QBManager

Laws of Motion

145852 When a metal wire of length ' $l$ ' is subjected to tensions $T_{1}$ and $T_{2}$ respectively its length changes to $l_{1}$ and $l_{2}$ , then the relation of ' $l$ ' is correctly given by

1

l \sqrt{l_{1} - l_{2}}

2

l = \frac{1}{2} (l_{1} + l_{2})

3

l = \frac{l_{1} T_{2} + l_{2} T_{1}}{T_{1} + T_{2}}

4

l = \frac{l_{1} T_{2} - l_{2} T_{1}}{T_{1} - T_{2}}

Explanation:

D Given,
Tension is $T_{1}$ when length is $l_{1}$ and when, length is $l_{2}$ , tension is $T_{2}$ .
Let natural length of wire is $l$
$∵$ Young's modules $(E) = \frac{Stress}{Strain}$
$E = \frac{T \times l}{A \times Δ l}$
(i) When length is $l_{1}$ and tension is $T_{1}$
$E = \frac{T_{1} \times l}{A (l_{1} - l)}$
(ii) When length is $l_{2}$ and tension is $T_{2}$
$E = \frac{T_{2} \times l}{A (l_{2} - l)}$
$∵ E$ is equal in both case,
$\frac{T_{1} \times l}{A (l_{1} - l)} = \frac{T_{2} \times l}{A (l_{2} - l)}$
$T_{1} l_{2} - T_{1} l = T_{2} l_{1} - T_{2} l$
$l (T_{2} - T_{1}) = T_{2} l_{1} - T_{1} l_{2}$
$l = \frac{T_{2} l_{1} - T_{1} l_{2}}{T_{2} - T_{1}}$

Tripura-2021

Laws of Motion

145853 The ratio of weights of a man inside a lift when it is stationary and when it is going down with a uniform acceleration ' $a$ ' is $3 : 2$ . The value of ' $a$ ' will be $(a < g, g =$ acceleration due to gravity)

1

\frac{3}{2} g

2

\frac{g}{3}

3

g

4

\frac{2}{3} g

Explanation:

B When the lift is stationary the weight of the $man$ is $w_{o} = mg$
The weight of the man when the lift is moving downward with acceleration a is $w = mg - ma$
$\frac{w_{o}}{w} = \frac{3}{2} (given)$
Therefore, $\frac{mg}{mg - ma} = \frac{3}{2}$
$\frac{m g}{m (g - a)} = \frac{3}{2}$
$2 g = 3 g - 3 a$
$g = 3 a$
$So, a = \frac{g}{3}$

MHT-CET 2020

Laws of Motion

145854 A block of mass $m$ is resting on a smooth horizontal surface. One end of a uniform rope of mass $(\frac{m}{3})$ is fixed to the block, which is pulled in the horizontal direction by applying force $F$ at the other end. The tension in the middle of the rope is

1

\frac{8}{7} F

2

\frac{1}{7} F

3

\frac{1}{8} F

4

\frac{1}{5} F

5

\frac{7}{8} F

Explanation:

E
$F = m_{1} a$
$a = (\frac{F}{m + \frac{m}{3}}) = \frac{3 F}{4 m}$
$∴ Now, Tension at middle of rope$
$∴ T = (m + \frac{m}{6}) (\frac{F}{m + \frac{m}{3}}) = \frac{7 m}{6} \times \frac{3 F}{4 m}$
$T = \frac{7 F}{8}$

BECEC - 2018

Laws of Motion

145855 A monkey of mass $20 kg$ is holding a vertical rope. The rope will not break, when a mass of $25 kg$ is suspended from it but will break, if the mass exceeds $25 kg$ . What is the maximum acceleration with which the monkey can climb up along the rope? $($ Take $g = 10 m / s^{2}$ )

1

25 m / s^{2}

2

2.5 m / s^{2}

3

5 m / s^{2}

4

10 m / s^{2}

Explanation:

B Case I:- Maximum mass $25 kg$ is applied.

$∴ T_{max} = 25 \times 10 = 250 N$
(breaking force)
Case I: mass of monkey of $20 kg$ moving upward with acceleration a

We know,
$T_{max} = mg + ma$
$250 = 20 g + 20 a$
$250 = 200 + 20 a$
$20 a = 250 - 200$
$a = \frac{50}{20}$
$a = 2.5 m / s^{2}$

AIPMT- 2003

Laws of Motion

145856 A lift of mass $1000 kg$ is moving upwards with an acceleration of $1 m / s^{2}$ . The tension developed in the string, which is connected to lift is
$(g = 9.8 m / s^{2})$

1

9800 N

2

10800 N

3

11000 N

4

10000 N

Explanation:

B Given that,
$m = 1000 kg, a = 1 m / s^{2}$

String connected to the lift (upward) $T = mg + ma$
$T = m (g + a)$
$T = 1000 (9.8 + 1)$
$T = 10800 N$

AIPMT- 2002

Laws of Motion

145852 When a metal wire of length ' $l$ ' is subjected to tensions $T_{1}$ and $T_{2}$ respectively its length changes to $l_{1}$ and $l_{2}$ , then the relation of ' $l$ ' is correctly given by

1

l \sqrt{l_{1} - l_{2}}

2

l = \frac{1}{2} (l_{1} + l_{2})

3

l = \frac{l_{1} T_{2} + l_{2} T_{1}}{T_{1} + T_{2}}

4

l = \frac{l_{1} T_{2} - l_{2} T_{1}}{T_{1} - T_{2}}

Explanation:

D Given,
Tension is $T_{1}$ when length is $l_{1}$ and when, length is $l_{2}$ , tension is $T_{2}$ .
Let natural length of wire is $l$
$∵$ Young's modules $(E) = \frac{Stress}{Strain}$
$E = \frac{T \times l}{A \times Δ l}$
(i) When length is $l_{1}$ and tension is $T_{1}$
$E = \frac{T_{1} \times l}{A (l_{1} - l)}$
(ii) When length is $l_{2}$ and tension is $T_{2}$
$E = \frac{T_{2} \times l}{A (l_{2} - l)}$
$∵ E$ is equal in both case,
$\frac{T_{1} \times l}{A (l_{1} - l)} = \frac{T_{2} \times l}{A (l_{2} - l)}$
$T_{1} l_{2} - T_{1} l = T_{2} l_{1} - T_{2} l$
$l (T_{2} - T_{1}) = T_{2} l_{1} - T_{1} l_{2}$
$l = \frac{T_{2} l_{1} - T_{1} l_{2}}{T_{2} - T_{1}}$

Tripura-2021

Laws of Motion

145853 The ratio of weights of a man inside a lift when it is stationary and when it is going down with a uniform acceleration ' $a$ ' is $3 : 2$ . The value of ' $a$ ' will be $(a < g, g =$ acceleration due to gravity)

1

\frac{3}{2} g

2

\frac{g}{3}

3

g

4

\frac{2}{3} g

Explanation:

B When the lift is stationary the weight of the $man$ is $w_{o} = mg$
The weight of the man when the lift is moving downward with acceleration a is $w = mg - ma$
$\frac{w_{o}}{w} = \frac{3}{2} (given)$
Therefore, $\frac{mg}{mg - ma} = \frac{3}{2}$
$\frac{m g}{m (g - a)} = \frac{3}{2}$
$2 g = 3 g - 3 a$
$g = 3 a$
$So, a = \frac{g}{3}$

MHT-CET 2020

Laws of Motion

145854 A block of mass $m$ is resting on a smooth horizontal surface. One end of a uniform rope of mass $(\frac{m}{3})$ is fixed to the block, which is pulled in the horizontal direction by applying force $F$ at the other end. The tension in the middle of the rope is

1

\frac{8}{7} F

2

\frac{1}{7} F

3

\frac{1}{8} F

4

\frac{1}{5} F

5

\frac{7}{8} F

Explanation:

E
$F = m_{1} a$
$a = (\frac{F}{m + \frac{m}{3}}) = \frac{3 F}{4 m}$
$∴ Now, Tension at middle of rope$
$∴ T = (m + \frac{m}{6}) (\frac{F}{m + \frac{m}{3}}) = \frac{7 m}{6} \times \frac{3 F}{4 m}$
$T = \frac{7 F}{8}$

