QBManager

Laws of Motion

270246 A balloon of mass $M$ is descending at a constant acceleration $α$ . When a mass $m$ is released from the balloon it starts rising with the same acceleration $α$ . Assuming that its volume does not change, what is the value of $m$ ?

1

\frac{α}{α + g} M

2

\frac{2 α}{α + g} M

3

\frac{α + g}{α} M

4

\frac{α + g}{2 α} M

Explanation:

While descending, $M g - F_{B} = M α$
While ascending $F_{B} - (M - m) g = (M - m) α$
Where ' $F_{B}$ ' is the buoyant force

Laws of Motion

270247 A monkey of mass $40 kg$ climbs on a massless rope of breaking strength $600 N$ . The rope will break if the m onkey. (Take $g = 10 m / s^{2}$ )

1 climbs up with a uniform speed of

6 m / s

2 climbs up with an acceleration of

6 m / s^{2}

3 climbs down with an acceleration of

4 m / s^{2}

4 climbs down with a uniform speed of

5 m / s

Explanation:

To move up with an acceleration $a$ the monkey will push the rope downwards with a force of ma.
$T_{max} = m g + m a_{max}$

Laws of Motion

270248 Two persons are holding a rope of negligible weight tightly at its ends so that it is horizontal. A $15 kg$ weight is attached to rope at the midpoint which now no more remains horizontal. The minimum tension required to completely straighten the rope is

1

150 N

2

75 N

3

50 N

4 Infinitely large

Explanation:

$T = F, 2 T \cos θ = m g$
To make rope straight, $θ = 90^{\circ}$

Laws of Motion

270249 A straight rope of length ' $L$ ' is kept on a frictionless horizontal surface and a force ' $F$ ' is applied to one end of the rope in the direction of its length and away from that end. The tension in the rope at a distance ' $I$ ' from that end is

1

\frac{F}{l}

2

\frac{L F}{l}

3

\to - \frac{l}{L} θ_{F}

4

G 1 + \frac{l}{L} θ_{F}

Explanation:

$F - T = m_{1} a, T = m_{2} a$

Laws of Motion

270246 A balloon of mass $M$ is descending at a constant acceleration $α$ . When a mass $m$ is released from the balloon it starts rising with the same acceleration $α$ . Assuming that its volume does not change, what is the value of $m$ ?

1

\frac{α}{α + g} M

2

\frac{2 α}{α + g} M

3

\frac{α + g}{α} M

4

\frac{α + g}{2 α} M

Explanation:

While descending, $M g - F_{B} = M α$
While ascending $F_{B} - (M - m) g = (M - m) α$
Where ' $F_{B}$ ' is the buoyant force

Laws of Motion

270247 A monkey of mass $40 kg$ climbs on a massless rope of breaking strength $600 N$ . The rope will break if the m onkey. (Take $g = 10 m / s^{2}$ )

1 climbs up with a uniform speed of

6 m / s

2 climbs up with an acceleration of

6 m / s^{2}

3 climbs down with an acceleration of

4 m / s^{2}

4 climbs down with a uniform speed of

5 m / s

Explanation:

To move up with an acceleration $a$ the monkey will push the rope downwards with a force of ma.
$T_{max} = m g + m a_{max}$

Laws of Motion

270248 Two persons are holding a rope of negligible weight tightly at its ends so that it is horizontal. A $15 kg$ weight is attached to rope at the midpoint which now no more remains horizontal. The minimum tension required to completely straighten the rope is

1

150 N

2

75 N

3

50 N

4 Infinitely large

Explanation:

$T = F, 2 T \cos θ = m g$
To make rope straight, $θ = 90^{\circ}$

Laws of Motion

270249 A straight rope of length ' $L$ ' is kept on a frictionless horizontal surface and a force ' $F$ ' is applied to one end of the rope in the direction of its length and away from that end. The tension in the rope at a distance ' $I$ ' from that end is

1

\frac{F}{l}

2

\frac{L F}{l}

3

\to - \frac{l}{L} θ_{F}

4

G 1 + \frac{l}{L} θ_{F}

Explanation:

$F - T = m_{1} a, T = m_{2} a$

Laws of Motion

270246 A balloon of mass $M$ is descending at a constant acceleration $α$ . When a mass $m$ is released from the balloon it starts rising with the same acceleration $α$ . Assuming that its volume does not change, what is the value of $m$ ?

