QBManager

LAWS OF MOTION (ADDITIONAL)

371735 A lift is tied with thick iron ropes having mass ' $M$ '. The maximum acceleration of the lift is ' $a$ ' $m / s^{2}$ and maximum safe stress is ' $S$ ' $N / m^{2}$ . The minimum diameter of the rope is ( $g$ = acceleration due to gravity)

1

{[\frac{2 M (g + a)}{π S}]}^{\frac{1}{2}}

2

{[\frac{2 M (g - a)}{π S}]}^{\frac{1}{2}}

3

{[\frac{4 M (g + a)}{π S}]}^{\frac{1}{2}}

4

{[\frac{4 M (g - a)}{π S}]}^{\frac{1}{2}}

Explanation:

C We know that,
Force, $F = m (g + a)$
Stress $\times$ area $= M (g + a) [∵$ Stress $= \frac{Force}{area}]$
$S \times π \frac{d^{2}}{4} = M (g + a)$
$d^{2} = \frac{4 M (g + a)}{S π}$
$d = {(\frac{4 M (g + a)}{S π})}^{\frac{1}{2}}$

MHT-CET 2020

LAWS OF MOTION (ADDITIONAL)

371736 A mass of $1 kg$ is suspended by a string. It is first lifted up with an acceleration of $4.9 m / s^{2}$ and then lowered down with same acceleration. The ratio of tensions in the string in the two cases, respectively is $g = 9.8 m / s^{2}$

1

1 : 3

2

2 : 1

3

3 : 1

4

1 : 2

Explanation:

C We know,

${ma}_{1} = T_{1} - mg$
$a_{1} = a_{2} = 4.9 m / s^{2}$
$g = 9.8 m / s^{2}$
$m = 1 kg$
${ma}_{2} = mg - T_{2}$
Case-I
$T_{1} - mg = {ma}_{1}$
$T_{1} = m (g + a_{1})$
$T_{1} = 1 (9.8 + 4.9)$
$T_{1} = 14.7 N$
Case-II
${ma}_{2} = mg - T_{2}$
$T_{2} = m (g - a_{2})$
$T_{2} = 1 (9.8 - 4.9)$
$T_{2} = 4.9 N$
$∴$ Ratio $= \frac{T_{1}}{T_{2}} = \frac{14.7}{4.9} = \frac{3}{1}$
$T_{1} : T_{2} = 3 : 1$

MHT-CET 2020

LAWS OF MOTION (ADDITIONAL)

371737 A lift of mass ' $m$ ' is ascending with an acceleration ' $a$ ' $(a < g)$ . The tension in the cable of the lift is ( $g =$ acceleration due to gravity)

1

m (a - g)

2

m (g - a)

3

m (2 g + a)

4

m (g + a)

Explanation:

D Given:
Mass of lift $= m$
Acceleration $= a$

Where, $a < g$
So,
Tension on cable of the lift will be
$T - mg = ma$
(Where ' $g$ ' is the acceleration due to gravity)
$\overset{\circ}{T} = mg + ma$
$∴ T = m (g + a)$

MHT-CET 2020

LAWS OF MOTION (ADDITIONAL)

371738 Pick out the correct statement

1 Second law of motion is a vector equation

2 Second law of motion is applicable to a particle and not to the system of particles

3 Force is always in the direction of motion

4 If external force on a body is zero, it does not mean the acceleration is zero

5 Acceleration at an instant depends on the history of the motion of the particle

Explanation:

A $\vec{F} = ma$ (Second law of motion)
In this case, force $\vec{F}$ refers to the total external force on the system (internal forces in the system are not
included in $\vec{F}$ ) and a refers to the acceleration of the whole system, i.e., acceleration of the centre of mass of the system.

Kerala CEE 2020

LAWS OF MOTION (ADDITIONAL)

371739 A boy is standing on a weighing machine inside a lift. When the lift goes upwards with acceleration $\frac{g}{4}$ , the machine shows the reading $50 kg$ . wt. When the lift goes downward with acceleration $\frac{g}{4}$ , the reading of the machine in kg. wt. would be

1 50

2 30

3 45.5

4 62.5

5 14

Explanation:

B $∙$ Lift goes upwards,

$\Rightarrow 50 \times g = m (g + \frac{g}{4})$
$50 g = m (\frac{5 g}{4})$
$m = 40 kg$
$- Lift goes downward,$
$\Rightarrow F_{net} = m (g - \frac{g}{4})$
$= 40 (\frac{3 g}{4})$
$= 30 N$

Kerala CEE 2020

LAWS OF MOTION (ADDITIONAL)

371735 A lift is tied with thick iron ropes having mass ' $M$ '. The maximum acceleration of the lift is ' $a$ ' $m / s^{2}$ and maximum safe stress is ' $S$ ' $N / m^{2}$ . The minimum diameter of the rope is ( $g$ = acceleration due to gravity)

1

{[\frac{2 M (g + a)}{π S}]}^{\frac{1}{2}}

2

{[\frac{2 M (g - a)}{π S}]}^{\frac{1}{2}}

3

{[\frac{4 M (g + a)}{π S}]}^{\frac{1}{2}}

4

{[\frac{4 M (g - a)}{π S}]}^{\frac{1}{2}}

Explanation:

C We know that,
Force, $F = m (g + a)$
Stress $\times$ area $= M (g + a) [∵$ Stress $= \frac{Force}{area}]$
$S \times π \frac{d^{2}}{4} = M (g + a)$
$d^{2} = \frac{4 M (g + a)}{S π}$
$d = {(\frac{4 M (g + a)}{S π})}^{\frac{1}{2}}$

MHT-CET 2020

LAWS OF MOTION (ADDITIONAL)

371736 A mass of $1 kg$ is suspended by a string. It is first lifted up with an acceleration of $4.9 m / s^{2}$ and then lowered down with same acceleration. The ratio of tensions in the string in the two cases, respectively is $g = 9.8 m / s^{2}$

1

1 : 3

2

2 : 1

3

3 : 1

4

1 : 2

Explanation:

C We know,

${ma}_{1} = T_{1} - mg$
$a_{1} = a_{2} = 4.9 m / s^{2}$
$g = 9.8 m / s^{2}$
$m = 1 kg$
${ma}_{2} = mg - T_{2}$
Case-I
$T_{1} - mg = {ma}_{1}$
$T_{1} = m (g + a_{1})$
$T_{1} = 1 (9.8 + 4.9)$
$T_{1} = 14.7 N$
Case-II
${ma}_{2} = mg - T_{2}$
$T_{2} = m (g - a_{2})$
$T_{2} = 1 (9.8 - 4.9)$
$T_{2} = 4.9 N$
$∴$ Ratio $= \frac{T_{1}}{T_{2}} = \frac{14.7}{4.9} = \frac{3}{1}$
$T_{1} : T_{2} = 3 : 1$

MHT-CET 2020

LAWS OF MOTION (ADDITIONAL)

371737 A lift of mass ' $m$ ' is ascending with an acceleration ' $a$ ' $(a < g)$ . The tension in the cable of the lift is ( $g =$ acceleration due to gravity)

1

m (a - g)

2

m (g - a)

3

m (2 g + a)

4

m (g + a)

Explanation:

D Given:
Mass of lift $= m$
Acceleration $= a$

Where, $a < g$
So,
Tension on cable of the lift will be
$T - mg = ma$
(Where ' $g$ ' is the acceleration due to gravity)
$\overset{\circ}{T} = mg + ma$
$∴ T = m (g + a)$

MHT-CET 2020

LAWS OF MOTION (ADDITIONAL)

371738 Pick out the correct statement

1 Second law of motion is a vector equation

2 Second law of motion is applicable to a particle and not to the system of particles

3 Force is always in the direction of motion

4 If external force on a body is zero, it does not mean the acceleration is zero

5 Acceleration at an instant depends on the history of the motion of the particle

Explanation:

A $\vec{F} = ma$ (Second law of motion)
In this case, force $\vec{F}$ refers to the total external force on the system (internal forces in the system are not
included in $\vec{F}$ ) and a refers to the acceleration of the whole system, i.e., acceleration of the centre of mass of the system.

Kerala CEE 2020

LAWS OF MOTION (ADDITIONAL)

371739 A boy is standing on a weighing machine inside a lift. When the lift goes upwards with acceleration $\frac{g}{4}$ , the machine shows the reading $50 kg$ . wt. When the lift goes downward with acceleration $\frac{g}{4}$ , the reading of the machine in kg. wt. would be

1 50

2 30

3 45.5

4 62.5

5 14

Explanation:

B $∙$ Lift goes upwards,

$\Rightarrow 50 \times g = m (g + \frac{g}{4})$
$50 g = m (\frac{5 g}{4})$
$m = 40 kg$
$- Lift goes downward,$
$\Rightarrow F_{net} = m (g - \frac{g}{4})$
$= 40 (\frac{3 g}{4})$
$= 30 N$