BECEC - 2018

Laws of Motion

145855 A monkey of mass $20 kg$ is holding a vertical rope. The rope will not break, when a mass of $25 kg$ is suspended from it but will break, if the mass exceeds $25 kg$ . What is the maximum acceleration with which the monkey can climb up along the rope? $($ Take $g = 10 m / s^{2}$ )

1

25 m / s^{2}

2

2.5 m / s^{2}

3

5 m / s^{2}

4

10 m / s^{2}

Explanation:

B Case I:- Maximum mass $25 kg$ is applied.

$∴ T_{max} = 25 \times 10 = 250 N$
(breaking force)
Case I: mass of monkey of $20 kg$ moving upward with acceleration a

We know,
$T_{max} = mg + ma$
$250 = 20 g + 20 a$
$250 = 200 + 20 a$
$20 a = 250 - 200$
$a = \frac{50}{20}$
$a = 2.5 m / s^{2}$

AIPMT- 2003

Laws of Motion

145856 A lift of mass $1000 kg$ is moving upwards with an acceleration of $1 m / s^{2}$ . The tension developed in the string, which is connected to lift is
$(g = 9.8 m / s^{2})$

1

9800 N

2

10800 N

3

11000 N

4

10000 N

Explanation:

B Given that,
$m = 1000 kg, a = 1 m / s^{2}$

String connected to the lift (upward) $T = mg + ma$
$T = m (g + a)$
$T = 1000 (9.8 + 1)$
$T = 10800 N$

AIPMT- 2002

Laws of Motion

145852 When a metal wire of length ' $l$ ' is subjected to tensions $T_{1}$ and $T_{2}$ respectively its length changes to $l_{1}$ and $l_{2}$ , then the relation of ' $l$ ' is correctly given by

1

l \sqrt{l_{1} - l_{2}}

2

l = \frac{1}{2} (l_{1} + l_{2})

3

l = \frac{l_{1} T_{2} + l_{2} T_{1}}{T_{1} + T_{2}}

4

l = \frac{l_{1} T_{2} - l_{2} T_{1}}{T_{1} - T_{2}}

Explanation:

D Given,
Tension is $T_{1}$ when length is $l_{1}$ and when, length is $l_{2}$ , tension is $T_{2}$ .
Let natural length of wire is $l$
$∵$ Young's modules $(E) = \frac{Stress}{Strain}$
$E = \frac{T \times l}{A \times Δ l}$
(i) When length is $l_{1}$ and tension is $T_{1}$
$E = \frac{T_{1} \times l}{A (l_{1} - l)}$
(ii) When length is $l_{2}$ and tension is $T_{2}$
$E = \frac{T_{2} \times l}{A (l_{2} - l)}$
$∵ E$ is equal in both case,
$\frac{T_{1} \times l}{A (l_{1} - l)} = \frac{T_{2} \times l}{A (l_{2} - l)}$
$T_{1} l_{2} - T_{1} l = T_{2} l_{1} - T_{2} l$
$l (T_{2} - T_{1}) = T_{2} l_{1} - T_{1} l_{2}$
$l = \frac{T_{2} l_{1} - T_{1} l_{2}}{T_{2} - T_{1}}$

Tripura-2021

Laws of Motion

145853 The ratio of weights of a man inside a lift when it is stationary and when it is going down with a uniform acceleration ' $a$ ' is $3 : 2$ . The value of ' $a$ ' will be $(a < g, g =$ acceleration due to gravity)

1

\frac{3}{2} g

2

\frac{g}{3}

3

g

4

\frac{2}{3} g

Explanation:

B When the lift is stationary the weight of the $man$ is $w_{o} = mg$
The weight of the man when the lift is moving downward with acceleration a is $w = mg - ma$
$\frac{w_{o}}{w} = \frac{3}{2} (given)$
Therefore, $\frac{mg}{mg - ma} = \frac{3}{2}$
$\frac{m g}{m (g - a)} = \frac{3}{2}$
$2 g = 3 g - 3 a$
$g = 3 a$
$So, a = \frac{g}{3}$

MHT-CET 2020

Laws of Motion

145854 A block of mass $m$ is resting on a smooth horizontal surface. One end of a uniform rope of mass $(\frac{m}{3})$ is fixed to the block, which is pulled in the horizontal direction by applying force $F$ at the other end. The tension in the middle of the rope is

1

\frac{8}{7} F

2

\frac{1}{7} F

3

\frac{1}{8} F

4

\frac{1}{5} F

5

\frac{7}{8} F

Explanation:

E
$F = m_{1} a$
$a = (\frac{F}{m + \frac{m}{3}}) = \frac{3 F}{4 m}$
$∴ Now, Tension at middle of rope$
$∴ T = (m + \frac{m}{6}) (\frac{F}{m + \frac{m}{3}}) = \frac{7 m}{6} \times \frac{3 F}{4 m}$
$T = \frac{7 F}{8}$

BECEC - 2018

Laws of Motion

145855 A monkey of mass $20 kg$ is holding a vertical rope. The rope will not break, when a mass of $25 kg$ is suspended from it but will break, if the mass exceeds $25 kg$ . What is the maximum acceleration with which the monkey can climb up along the rope? $($ Take $g = 10 m / s^{2}$ )

1

25 m / s^{2}

2

2.5 m / s^{2}

3

5 m / s^{2}

4

10 m / s^{2}

Explanation:

B Case I:- Maximum mass $25 kg$ is applied.

$∴ T_{max} = 25 \times 10 = 250 N$
(breaking force)
Case I: mass of monkey of $20 kg$ moving upward with acceleration a

We know,
$T_{max} = mg + ma$
$250 = 20 g + 20 a$
$250 = 200 + 20 a$
$20 a = 250 - 200$
$a = \frac{50}{20}$
$a = 2.5 m / s^{2}$

AIPMT- 2003

Laws of Motion

145856 A lift of mass $1000 kg$ is moving upwards with an acceleration of $1 m / s^{2}$ . The tension developed in the string, which is connected to lift is
$(g = 9.8 m / s^{2})$

1

9800 N

2

10800 N

3

11000 N

4

10000 N

Explanation:

B Given that,
$m = 1000 kg, a = 1 m / s^{2}$

String connected to the lift (upward) $T = mg + ma$
$T = m (g + a)$
$T = 1000 (9.8 + 1)$
$T = 10800 N$

AIPMT- 2002

Laws of Motion

145852 When a metal wire of length ' $l$ ' is subjected to tensions $T_{1}$ and $T_{2}$ respectively its length changes to $l_{1}$ and $l_{2}$ , then the relation of ' $l$ ' is correctly given by

1

l \sqrt{l_{1} - l_{2}}

2

l = \frac{1}{2} (l_{1} + l_{2})

3

l = \frac{l_{1} T_{2} + l_{2} T_{1}}{T_{1} + T_{2}}

4

l = \frac{l_{1} T_{2} - l_{2} T_{1}}{T_{1} - T_{2}}

Explanation:

D Given,
Tension is $T_{1}$ when length is $l_{1}$ and when, length is $l_{2}$ , tension is $T_{2}$ .
Let natural length of wire is $l$
$∵$ Young's modules $(E) = \frac{Stress}{Strain}$
$E = \frac{T \times l}{A \times Δ l}$
(i) When length is $l_{1}$ and tension is $T_{1}$
$E = \frac{T_{1} \times l}{A (l_{1} - l)}$
(ii) When length is $l_{2}$ and tension is $T_{2}$
$E = \frac{T_{2} \times l}{A (l_{2} - l)}$
$∵ E$ is equal in both case,
$\frac{T_{1} \times l}{A (l_{1} - l)} = \frac{T_{2} \times l}{A (l_{2} - l)}$
$T_{1} l_{2} - T_{1} l = T_{2} l_{1} - T_{2} l$
$l (T_{2} - T_{1}) = T_{2} l_{1} - T_{1} l_{2}$
$l = \frac{T_{2} l_{1} - T_{1} l_{2}}{T_{2} - T_{1}}$

