be pushed along the floor. The force required will be minimum when the surface in contact with ground is

1

8 cm \times 15 cm

surface

2

5 cm \times 15 cm

surface

3

8 cm \times 5 cm

surface

4 Force is same for all surfaces

Explanation:

D Friction force is independent of the area of contact between the bodies surfaces. So the friction will be same for different areas in contact, its depend on roughness of surface of contact.
The force $(F)$ required to push the iron block should be more than static friction $f_{s} = μ_{s} N$
$F > f_{s}$

EAMCET-1998

LAWS OF MOTION (ADDITIONAL)

372132 To determine the coefficient of friction between a rough surface and a block, the surface is kept inclined at $45^{\circ}$ and the block is released from rest. The block takes a time $t$ in moving a distance $d$ . The rough surface is then replaced by a smooth surface and the same experiment is repeated. The block now takes a time $t / 2$ in moving down the same distance $d$ . The coefficient of friction is

1

3 / 4

2

5 / 4

3

1 / 2

4

1 / \sqrt{2}

Explanation:

A Given, $θ = 45^{\circ}$
Time takes moving distance ' $d$ ' on rough surface $= t$
Time takes moving distance ' $d$ ' on smooth surface $= t / 2$

Case (I) when plane is rough,
$N = mg \cos θ$
$f_{s} = μ N$
$f_{s} = μ mg \cos θ$
$m g \sin θ - f_{s} = m a$
$mg \sin θ - μ mg \cos θ = ma$
$a = g \sin θ - μ g \cos θ$
$∴ s = ut + \frac{1}{2} {at}^{2} ($ where, $u = 0)$
$d = 0 \times t_{1} + \frac{1}{2} (g \sin θ - μ g \cos θ) t_{1}^{2}$ $\sqrt{\frac{2 d}{g \sin θ - μ g \cos θ}} = t_{1}$
Case (II), When $t_{2} = t / 2$
$s = ut + \frac{1}{2} {at}^{2}$
where, $u = 0, s = - d$
$a = - g \sin θ$ (plane is smooth)
$∴ d = \frac{1}{2} g \sin θ t_{2}^{2}$
$\frac{2 d}{g \sin θ} = t_{2}^{2}$
$t_{2} = \sqrt{\frac{2 d}{g \sin θ}}$
Given
$\frac{t_{1}}{2} = t_{2}$
$\frac{1}{2} \sqrt{\frac{2 d}{g \sin θ - μ g \cos θ}} = \sqrt{\frac{2 d}{g \sin θ}}$
$\frac{1}{4} \times \frac{2 d}{g \sin θ - μ g \cos θ} = \frac{2 d}{g \sin θ}$
$\frac{1}{4 \sin θ - 4 μ \cos θ} = \frac{1}{\sin θ}$
$4 \sin θ - 4 μ \cos θ = \sin θ$
$3 \sin θ = 4 μ \cos θ$
$\frac{3}{4} \tan θ = μ$
$μ = \frac{3}{4} \tan 45^{\circ} (∵ θ = 45^{\circ})$
$∴ μ = 3 / 4$

EAMCET-2008

LAWS OF MOTION (ADDITIONAL)

372128 A uniform chain of length $L$ hangs partially from table and held in equilibrium by friction. If greatest length of chain that hangs without slipping is $l$ then the coefficient of friction between chain and table is

1

\frac{l}{2}

2

\frac{l}{L + l}

3

\frac{l}{L - l}

4

\frac{l}{L + 1}

Explanation:

C
Let, mass of chain $= m kg$
Per unit mass $= λ kg / m$
So, mass of hanging part $= λ l kg$
mass of upper part of chain $= λ (L - l) kg$
For hanging part,
$T = λ l g$
For part on table,
$N = (L - l) g λ$
$T = f \Rightarrow T = μ (L - l) g λ$
$\lg λ = μ (L - l) g λ$
$μ = \frac{l}{L - l}$

EAMCET-1995

LAWS OF MOTION (ADDITIONAL)

372129 A body of weight $50 N$ is placed on a smooth surface. If the force required to move the body on the rough surface is $30 N$ the coefficient of friction is