Kerala CEE 2020

LAWS OF MOTION (ADDITIONAL)

371735 A lift is tied with thick iron ropes having mass ' $M$ '. The maximum acceleration of the lift is ' $a$ ' $m / s^{2}$ and maximum safe stress is ' $S$ ' $N / m^{2}$ . The minimum diameter of the rope is ( $g$ = acceleration due to gravity)

1

{[\frac{2 M (g + a)}{π S}]}^{\frac{1}{2}}

2

{[\frac{2 M (g - a)}{π S}]}^{\frac{1}{2}}

3

{[\frac{4 M (g + a)}{π S}]}^{\frac{1}{2}}

4

{[\frac{4 M (g - a)}{π S}]}^{\frac{1}{2}}

Explanation:

C We know that,
Force, $F = m (g + a)$
Stress $\times$ area $= M (g + a) [∵$ Stress $= \frac{Force}{area}]$
$S \times π \frac{d^{2}}{4} = M (g + a)$
$d^{2} = \frac{4 M (g + a)}{S π}$
$d = {(\frac{4 M (g + a)}{S π})}^{\frac{1}{2}}$

MHT-CET 2020

LAWS OF MOTION (ADDITIONAL)

371736 A mass of $1 kg$ is suspended by a string. It is first lifted up with an acceleration of $4.9 m / s^{2}$ and then lowered down with same acceleration. The ratio of tensions in the string in the two cases, respectively is $g = 9.8 m / s^{2}$

1

1 : 3

2

2 : 1

3

3 : 1

4

1 : 2

Explanation:

C We know,

${ma}_{1} = T_{1} - mg$
$a_{1} = a_{2} = 4.9 m / s^{2}$
$g = 9.8 m / s^{2}$
$m = 1 kg$
${ma}_{2} = mg - T_{2}$
Case-I
$T_{1} - mg = {ma}_{1}$
$T_{1} = m (g + a_{1})$
$T_{1} = 1 (9.8 + 4.9)$
$T_{1} = 14.7 N$
Case-II
${ma}_{2} = mg - T_{2}$
$T_{2} = m (g - a_{2})$
$T_{2} = 1 (9.8 - 4.9)$
$T_{2} = 4.9 N$
$∴$ Ratio $= \frac{T_{1}}{T_{2}} = \frac{14.7}{4.9} = \frac{3}{1}$
$T_{1} : T_{2} = 3 : 1$

MHT-CET 2020

LAWS OF MOTION (ADDITIONAL)

371737 A lift of mass ' $m$ ' is ascending with an acceleration ' $a$ ' $(a < g)$ . The tension in the cable of the lift is ( $g =$ acceleration due to gravity)

1

m (a - g)

2

m (g - a)

3

m (2 g + a)

4

m (g + a)

Explanation:

D Given:
Mass of lift $= m$
Acceleration $= a$

Where, $a < g$
So,
Tension on cable of the lift will be
$T - mg = ma$
(Where ' $g$ ' is the acceleration due to gravity)
$\overset{\circ}{T} = mg + ma$
$∴ T = m (g + a)$

MHT-CET 2020

LAWS OF MOTION (ADDITIONAL)

371738 Pick out the correct statement

1 Second law of motion is a vector equation

2 Second law of motion is applicable to a particle and not to the system of particles

3 Force is always in the direction of motion

4 If external force on a body is zero, it does not mean the acceleration is zero

5 Acceleration at an instant depends on the history of the motion of the particle

Explanation:

A $\vec{F} = ma$ (Second law of motion)
In this case, force $\vec{F}$ refers to the total external force on the system (internal forces in the system are not
included in $\vec{F}$ ) and a refers to the acceleration of the whole system, i.e., acceleration of the centre of mass of the system.

Kerala CEE 2020

LAWS OF MOTION (ADDITIONAL)

371739 A boy is standing on a weighing machine inside a lift. When the lift goes upwards with acceleration $\frac{g}{4}$ , the machine shows the reading $50 kg$ . wt. When the lift goes downward with acceleration $\frac{g}{4}$ , the reading of the machine in kg. wt. would be

1 50

2 30

3 45.5

4 62.5

5 14

Explanation:

B $∙$ Lift goes upwards,

$\Rightarrow 50 \times g = m (g + \frac{g}{4})$
$50 g = m (\frac{5 g}{4})$
$m = 40 kg$
$- Lift goes downward,$
$\Rightarrow F_{net} = m (g - \frac{g}{4})$
$= 40 (\frac{3 g}{4})$
$= 30 N$

Kerala CEE 2020

LAWS OF MOTION (ADDITIONAL)

371735 A lift is tied with thick iron ropes having mass ' $M$ '. The maximum acceleration of the lift is ' $a$ ' $m / s^{2}$ and maximum safe stress is ' $S$ ' $N / m^{2}$ . The minimum diameter of the rope is ( $g$ = acceleration due to gravity)

1

{[\frac{2 M (g + a)}{π S}]}^{\frac{1}{2}}

2

{[\frac{2 M (g - a)}{π S}]}^{\frac{1}{2}}

3

{[\frac{4 M (g + a)}{π S}]}^{\frac{1}{2}}

4

{[\frac{4 M (g - a)}{π S}]}^{\frac{1}{2}}

Explanation:

C We know that,
Force, $F = m (g + a)$
Stress $\times$ area $= M (g + a) [∵$ Stress $= \frac{Force}{area}]$
$S \times π \frac{d^{2}}{4} = M (g + a)$
$d^{2} = \frac{4 M (g + a)}{S π}$
$d = {(\frac{4 M (g + a)}{S π})}^{\frac{1}{2}}$

MHT-CET 2020

LAWS OF MOTION (ADDITIONAL)

371736 A mass of $1 kg$ is suspended by a string. It is first lifted up with an acceleration of $4.9 m / s^{2}$ and then lowered down with same acceleration. The ratio of tensions in the string in the two cases, respectively is $g = 9.8 m / s^{2}$

1

1 : 3

2

2 : 1

3

3 : 1

4

1 : 2

Explanation:

C We know,

${ma}_{1} = T_{1} - mg$
$a_{1} = a_{2} = 4.9 m / s^{2}$
$g = 9.8 m / s^{2}$
$m = 1 kg$
${ma}_{2} = mg - T_{2}$
Case-I
$T_{1} - mg = {ma}_{1}$
$T_{1} = m (g + a_{1})$
$T_{1} = 1 (9.8 + 4.9)$
$T_{1} = 14.7 N$
Case-II
${ma}_{2} = mg - T_{2}$
$T_{2} = m (g - a_{2})$
$T_{2} = 1 (9.8 - 4.9)$
$T_{2} = 4.9 N$
$∴$ Ratio $= \frac{T_{1}}{T_{2}} = \frac{14.7}{4.9} = \frac{3}{1}$
$T_{1} : T_{2} = 3 : 1$

MHT-CET 2020

LAWS OF MOTION (ADDITIONAL)

371737 A lift of mass ' $m$ ' is ascending with an acceleration ' $a$ ' $(a < g)$ . The tension in the cable of the lift is ( $g =$ acceleration due to gravity)

1

m (a - g)

2

m (g - a)

3

m (2 g + a)

4

m (g + a)

Explanation:

D Given:
Mass of lift $= m$
Acceleration $= a$

Where, $a < g$
So,
Tension on cable of the lift will be
$T - mg = ma$
(Where ' $g$ ' is the acceleration due to gravity)
$\overset{\circ}{T} = mg + ma$
$∴ T = m (g + a)$

MHT-CET 2020

LAWS OF MOTION (ADDITIONAL)

371738 Pick out the correct statement

1 Second law of motion is a vector equation

2 Second law of motion is applicable to a particle and not to the system of particles

3 Force is always in the direction of motion

4 If external force on a body is zero, it does not mean the acceleration is zero

5 Acceleration at an instant depends on the history of the motion of the particle

Explanation:

A $\vec{F} = ma$ (Second law of motion)
In this case, force $\vec{F}$ refers to the total external force on the system (internal forces in the system are not
included in $\vec{F}$ ) and a refers to the acceleration of the whole system, i.e., acceleration of the centre of mass of the system.

Kerala CEE 2020

LAWS OF MOTION (ADDITIONAL)

371739 A boy is standing on a weighing machine inside a lift. When the lift goes upwards with acceleration $\frac{g}{4}$ , the machine shows the reading $50 kg$ . wt. When the lift goes downward with acceleration $\frac{g}{4}$ , the reading of the machine in kg. wt. would be

1 50

2 30

3 45.5

4 62.5

5 14

Explanation:

B $∙$ Lift goes upwards,

$\Rightarrow 50 \times g = m (g + \frac{g}{4})$
$50 g = m (\frac{5 g}{4})$
$m = 40 kg$
$- Lift goes downward,$
$\Rightarrow F_{net} = m (g - \frac{g}{4})$
$= 40 (\frac{3 g}{4})$
$= 30 N$

