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PHXI05:LAWS OF MOTION

363113 A force of $F$ is applied as shown in figure. Find the tension in the string between $B C$ , if the friction force is negligible.
supporting img

1

F / 3

2

F / 2

3

F / 4

4

F / 5

Explanation:

Common acceleration of the system
supporting img
$a = \frac{F}{M_{1} + M_{2} + M_{3}} = \frac{F}{8 + 8 + 8} = \frac{F}{24}$
$T_{2} = \frac{M_{3} F}{M_{1} + M_{2} + M_{3}} = \frac{8 \times F}{24} = \frac{F}{3}$

PHXI05:LAWS OF MOTION

363114 Three masses are connected as shown in figure on a horizontal frictionless surface and pulled by a force of $60 N$ . The tensions $T_{1}$ and $T_{2}$ are in the ratio
supporting img

1

1 : 1

2

1 : 5

3

1 : 4

4

4 : 5

Explanation:

Horizontal component of force
$= 60 \cos 30^{\circ} = 30 \sqrt{3} N$
Acceleration of the system,
$a = \frac{f o r c e}{(3 + 12 + 15)} = \frac{30 \sqrt{3}}{30} = \sqrt{3} m / s^{2}$
$∴ T_{2} =$ mass pulled $\times$ acceleration
$= (3 + 12) \times \sqrt{3} = 15 \sqrt{3} N$ and
$T_{1} = 3 \times \sqrt{3} = 3 \sqrt{3} N$
$∴ \frac{T_{1}}{T_{2}} = \frac{3 \sqrt{3}}{15 \sqrt{3}} = \frac{1}{5}$

PHXI05:LAWS OF MOTION

363115 A block of mass 10 $k g$ is suspended through two light spring balances as shown in figure.
supporting img

1 Both the scales will read

10 k g

2 Both the scales will read

5 k g

3 The upper scale will read

10 k g

and the lower zero

4 The readings may be anything but their sum will be

10 k g

PHXI05:LAWS OF MOTION

363116 Two masses $M_{1}$ and $M_{2}$ are accelerated uniformly on a frictionless surface. The ratio of the tensions $T_{1} / T_{2}$ is
supporting img

1

\frac{M_{1}}{M_{1} + M_{2}}

2

\frac{M_{2}}{M_{1} + M_{2}}

3

\frac{M_{2}}{M_{1}}

4

\frac{M_{1} M_{2}}{M_{1} + M_{2}}

Explanation:

$T_{2} - T_{1} = M_{2} a (1)$
$T_{1} = M_{1} a (2)$
$T_{2} - T_{1} = M_{2} \times \frac{T_{1}}{M_{1}} ($ from eq (2))
$T_{2} = \frac{T_{1} M_{2}}{M_{1}} + T_{1} \Rightarrow T_{2} = T_{1} (1 + \frac{M_{2}}{M_{1}})$
$\frac{T_{1}}{T_{2}} = \frac{M_{1}}{M_{1} + M_{2}}$

PHXI05:LAWS OF MOTION

363113 A force of $F$ is applied as shown in figure. Find the tension in the string between $B C$ , if the friction force is negligible.
supporting img

1

F / 3

2

F / 2

3

F / 4

4

F / 5

Explanation:

Common acceleration of the system
supporting img
$a = \frac{F}{M_{1} + M_{2} + M_{3}} = \frac{F}{8 + 8 + 8} = \frac{F}{24}$
$T_{2} = \frac{M_{3} F}{M_{1} + M_{2} + M_{3}} = \frac{8 \times F}{24} = \frac{F}{3}$

PHXI05:LAWS OF MOTION

363114 Three masses are connected as shown in figure on a horizontal frictionless surface and pulled by a force of $60 N$ . The tensions $T_{1}$ and $T_{2}$ are in the ratio
supporting img

1

1 : 1

2

1 : 5

3

1 : 4

4

4 : 5

Explanation:

Horizontal component of force
$= 60 \cos 30^{\circ} = 30 \sqrt{3} N$
Acceleration of the system,
$a = \frac{f o r c e}{(3 + 12 + 15)} = \frac{30 \sqrt{3}}{30} = \sqrt{3} m / s^{2}$
$∴ T_{2} =$ mass pulled $\times$ acceleration
$= (3 + 12) \times \sqrt{3} = 15 \sqrt{3} N$ and
$T_{1} = 3 \times \sqrt{3} = 3 \sqrt{3} N$
$∴ \frac{T_{1}}{T_{2}} = \frac{3 \sqrt{3}}{15 \sqrt{3}} = \frac{1}{5}$

PHXI05:LAWS OF MOTION

363115 A block of mass 10 $k g$ is suspended through two light spring balances as shown in figure.
supporting img

1 Both the scales will read

10 k g

2 Both the scales will read

5 k g

3 The upper scale will read

10 k g

and the lower zero

4 The readings may be anything but their sum will be

10 k g

PHXI05:LAWS OF MOTION

363116 Two masses $M_{1}$ and $M_{2}$ are accelerated uniformly on a frictionless surface. The ratio of the tensions $T_{1} / T_{2}$ is
supporting img

1

\frac{M_{1}}{M_{1} + M_{2}}

2

\frac{M_{2}}{M_{1} + M_{2}}

3

\frac{M_{2}}{M_{1}}

4

\frac{M_{1} M_{2}}{M_{1} + M_{2}}

Explanation:

$T_{2} - T_{1} = M_{2} a (1)$
$T_{1} = M_{1} a (2)$
$T_{2} - T_{1} = M_{2} \times \frac{T_{1}}{M_{1}} ($ from eq (2))
$T_{2} = \frac{T_{1} M_{2}}{M_{1}} + T_{1} \Rightarrow T_{2} = T_{1} (1 + \frac{M_{2}}{M_{1}})$
$\frac{T_{1}}{T_{2}} = \frac{M_{1}}{M_{1} + M_{2}}$

PHXI05:LAWS OF MOTION

363113 A force of $F$ is applied as shown in figure. Find the tension in the string between $B C$ , if the friction force is negligible.
supporting img

1

F / 3

2

F / 2

3

F / 4

4

F / 5

Explanation:

Common acceleration of the system
supporting img
$a = \frac{F}{M_{1} + M_{2} + M_{3}} = \frac{F}{8 + 8 + 8} = \frac{F}{24}$
$T_{2} = \frac{M_{3} F}{M_{1} + M_{2} + M_{3}} = \frac{8 \times F}{24} = \frac{F}{3}$

PHXI05:LAWS OF MOTION

363114 Three masses are connected as shown in figure on a horizontal frictionless surface and pulled by a force of $60 N$ . The tensions $T_{1}$ and $T_{2}$ are in the ratio
supporting img

1

1 : 1

2

1 : 5

3

1 : 4

4

4 : 5

Explanation:

Horizontal component of force
$= 60 \cos 30^{\circ} = 30 \sqrt{3} N$
Acceleration of the system,
$a = \frac{f o r c e}{(3 + 12 + 15)} = \frac{30 \sqrt{3}}{30} = \sqrt{3} m / s^{2}$
$∴ T_{2} =$ mass pulled $\times$ acceleration
$= (3 + 12) \times \sqrt{3} = 15 \sqrt{3} N$ and
$T_{1} = 3 \times \sqrt{3} = 3 \sqrt{3} N$
$∴ \frac{T_{1}}{T_{2}} = \frac{3 \sqrt{3}}{15 \sqrt{3}} = \frac{1}{5}$

PHXI05:LAWS OF MOTION

363115 A block of mass 10 $k g$ is suspended through two light spring balances as shown in figure.
supporting img

1 Both the scales will read

10 k g

2 Both the scales will read

5 k g

3 The upper scale will read

10 k g

and the lower zero

4 The readings may be anything but their sum will be

10 k g

PHXI05:LAWS OF MOTION

363116 Two masses $M_{1}$ and $M_{2}$ are accelerated uniformly on a frictionless surface. The ratio of the tensions $T_{1} / T_{2}$ is
supporting img

