QBManager

Laws of Motion

146009 A block of mass $8 kg$ is suspended by a rope of length $3 m$ from the ceiling. A force of $40 N$ is applied horizontally to the block. Then the angle that the rope makes with the vertical in equilibrium is
(acceleration due to gravity $= 10 {ms}^{- 2}$ , neglect the mass of the rope)

1

\sin^{- 1} (\frac{1}{2})

2

\tan^{- 1} (\frac{1}{2})

3

\sin^{- 1} (\frac{1}{3})

4

\tan^{- 1} (\frac{1}{3})

Explanation:

B Given, mass (m) $= 8 kg = 80 N$
Length of the rope $= 3 m$

Let $θ$ angle make by the rope with vertical for equilibrium-
From the figure,
$\sum F_{x} = 0$
$T \sin θ = 40$
And
$\sum F_{y} = 0$
$T \cos θ = mg$
$T \cos θ = 80$
From equation (i) / (ii), we get-
$\frac{T \sin θ}{T \cos θ} = \frac{40}{80}$
$\tan θ = \frac{40}{80}$
$θ = \tan^{- 1} (1 / 2)$

AP EAMCET-08.07.2022

Laws of Motion

146010 Three masses $M = 100 kg, m_{1} = 10 kg$ , and $m_{2} =$ $20 kg$ are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force $F$ is applied on the system so that the mass $m_{2}$ moves upward with an acceleration of $2 {ms}^{- 2}$ . The value of $F$ is :
$($ Take $g = 10 {ms}^{- 2}$ )

1

3360 N

2

3380 N

3

3120 N

4

3240 N

Explanation:

C In frame of block of mass M moving with acceleration a.
FBD of $10 kg$ block w.r.t. $100 kg$

$m_{1} a - T = 2 m_{1}$
$10 a - T = 20$
Now, FBD of $20 kg$ block w.r.t. $100 kg$ -

$T - m_{2} g = m_{2} \cdot 2$
$T - 20 \times 10 = 20 \times 2$
$T - 200 = 40$
$T = 240$
From equation (i) and (ii)
$10 a - 240 = 20$
$10 a = 260$
$a = 26 m / s^{2}$
For block $m_{2}$ -
$F = (M + m_{2}) a = 120 \times 26$
$F = 3120 N$

JEE Main-26.07.2022

Laws of Motion

146011 A block of mass $40 kg$ sliders over a surface, when a mass of $4 kg$ is suspended through an inextensible massless string passing over frictionless pulley as shown below. The coefficient of kinetic friction between the surface and block is $0.02$ . The acceleration of block is. (Given $g = 10 {ms}^{- 2}$ )

1

1 {ms}^{- 2}

2

1 / 5 {ms}^{- 2}

3

4 / 5 {ms}^{- 2}

4

8 / 11 {ms}^{- 2}

Explanation:

D Given, $μ = 0.02$
For $4 kg$ block,

$mg - T = ma$
$4 g - T = 4 a$
For $40 kg$ block,

$T - f = ma$
$T - μ N = ma$
$T - μ mg = ma$
$T - 0.02 \times 40 g = 40 a$
On adding equation (i) $&$ equation (ii), we get-
$40 - 8 = 44 a$
$44 a = 32$
$a = \frac{8}{11} m / s^{2}$

JEE Main-29.06.2022

Laws of Motion

146012 Two masses $M_{1}$ and $M_{2}$ are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass $M_{2}$ is twice that of $M_{1}$ , the acceleration of the system is $a_{1}$ . When the mass $M_{2}$ is thrice that of $M_{1}$ . The acceleration of the system is $a_{2}$ . The ratio $\frac{a_{1}}{a_{2}}$ will be:

1

\frac{1}{3}

2

\frac{2}{3}

3

\frac{3}{2}

4

\frac{1}{2}

Explanation:

B According to the question -
Case-I :
When, $M_{2} = 2 M_{1}$
Then, Acceleration $= a_{1}$

