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PHXI05:LAWS OF MOTION

363083 The tension in the string in the pulley system shown in the figure is
supporting img

1

75 N

2

80 N

3

7.5 N

4

30 N

Explanation:

Equations for the given system For $10 k g$
mass,
$10 g - T = 10 a (1)$
For $6 k g m a s s, T - 6 g = 6 a (2)$

supporting img

From Eqs. (1) and (2), we get
$T = 75 N$

PHXI05:LAWS OF MOTION

363084 A light unstretchable string passing over a smooth light pulley connects two blocks of masses $m_{1}$ and $m_{2}$ . If the acceleration of the system is $\frac{g}{8}$ , then the ratio of the masses $\frac{m_{2}}{m_{1}}$ is

1

5 : 3

2

8 : 1

3

9 : 7

4

4 : 3

Explanation:

supporting img

From free body diagram $m_{2} g - T = m_{2} a$ (1)
$T - m_{2} g = m_{2} a$ $(2)$
From equation (1) and (2), we get
$\begin{aligned} (m_{2} - m_{1}) g = (m_{1} + m_{2}) a \\ or a = \frac{m_{2} - m_{1}}{m_{1} + m_{2}} g \Rightarrow \frac{g}{8} = \frac{(m_{2} - m_{1}) g}{m_{1} + m_{2}} \\ or m_{1} + m_{2} = 8 m_{2} - 8 m_{1} \\ \Rightarrow 9 m_{1} = 7 m_{2} \end{aligned}$

JEE - 2024

PHXI05:LAWS OF MOTION

363085 Two monkeys $M_{1}$ and $M_{2}$ of equal mass ' $m$ ' can climb strings of a pulley arrangement as shown in figure. Find magnitude of acceleration (in $m / s^{2}$ ) of $M_{1}$ with respect to rope so that block remains stationary. It is given that $M_{2}$ is just holding the string. Assume pulley is frictionless and string is massless and inextensible.
supporting img

1

210 m / s^{2}

2

110 m / s^{2}

3

320 m / s^{2}

4

160 m / s^{2}

Explanation:

$T = 5 m g$
$a_{2} = \frac{5 m g - m g}{m} = 4 g$
$T^{'} = 2 T = 10 m g$
$a_{1} = \frac{10 m g - m g}{m} = 9 g$
Acceleration of the rope is equal to the acceleration of the moving pulley. Hence acceleration of $M_{1}$ with respect to rope
supporting img
$a_{M_{1}, R o p e} = 9 g - (- 2 g) = 11 g = 110 m / s^{2}$

PHXI05:LAWS OF MOTION

363086 A light rope passes over a pulley. One section of the rope is held by a child and the other section by a man, then
supporting img

1 The man and the child have same vector acceleration

2 The man and the child have same magnitude of acceleration but in opposite direction

3 The man and the child have different magnitude of acceleration

4 The man and the child have accelerations which keep on interchanging with each other

PHXI05:LAWS OF MOTION

363083 The tension in the string in the pulley system shown in the figure is
supporting img

1

75 N

2

80 N

3

7.5 N

4

30 N

Explanation:

Equations for the given system For $10 k g$
mass,
$10 g - T = 10 a (1)$
For $6 k g m a s s, T - 6 g = 6 a (2)$

supporting img

From Eqs. (1) and (2), we get
$T = 75 N$

PHXI05:LAWS OF MOTION

363084 A light unstretchable string passing over a smooth light pulley connects two blocks of masses $m_{1}$ and $m_{2}$ . If the acceleration of the system is $\frac{g}{8}$ , then the ratio of the masses $\frac{m_{2}}{m_{1}}$ is

1

5 : 3

2

8 : 1

3

9 : 7

4

4 : 3

Explanation:

supporting img

From free body diagram $m_{2} g - T = m_{2} a$ (1)
$T - m_{2} g = m_{2} a$ $(2)$
From equation (1) and (2), we get
$\begin{aligned} (m_{2} - m_{1}) g = (m_{1} + m_{2}) a \\ or a = \frac{m_{2} - m_{1}}{m_{1} + m_{2}} g \Rightarrow \frac{g}{8} = \frac{(m_{2} - m_{1}) g}{m_{1} + m_{2}} \\ or m_{1} + m_{2} = 8 m_{2} - 8 m_{1} \\ \Rightarrow 9 m_{1} = 7 m_{2} \end{aligned}$

JEE - 2024

PHXI05:LAWS OF MOTION

363085 Two monkeys $M_{1}$ and $M_{2}$ of equal mass ' $m$ ' can climb strings of a pulley arrangement as shown in figure. Find magnitude of acceleration (in $m / s^{2}$ ) of $M_{1}$ with respect to rope so that block remains stationary. It is given that $M_{2}$ is just holding the string. Assume pulley is frictionless and string is massless and inextensible.
supporting img

1

210 m / s^{2}

2

110 m / s^{2}

3

320 m / s^{2}

4

160 m / s^{2}

Explanation:

$T = 5 m g$
$a_{2} = \frac{5 m g - m g}{m} = 4 g$
$T^{'} = 2 T = 10 m g$
$a_{1} = \frac{10 m g - m g}{m} = 9 g$
Acceleration of the rope is equal to the acceleration of the moving pulley. Hence acceleration of $M_{1}$ with respect to rope
supporting img
$a_{M_{1}, R o p e} = 9 g - (- 2 g) = 11 g = 110 m / s^{2}$

PHXI05:LAWS OF MOTION

363086 A light rope passes over a pulley. One section of the rope is held by a child and the other section by a man, then
supporting img

1 The man and the child have same vector acceleration

2 The man and the child have same magnitude of acceleration but in opposite direction

3 The man and the child have different magnitude of acceleration

4 The man and the child have accelerations which keep on interchanging with each other

PHXI05:LAWS OF MOTION

363083 The tension in the string in the pulley system shown in the figure is
supporting img

1

75 N

2

80 N

3

7.5 N

4

30 N

Explanation:

Equations for the given system For $10 k g$
mass,
$10 g - T = 10 a (1)$
For $6 k g m a s s, T - 6 g = 6 a (2)$

supporting img

From Eqs. (1) and (2), we get
$T = 75 N$

PHXI05:LAWS OF MOTION

363084 A light unstretchable string passing over a smooth light pulley connects two blocks of masses $m_{1}$ and $m_{2}$ . If the acceleration of the system is $\frac{g}{8}$ , then the ratio of the masses $\frac{m_{2}}{m_{1}}$ is

1

5 : 3

2

8 : 1

3

9 : 7

4

4 : 3

Explanation:

supporting img

From free body diagram $m_{2} g - T = m_{2} a$ (1)
$T - m_{2} g = m_{2} a$ $(2)$
From equation (1) and (2), we get
$\begin{aligned} (m_{2} - m_{1}) g = (m_{1} + m_{2}) a \\ or a = \frac{m_{2} - m_{1}}{m_{1} + m_{2}} g \Rightarrow \frac{g}{8} = \frac{(m_{2} - m_{1}) g}{m_{1} + m_{2}} \\ or m_{1} + m_{2} = 8 m_{2} - 8 m_{1} \\ \Rightarrow 9 m_{1} = 7 m_{2} \end{aligned}$

JEE - 2024

PHXI05:LAWS OF MOTION

363085 Two monkeys $M_{1}$ and $M_{2}$ of equal mass ' $m$ ' can climb strings of a pulley arrangement as shown in figure. Find magnitude of acceleration (in $m / s^{2}$ ) of $M_{1}$ with respect to rope so that block remains stationary. It is given that $M_{2}$ is just holding the string. Assume pulley is frictionless and string is massless and inextensible.
supporting img