1 0.6

2 1.2

3 0.3

4 1.67

Explanation:

A Given,
Weight of body $(w) = 50 N$
Force $(F) = 30 N$
Coefficient of friction $(μ) = \frac{F}{N} [∵ N = w]$
$μ = \frac{30}{50}$
$μ = 0.6$

EAMCET-1996

LAWS OF MOTION (ADDITIONAL)

372130 A body of mass $10 kg$ lies on a rough horizontal surface. When a horizontal force $F$ Newton acts on it, it gets an acceleration of $5 {ms}^{- 2}$ and when the horizontal force is doubled, it gets an acceleration of $18 {ms}^{- 2}$ . Then, the coefficient of friction between the body and the horizontal surface is (assume $g = 10 {m s}^{- 2}$ )

1 0.2

2 0.8

3 0.4

4 0.6

Explanation:

B $Mass (m) = 10 kg$
Friction force $(f) = μ . mg$
When force is F Newton is acts,
acceleration $(a) = \frac{F - f}{m} = 5 m / s^{2}$
$F - f = 50 N$
When force is $2 F$ is act,
Then, acceleration $(a) = \frac{2 F - f}{m} = 18 m / s^{2}$
$2 F - f = 180 N$
From equation (i) and (ii), we get
$F = 130 N$
Put this value in equation (i), we get
$f = 80 N$
$μ N = 80 \Rightarrow μ = \frac{80}{N} = \frac{80}{mg} = \frac{80}{10 \times 10} = 0.8$

EAMCET-1997

LAWS OF MOTION (ADDITIONAL)

372131 An iron block of sides $50 cm \times 8 cm \times 15 cm$ has to be pushed along the floor. The force required will be minimum when the surface in contact with ground is

1

8 cm \times 15 cm

surface

2

5 cm \times 15 cm

surface

3

8 cm \times 5 cm

surface

4 Force is same for all surfaces

Explanation:

D Friction force is independent of the area of contact between the bodies surfaces. So the friction will be same for different areas in contact, its depend on roughness of surface of contact.
The force $(F)$ required to push the iron block should be more than static friction $f_{s} = μ_{s} N$
$F > f_{s}$

EAMCET-1998

LAWS OF MOTION (ADDITIONAL)

372132 To determine the coefficient of friction between a rough surface and a block, the surface is kept inclined at $45^{\circ}$ and the block is released from rest. The block takes a time $t$ in moving a distance $d$ . The rough surface is then replaced by a smooth surface and the same experiment is repeated. The block now takes a time $t / 2$ in moving down the same distance $d$ . The coefficient of friction is

1

3 / 4

2

5 / 4

3

1 / 2

4

1 / \sqrt{2}

Explanation:

A Given, $θ = 45^{\circ}$
Time takes moving distance ' $d$ ' on rough surface $= t$
Time takes moving distance ' $d$ ' on smooth surface $= t / 2$

Case (I) when plane is rough,
$N = mg \cos θ$
$f_{s} = μ N$
$f_{s} = μ mg \cos θ$
$m g \sin θ - f_{s} = m a$
$mg \sin θ - μ mg \cos θ = ma$
$a = g \sin θ - μ g \cos θ$
$∴ s = ut + \frac{1}{2} {at}^{2} ($ where, $u = 0)$
$d = 0 \times t_{1} + \frac{1}{2} (g \sin θ - μ g \cos θ) t_{1}^{2}$ $\sqrt{\frac{2 d}{g \sin θ - μ g \cos θ}} = t_{1}$
Case (II), When $t_{2} = t / 2$
$s = ut + \frac{1}{2} {at}^{2}$
where, $u = 0, s = - d$
$a = - g \sin θ$ (plane is smooth)
$∴ d = \frac{1}{2} g \sin θ t_{2}^{2}$
$\frac{2 d}{g \sin θ} = t_{2}^{2}$
$t_{2} = \sqrt{\frac{2 d}{g \sin θ}}$
Given
$\frac{t_{1}}{2} = t_{2}$
$\frac{1}{2} \sqrt{\frac{2 d}{g \sin θ - μ g \cos θ}} = \sqrt{\frac{2 d}{g \sin θ}}$
$\frac{1}{4} \times \frac{2 d}{g \sin θ - μ g \cos θ} = \frac{2 d}{g \sin θ}$
$\frac{1}{4 \sin θ - 4 μ \cos θ} = \frac{1}{\sin θ}$
$4 \sin θ - 4 μ \cos θ = \sin θ$
$3 \sin θ = 4 μ \cos θ$
$\frac{3}{4} \tan θ = μ$
$μ = \frac{3}{4} \tan 45^{\circ} (∵ θ = 45^{\circ})$
$∴ μ = 3 / 4$