So, for $M_{1}$ block,
$T - M_{1} g = M_{1} a_{1}$
For $M_{2}$ block,
$M_{2} g - T = M_{2} a_{1}$
$2 M_{1} g - T = 2 M_{1} a_{1}$
On adding equation (i) \equation (ii), we get-
$M_{1} g = 3 M_{1} a_{1}$
$a_{1} = g / 3$
Case-II :
When, $M_{2} = 3 M_{1}$
Then, Acceleration $= a_{2}$
So, for block $M_{1}$ ,
For block $M_{2}$
$T - M_{1} g = M_{1} a_{2}$
$M_{2} g - T = M_{2} a_{2}$
$3 M_{1} g - T = 3 M_{1} a_{2}$
From eq $^{n}$ (i) \eq $^{n}$ (ii), we get-
$2 M_{1} g = 4 M_{1} a_{2}$
$a_{2} = g / 2$
Hence, $\frac{a_{1}}{a_{2}} = \frac{g / 3}{g / 2} = \frac{2}{3}$

JEE Main-26.07.2022

Laws of Motion

146013 A uniform metal chain of mass $M$ and length ' $L$ ' passes over a massless and frictionless pulley. It is released from rest with a part of its length ' $l$ ' is hanging on one side and rest of its length ' $L - l$ ' is hanging on the other side of the pulley. At a certain point of time, when $ℓ = \frac{L}{x}$ , the acceleration of the chain is $\frac{g}{2}$ . The value of $x$ is.....

1 6

2 2

3 1.5

4 4

Explanation:

D
We know,
Acceleration of chain is-
$a = \frac{(m_{2} - m_{1})}{(m_{2} + m_{1})} g$
Given, $a = \frac{g}{2}$
So, $\frac{g}{2} = \frac{(m_{2} - m_{1})}{(m_{2} + m_{1})} g$
$2 m_{2} - 2 m_{1} = m_{2} + m_{1}$
$3 m_{1} = m_{2}$
Now, from figure,
length)
$m_{2} = (L - l) λ ($ where, $λ =$ mass per unit
$m_{1} = l λ$
From equation (i), we get-
$3 l λ = (L - l) λ$
$4 l = L$
$l = \frac{L}{4}$
Compare to $l = \frac{L}{x}$ from question
We get, $x = 4$

JEE Main-28.07.2022

Laws of Motion

146009 A block of mass $8 kg$ is suspended by a rope of length $3 m$ from the ceiling. A force of $40 N$ is applied horizontally to the block. Then the angle that the rope makes with the vertical in equilibrium is
(acceleration due to gravity $= 10 {ms}^{- 2}$ , neglect the mass of the rope)

1

\sin^{- 1} (\frac{1}{2})

2

\tan^{- 1} (\frac{1}{2})

3

\sin^{- 1} (\frac{1}{3})

4

\tan^{- 1} (\frac{1}{3})

Explanation:

B Given, mass (m) $= 8 kg = 80 N$
Length of the rope $= 3 m$

Let $θ$ angle make by the rope with vertical for equilibrium-
From the figure,
$\sum F_{x} = 0$
$T \sin θ = 40$
And
$\sum F_{y} = 0$
$T \cos θ = mg$
$T \cos θ = 80$
From equation (i) / (ii), we get-
$\frac{T \sin θ}{T \cos θ} = \frac{40}{80}$
$\tan θ = \frac{40}{80}$
$θ = \tan^{- 1} (1 / 2)$

AP EAMCET-08.07.2022

Laws of Motion

146010 Three masses $M = 100 kg, m_{1} = 10 kg$ , and $m_{2} =$ $20 kg$ are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force $F$ is applied on the system so that the mass $m_{2}$ moves upward with an acceleration of $2 {ms}^{- 2}$ . The value of $F$ is :
$($ Take $g = 10 {ms}^{- 2}$ )