1

210 m / s^{2}

2

110 m / s^{2}

3

320 m / s^{2}

4

160 m / s^{2}

Explanation:

$T = 5 m g$
$a_{2} = \frac{5 m g - m g}{m} = 4 g$
$T^{'} = 2 T = 10 m g$
$a_{1} = \frac{10 m g - m g}{m} = 9 g$
Acceleration of the rope is equal to the acceleration of the moving pulley. Hence acceleration of $M_{1}$ with respect to rope
supporting img
$a_{M_{1}, R o p e} = 9 g - (- 2 g) = 11 g = 110 m / s^{2}$

PHXI05:LAWS OF MOTION

363086 A light rope passes over a pulley. One section of the rope is held by a child and the other section by a man, then
supporting img

1 The man and the child have same vector acceleration

2 The man and the child have same magnitude of acceleration but in opposite direction

3 The man and the child have different magnitude of acceleration

4 The man and the child have accelerations which keep on interchanging with each other

PHXI05:LAWS OF MOTION

363083 The tension in the string in the pulley system shown in the figure is
supporting img

1

75 N

2

80 N

3

7.5 N

4

30 N

Explanation:

Equations for the given system For $10 k g$
mass,
$10 g - T = 10 a (1)$
For $6 k g m a s s, T - 6 g = 6 a (2)$

supporting img

From Eqs. (1) and (2), we get
$T = 75 N$

PHXI05:LAWS OF MOTION

363084 A light unstretchable string passing over a smooth light pulley connects two blocks of masses $m_{1}$ and $m_{2}$ . If the acceleration of the system is $\frac{g}{8}$ , then the ratio of the masses $\frac{m_{2}}{m_{1}}$ is

1

5 : 3

2

8 : 1

3

9 : 7

4

4 : 3

Explanation:

supporting img

From free body diagram $m_{2} g - T = m_{2} a$ (1)
$T - m_{2} g = m_{2} a$ $(2)$
From equation (1) and (2), we get
$\begin{aligned} (m_{2} - m_{1}) g = (m_{1} + m_{2}) a \\ or a = \frac{m_{2} - m_{1}}{m_{1} + m_{2}} g \Rightarrow \frac{g}{8} = \frac{(m_{2} - m_{1}) g}{m_{1} + m_{2}} \\ or m_{1} + m_{2} = 8 m_{2} - 8 m_{1} \\ \Rightarrow 9 m_{1} = 7 m_{2} \end{aligned}$

JEE - 2024

PHXI05:LAWS OF MOTION

363085 Two monkeys $M_{1}$ and $M_{2}$ of equal mass ' $m$ ' can climb strings of a pulley arrangement as shown in figure. Find magnitude of acceleration (in $m / s^{2}$ ) of $M_{1}$ with respect to rope so that block remains stationary. It is given that $M_{2}$ is just holding the string. Assume pulley is frictionless and string is massless and inextensible.
supporting img

1

210 m / s^{2}

2

110 m / s^{2}

3

320 m / s^{2}

4

160 m / s^{2}

Explanation:

$T = 5 m g$
$a_{2} = \frac{5 m g - m g}{m} = 4 g$
$T^{'} = 2 T = 10 m g$
$a_{1} = \frac{10 m g - m g}{m} = 9 g$
Acceleration of the rope is equal to the acceleration of the moving pulley. Hence acceleration of $M_{1}$ with respect to rope
supporting img
$a_{M_{1}, R o p e} = 9 g - (- 2 g) = 11 g = 110 m / s^{2}$

PHXI05:LAWS OF MOTION

363086 A light rope passes over a pulley. One section of the rope is held by a child and the other section by a man, then
supporting img

1 The man and the child have same vector acceleration

2 The man and the child have same magnitude of acceleration but in opposite direction

3 The man and the child have different magnitude of acceleration

4 The man and the child have accelerations which keep on interchanging with each other

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