EAMCET-2008

LAWS OF MOTION (ADDITIONAL)

372128 A uniform chain of length $L$ hangs partially from table and held in equilibrium by friction. If greatest length of chain that hangs without slipping is $l$ then the coefficient of friction between chain and table is

1

\frac{l}{2}

2

\frac{l}{L + l}

3

\frac{l}{L - l}

4

\frac{l}{L + 1}

Explanation:

C
Let, mass of chain $= m kg$
Per unit mass $= λ kg / m$
So, mass of hanging part $= λ l kg$
mass of upper part of chain $= λ (L - l) kg$
For hanging part,
$T = λ l g$
For part on table,
$N = (L - l) g λ$
$T = f \Rightarrow T = μ (L - l) g λ$
$\lg λ = μ (L - l) g λ$
$μ = \frac{l}{L - l}$

EAMCET-1995

LAWS OF MOTION (ADDITIONAL)

372129 A body of weight $50 N$ is placed on a smooth surface. If the force required to move the body on the rough surface is $30 N$ the coefficient of friction is

1 0.6

2 1.2

3 0.3

4 1.67

Explanation:

A Given,
Weight of body $(w) = 50 N$
Force $(F) = 30 N$
Coefficient of friction $(μ) = \frac{F}{N} [∵ N = w]$
$μ = \frac{30}{50}$
$μ = 0.6$

EAMCET-1996

LAWS OF MOTION (ADDITIONAL)

372130 A body of mass $10 kg$ lies on a rough horizontal surface. When a horizontal force $F$ Newton acts on it, it gets an acceleration of $5 {ms}^{- 2}$ and when the horizontal force is doubled, it gets an acceleration of $18 {ms}^{- 2}$ . Then, the coefficient of friction between the body and the horizontal surface is (assume $g = 10 {m s}^{- 2}$ )

1 0.2

2 0.8

3 0.4

4 0.6

Explanation:

B $Mass (m) = 10 kg$
Friction force $(f) = μ . mg$
When force is F Newton is acts,
acceleration $(a) = \frac{F - f}{m} = 5 m / s^{2}$
$F - f = 50 N$
When force is $2 F$ is act,
Then, acceleration $(a) = \frac{2 F - f}{m} = 18 m / s^{2}$
$2 F - f = 180 N$
From equation (i) and (ii), we get
$F = 130 N$
Put this value in equation (i), we get
$f = 80 N$
$μ N = 80 \Rightarrow μ = \frac{80}{N} = \frac{80}{mg} = \frac{80}{10 \times 10} = 0.8$

EAMCET-1997

LAWS OF MOTION (ADDITIONAL)

372131 An iron block of sides $50 cm \times 8 cm \times 15 cm$ has to be pushed along the floor. The force required will be minimum when the surface in contact with ground is

1

8 cm \times 15 cm

surface

2

5 cm \times 15 cm

surface

3

8 cm \times 5 cm

surface

4 Force is same for all surfaces

Explanation:

D Friction force is independent of the area of contact between the bodies surfaces. So the friction will be same for different areas in contact, its depend on roughness of surface of contact.
The force $(F)$ required to push the iron block should be more than static friction $f_{s} = μ_{s} N$
$F > f_{s}$

EAMCET-1998

LAWS OF MOTION (ADDITIONAL)

372132 To determine the coefficient of friction between a rough surface and a block, the surface is kept inclined at $45^{\circ}$ and the block is released from rest. The block takes a time $t$ in moving a distance $d$ . The rough surface is then replaced by a smooth surface and the same experiment is repeated. The block now takes a time $t / 2$ in moving down the same distance $d$ . The coefficient of friction is

1

3 / 4

2

5 / 4

3

1 / 2

4

1 / \sqrt{2}

Explanation:

A Given, $θ = 45^{\circ}$
Time takes moving distance ' $d$ ' on rough surface $= t$
Time takes moving distance ' $d$ ' on smooth surface $= t / 2$

Case (I) when plane is rough,
$N = mg \cos θ$
$f_{s} = μ N$
$f_{s} = μ mg \cos θ$
$m g \sin θ - f_{s} = m a$
$mg \sin θ - μ mg \cos θ = ma$
$a = g \sin θ - μ g \cos θ$
$∴ s = ut + \frac{1}{2} {at}^{2} ($ where, $u = 0)$
$d = 0 \times t_{1} + \frac{1}{2} (g \sin θ - μ g \cos θ) t_{1}^{2}$ $\sqrt{\frac{2 d}{g \sin θ - μ g \cos θ}} = t_{1}$
Case (II), When $t_{2} = t / 2$
$s = ut + \frac{1}{2} {at}^{2}$
where, $u = 0, s = - d$
$a = - g \sin θ$ (plane is smooth)
$∴ d = \frac{1}{2} g \sin θ t_{2}^{2}$
$\frac{2 d}{g \sin θ} = t_{2}^{2}$
$t_{2} = \sqrt{\frac{2 d}{g \sin θ}}$
Given
$\frac{t_{1}}{2} = t_{2}$
$\frac{1}{2} \sqrt{\frac{2 d}{g \sin θ - μ g \cos θ}} = \sqrt{\frac{2 d}{g \sin θ}}$
$\frac{1}{4} \times \frac{2 d}{g \sin θ - μ g \cos θ} = \frac{2 d}{g \sin θ}$
$\frac{1}{4 \sin θ - 4 μ \cos θ} = \frac{1}{\sin θ}$
$4 \sin θ - 4 μ \cos θ = \sin θ$
$3 \sin θ = 4 μ \cos θ$
$\frac{3}{4} \tan θ = μ$
$μ = \frac{3}{4} \tan 45^{\circ} (∵ θ = 45^{\circ})$
$∴ μ = 3 / 4$

EAMCET-2008

LAWS OF MOTION (ADDITIONAL)

372128 A uniform chain of length $L$ hangs partially from table and held in equilibrium by friction. If greatest length of chain that hangs without slipping is $l$ then the coefficient of friction between chain and table is

1

\frac{l}{2}

2

\frac{l}{L + l}

3

\frac{l}{L - l}

4

\frac{l}{L + 1}

Explanation:

C
Let, mass of chain $= m kg$
Per unit mass $= λ kg / m$
So, mass of hanging part $= λ l kg$
mass of upper part of chain $= λ (L - l) kg$
For hanging part,
$T = λ l g$
For part on table,
$N = (L - l) g λ$
$T = f \Rightarrow T = μ (L - l) g λ$
$\lg λ = μ (L - l) g λ$
$μ = \frac{l}{L - l}$

EAMCET-1995

LAWS OF MOTION (ADDITIONAL)

372129 A body of weight $50 N$ is placed on a smooth surface. If the force required to move the body on the rough surface is $30 N$ the coefficient of friction is

1 0.6

2 1.2

3 0.3

4 1.67

Explanation:

A Given,
Weight of body $(w) = 50 N$
Force $(F) = 30 N$
Coefficient of friction $(μ) = \frac{F}{N} [∵ N = w]$
$μ = \frac{30}{50}$
$μ = 0.6$

EAMCET-1996

LAWS OF MOTION (ADDITIONAL)

372130 A body of mass $10 kg$ lies on a rough horizontal surface. When a horizontal force $F$ Newton acts on it, it gets an acceleration of $5 {ms}^{- 2}$ and when the horizontal force is doubled, it gets an acceleration of $18 {ms}^{- 2}$ . Then, the coefficient of friction between the body and the horizontal surface is (assume $g = 10 {m s}^{- 2}$ )