1

3360 N

2

3380 N

3

3120 N

4

3240 N

Explanation:

C In frame of block of mass M moving with acceleration a.
FBD of $10 kg$ block w.r.t. $100 kg$

$m_{1} a - T = 2 m_{1}$
$10 a - T = 20$
Now, FBD of $20 kg$ block w.r.t. $100 kg$ -

$T - m_{2} g = m_{2} \cdot 2$
$T - 20 \times 10 = 20 \times 2$
$T - 200 = 40$
$T = 240$
From equation (i) and (ii)
$10 a - 240 = 20$
$10 a = 260$
$a = 26 m / s^{2}$
For block $m_{2}$ -
$F = (M + m_{2}) a = 120 \times 26$
$F = 3120 N$

JEE Main-26.07.2022

Laws of Motion

146011 A block of mass $40 kg$ sliders over a surface, when a mass of $4 kg$ is suspended through an inextensible massless string passing over frictionless pulley as shown below. The coefficient of kinetic friction between the surface and block is $0.02$ . The acceleration of block is. (Given $g = 10 {ms}^{- 2}$ )

1

1 {ms}^{- 2}

2

1 / 5 {ms}^{- 2}

3

4 / 5 {ms}^{- 2}

4

8 / 11 {ms}^{- 2}

Explanation:

D Given, $μ = 0.02$
For $4 kg$ block,

$mg - T = ma$
$4 g - T = 4 a$
For $40 kg$ block,

$T - f = ma$
$T - μ N = ma$
$T - μ mg = ma$
$T - 0.02 \times 40 g = 40 a$
On adding equation (i) $&$ equation (ii), we get-
$40 - 8 = 44 a$
$44 a = 32$
$a = \frac{8}{11} m / s^{2}$

JEE Main-29.06.2022

Laws of Motion

146012 Two masses $M_{1}$ and $M_{2}$ are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass $M_{2}$ is twice that of $M_{1}$ , the acceleration of the system is $a_{1}$ . When the mass $M_{2}$ is thrice that of $M_{1}$ . The acceleration of the system is $a_{2}$ . The ratio $\frac{a_{1}}{a_{2}}$ will be:

1

\frac{1}{3}

2

\frac{2}{3}

3

\frac{3}{2}

4

\frac{1}{2}

Explanation:

B According to the question -
Case-I :
When, $M_{2} = 2 M_{1}$
Then, Acceleration $= a_{1}$

So, for $M_{1}$ block,
$T - M_{1} g = M_{1} a_{1}$
For $M_{2}$ block,
$M_{2} g - T = M_{2} a_{1}$
$2 M_{1} g - T = 2 M_{1} a_{1}$
On adding equation (i) \equation (ii), we get-
$M_{1} g = 3 M_{1} a_{1}$
$a_{1} = g / 3$
Case-II :
When, $M_{2} = 3 M_{1}$
Then, Acceleration $= a_{2}$
So, for block $M_{1}$ ,
For block $M_{2}$
$T - M_{1} g = M_{1} a_{2}$
$M_{2} g - T = M_{2} a_{2}$
$3 M_{1} g - T = 3 M_{1} a_{2}$
From eq $^{n}$ (i) \eq $^{n}$ (ii), we get-
$2 M_{1} g = 4 M_{1} a_{2}$
$a_{2} = g / 2$
Hence, $\frac{a_{1}}{a_{2}} = \frac{g / 3}{g / 2} = \frac{2}{3}$

JEE Main-26.07.2022

Laws of Motion

146013 A uniform metal chain of mass $M$ and length ' $L$ ' passes over a massless and frictionless pulley. It is released from rest with a part of its length ' $l$ ' is hanging on one side and rest of its length ' $L - l$ ' is hanging on the other side of the pulley. At a certain point of time, when $ℓ = \frac{L}{x}$ , the acceleration of the chain is $\frac{g}{2}$ . The value of $x$ is.....

