QBManager

LAWS OF MOTION (ADDITIONAL)

372227 Two masses $M_{1}$ and $M_{2}$ are accelerated uniformly on frictionless surface as shown in figure. The ratio of the tensions $\frac{T_{1}}{T_{2}}$ is
$⟶ \bar{a}$

1

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

2

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

3

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

4

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

Explanation:

A From the above figure.
$T_{1} = M_{1} \cdot a$
Where $a \to$ acceleration
And, $T_{2} - T_{1} = M_{2}$ . a
From eq. (i) and (ii)
Hence,
$T_{2} - M_{1} a = M_{2} a$
$T_{2} = (M_{1} + M_{2}) a$
$∴ \frac{T_{1}}{T_{2}} = \frac{M_{1} a}{(M_{1} + M_{2}) a} = \frac{M_{1}}{M_{1} + M_{2}}$

MHT-CET 2020

LAWS OF MOTION (ADDITIONAL)

372228 Three blocks are connected by massless strings on a frictionless inclined plane of $30^{\circ}$ as shown in the figure. A force of $104 N$ is applied upward along the incline to mass $m_{3}$ causing an upward motion of the blocks. What is the acceleration of the blocks? (Assume, acceleration due to gravitv, $g = 10 m / s^{2}$ )
original image

1

6.0 m / s^{2}

2

4.5 m / s^{2}

3

3.0 m / s^{2}

4

1.5 m / s^{2}

Explanation:

D
$F - (m_{1} g \sin θ + m_{2} g \sin θ + m_{3} g \sin θ) = (m_{1} + m_{2} + m_{3}) \times$ a
$104 - (3 g \sin 30^{\circ} + 5 g \sin 30^{\circ} + 8 g \sin 30^{\circ}) = (3 + 5 + 8) \times a$
$104 - (3 + 5 + 8) g \sin 30 = 16 a$
$104 - 16 \times 10 \times \frac{1}{2} = 16 a$
$∴ a = \frac{24}{16}, a = 1.5 m / s^{2}$

TS- EAMCET-03.05.2019

LAWS OF MOTION (ADDITIONAL)

372229 Two masses connected in series with two massless strings are hanging from a support as shown in the figure. Find the tension in the upper string

1

m_{1} g

2

(m_{1} - m_{2}) g

3

m_{2} g

4

(m_{1} + m_{2}) g

5

(m_{1} \times m_{2}) g

Explanation:

D Two masses $m_{1} m a t h r m m_{2}$ are hanging from a support as shown in figure-

The tension is in string $QR$ ,
$T_{1} =$ Total mass hanging on string $\times g = m_{2} g$ Similarly,
The tension is in string OP,
$T_{2} =$ Total mass hanging on string $OP \times g = (m_{1} + m_{2}) g$ $T_{2} = (m_{1} + m_{2}) g$
Hence the tension in upper string is,
$T = (m_{1} + m_{2}) g$

Kerala CEE-2019

LAWS OF MOTION (ADDITIONAL)

372230 A $1 kg$ block and a $0.5 kg$ block move together on a horizontal frictionless surface. Each block exerts a force of $6 N$ on the other. The block move with a uniform acceleration of
original image

1

3 {ms}^{- 2}

2

6 {ms}^{- 2}

3

9 {ms}^{- 2}

4

12 {ms}^{- 2}

AIIMS-27.05.2018(E)

LAWS OF MOTION (ADDITIONAL)

372227 Two masses $M_{1}$ and $M_{2}$ are accelerated uniformly on frictionless surface as shown in figure. The ratio of the tensions $\frac{T_{1}}{T_{2}}$ is
$⟶ \bar{a}$

1

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

2

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

3

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

4

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

Explanation:

A From the above figure.
$T_{1} = M_{1} \cdot a$
Where $a \to$ acceleration
And, $T_{2} - T_{1} = M_{2}$ . a
From eq. (i) and (ii)
Hence,
$T_{2} - M_{1} a = M_{2} a$
$T_{2} = (M_{1} + M_{2}) a$
$∴ \frac{T_{1}}{T_{2}} = \frac{M_{1} a}{(M_{1} + M_{2}) a} = \frac{M_{1}}{M_{1} + M_{2}}$

MHT-CET 2020

LAWS OF MOTION (ADDITIONAL)

372228 Three blocks are connected by massless strings on a frictionless inclined plane of $30^{\circ}$ as shown in the figure. A force of $104 N$ is applied upward along the incline to mass $m_{3}$ causing an upward motion of the blocks. What is the acceleration of the blocks? (Assume, acceleration due to gravitv, $g = 10 m / s^{2}$ )
original image

1

6.0 m / s^{2}

2

4.5 m / s^{2}

3

3.0 m / s^{2}

4

1.5 m / s^{2}

Explanation:

D
$F - (m_{1} g \sin θ + m_{2} g \sin θ + m_{3} g \sin θ) = (m_{1} + m_{2} + m_{3}) \times$ a
$104 - (3 g \sin 30^{\circ} + 5 g \sin 30^{\circ} + 8 g \sin 30^{\circ}) = (3 + 5 + 8) \times a$
$104 - (3 + 5 + 8) g \sin 30 = 16 a$
$104 - 16 \times 10 \times \frac{1}{2} = 16 a$
$∴ a = \frac{24}{16}, a = 1.5 m / s^{2}$

TS- EAMCET-03.05.2019

LAWS OF MOTION (ADDITIONAL)

372229 Two masses connected in series with two massless strings are hanging from a support as shown in the figure. Find the tension in the upper string

1

m_{1} g

2

(m_{1} - m_{2}) g

3

m_{2} g

4

(m_{1} + m_{2}) g

5

(m_{1} \times m_{2}) g

Explanation:

D Two masses $m_{1} m a t h r m m_{2}$ are hanging from a support as shown in figure-

The tension is in string $QR$ ,
$T_{1} =$ Total mass hanging on string $\times g = m_{2} g$ Similarly,
The tension is in string OP,
$T_{2} =$ Total mass hanging on string $OP \times g = (m_{1} + m_{2}) g$ $T_{2} = (m_{1} + m_{2}) g$
Hence the tension in upper string is,
$T = (m_{1} + m_{2}) g$

Kerala CEE-2019

LAWS OF MOTION (ADDITIONAL)

372230 A $1 kg$ block and a $0.5 kg$ block move together on a horizontal frictionless surface. Each block exerts a force of $6 N$ on the other. The block move with a uniform acceleration of
original image

1

3 {ms}^{- 2}

2

6 {ms}^{- 2}

3

9 {ms}^{- 2}

4

12 {ms}^{- 2}

AIIMS-27.05.2018(E)

LAWS OF MOTION (ADDITIONAL)

372227 Two masses $M_{1}$ and $M_{2}$ are accelerated uniformly on frictionless surface as shown in figure. The ratio of the tensions $\frac{T_{1}}{T_{2}}$ is
$⟶ \bar{a}$

1

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

2

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

3

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

4

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

Explanation:

A From the above figure.
$T_{1} = M_{1} \cdot a$
Where $a \to$ acceleration
And, $T_{2} - T_{1} = M_{2}$ . a
From eq. (i) and (ii)
Hence,
$T_{2} - M_{1} a = M_{2} a$
$T_{2} = (M_{1} + M_{2}) a$
$∴ \frac{T_{1}}{T_{2}} = \frac{M_{1} a}{(M_{1} + M_{2}) a} = \frac{M_{1}}{M_{1} + M_{2}}$

MHT-CET 2020

LAWS OF MOTION (ADDITIONAL)

372228 Three blocks are connected by massless strings on a frictionless inclined plane of $30^{\circ}$ as shown in the figure. A force of $104 N$ is applied upward along the incline to mass $m_{3}$ causing an upward motion of the blocks. What is the acceleration of the blocks? (Assume, acceleration due to gravitv, $g = 10 m / s^{2}$ )
original image

1

6.0 m / s^{2}

2

4.5 m / s^{2}

3

3.0 m / s^{2}

4

1.5 m / s^{2}

Explanation:

D
$F - (m_{1} g \sin θ + m_{2} g \sin θ + m_{3} g \sin θ) = (m_{1} + m_{2} + m_{3}) \times$ a
$104 - (3 g \sin 30^{\circ} + 5 g \sin 30^{\circ} + 8 g \sin 30^{\circ}) = (3 + 5 + 8) \times a$
$104 - (3 + 5 + 8) g \sin 30 = 16 a$
$104 - 16 \times 10 \times \frac{1}{2} = 16 a$
$∴ a = \frac{24}{16}, a = 1.5 m / s^{2}$

TS- EAMCET-03.05.2019

LAWS OF MOTION (ADDITIONAL)

372229 Two masses connected in series with two massless strings are hanging from a support as shown in the figure. Find the tension in the upper string

1

m_{1} g

2

(m_{1} - m_{2}) g

3

m_{2} g

4

(m_{1} + m_{2}) g

5

(m_{1} \times m_{2}) g

Explanation:

D Two masses $m_{1} m a t h r m m_{2}$ are hanging from a support as shown in figure-

The tension is in string $QR$ ,
$T_{1} =$ Total mass hanging on string $\times g = m_{2} g$ Similarly,
The tension is in string OP,
$T_{2} =$ Total mass hanging on string $OP \times g = (m_{1} + m_{2}) g$ $T_{2} = (m_{1} + m_{2}) g$
Hence the tension in upper string is,
$T = (m_{1} + m_{2}) g$

Kerala CEE-2019

LAWS OF MOTION (ADDITIONAL)

372230 A $1 kg$ block and a $0.5 kg$ block move together on a horizontal frictionless surface. Each block exerts a force of $6 N$ on the other. The block move with a uniform acceleration of
original image

1

3 {ms}^{- 2}

2

6 {ms}^{- 2}

3

9 {ms}^{- 2}

4

12 {ms}^{- 2}

AIIMS-27.05.2018(E)

LAWS OF MOTION (ADDITIONAL)

372227 Two masses $M_{1}$ and $M_{2}$ are accelerated uniformly on frictionless surface as shown in figure. The ratio of the tensions $\frac{T_{1}}{T_{2}}$ is
$⟶ \bar{a}$

1

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

2

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

3

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

4

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

Explanation:

A From the above figure.
$T_{1} = M_{1} \cdot a$
Where $a \to$ acceleration
And, $T_{2} - T_{1} = M_{2}$ . a
From eq. (i) and (ii)
Hence,
$T_{2} - M_{1} a = M_{2} a$
$T_{2} = (M_{1} + M_{2}) a$
$∴ \frac{T_{1}}{T_{2}} = \frac{M_{1} a}{(M_{1} + M_{2}) a} = \frac{M_{1}}{M_{1} + M_{2}}$

MHT-CET 2020

LAWS OF MOTION (ADDITIONAL)

372228 Three blocks are connected by massless strings on a frictionless inclined plane of $30^{\circ}$ as shown in the figure. A force of $104 N$ is applied upward along the incline to mass $m_{3}$ causing an upward motion of the blocks. What is the acceleration of the blocks? (Assume, acceleration due to gravitv, $g = 10 m / s^{2}$ )
original image

1

6.0 m / s^{2}

2

4.5 m / s^{2}

3

3.0 m / s^{2}

4

1.5 m / s^{2}

Explanation:

D
$F - (m_{1} g \sin θ + m_{2} g \sin θ + m_{3} g \sin θ) = (m_{1} + m_{2} + m_{3}) \times$ a
$104 - (3 g \sin 30^{\circ} + 5 g \sin 30^{\circ} + 8 g \sin 30^{\circ}) = (3 + 5 + 8) \times a$
$104 - (3 + 5 + 8) g \sin 30 = 16 a$
$104 - 16 \times 10 \times \frac{1}{2} = 16 a$
$∴ a = \frac{24}{16}, a = 1.5 m / s^{2}$

TS- EAMCET-03.05.2019

LAWS OF MOTION (ADDITIONAL)

372229 Two masses connected in series with two massless strings are hanging from a support as shown in the figure. Find the tension in the upper string

1

m_{1} g

2

(m_{1} - m_{2}) g

3

m_{2} g

4

(m_{1} + m_{2}) g

5

(m_{1} \times m_{2}) g

Explanation:

D Two masses $m_{1} m a t h r m m_{2}$ are hanging from a support as shown in figure-

The tension is in string $QR$ ,
$T_{1} =$ Total mass hanging on string $\times g = m_{2} g$ Similarly,
The tension is in string OP,
$T_{2} =$ Total mass hanging on string $OP \times g = (m_{1} + m_{2}) g$ $T_{2} = (m_{1} + m_{2}) g$
Hence the tension in upper string is,
$T = (m_{1} + m_{2}) g$

Kerala CEE-2019

LAWS OF MOTION (ADDITIONAL)

372230 A $1 kg$ block and a $0.5 kg$ block move together on a horizontal frictionless surface. Each block exerts a force of $6 N$ on the other. The block move with a uniform acceleration of
original image

1

3 {ms}^{- 2}

2

6 {ms}^{- 2}

3

9 {ms}^{- 2}

4

12 {ms}^{- 2}

AIIMS-27.05.2018(E)

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