1 0.2

2 0.8

3 0.4

4 0.6

Explanation:

B $Mass (m) = 10 kg$
Friction force $(f) = μ . mg$
When force is F Newton is acts,
acceleration $(a) = \frac{F - f}{m} = 5 m / s^{2}$
$F - f = 50 N$
When force is $2 F$ is act,
Then, acceleration $(a) = \frac{2 F - f}{m} = 18 m / s^{2}$
$2 F - f = 180 N$
From equation (i) and (ii), we get
$F = 130 N$
Put this value in equation (i), we get
$f = 80 N$
$μ N = 80 \Rightarrow μ = \frac{80}{N} = \frac{80}{mg} = \frac{80}{10 \times 10} = 0.8$

EAMCET-1997

LAWS OF MOTION (ADDITIONAL)

372131 An iron block of sides $50 cm \times 8 cm \times 15 cm$ has to be pushed along the floor. The force required will be minimum when the surface in contact with ground is

1

8 cm \times 15 cm

surface

2

5 cm \times 15 cm

surface

3

8 cm \times 5 cm

surface

4 Force is same for all surfaces

Explanation:

D Friction force is independent of the area of contact between the bodies surfaces. So the friction will be same for different areas in contact, its depend on roughness of surface of contact.
The force $(F)$ required to push the iron block should be more than static friction $f_{s} = μ_{s} N$
$F > f_{s}$

EAMCET-1998

LAWS OF MOTION (ADDITIONAL)

372132 To determine the coefficient of friction between a rough surface and a block, the surface is kept inclined at $45^{\circ}$ and the block is released from rest. The block takes a time $t$ in moving a distance $d$ . The rough surface is then replaced by a smooth surface and the same experiment is repeated. The block now takes a time $t / 2$ in moving down the same distance $d$ . The coefficient of friction is

1

3 / 4

2

5 / 4

3

1 / 2

4

1 / \sqrt{2}

Explanation:

A Given, $θ = 45^{\circ}$
Time takes moving distance ' $d$ ' on rough surface $= t$
Time takes moving distance ' $d$ ' on smooth surface $= t / 2$

Case (I) when plane is rough,
$N = mg \cos θ$
$f_{s} = μ N$
$f_{s} = μ mg \cos θ$
$m g \sin θ - f_{s} = m a$
$mg \sin θ - μ mg \cos θ = ma$
$a = g \sin θ - μ g \cos θ$
$∴ s = ut + \frac{1}{2} {at}^{2} ($ where, $u = 0)$
$d = 0 \times t_{1} + \frac{1}{2} (g \sin θ - μ g \cos θ) t_{1}^{2}$ $\sqrt{\frac{2 d}{g \sin θ - μ g \cos θ}} = t_{1}$
Case (II), When $t_{2} = t / 2$
$s = ut + \frac{1}{2} {at}^{2}$
where, $u = 0, s = - d$
$a = - g \sin θ$ (plane is smooth)
$∴ d = \frac{1}{2} g \sin θ t_{2}^{2}$
$\frac{2 d}{g \sin θ} = t_{2}^{2}$
$t_{2} = \sqrt{\frac{2 d}{g \sin θ}}$
Given
$\frac{t_{1}}{2} = t_{2}$
$\frac{1}{2} \sqrt{\frac{2 d}{g \sin θ - μ g \cos θ}} = \sqrt{\frac{2 d}{g \sin θ}}$
$\frac{1}{4} \times \frac{2 d}{g \sin θ - μ g \cos θ} = \frac{2 d}{g \sin θ}$
$\frac{1}{4 \sin θ - 4 μ \cos θ} = \frac{1}{\sin θ}$
$4 \sin θ - 4 μ \cos θ = \sin θ$
$3 \sin θ = 4 μ \cos θ$
$\frac{3}{4} \tan θ = μ$
$μ = \frac{3}{4} \tan 45^{\circ} (∵ θ = 45^{\circ})$
$∴ μ = 3 / 4$

EAMCET-2008

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Question Bank Series

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