Tripura-2021

Laws of Motion

145853 The ratio of weights of a man inside a lift when it is stationary and when it is going down with a uniform acceleration ' $a$ ' is $3 : 2$ . The value of ' $a$ ' will be $(a < g, g =$ acceleration due to gravity)

1

\frac{3}{2} g

2

\frac{g}{3}

3

g

4

\frac{2}{3} g

Explanation:

B When the lift is stationary the weight of the $man$ is $w_{o} = mg$
The weight of the man when the lift is moving downward with acceleration a is $w = mg - ma$
$\frac{w_{o}}{w} = \frac{3}{2} (given)$
Therefore, $\frac{mg}{mg - ma} = \frac{3}{2}$
$\frac{m g}{m (g - a)} = \frac{3}{2}$
$2 g = 3 g - 3 a$
$g = 3 a$
$So, a = \frac{g}{3}$

MHT-CET 2020

Laws of Motion

145854 A block of mass $m$ is resting on a smooth horizontal surface. One end of a uniform rope of mass $(\frac{m}{3})$ is fixed to the block, which is pulled in the horizontal direction by applying force $F$ at the other end. The tension in the middle of the rope is

1

\frac{8}{7} F

2

\frac{1}{7} F

3

\frac{1}{8} F

4

\frac{1}{5} F

5

\frac{7}{8} F

Explanation:

E
$F = m_{1} a$
$a = (\frac{F}{m + \frac{m}{3}}) = \frac{3 F}{4 m}$
$∴ Now, Tension at middle of rope$
$∴ T = (m + \frac{m}{6}) (\frac{F}{m + \frac{m}{3}}) = \frac{7 m}{6} \times \frac{3 F}{4 m}$
$T = \frac{7 F}{8}$

BECEC - 2018

Laws of Motion

145855 A monkey of mass $20 kg$ is holding a vertical rope. The rope will not break, when a mass of $25 kg$ is suspended from it but will break, if the mass exceeds $25 kg$ . What is the maximum acceleration with which the monkey can climb up along the rope? $($ Take $g = 10 m / s^{2}$ )

1

25 m / s^{2}

2

2.5 m / s^{2}

3

5 m / s^{2}

4

10 m / s^{2}

Explanation:

B Case I:- Maximum mass $25 kg$ is applied.

$∴ T_{max} = 25 \times 10 = 250 N$
(breaking force)
Case I: mass of monkey of $20 kg$ moving upward with acceleration a

We know,
$T_{max} = mg + ma$
$250 = 20 g + 20 a$
$250 = 200 + 20 a$
$20 a = 250 - 200$
$a = \frac{50}{20}$
$a = 2.5 m / s^{2}$

AIPMT- 2003

Laws of Motion

145856 A lift of mass $1000 kg$ is moving upwards with an acceleration of $1 m / s^{2}$ . The tension developed in the string, which is connected to lift is
$(g = 9.8 m / s^{2})$

1

9800 N

2

10800 N

3

11000 N

4

10000 N

Explanation:

B Given that,
$m = 1000 kg, a = 1 m / s^{2}$

String connected to the lift (upward) $T = mg + ma$
$T = m (g + a)$
$T = 1000 (9.8 + 1)$
$T = 10800 N$

AIPMT- 2002

Laws of Motion

145852 When a metal wire of length ' $l$ ' is subjected to tensions $T_{1}$ and $T_{2}$ respectively its length changes to $l_{1}$ and $l_{2}$ , then the relation of ' $l$ ' is correctly given by

1

l \sqrt{l_{1} - l_{2}}

2

l = \frac{1}{2} (l_{1} + l_{2})

3

l = \frac{l_{1} T_{2} + l_{2} T_{1}}{T_{1} + T_{2}}

4

l = \frac{l_{1} T_{2} - l_{2} T_{1}}{T_{1} - T_{2}}

Explanation:

D Given,
Tension is $T_{1}$ when length is $l_{1}$ and when, length is $l_{2}$ , tension is $T_{2}$ .
Let natural length of wire is $l$
$∵$ Young's modules $(E) = \frac{Stress}{Strain}$
$E = \frac{T \times l}{A \times Δ l}$
(i) When length is $l_{1}$ and tension is $T_{1}$
$E = \frac{T_{1} \times l}{A (l_{1} - l)}$
(ii) When length is $l_{2}$ and tension is $T_{2}$
$E = \frac{T_{2} \times l}{A (l_{2} - l)}$
$∵ E$ is equal in both case,
$\frac{T_{1} \times l}{A (l_{1} - l)} = \frac{T_{2} \times l}{A (l_{2} - l)}$
$T_{1} l_{2} - T_{1} l = T_{2} l_{1} - T_{2} l$
$l (T_{2} - T_{1}) = T_{2} l_{1} - T_{1} l_{2}$
$l = \frac{T_{2} l_{1} - T_{1} l_{2}}{T_{2} - T_{1}}$

Tripura-2021

Laws of Motion

145853 The ratio of weights of a man inside a lift when it is stationary and when it is going down with a uniform acceleration ' $a$ ' is $3 : 2$ . The value of ' $a$ ' will be $(a < g, g =$ acceleration due to gravity)

1

\frac{3}{2} g

2

\frac{g}{3}

3

g

4

\frac{2}{3} g

Explanation:

B When the lift is stationary the weight of the $man$ is $w_{o} = mg$
The weight of the man when the lift is moving downward with acceleration a is $w = mg - ma$
$\frac{w_{o}}{w} = \frac{3}{2} (given)$
Therefore, $\frac{mg}{mg - ma} = \frac{3}{2}$
$\frac{m g}{m (g - a)} = \frac{3}{2}$
$2 g = 3 g - 3 a$
$g = 3 a$
$So, a = \frac{g}{3}$

MHT-CET 2020

Laws of Motion

145854 A block of mass $m$ is resting on a smooth horizontal surface. One end of a uniform rope of mass $(\frac{m}{3})$ is fixed to the block, which is pulled in the horizontal direction by applying force $F$ at the other end. The tension in the middle of the rope is

1

\frac{8}{7} F

2

\frac{1}{7} F

3

\frac{1}{8} F

4

\frac{1}{5} F

5

\frac{7}{8} F

Explanation:

E
$F = m_{1} a$
$a = (\frac{F}{m + \frac{m}{3}}) = \frac{3 F}{4 m}$
$∴ Now, Tension at middle of rope$
$∴ T = (m + \frac{m}{6}) (\frac{F}{m + \frac{m}{3}}) = \frac{7 m}{6} \times \frac{3 F}{4 m}$
$T = \frac{7 F}{8}$

BECEC - 2018

Laws of Motion

145855 A monkey of mass $20 kg$ is holding a vertical rope. The rope will not break, when a mass of $25 kg$ is suspended from it but will break, if the mass exceeds $25 kg$ . What is the maximum acceleration with which the monkey can climb up along the rope? $($ Take $g = 10 m / s^{2}$ )

1

25 m / s^{2}

2

2.5 m / s^{2}

3

5 m / s^{2}

4

10 m / s^{2}

Explanation:

B Case I:- Maximum mass $25 kg$ is applied.

$∴ T_{max} = 25 \times 10 = 250 N$
(breaking force)
Case I: mass of monkey of $20 kg$ moving upward with acceleration a

We know,
$T_{max} = mg + ma$
$250 = 20 g + 20 a$
$250 = 200 + 20 a$
$20 a = 250 - 200$
$a = \frac{50}{20}$
$a = 2.5 m / s^{2}$

AIPMT- 2003

Laws of Motion

145856 A lift of mass $1000 kg$ is moving upwards with an acceleration of $1 m / s^{2}$ . The tension developed in the string, which is connected to lift is
$(g = 9.8 m / s^{2})$

1

9800 N

2

10800 N

3

11000 N

4

10000 N

Explanation:

B Given that,
$m = 1000 kg, a = 1 m / s^{2}$

String connected to the lift (upward) $T = mg + ma$
$T = m (g + a)$
$T = 1000 (9.8 + 1)$
$T = 10800 N$

AIPMT- 2002