1 6

2 2

3 1.5

4 4

Explanation:

D
We know,
Acceleration of chain is-
$a = \frac{(m_{2} - m_{1})}{(m_{2} + m_{1})} g$
Given, $a = \frac{g}{2}$
So, $\frac{g}{2} = \frac{(m_{2} - m_{1})}{(m_{2} + m_{1})} g$
$2 m_{2} - 2 m_{1} = m_{2} + m_{1}$
$3 m_{1} = m_{2}$
Now, from figure,
length)
$m_{2} = (L - l) λ ($ where, $λ =$ mass per unit
$m_{1} = l λ$
From equation (i), we get-
$3 l λ = (L - l) λ$
$4 l = L$
$l = \frac{L}{4}$
Compare to $l = \frac{L}{x}$ from question
We get, $x = 4$

JEE Main-28.07.2022

Laws of Motion

146009 A block of mass $8 kg$ is suspended by a rope of length $3 m$ from the ceiling. A force of $40 N$ is applied horizontally to the block. Then the angle that the rope makes with the vertical in equilibrium is
(acceleration due to gravity $= 10 {ms}^{- 2}$ , neglect the mass of the rope)

1

\sin^{- 1} (\frac{1}{2})

2

\tan^{- 1} (\frac{1}{2})

3

\sin^{- 1} (\frac{1}{3})

4

\tan^{- 1} (\frac{1}{3})

Explanation:

B Given, mass (m) $= 8 kg = 80 N$
Length of the rope $= 3 m$

Let $θ$ angle make by the rope with vertical for equilibrium-
From the figure,
$\sum F_{x} = 0$
$T \sin θ = 40$
And
$\sum F_{y} = 0$
$T \cos θ = mg$
$T \cos θ = 80$
From equation (i) / (ii), we get-
$\frac{T \sin θ}{T \cos θ} = \frac{40}{80}$
$\tan θ = \frac{40}{80}$
$θ = \tan^{- 1} (1 / 2)$

AP EAMCET-08.07.2022

Laws of Motion

146010 Three masses $M = 100 kg, m_{1} = 10 kg$ , and $m_{2} =$ $20 kg$ are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force $F$ is applied on the system so that the mass $m_{2}$ moves upward with an acceleration of $2 {ms}^{- 2}$ . The value of $F$ is :
$($ Take $g = 10 {ms}^{- 2}$ )

1

3360 N

2

3380 N

3

3120 N

4

3240 N

Explanation:

C In frame of block of mass M moving with acceleration a.
FBD of $10 kg$ block w.r.t. $100 kg$

$m_{1} a - T = 2 m_{1}$
$10 a - T = 20$
Now, FBD of $20 kg$ block w.r.t. $100 kg$ -

$T - m_{2} g = m_{2} \cdot 2$
$T - 20 \times 10 = 20 \times 2$
$T - 200 = 40$
$T = 240$
From equation (i) and (ii)
$10 a - 240 = 20$
$10 a = 260$
$a = 26 m / s^{2}$
For block $m_{2}$ -
$F = (M + m_{2}) a = 120 \times 26$
$F = 3120 N$

JEE Main-26.07.2022

Laws of Motion

146011 A block of mass $40 kg$ sliders over a surface, when a mass of $4 kg$ is suspended through an inextensible massless string passing over frictionless pulley as shown below. The coefficient of kinetic friction between the surface and block is $0.02$ . The acceleration of block is. (Given $g = 10 {ms}^{- 2}$ )

1

1 {ms}^{- 2}

2

1 / 5 {ms}^{- 2}

3

4 / 5 {ms}^{- 2}

4

8 / 11 {ms}^{- 2}

Explanation:

D Given, $μ = 0.02$
For $4 kg$ block,

$mg - T = ma$
$4 g - T = 4 a$
For $40 kg$ block,

$T - f = ma$
$T - μ N = ma$
$T - μ mg = ma$
$T - 0.02 \times 40 g = 40 a$
On adding equation (i) $&$ equation (ii), we get-
$40 - 8 = 44 a$
$44 a = 32$
$a = \frac{8}{11} m / s^{2}$

JEE Main-29.06.2022

Laws of Motion

146012 Two masses $M_{1}$ and $M_{2}$ are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass $M_{2}$ is twice that of $M_{1}$ , the acceleration of the system is $a_{1}$ . When the mass $M_{2}$ is thrice that of $M_{1}$ . The acceleration of the system is $a_{2}$ . The ratio $\frac{a_{1}}{a_{2}}$ will be:

1

\frac{1}{3}

2

\frac{2}{3}

3

\frac{3}{2}

4

\frac{1}{2}

Explanation:

B According to the question -
Case-I :
When, $M_{2} = 2 M_{1}$
Then, Acceleration $= a_{1}$

So, for $M_{1}$ block,
$T - M_{1} g = M_{1} a_{1}$
For $M_{2}$ block,
$M_{2} g - T = M_{2} a_{1}$
$2 M_{1} g - T = 2 M_{1} a_{1}$
On adding equation (i) \equation (ii), we get-
$M_{1} g = 3 M_{1} a_{1}$
$a_{1} = g / 3$
Case-II :
When, $M_{2} = 3 M_{1}$
Then, Acceleration $= a_{2}$
So, for block $M_{1}$ ,
For block $M_{2}$
$T - M_{1} g = M_{1} a_{2}$
$M_{2} g - T = M_{2} a_{2}$
$3 M_{1} g - T = 3 M_{1} a_{2}$
From eq $^{n}$ (i) \eq $^{n}$ (ii), we get-
$2 M_{1} g = 4 M_{1} a_{2}$
$a_{2} = g / 2$
Hence, $\frac{a_{1}}{a_{2}} = \frac{g / 3}{g / 2} = \frac{2}{3}$

JEE Main-26.07.2022

Laws of Motion

146013 A uniform metal chain of mass $M$ and length ' $L$ ' passes over a massless and frictionless pulley. It is released from rest with a part of its length ' $l$ ' is hanging on one side and rest of its length ' $L - l$ ' is hanging on the other side of the pulley. At a certain point of time, when $ℓ = \frac{L}{x}$ , the acceleration of the chain is $\frac{g}{2}$ . The value of $x$ is.....

1 6

2 2

3 1.5

4 4

Explanation:

D
We know,
Acceleration of chain is-
$a = \frac{(m_{2} - m_{1})}{(m_{2} + m_{1})} g$
Given, $a = \frac{g}{2}$
So, $\frac{g}{2} = \frac{(m_{2} - m_{1})}{(m_{2} + m_{1})} g$
$2 m_{2} - 2 m_{1} = m_{2} + m_{1}$
$3 m_{1} = m_{2}$
Now, from figure,
length)
$m_{2} = (L - l) λ ($ where, $λ =$ mass per unit
$m_{1} = l λ$
From equation (i), we get-
$3 l λ = (L - l) λ$
$4 l = L$
$l = \frac{L}{4}$
Compare to $l = \frac{L}{x}$ from question
We get, $x = 4$

JEE Main-28.07.2022

Laws of Motion

146009 A block of mass $8 kg$ is suspended by a rope of length $3 m$ from the ceiling. A force of $40 N$ is applied horizontally to the block. Then the angle that the rope makes with the vertical in equilibrium is
(acceleration due to gravity $= 10 {ms}^{- 2}$ , neglect the mass of the rope)

1

\sin^{- 1} (\frac{1}{2})

2

\tan^{- 1} (\frac{1}{2})

3

\sin^{- 1} (\frac{1}{3})

4

\tan^{- 1} (\frac{1}{3})

Explanation:

B Given, mass (m) $= 8 kg = 80 N$
Length of the rope $= 3 m$

Let $θ$ angle make by the rope with vertical for equilibrium-
From the figure,
$\sum F_{x} = 0$
$T \sin θ = 40$
And
$\sum F_{y} = 0$
$T \cos θ = mg$
$T \cos θ = 80$
From equation (i) / (ii), we get-
$\frac{T \sin θ}{T \cos θ} = \frac{40}{80}$
$\tan θ = \frac{40}{80}$
$θ = \tan^{- 1} (1 / 2)$

AP EAMCET-08.07.2022

Laws of Motion

146010 Three masses $M = 100 kg, m_{1} = 10 kg$ , and $m_{2} =$ $20 kg$ are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force $F$ is applied on the system so that the mass $m_{2}$ moves upward with an acceleration of $2 {ms}^{- 2}$ . The value of $F$ is :
$($ Take $g = 10 {ms}^{- 2}$ )

1

3360 N

2

3380 N

3

3120 N

4

3240 N

Explanation:

C In frame of block of mass M moving with acceleration a.
FBD of $10 kg$ block w.r.t. $100 kg$

$m_{1} a - T = 2 m_{1}$
$10 a - T = 20$
Now, FBD of $20 kg$ block w.r.t. $100 kg$ -

$T - m_{2} g = m_{2} \cdot 2$
$T - 20 \times 10 = 20 \times 2$
$T - 200 = 40$
$T = 240$
From equation (i) and (ii)
$10 a - 240 = 20$
$10 a = 260$
$a = 26 m / s^{2}$
For block $m_{2}$ -
$F = (M + m_{2}) a = 120 \times 26$
$F = 3120 N$

JEE Main-26.07.2022

Laws of Motion

146011 A block of mass $40 kg$ sliders over a surface, when a mass of $4 kg$ is suspended through an inextensible massless string passing over frictionless pulley as shown below. The coefficient of kinetic friction between the surface and block is $0.02$ . The acceleration of block is. (Given $g = 10 {ms}^{- 2}$ )

1

1 {ms}^{- 2}

2

1 / 5 {ms}^{- 2}

3

4 / 5 {ms}^{- 2}

4

8 / 11 {ms}^{- 2}

Explanation:

D Given, $μ = 0.02$
For $4 kg$ block,

$mg - T = ma$
$4 g - T = 4 a$
For $40 kg$ block,

$T - f = ma$
$T - μ N = ma$
$T - μ mg = ma$
$T - 0.02 \times 40 g = 40 a$
On adding equation (i) $&$ equation (ii), we get-
$40 - 8 = 44 a$
$44 a = 32$
$a = \frac{8}{11} m / s^{2}$

JEE Main-29.06.2022

Laws of Motion

146012 Two masses $M_{1}$ and $M_{2}$ are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass $M_{2}$ is twice that of $M_{1}$ , the acceleration of the system is $a_{1}$ . When the mass $M_{2}$ is thrice that of $M_{1}$ . The acceleration of the system is $a_{2}$ . The ratio $\frac{a_{1}}{a_{2}}$ will be:

1

\frac{1}{3}

2

\frac{2}{3}

3

\frac{3}{2}

4

\frac{1}{2}

Explanation:

B According to the question -
Case-I :
When, $M_{2} = 2 M_{1}$
Then, Acceleration $= a_{1}$

So, for $M_{1}$ block,
$T - M_{1} g = M_{1} a_{1}$
For $M_{2}$ block,
$M_{2} g - T = M_{2} a_{1}$
$2 M_{1} g - T = 2 M_{1} a_{1}$
On adding equation (i) \equation (ii), we get-
$M_{1} g = 3 M_{1} a_{1}$
$a_{1} = g / 3$
Case-II :
When, $M_{2} = 3 M_{1}$
Then, Acceleration $= a_{2}$
So, for block $M_{1}$ ,
For block $M_{2}$
$T - M_{1} g = M_{1} a_{2}$
$M_{2} g - T = M_{2} a_{2}$
$3 M_{1} g - T = 3 M_{1} a_{2}$
From eq $^{n}$ (i) \eq $^{n}$ (ii), we get-
$2 M_{1} g = 4 M_{1} a_{2}$
$a_{2} = g / 2$
Hence, $\frac{a_{1}}{a_{2}} = \frac{g / 3}{g / 2} = \frac{2}{3}$