1

\frac{M_{1}}{M_{1} + M_{2}}

2

\frac{M_{2}}{M_{1} + M_{2}}

3

\frac{M_{2}}{M_{1}}

4

\frac{M_{1} M_{2}}{M_{1} + M_{2}}

Explanation:

$T_{2} - T_{1} = M_{2} a (1)$
$T_{1} = M_{1} a (2)$
$T_{2} - T_{1} = M_{2} \times \frac{T_{1}}{M_{1}} ($ from eq (2))
$T_{2} = \frac{T_{1} M_{2}}{M_{1}} + T_{1} \Rightarrow T_{2} = T_{1} (1 + \frac{M_{2}}{M_{1}})$
$\frac{T_{1}}{T_{2}} = \frac{M_{1}}{M_{1} + M_{2}}$

PHXI05:LAWS OF MOTION

363113 A force of $F$ is applied as shown in figure. Find the tension in the string between $B C$ , if the friction force is negligible.
supporting img

1

F / 3

2

F / 2

3

F / 4

4

F / 5

Explanation:

Common acceleration of the system
supporting img
$a = \frac{F}{M_{1} + M_{2} + M_{3}} = \frac{F}{8 + 8 + 8} = \frac{F}{24}$
$T_{2} = \frac{M_{3} F}{M_{1} + M_{2} + M_{3}} = \frac{8 \times F}{24} = \frac{F}{3}$

PHXI05:LAWS OF MOTION

363114 Three masses are connected as shown in figure on a horizontal frictionless surface and pulled by a force of $60 N$ . The tensions $T_{1}$ and $T_{2}$ are in the ratio
supporting img

1

1 : 1

2

1 : 5

3

1 : 4

4

4 : 5

Explanation:

Horizontal component of force
$= 60 \cos 30^{\circ} = 30 \sqrt{3} N$
Acceleration of the system,
$a = \frac{f o r c e}{(3 + 12 + 15)} = \frac{30 \sqrt{3}}{30} = \sqrt{3} m / s^{2}$
$∴ T_{2} =$ mass pulled $\times$ acceleration
$= (3 + 12) \times \sqrt{3} = 15 \sqrt{3} N$ and
$T_{1} = 3 \times \sqrt{3} = 3 \sqrt{3} N$
$∴ \frac{T_{1}}{T_{2}} = \frac{3 \sqrt{3}}{15 \sqrt{3}} = \frac{1}{5}$

PHXI05:LAWS OF MOTION

363115 A block of mass 10 $k g$ is suspended through two light spring balances as shown in figure.
supporting img

1 Both the scales will read

10 k g

2 Both the scales will read

5 k g

3 The upper scale will read

10 k g

and the lower zero

4 The readings may be anything but their sum will be

10 k g

PHXI05:LAWS OF MOTION

363116 Two masses $M_{1}$ and $M_{2}$ are accelerated uniformly on a frictionless surface. The ratio of the tensions $T_{1} / T_{2}$ is
supporting img

1

\frac{M_{1}}{M_{1} + M_{2}}

2

\frac{M_{2}}{M_{1} + M_{2}}

3

\frac{M_{2}}{M_{1}}

4

\frac{M_{1} M_{2}}{M_{1} + M_{2}}

Explanation:

$T_{2} - T_{1} = M_{2} a (1)$
$T_{1} = M_{1} a (2)$
$T_{2} - T_{1} = M_{2} \times \frac{T_{1}}{M_{1}} ($ from eq (2))
$T_{2} = \frac{T_{1} M_{2}}{M_{1}} + T_{1} \Rightarrow T_{2} = T_{1} (1 + \frac{M_{2}}{M_{1}})$
$\frac{T_{1}}{T_{2}} = \frac{M_{1}}{M_{1} + M_{2}}$

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