Kerala CEE 2020

LAWS OF MOTION (ADDITIONAL)

371735 A lift is tied with thick iron ropes having mass ' $M$ '. The maximum acceleration of the lift is ' $a$ ' $m / s^{2}$ and maximum safe stress is ' $S$ ' $N / m^{2}$ . The minimum diameter of the rope is ( $g$ = acceleration due to gravity)

1

{[\frac{2 M (g + a)}{π S}]}^{\frac{1}{2}}

2

{[\frac{2 M (g - a)}{π S}]}^{\frac{1}{2}}

3

{[\frac{4 M (g + a)}{π S}]}^{\frac{1}{2}}

4

{[\frac{4 M (g - a)}{π S}]}^{\frac{1}{2}}

Explanation:

C We know that,
Force, $F = m (g + a)$
Stress $\times$ area $= M (g + a) [∵$ Stress $= \frac{Force}{area}]$
$S \times π \frac{d^{2}}{4} = M (g + a)$
$d^{2} = \frac{4 M (g + a)}{S π}$
$d = {(\frac{4 M (g + a)}{S π})}^{\frac{1}{2}}$

MHT-CET 2020

LAWS OF MOTION (ADDITIONAL)

371736 A mass of $1 kg$ is suspended by a string. It is first lifted up with an acceleration of $4.9 m / s^{2}$ and then lowered down with same acceleration. The ratio of tensions in the string in the two cases, respectively is $g = 9.8 m / s^{2}$

1

1 : 3

2

2 : 1

3

3 : 1

4

1 : 2

Explanation:

C We know,

${ma}_{1} = T_{1} - mg$
$a_{1} = a_{2} = 4.9 m / s^{2}$
$g = 9.8 m / s^{2}$
$m = 1 kg$
${ma}_{2} = mg - T_{2}$
Case-I
$T_{1} - mg = {ma}_{1}$
$T_{1} = m (g + a_{1})$
$T_{1} = 1 (9.8 + 4.9)$
$T_{1} = 14.7 N$
Case-II
${ma}_{2} = mg - T_{2}$
$T_{2} = m (g - a_{2})$
$T_{2} = 1 (9.8 - 4.9)$
$T_{2} = 4.9 N$
$∴$ Ratio $= \frac{T_{1}}{T_{2}} = \frac{14.7}{4.9} = \frac{3}{1}$
$T_{1} : T_{2} = 3 : 1$

MHT-CET 2020

LAWS OF MOTION (ADDITIONAL)

371737 A lift of mass ' $m$ ' is ascending with an acceleration ' $a$ ' $(a < g)$ . The tension in the cable of the lift is ( $g =$ acceleration due to gravity)

1

m (a - g)

2

m (g - a)

3

m (2 g + a)

4

m (g + a)

Explanation:

D Given:
Mass of lift $= m$
Acceleration $= a$

Where, $a < g$
So,
Tension on cable of the lift will be
$T - mg = ma$
(Where ' $g$ ' is the acceleration due to gravity)
$\overset{\circ}{T} = mg + ma$
$∴ T = m (g + a)$

MHT-CET 2020

LAWS OF MOTION (ADDITIONAL)

371738 Pick out the correct statement

1 Second law of motion is a vector equation

2 Second law of motion is applicable to a particle and not to the system of particles

3 Force is always in the direction of motion

4 If external force on a body is zero, it does not mean the acceleration is zero

5 Acceleration at an instant depends on the history of the motion of the particle

Explanation:

A $\vec{F} = ma$ (Second law of motion)
In this case, force $\vec{F}$ refers to the total external force on the system (internal forces in the system are not
included in $\vec{F}$ ) and a refers to the acceleration of the whole system, i.e., acceleration of the centre of mass of the system.

Kerala CEE 2020

LAWS OF MOTION (ADDITIONAL)

371739 A boy is standing on a weighing machine inside a lift. When the lift goes upwards with acceleration $\frac{g}{4}$ , the machine shows the reading $50 kg$ . wt. When the lift goes downward with acceleration $\frac{g}{4}$ , the reading of the machine in kg. wt. would be

1 50

2 30

3 45.5

4 62.5

5 14

Explanation:

B $∙$ Lift goes upwards,

$\Rightarrow 50 \times g = m (g + \frac{g}{4})$
$50 g = m (\frac{5 g}{4})$
$m = 40 kg$
$- Lift goes downward,$
$\Rightarrow F_{net} = m (g - \frac{g}{4})$
$= 40 (\frac{3 g}{4})$
$= 30 N$

Kerala CEE 2020