JEE Main-26.07.2022

Laws of Motion

146013 A uniform metal chain of mass $M$ and length ' $L$ ' passes over a massless and frictionless pulley. It is released from rest with a part of its length ' $l$ ' is hanging on one side and rest of its length ' $L - l$ ' is hanging on the other side of the pulley. At a certain point of time, when $ℓ = \frac{L}{x}$ , the acceleration of the chain is $\frac{g}{2}$ . The value of $x$ is.....

1 6

2 2

3 1.5

4 4

Explanation:

D
We know,
Acceleration of chain is-
$a = \frac{(m_{2} - m_{1})}{(m_{2} + m_{1})} g$
Given, $a = \frac{g}{2}$
So, $\frac{g}{2} = \frac{(m_{2} - m_{1})}{(m_{2} + m_{1})} g$
$2 m_{2} - 2 m_{1} = m_{2} + m_{1}$
$3 m_{1} = m_{2}$
Now, from figure,
length)
$m_{2} = (L - l) λ ($ where, $λ =$ mass per unit
$m_{1} = l λ$
From equation (i), we get-
$3 l λ = (L - l) λ$
$4 l = L$
$l = \frac{L}{4}$
Compare to $l = \frac{L}{x}$ from question
We get, $x = 4$

JEE Main-28.07.2022

Laws of Motion

146009 A block of mass $8 kg$ is suspended by a rope of length $3 m$ from the ceiling. A force of $40 N$ is applied horizontally to the block. Then the angle that the rope makes with the vertical in equilibrium is
(acceleration due to gravity $= 10 {ms}^{- 2}$ , neglect the mass of the rope)

1

\sin^{- 1} (\frac{1}{2})

2

\tan^{- 1} (\frac{1}{2})

3

\sin^{- 1} (\frac{1}{3})

4

\tan^{- 1} (\frac{1}{3})

Explanation:

B Given, mass (m) $= 8 kg = 80 N$
Length of the rope $= 3 m$

Let $θ$ angle make by the rope with vertical for equilibrium-
From the figure,
$\sum F_{x} = 0$
$T \sin θ = 40$
And
$\sum F_{y} = 0$
$T \cos θ = mg$
$T \cos θ = 80$
From equation (i) / (ii), we get-
$\frac{T \sin θ}{T \cos θ} = \frac{40}{80}$
$\tan θ = \frac{40}{80}$
$θ = \tan^{- 1} (1 / 2)$

AP EAMCET-08.07.2022

Laws of Motion

146010 Three masses $M = 100 kg, m_{1} = 10 kg$ , and $m_{2} =$ $20 kg$ are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force $F$ is applied on the system so that the mass $m_{2}$ moves upward with an acceleration of $2 {ms}^{- 2}$ . The value of $F$ is :
$($ Take $g = 10 {ms}^{- 2}$ )

1

3360 N

2

3380 N

3

3120 N

4

3240 N

Explanation:

C In frame of block of mass M moving with acceleration a.
FBD of $10 kg$ block w.r.t. $100 kg$

$m_{1} a - T = 2 m_{1}$
$10 a - T = 20$
Now, FBD of $20 kg$ block w.r.t. $100 kg$ -

$T - m_{2} g = m_{2} \cdot 2$
$T - 20 \times 10 = 20 \times 2$
$T - 200 = 40$
$T = 240$
From equation (i) and (ii)
$10 a - 240 = 20$
$10 a = 260$
$a = 26 m / s^{2}$
For block $m_{2}$ -
$F = (M + m_{2}) a = 120 \times 26$
$F = 3120 N$

JEE Main-26.07.2022

Laws of Motion

146011 A block of mass $40 kg$ sliders over a surface, when a mass of $4 kg$ is suspended through an inextensible massless string passing over frictionless pulley as shown below. The coefficient of kinetic friction between the surface and block is $0.02$ . The acceleration of block is. (Given $g = 10 {ms}^{- 2}$ )

1

1 {ms}^{- 2}

2

1 / 5 {ms}^{- 2}

3

4 / 5 {ms}^{- 2}

4

8 / 11 {ms}^{- 2}

Explanation:

D Given, $μ = 0.02$
For $4 kg$ block,

$mg - T = ma$
$4 g - T = 4 a$
For $40 kg$ block,

$T - f = ma$
$T - μ N = ma$
$T - μ mg = ma$
$T - 0.02 \times 40 g = 40 a$
On adding equation (i) $&$ equation (ii), we get-
$40 - 8 = 44 a$
$44 a = 32$
$a = \frac{8}{11} m / s^{2}$

JEE Main-29.06.2022

Laws of Motion

146012 Two masses $M_{1}$ and $M_{2}$ are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass $M_{2}$ is twice that of $M_{1}$ , the acceleration of the system is $a_{1}$ . When the mass $M_{2}$ is thrice that of $M_{1}$ . The acceleration of the system is $a_{2}$ . The ratio $\frac{a_{1}}{a_{2}}$ will be:

1

\frac{1}{3}

2

\frac{2}{3}

3

\frac{3}{2}

4

\frac{1}{2}

Explanation:

B According to the question -
Case-I :
When, $M_{2} = 2 M_{1}$
Then, Acceleration $= a_{1}$

So, for $M_{1}$ block,
$T - M_{1} g = M_{1} a_{1}$
For $M_{2}$ block,
$M_{2} g - T = M_{2} a_{1}$
$2 M_{1} g - T = 2 M_{1} a_{1}$
On adding equation (i) \equation (ii), we get-
$M_{1} g = 3 M_{1} a_{1}$
$a_{1} = g / 3$
Case-II :
When, $M_{2} = 3 M_{1}$
Then, Acceleration $= a_{2}$
So, for block $M_{1}$ ,
For block $M_{2}$
$T - M_{1} g = M_{1} a_{2}$
$M_{2} g - T = M_{2} a_{2}$
$3 M_{1} g - T = 3 M_{1} a_{2}$
From eq $^{n}$ (i) \eq $^{n}$ (ii), we get-
$2 M_{1} g = 4 M_{1} a_{2}$
$a_{2} = g / 2$
Hence, $\frac{a_{1}}{a_{2}} = \frac{g / 3}{g / 2} = \frac{2}{3}$

JEE Main-26.07.2022

Laws of Motion

146013 A uniform metal chain of mass $M$ and length ' $L$ ' passes over a massless and frictionless pulley. It is released from rest with a part of its length ' $l$ ' is hanging on one side and rest of its length ' $L - l$ ' is hanging on the other side of the pulley. At a certain point of time, when $ℓ = \frac{L}{x}$ , the acceleration of the chain is $\frac{g}{2}$ . The value of $x$ is.....

1 6

2 2

3 1.5

4 4

Explanation:

D
We know,
Acceleration of chain is-
$a = \frac{(m_{2} - m_{1})}{(m_{2} + m_{1})} g$
Given, $a = \frac{g}{2}$
So, $\frac{g}{2} = \frac{(m_{2} - m_{1})}{(m_{2} + m_{1})} g$
$2 m_{2} - 2 m_{1} = m_{2} + m_{1}$
$3 m_{1} = m_{2}$
Now, from figure,
length)
$m_{2} = (L - l) λ ($ where, $λ =$ mass per unit
$m_{1} = l λ$
From equation (i), we get-
$3 l λ = (L - l) λ$
$4 l = L$
$l = \frac{L}{4}$
Compare to $l = \frac{L}{x}$ from question
We get, $x = 4$

JEE Main-28.07.2022