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LAWS OF MOTION (ADDITIONAL)

371885 In the figure shown, acceleration with which the mass $m$ falls down when released is (consider the string to be massless, gacceleration due to gravity)

1

\frac{2 g}{3}

2

\frac{g}{2}

3

\frac{5 g}{6}

4

g

Explanation:

B Let $m$ be the mass \' $R$ ' be the radius of hollow cylinder.

When the system is released, torque produced in hollow cylinder is
$τ = I α,$
$= {MR}^{2} \cdot \frac{a}{R} {∵ I = {MR}^{2} & a = α R}$
$τ = MRa$
Also, we know $τ = FR = TR$
Equating (i) \& (ii) we get
$TR = M . R a.$
$T = Ma$
Now, from F.B.D.
$Mg - T = Ma$
$Mg - Ma = Ma {using eq$
$g = 2 a$
$a = \frac{g}{2}$

AP EAMCET (20.04.2019) Shift-1

LAWS OF MOTION (ADDITIONAL)

371886 One end of a light string is fixed to a clamp on the ground and the other end passes over a fixed frictionless pulley as shown in the figure. It makes an angle of $30^{\circ}$ with the ground. The clamp can tolerate a vertical force of $40 N$ . If a monkey of mass $5 kg$ were to climb up the rope, then the maximum acceleration in the upward direction with which it can climb safely is $(g =$ $10 {ms}^{- 2}$ )

1

2 {ms}^{- 2}

2

4 {ms}^{- 2}

3

6 {ms}^{- 2}

4

8 {ms}^{- 2}

Explanation:

C Let ' $T$ ' be the tension in the string and ' $a$ ' be the acceleration in the upward direction.

The upward force created on damp is,
$T \sin 30^{\circ} = \frac{T}{2}$
$\frac{T}{2} = 40 N$
(Given)
$T = 80 N$
Now, for monkey to climb up,
$T - mg = ma$
$a = \frac{T - mg}{m}$
$= \frac{80 - 5 \times 10}{5}$
$a = 6 m / s^{2}$

AP EAMCET (20.04.2019) Shift-1

LAWS OF MOTION (ADDITIONAL)

371887 Four blocks A, B, C and D of masses $6 kg, 3 kg$ , $6 kg$ and $1 kg$ respectively are connected by light strings passing over frictionless pulleys as shown in the figure. The strings $P$ and $Q$ are horizontal. The coefficient of friction between the horizontal surface and the block $B$ is 0.2 and the blocks $A$ and $B$ move together. If the system is released from rest.
Then the tension in the string $Q$ is
(Acceleration due to gravity, $g = 10 {m s}^{- 2}$ )

1

48 N

2

24 N

3

12 N

4

6 N

Explanation:

C
From F.B.D. we have,
$T_{Q} - 1 g = 1 \times a$
$T_{Q} = a + g$
$(m_{A} + m_{B} + m_{C} + m_{D}) a = m_{C} g - m_{D} g - f_{s}$
$a = \frac{[m_{C} - m_{D} - μ (m_{A} + m_{B})]}{m_{A} + m_{B} + m_{C} + m_{D}} g {∴ f_{s} = μ (m_{A} + m_{B}) g}$
$a = \frac{[6 - 1 - 0.2 \times 9]}{6 + 3 + 6 + 1} \times 10$
$= \frac{32}{16}$
$a = 2 m / s^{2}$
$T_{Q} = 1 kg \times \frac{2 m}{s^{2}} + 1 kg \times 10 \frac{m}{s^{2}}$
$= 12 kg m / s^{2} = 12 N$

AP EAMCET (23.04.2019) Shift-I

LAWS OF MOTION (ADDITIONAL)

371888 A system containing masses and pulleys connected on an inclined plane is shown in the figure. If the system is in equilibrium then the value of $m$ is

1

1 kg

2

0.5 kg

3

0.75 kg

4

0.25 kg

Explanation:

B
From the F.B.D. we have -
Since the system is in equilibrium.
$T = mg$
$T = (2 g - T_{1}) \sin 30^{\circ}$
$T_{1} = g$
From eq $^{n}$ . (ii) $i i i$ we have
$T = (2 g - g) \sin 30^{\circ} = \frac{g}{2}$
From eq $^{n}$ . (i) $i v$ , we get
$\frac{g}{2} = mg$
$m = \frac{1}{2} kg$
$m = 0.5 kg$

AP EAMCET (22.04.2019) Shift-I

LAWS OF MOTION (ADDITIONAL)

371885 In the figure shown, acceleration with which the mass $m$ falls down when released is (consider the string to be massless, gacceleration due to gravity)

1

\frac{2 g}{3}

2

\frac{g}{2}

3

\frac{5 g}{6}

4

g

Explanation:

B Let $m$ be the mass \' $R$ ' be the radius of hollow cylinder.

When the system is released, torque produced in hollow cylinder is
$τ = I α,$
$= {MR}^{2} \cdot \frac{a}{R} {∵ I = {MR}^{2} & a = α R}$
$τ = MRa$
Also, we know $τ = FR = TR$
Equating (i) \& (ii) we get
$TR = M . R a.$
$T = Ma$
Now, from F.B.D.
$Mg - T = Ma$
$Mg - Ma = Ma {using eq$
$g = 2 a$
$a = \frac{g}{2}$

AP EAMCET (20.04.2019) Shift-1

LAWS OF MOTION (ADDITIONAL)

371886 One end of a light string is fixed to a clamp on the ground and the other end passes over a fixed frictionless pulley as shown in the figure. It makes an angle of $30^{\circ}$ with the ground. The clamp can tolerate a vertical force of $40 N$ . If a monkey of mass $5 kg$ were to climb up the rope, then the maximum acceleration in the upward direction with which it can climb safely is $(g =$ $10 {ms}^{- 2}$ )

1

2 {ms}^{- 2}

2

4 {ms}^{- 2}

3

6 {ms}^{- 2}

4

8 {ms}^{- 2}

Explanation:

C Let ' $T$ ' be the tension in the string and ' $a$ ' be the acceleration in the upward direction.

The upward force created on damp is,
$T \sin 30^{\circ} = \frac{T}{2}$
$\frac{T}{2} = 40 N$
(Given)
$T = 80 N$
Now, for monkey to climb up,
$T - mg = ma$
$a = \frac{T - mg}{m}$
$= \frac{80 - 5 \times 10}{5}$
$a = 6 m / s^{2}$

AP EAMCET (20.04.2019) Shift-1

LAWS OF MOTION (ADDITIONAL)

371887 Four blocks A, B, C and D of masses $6 kg, 3 kg$ , $6 kg$ and $1 kg$ respectively are connected by light strings passing over frictionless pulleys as shown in the figure. The strings $P$ and $Q$ are horizontal. The coefficient of friction between the horizontal surface and the block $B$ is 0.2 and the blocks $A$ and $B$ move together. If the system is released from rest.
Then the tension in the string $Q$ is
(Acceleration due to gravity, $g = 10 {m s}^{- 2}$ )

1

48 N

2

24 N

3

12 N

4

6 N

Explanation:

C
From F.B.D. we have,
$T_{Q} - 1 g = 1 \times a$
$T_{Q} = a + g$
$(m_{A} + m_{B} + m_{C} + m_{D}) a = m_{C} g - m_{D} g - f_{s}$
$a = \frac{[m_{C} - m_{D} - μ (m_{A} + m_{B})]}{m_{A} + m_{B} + m_{C} + m_{D}} g {∴ f_{s} = μ (m_{A} + m_{B}) g}$
$a = \frac{[6 - 1 - 0.2 \times 9]}{6 + 3 + 6 + 1} \times 10$
$= \frac{32}{16}$
$a = 2 m / s^{2}$
$T_{Q} = 1 kg \times \frac{2 m}{s^{2}} + 1 kg \times 10 \frac{m}{s^{2}}$
$= 12 kg m / s^{2} = 12 N$

AP EAMCET (23.04.2019) Shift-I

LAWS OF MOTION (ADDITIONAL)

371888 A system containing masses and pulleys connected on an inclined plane is shown in the figure. If the system is in equilibrium then the value of $m$ is

1

1 kg

2

0.5 kg

3

0.75 kg

4

0.25 kg

Explanation:

B
From the F.B.D. we have -
Since the system is in equilibrium.
$T = mg$
$T = (2 g - T_{1}) \sin 30^{\circ}$
$T_{1} = g$
From eq $^{n}$ . (ii) $i i i$ we have
$T = (2 g - g) \sin 30^{\circ} = \frac{g}{2}$
From eq $^{n}$ . (i) $i v$ , we get
$\frac{g}{2} = mg$
$m = \frac{1}{2} kg$
$m = 0.5 kg$

AP EAMCET (22.04.2019) Shift-I

LAWS OF MOTION (ADDITIONAL)

371885 In the figure shown, acceleration with which the mass $m$ falls down when released is (consider the string to be massless, gacceleration due to gravity)

1

\frac{2 g}{3}

2

\frac{g}{2}

3

\frac{5 g}{6}

4

g

Explanation:

B Let $m$ be the mass \' $R$ ' be the radius of hollow cylinder.

When the system is released, torque produced in hollow cylinder is
$τ = I α,$
$= {MR}^{2} \cdot \frac{a}{R} {∵ I = {MR}^{2} & a = α R}$
$τ = MRa$
Also, we know $τ = FR = TR$
Equating (i) \& (ii) we get
$TR = M . R a.$
$T = Ma$
Now, from F.B.D.
$Mg - T = Ma$
$Mg - Ma = Ma {using eq$
$g = 2 a$
$a = \frac{g}{2}$

AP EAMCET (20.04.2019) Shift-1

LAWS OF MOTION (ADDITIONAL)

371886 One end of a light string is fixed to a clamp on the ground and the other end passes over a fixed frictionless pulley as shown in the figure. It makes an angle of $30^{\circ}$ with the ground. The clamp can tolerate a vertical force of $40 N$ . If a monkey of mass $5 kg$ were to climb up the rope, then the maximum acceleration in the upward direction with which it can climb safely is $(g =$ $10 {ms}^{- 2}$ )

1

2 {ms}^{- 2}

2

4 {ms}^{- 2}

3

6 {ms}^{- 2}

4

8 {ms}^{- 2}

Explanation:

C Let ' $T$ ' be the tension in the string and ' $a$ ' be the acceleration in the upward direction.

The upward force created on damp is,
$T \sin 30^{\circ} = \frac{T}{2}$
$\frac{T}{2} = 40 N$
(Given)
$T = 80 N$
Now, for monkey to climb up,
$T - mg = ma$
$a = \frac{T - mg}{m}$
$= \frac{80 - 5 \times 10}{5}$
$a = 6 m / s^{2}$

AP EAMCET (20.04.2019) Shift-1

LAWS OF MOTION (ADDITIONAL)

371887 Four blocks A, B, C and D of masses $6 kg, 3 kg$ , $6 kg$ and $1 kg$ respectively are connected by light strings passing over frictionless pulleys as shown in the figure. The strings $P$ and $Q$ are horizontal. The coefficient of friction between the horizontal surface and the block $B$ is 0.2 and the blocks $A$ and $B$ move together. If the system is released from rest.
Then the tension in the string $Q$ is
(Acceleration due to gravity, $g = 10 {m s}^{- 2}$ )

1

48 N

2

24 N

3

12 N

4

6 N

Explanation:

C
From F.B.D. we have,
$T_{Q} - 1 g = 1 \times a$
$T_{Q} = a + g$
$(m_{A} + m_{B} + m_{C} + m_{D}) a = m_{C} g - m_{D} g - f_{s}$
$a = \frac{[m_{C} - m_{D} - μ (m_{A} + m_{B})]}{m_{A} + m_{B} + m_{C} + m_{D}} g {∴ f_{s} = μ (m_{A} + m_{B}) g}$
$a = \frac{[6 - 1 - 0.2 \times 9]}{6 + 3 + 6 + 1} \times 10$
$= \frac{32}{16}$
$a = 2 m / s^{2}$
$T_{Q} = 1 kg \times \frac{2 m}{s^{2}} + 1 kg \times 10 \frac{m}{s^{2}}$
$= 12 kg m / s^{2} = 12 N$

AP EAMCET (23.04.2019) Shift-I

LAWS OF MOTION (ADDITIONAL)

371888 A system containing masses and pulleys connected on an inclined plane is shown in the figure. If the system is in equilibrium then the value of $m$ is

1

1 kg

2

0.5 kg

3

0.75 kg

4

0.25 kg

Explanation:

B
From the F.B.D. we have -
Since the system is in equilibrium.
$T = mg$
$T = (2 g - T_{1}) \sin 30^{\circ}$
$T_{1} = g$
From eq $^{n}$ . (ii) $i i i$ we have
$T = (2 g - g) \sin 30^{\circ} = \frac{g}{2}$
From eq $^{n}$ . (i) $i v$ , we get
$\frac{g}{2} = mg$
$m = \frac{1}{2} kg$
$m = 0.5 kg$

AP EAMCET (22.04.2019) Shift-I

LAWS OF MOTION (ADDITIONAL)

371885 In the figure shown, acceleration with which the mass $m$ falls down when released is (consider the string to be massless, gacceleration due to gravity)

1

\frac{2 g}{3}

2

\frac{g}{2}

3

\frac{5 g}{6}

4

g

Explanation:

B Let $m$ be the mass \' $R$ ' be the radius of hollow cylinder.

When the system is released, torque produced in hollow cylinder is
$τ = I α,$
$= {MR}^{2} \cdot \frac{a}{R} {∵ I = {MR}^{2} & a = α R}$
$τ = MRa$
Also, we know $τ = FR = TR$
Equating (i) \& (ii) we get
$TR = M . R a.$
$T = Ma$
Now, from F.B.D.
$Mg - T = Ma$
$Mg - Ma = Ma {using eq$
$g = 2 a$
$a = \frac{g}{2}$

AP EAMCET (20.04.2019) Shift-1

LAWS OF MOTION (ADDITIONAL)

371886 One end of a light string is fixed to a clamp on the ground and the other end passes over a fixed frictionless pulley as shown in the figure. It makes an angle of $30^{\circ}$ with the ground. The clamp can tolerate a vertical force of $40 N$ . If a monkey of mass $5 kg$ were to climb up the rope, then the maximum acceleration in the upward direction with which it can climb safely is $(g =$ $10 {ms}^{- 2}$ )

1

2 {ms}^{- 2}

2

4 {ms}^{- 2}

3

6 {ms}^{- 2}

4

8 {ms}^{- 2}

Explanation:

C Let ' $T$ ' be the tension in the string and ' $a$ ' be the acceleration in the upward direction.

The upward force created on damp is,
$T \sin 30^{\circ} = \frac{T}{2}$
$\frac{T}{2} = 40 N$
(Given)
$T = 80 N$
Now, for monkey to climb up,
$T - mg = ma$
$a = \frac{T - mg}{m}$
$= \frac{80 - 5 \times 10}{5}$
$a = 6 m / s^{2}$

AP EAMCET (20.04.2019) Shift-1

LAWS OF MOTION (ADDITIONAL)

371887 Four blocks A, B, C and D of masses $6 kg, 3 kg$ , $6 kg$ and $1 kg$ respectively are connected by light strings passing over frictionless pulleys as shown in the figure. The strings $P$ and $Q$ are horizontal. The coefficient of friction between the horizontal surface and the block $B$ is 0.2 and the blocks $A$ and $B$ move together. If the system is released from rest.
Then the tension in the string $Q$ is
(Acceleration due to gravity, $g = 10 {m s}^{- 2}$ )

1

48 N

2

24 N

3

12 N

4

6 N

Explanation:

C
From F.B.D. we have,
$T_{Q} - 1 g = 1 \times a$
$T_{Q} = a + g$
$(m_{A} + m_{B} + m_{C} + m_{D}) a = m_{C} g - m_{D} g - f_{s}$
$a = \frac{[m_{C} - m_{D} - μ (m_{A} + m_{B})]}{m_{A} + m_{B} + m_{C} + m_{D}} g {∴ f_{s} = μ (m_{A} + m_{B}) g}$
$a = \frac{[6 - 1 - 0.2 \times 9]}{6 + 3 + 6 + 1} \times 10$
$= \frac{32}{16}$
$a = 2 m / s^{2}$
$T_{Q} = 1 kg \times \frac{2 m}{s^{2}} + 1 kg \times 10 \frac{m}{s^{2}}$
$= 12 kg m / s^{2} = 12 N$

AP EAMCET (23.04.2019) Shift-I

LAWS OF MOTION (ADDITIONAL)

371888 A system containing masses and pulleys connected on an inclined plane is shown in the figure. If the system is in equilibrium then the value of $m$ is

1

1 kg

2

0.5 kg

3

0.75 kg

4

0.25 kg

Explanation:

B
From the F.B.D. we have -
Since the system is in equilibrium.
$T = mg$
$T = (2 g - T_{1}) \sin 30^{\circ}$
$T_{1} = g$
From eq $^{n}$ . (ii) $i i i$ we have
$T = (2 g - g) \sin 30^{\circ} = \frac{g}{2}$
From eq $^{n}$ . (i) $i v$ , we get
$\frac{g}{2} = mg$
$m = \frac{1}{2} kg$
$m = 0.5 kg$

AP EAMCET (22.04.2019) Shift-I