1

\frac{α}{α + g} M

2

\frac{2 α}{α + g} M

3

\frac{α + g}{α} M

4

\frac{α + g}{2 α} M

Explanation:

While descending, $M g - F_{B} = M α$
While ascending $F_{B} - (M - m) g = (M - m) α$
Where ' $F_{B}$ ' is the buoyant force

Laws of Motion

270247 A monkey of mass $40 kg$ climbs on a massless rope of breaking strength $600 N$ . The rope will break if the m onkey. (Take $g = 10 m / s^{2}$ )

1 climbs up with a uniform speed of

6 m / s

2 climbs up with an acceleration of

6 m / s^{2}

3 climbs down with an acceleration of

4 m / s^{2}

4 climbs down with a uniform speed of

5 m / s

Explanation:

To move up with an acceleration $a$ the monkey will push the rope downwards with a force of ma.
$T_{max} = m g + m a_{max}$

Laws of Motion

270248 Two persons are holding a rope of negligible weight tightly at its ends so that it is horizontal. A $15 kg$ weight is attached to rope at the midpoint which now no more remains horizontal. The minimum tension required to completely straighten the rope is

1

150 N

2

75 N

3

50 N

4 Infinitely large

Explanation:

$T = F, 2 T \cos θ = m g$
To make rope straight, $θ = 90^{\circ}$

Laws of Motion

270249 A straight rope of length ' $L$ ' is kept on a frictionless horizontal surface and a force ' $F$ ' is applied to one end of the rope in the direction of its length and away from that end. The tension in the rope at a distance ' $I$ ' from that end is

1

\frac{F}{l}

2

\frac{L F}{l}

3

\to - \frac{l}{L} θ_{F}

4

G 1 + \frac{l}{L} θ_{F}

Explanation:

$F - T = m_{1} a, T = m_{2} a$

Laws of Motion

270246 A balloon of mass $M$ is descending at a constant acceleration $α$ . When a mass $m$ is released from the balloon it starts rising with the same acceleration $α$ . Assuming that its volume does not change, what is the value of $m$ ?

1

\frac{α}{α + g} M

2

\frac{2 α}{α + g} M

3

\frac{α + g}{α} M

4

\frac{α + g}{2 α} M

Explanation:

While descending, $M g - F_{B} = M α$
While ascending $F_{B} - (M - m) g = (M - m) α$
Where ' $F_{B}$ ' is the buoyant force

Laws of Motion

270247 A monkey of mass $40 kg$ climbs on a massless rope of breaking strength $600 N$ . The rope will break if the m onkey. (Take $g = 10 m / s^{2}$ )

1 climbs up with a uniform speed of

6 m / s

2 climbs up with an acceleration of

6 m / s^{2}

3 climbs down with an acceleration of

4 m / s^{2}

4 climbs down with a uniform speed of

5 m / s

Explanation:

To move up with an acceleration $a$ the monkey will push the rope downwards with a force of ma.
$T_{max} = m g + m a_{max}$

Laws of Motion

270248 Two persons are holding a rope of negligible weight tightly at its ends so that it is horizontal. A $15 kg$ weight is attached to rope at the midpoint which now no more remains horizontal. The minimum tension required to completely straighten the rope is

1

150 N

2

75 N

3

50 N

4 Infinitely large

Explanation:

$T = F, 2 T \cos θ = m g$
To make rope straight, $θ = 90^{\circ}$

Laws of Motion

270249 A straight rope of length ' $L$ ' is kept on a frictionless horizontal surface and a force ' $F$ ' is applied to one end of the rope in the direction of its length and away from that end. The tension in the rope at a distance ' $I$ ' from that end is

1

\frac{F}{l}

2

\frac{L F}{l}

3

\to - \frac{l}{L} θ_{F}

4

G 1 + \frac{l}{L} θ_{F}

Explanation:

$F - T = m_{1} a, T = m_{2} a$

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: