QBManager

LAWS OF MOTION (ADDITIONAL)

372231 Two blocks $A$ and $B$ of masses $3 m$ and $m$ respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of $A$ and B immediately after the string is cut, are respectively

1

g, \frac{g}{3}

2

\frac{g}{3}, g

3

g, g

4

\frac{g}{3}, \frac{g}{3}

Explanation:

B Case- I: Before the string is cut-

FBD before cutting the string
For block $A, S = T + 3 mg$
For block B, $T = mg$
Put the value $T = mg$ in equation (i), we get
$S = mg + 3 mg = 4 mg$
Case II : After the string is cut -

FBD after cutting the string
Using Newton's second law of motion for block B, we have-
$ma = mg$
$a = g$
Again using Newton's second law of motion for block A, we have .
$S - 3 mg = 3 {ma}^{'}$
By substituting $S = 4 mg$ in the above equation we get-
$4 {mg}^{-} - 3 mg = 3 {ma}^{'}$
$3 {ma}^{'} = mg$
$a^{'} = \frac{mg}{3 m}$
$a^{'} = \frac{g}{3}$
Therefore, the magnitudes of acceleration of block $A$ and B immediately after the sting is cut are $\frac{g}{3}$ (in upward direction) and $g$ (in downward direction) respectively.

NEET- 2017

LAWS OF MOTION (ADDITIONAL)

372232 Assertion: A man and a block rest on smooth horizontal surface. The man holds a rope which is connected to block. The man cannot move on the horizontal surface.

Reason: A man standing at rest on smooth horizontal surface cannot start walking due to absence of friction (The man is only in contact with floor as shown).

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 Assertion is incorrect, but reason is correct.

AIIMS-2017

LAWS OF MOTION (ADDITIONAL)

372233 A body of the mass $50 kg$ is suspended using a spring balance inside a lift at rest. If the lift starts falling freely, the reading of the spring balance is :

1

< 50 kg

2

= 50 kg

3

> 50 kg

4

= 0

Explanation:

D Given,
Mass of the body, $m = 50 kg$
As the lift is freely falling thus its acceleration in downward direction. (wightlessness)

Reading of the spring balance $=$ apparent weight of mass
$m (g - a) = 50 (g - g) = 0$

Karnataka CET-2017

LAWS OF MOTION (ADDITIONAL)

372234 Two blocks are connected over a mass less pulley as shown in fig. The mass of block $A$ is $10 kg$ and the coefficient of kinetic friction is 0.2. Block A slides down the incline at constant speed. The mass of block $B$ in $kg$ is:

1 3.5

2 3.3

3 3.0

4 2.5

Explanation:

B Mass of block $(m_{1}) = 10 kg$
Kinetic friction $(μ_{k}) = 0.2$
$a = 0$

For block of mass $m_{2}$
$T = m_{2} g = 10 m_{2}$
From the figure
Resolving the forces perpendicular to the plane
$N = m_{1} g \cos 30^{\circ}$
We know
$f_{k} = μ_{k} N = 0.2 \times m_{1} g \cos 30^{\circ} [{From eq}^{n} \cdot (ii)]$
$f_{k} = 0.2 \times 10 \times 10 \times \frac{\sqrt{3}}{2}$
$f_{k} = 10 \sqrt{3} N$
Resolving the forces parallel to the plane
$T + f_{k} = m_{1} g \sin 30^{\circ}$
Putting the value $f_{k}$ and $T$ in equation (iii)
$10 m_{2} + 10 \sqrt{3} = 10 \times 10 \times \frac{1}{2}$
$m_{2} + \sqrt{3} = 5$
$m_{2} = (5 - \sqrt{3})$
$m_{2} = 3.269 kg$
$m_{2} \approx 3.3 kg$

BITSAT-2010

LAWS OF MOTION (ADDITIONAL)

372235 Three blocks of masses $m_{1}, m_{2}$ and $m_{3}$ are connected by massless strings, as shown, on a frictionless table. They are pulled with a force $T_{3} = 40 N$ . If $m_{1} = 10 kg, m_{2} = 6 kg$ and $m_{3} = 4$ $kg$ , the tension $T_{2}$ will be

1

20 N

2

40 N

3

10 N

4

32 N

Explanation:

D Given $m_{1} = 10 kg, m_{2} = 6 kg, m_{3} = 4 kg, T_{3} =$ $40 N$

Total mass of system $= m_{1} + m_{2} + m_{3} = 20 kg$
$\begin{array}{r} ∴ a = \frac{F}{m} \\ a = \frac{40}{20} = 2 m / s^{2} \end{array}$
Because string $T_{2}$ pulls $m_{1}$ and $m_{2}$ block.
$T_{2} = (m_{1} + m_{2}) a$
$T_{2} = (10 + 6) \times 2$
$T_{2} = 32 N$

BITSAT-2011

LAWS OF MOTION (ADDITIONAL)

372231 Two blocks $A$ and $B$ of masses $3 m$ and $m$ respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of $A$ and B immediately after the string is cut, are respectively

1

g, \frac{g}{3}

2

\frac{g}{3}, g

3

g, g

4

\frac{g}{3}, \frac{g}{3}

Explanation:

B Case- I: Before the string is cut-

FBD before cutting the string
For block $A, S = T + 3 mg$
For block B, $T = mg$
Put the value $T = mg$ in equation (i), we get
$S = mg + 3 mg = 4 mg$
Case II : After the string is cut -

FBD after cutting the string
Using Newton's second law of motion for block B, we have-
$ma = mg$
$a = g$
Again using Newton's second law of motion for block A, we have .
$S - 3 mg = 3 {ma}^{'}$
By substituting $S = 4 mg$ in the above equation we get-
$4 {mg}^{-} - 3 mg = 3 {ma}^{'}$
$3 {ma}^{'} = mg$
$a^{'} = \frac{mg}{3 m}$
$a^{'} = \frac{g}{3}$
Therefore, the magnitudes of acceleration of block $A$ and B immediately after the sting is cut are $\frac{g}{3}$ (in upward direction) and $g$ (in downward direction) respectively.

NEET- 2017

LAWS OF MOTION (ADDITIONAL)

372232 Assertion: A man and a block rest on smooth horizontal surface. The man holds a rope which is connected to block. The man cannot move on the horizontal surface.

Reason: A man standing at rest on smooth horizontal surface cannot start walking due to absence of friction (The man is only in contact with floor as shown).

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 Assertion is incorrect, but reason is correct.

AIIMS-2017

LAWS OF MOTION (ADDITIONAL)

372233 A body of the mass $50 kg$ is suspended using a spring balance inside a lift at rest. If the lift starts falling freely, the reading of the spring balance is :

1

< 50 kg

2

= 50 kg

3

> 50 kg

4

= 0

Explanation:

D Given,
Mass of the body, $m = 50 kg$
As the lift is freely falling thus its acceleration in downward direction. (wightlessness)

Reading of the spring balance $=$ apparent weight of mass
$m (g - a) = 50 (g - g) = 0$

Karnataka CET-2017

LAWS OF MOTION (ADDITIONAL)

372234 Two blocks are connected over a mass less pulley as shown in fig. The mass of block $A$ is $10 kg$ and the coefficient of kinetic friction is 0.2. Block A slides down the incline at constant speed. The mass of block $B$ in $kg$ is:

1 3.5

2 3.3

3 3.0

4 2.5

Explanation:

B Mass of block $(m_{1}) = 10 kg$
Kinetic friction $(μ_{k}) = 0.2$
$a = 0$

For block of mass $m_{2}$
$T = m_{2} g = 10 m_{2}$
From the figure
Resolving the forces perpendicular to the plane
$N = m_{1} g \cos 30^{\circ}$
We know
$f_{k} = μ_{k} N = 0.2 \times m_{1} g \cos 30^{\circ} [{From eq}^{n} \cdot (ii)]$
$f_{k} = 0.2 \times 10 \times 10 \times \frac{\sqrt{3}}{2}$
$f_{k} = 10 \sqrt{3} N$
Resolving the forces parallel to the plane
$T + f_{k} = m_{1} g \sin 30^{\circ}$
Putting the value $f_{k}$ and $T$ in equation (iii)
$10 m_{2} + 10 \sqrt{3} = 10 \times 10 \times \frac{1}{2}$
$m_{2} + \sqrt{3} = 5$
$m_{2} = (5 - \sqrt{3})$
$m_{2} = 3.269 kg$
$m_{2} \approx 3.3 kg$

BITSAT-2010

LAWS OF MOTION (ADDITIONAL)

372235 Three blocks of masses $m_{1}, m_{2}$ and $m_{3}$ are connected by massless strings, as shown, on a frictionless table. They are pulled with a force $T_{3} = 40 N$ . If $m_{1} = 10 kg, m_{2} = 6 kg$ and $m_{3} = 4$ $kg$ , the tension $T_{2}$ will be

1

20 N

2

40 N

3

10 N

4

32 N

Explanation:

D Given $m_{1} = 10 kg, m_{2} = 6 kg, m_{3} = 4 kg, T_{3} =$ $40 N$

Total mass of system $= m_{1} + m_{2} + m_{3} = 20 kg$
$\begin{array}{r} ∴ a = \frac{F}{m} \\ a = \frac{40}{20} = 2 m / s^{2} \end{array}$
Because string $T_{2}$ pulls $m_{1}$ and $m_{2}$ block.
$T_{2} = (m_{1} + m_{2}) a$
$T_{2} = (10 + 6) \times 2$
$T_{2} = 32 N$

BITSAT-2011

LAWS OF MOTION (ADDITIONAL)

372231 Two blocks $A$ and $B$ of masses $3 m$ and $m$ respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of $A$ and B immediately after the string is cut, are respectively

1

g, \frac{g}{3}

2

\frac{g}{3}, g

3

g, g

4

\frac{g}{3}, \frac{g}{3}

Explanation:

B Case- I: Before the string is cut-

FBD before cutting the string
For block $A, S = T + 3 mg$
For block B, $T = mg$
Put the value $T = mg$ in equation (i), we get
$S = mg + 3 mg = 4 mg$
Case II : After the string is cut -

FBD after cutting the string
Using Newton's second law of motion for block B, we have-
$ma = mg$
$a = g$
Again using Newton's second law of motion for block A, we have .
$S - 3 mg = 3 {ma}^{'}$
By substituting $S = 4 mg$ in the above equation we get-
$4 {mg}^{-} - 3 mg = 3 {ma}^{'}$
$3 {ma}^{'} = mg$
$a^{'} = \frac{mg}{3 m}$
$a^{'} = \frac{g}{3}$
Therefore, the magnitudes of acceleration of block $A$ and B immediately after the sting is cut are $\frac{g}{3}$ (in upward direction) and $g$ (in downward direction) respectively.

NEET- 2017

LAWS OF MOTION (ADDITIONAL)

372232 Assertion: A man and a block rest on smooth horizontal surface. The man holds a rope which is connected to block. The man cannot move on the horizontal surface.

Reason: A man standing at rest on smooth horizontal surface cannot start walking due to absence of friction (The man is only in contact with floor as shown).

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 Assertion is incorrect, but reason is correct.

AIIMS-2017

LAWS OF MOTION (ADDITIONAL)

372233 A body of the mass $50 kg$ is suspended using a spring balance inside a lift at rest. If the lift starts falling freely, the reading of the spring balance is :

1

< 50 kg

2

= 50 kg

3

> 50 kg

4

= 0

Explanation:

D Given,
Mass of the body, $m = 50 kg$
As the lift is freely falling thus its acceleration in downward direction. (wightlessness)

Reading of the spring balance $=$ apparent weight of mass
$m (g - a) = 50 (g - g) = 0$

Karnataka CET-2017

LAWS OF MOTION (ADDITIONAL)

372234 Two blocks are connected over a mass less pulley as shown in fig. The mass of block $A$ is $10 kg$ and the coefficient of kinetic friction is 0.2. Block A slides down the incline at constant speed. The mass of block $B$ in $kg$ is:

1 3.5

2 3.3

3 3.0

4 2.5

Explanation:

B Mass of block $(m_{1}) = 10 kg$
Kinetic friction $(μ_{k}) = 0.2$
$a = 0$

For block of mass $m_{2}$
$T = m_{2} g = 10 m_{2}$
From the figure
Resolving the forces perpendicular to the plane
$N = m_{1} g \cos 30^{\circ}$
We know
$f_{k} = μ_{k} N = 0.2 \times m_{1} g \cos 30^{\circ} [{From eq}^{n} \cdot (ii)]$
$f_{k} = 0.2 \times 10 \times 10 \times \frac{\sqrt{3}}{2}$
$f_{k} = 10 \sqrt{3} N$
Resolving the forces parallel to the plane
$T + f_{k} = m_{1} g \sin 30^{\circ}$
Putting the value $f_{k}$ and $T$ in equation (iii)
$10 m_{2} + 10 \sqrt{3} = 10 \times 10 \times \frac{1}{2}$
$m_{2} + \sqrt{3} = 5$
$m_{2} = (5 - \sqrt{3})$
$m_{2} = 3.269 kg$
$m_{2} \approx 3.3 kg$

BITSAT-2010

LAWS OF MOTION (ADDITIONAL)

372235 Three blocks of masses $m_{1}, m_{2}$ and $m_{3}$ are connected by massless strings, as shown, on a frictionless table. They are pulled with a force $T_{3} = 40 N$ . If $m_{1} = 10 kg, m_{2} = 6 kg$ and $m_{3} = 4$ $kg$ , the tension $T_{2}$ will be

1

20 N

2

40 N

3

10 N

4

32 N

Explanation:

D Given $m_{1} = 10 kg, m_{2} = 6 kg, m_{3} = 4 kg, T_{3} =$ $40 N$

Total mass of system $= m_{1} + m_{2} + m_{3} = 20 kg$
$\begin{array}{r} ∴ a = \frac{F}{m} \\ a = \frac{40}{20} = 2 m / s^{2} \end{array}$
Because string $T_{2}$ pulls $m_{1}$ and $m_{2}$ block.
$T_{2} = (m_{1} + m_{2}) a$
$T_{2} = (10 + 6) \times 2$
$T_{2} = 32 N$

BITSAT-2011

LAWS OF MOTION (ADDITIONAL)

372231 Two blocks $A$ and $B$ of masses $3 m$ and $m$ respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of $A$ and B immediately after the string is cut, are respectively

1

g, \frac{g}{3}

2

\frac{g}{3}, g

3

g, g

4

\frac{g}{3}, \frac{g}{3}

Explanation:

B Case- I: Before the string is cut-

FBD before cutting the string
For block $A, S = T + 3 mg$
For block B, $T = mg$
Put the value $T = mg$ in equation (i), we get
$S = mg + 3 mg = 4 mg$
Case II : After the string is cut -

FBD after cutting the string
Using Newton's second law of motion for block B, we have-
$ma = mg$
$a = g$
Again using Newton's second law of motion for block A, we have .
$S - 3 mg = 3 {ma}^{'}$
By substituting $S = 4 mg$ in the above equation we get-
$4 {mg}^{-} - 3 mg = 3 {ma}^{'}$
$3 {ma}^{'} = mg$
$a^{'} = \frac{mg}{3 m}$
$a^{'} = \frac{g}{3}$
Therefore, the magnitudes of acceleration of block $A$ and B immediately after the sting is cut are $\frac{g}{3}$ (in upward direction) and $g$ (in downward direction) respectively.

NEET- 2017

LAWS OF MOTION (ADDITIONAL)

372232 Assertion: A man and a block rest on smooth horizontal surface. The man holds a rope which is connected to block. The man cannot move on the horizontal surface.

Reason: A man standing at rest on smooth horizontal surface cannot start walking due to absence of friction (The man is only in contact with floor as shown).

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 Assertion is incorrect, but reason is correct.

AIIMS-2017

LAWS OF MOTION (ADDITIONAL)

372233 A body of the mass $50 kg$ is suspended using a spring balance inside a lift at rest. If the lift starts falling freely, the reading of the spring balance is :

1

< 50 kg

2

= 50 kg

3

> 50 kg

4

= 0

Explanation:

D Given,
Mass of the body, $m = 50 kg$
As the lift is freely falling thus its acceleration in downward direction. (wightlessness)

Reading of the spring balance $=$ apparent weight of mass
$m (g - a) = 50 (g - g) = 0$

Karnataka CET-2017

LAWS OF MOTION (ADDITIONAL)

372234 Two blocks are connected over a mass less pulley as shown in fig. The mass of block $A$ is $10 kg$ and the coefficient of kinetic friction is 0.2. Block A slides down the incline at constant speed. The mass of block $B$ in $kg$ is:

1 3.5

2 3.3

3 3.0

4 2.5

Explanation:

B Mass of block $(m_{1}) = 10 kg$
Kinetic friction $(μ_{k}) = 0.2$
$a = 0$

For block of mass $m_{2}$
$T = m_{2} g = 10 m_{2}$
From the figure
Resolving the forces perpendicular to the plane
$N = m_{1} g \cos 30^{\circ}$
We know
$f_{k} = μ_{k} N = 0.2 \times m_{1} g \cos 30^{\circ} [{From eq}^{n} \cdot (ii)]$
$f_{k} = 0.2 \times 10 \times 10 \times \frac{\sqrt{3}}{2}$
$f_{k} = 10 \sqrt{3} N$
Resolving the forces parallel to the plane
$T + f_{k} = m_{1} g \sin 30^{\circ}$
Putting the value $f_{k}$ and $T$ in equation (iii)
$10 m_{2} + 10 \sqrt{3} = 10 \times 10 \times \frac{1}{2}$
$m_{2} + \sqrt{3} = 5$
$m_{2} = (5 - \sqrt{3})$
$m_{2} = 3.269 kg$
$m_{2} \approx 3.3 kg$

BITSAT-2010

LAWS OF MOTION (ADDITIONAL)

372235 Three blocks of masses $m_{1}, m_{2}$ and $m_{3}$ are connected by massless strings, as shown, on a frictionless table. They are pulled with a force $T_{3} = 40 N$ . If $m_{1} = 10 kg, m_{2} = 6 kg$ and $m_{3} = 4$ $kg$ , the tension $T_{2}$ will be

1

20 N

2

40 N

3

10 N

4

32 N

Explanation:

D Given $m_{1} = 10 kg, m_{2} = 6 kg, m_{3} = 4 kg, T_{3} =$ $40 N$

Total mass of system $= m_{1} + m_{2} + m_{3} = 20 kg$
$\begin{array}{r} ∴ a = \frac{F}{m} \\ a = \frac{40}{20} = 2 m / s^{2} \end{array}$
Because string $T_{2}$ pulls $m_{1}$ and $m_{2}$ block.
$T_{2} = (m_{1} + m_{2}) a$
$T_{2} = (10 + 6) \times 2$
$T_{2} = 32 N$

BITSAT-2011

LAWS OF MOTION (ADDITIONAL)

372231 Two blocks $A$ and $B$ of masses $3 m$ and $m$ respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of $A$ and B immediately after the string is cut, are respectively

1

g, \frac{g}{3}

2

\frac{g}{3}, g

3

g, g

4

\frac{g}{3}, \frac{g}{3}

Explanation:

B Case- I: Before the string is cut-

FBD before cutting the string
For block $A, S = T + 3 mg$
For block B, $T = mg$
Put the value $T = mg$ in equation (i), we get
$S = mg + 3 mg = 4 mg$
Case II : After the string is cut -

FBD after cutting the string
Using Newton's second law of motion for block B, we have-
$ma = mg$
$a = g$
Again using Newton's second law of motion for block A, we have .
$S - 3 mg = 3 {ma}^{'}$
By substituting $S = 4 mg$ in the above equation we get-
$4 {mg}^{-} - 3 mg = 3 {ma}^{'}$
$3 {ma}^{'} = mg$
$a^{'} = \frac{mg}{3 m}$
$a^{'} = \frac{g}{3}$
Therefore, the magnitudes of acceleration of block $A$ and B immediately after the sting is cut are $\frac{g}{3}$ (in upward direction) and $g$ (in downward direction) respectively.

NEET- 2017

LAWS OF MOTION (ADDITIONAL)

372232 Assertion: A man and a block rest on smooth horizontal surface. The man holds a rope which is connected to block. The man cannot move on the horizontal surface.

Reason: A man standing at rest on smooth horizontal surface cannot start walking due to absence of friction (The man is only in contact with floor as shown).

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 Assertion is incorrect, but reason is correct.

AIIMS-2017

LAWS OF MOTION (ADDITIONAL)

372233 A body of the mass $50 kg$ is suspended using a spring balance inside a lift at rest. If the lift starts falling freely, the reading of the spring balance is :

1

< 50 kg

2

= 50 kg

3

> 50 kg

4

= 0

Explanation:

D Given,
Mass of the body, $m = 50 kg$
As the lift is freely falling thus its acceleration in downward direction. (wightlessness)

Reading of the spring balance $=$ apparent weight of mass
$m (g - a) = 50 (g - g) = 0$

Karnataka CET-2017

LAWS OF MOTION (ADDITIONAL)

372234 Two blocks are connected over a mass less pulley as shown in fig. The mass of block $A$ is $10 kg$ and the coefficient of kinetic friction is 0.2. Block A slides down the incline at constant speed. The mass of block $B$ in $kg$ is:

1 3.5

2 3.3

3 3.0

4 2.5

Explanation:

B Mass of block $(m_{1}) = 10 kg$
Kinetic friction $(μ_{k}) = 0.2$
$a = 0$

For block of mass $m_{2}$
$T = m_{2} g = 10 m_{2}$
From the figure
Resolving the forces perpendicular to the plane
$N = m_{1} g \cos 30^{\circ}$
We know
$f_{k} = μ_{k} N = 0.2 \times m_{1} g \cos 30^{\circ} [{From eq}^{n} \cdot (ii)]$
$f_{k} = 0.2 \times 10 \times 10 \times \frac{\sqrt{3}}{2}$
$f_{k} = 10 \sqrt{3} N$
Resolving the forces parallel to the plane
$T + f_{k} = m_{1} g \sin 30^{\circ}$
Putting the value $f_{k}$ and $T$ in equation (iii)
$10 m_{2} + 10 \sqrt{3} = 10 \times 10 \times \frac{1}{2}$
$m_{2} + \sqrt{3} = 5$
$m_{2} = (5 - \sqrt{3})$
$m_{2} = 3.269 kg$
$m_{2} \approx 3.3 kg$

BITSAT-2010

LAWS OF MOTION (ADDITIONAL)

372235 Three blocks of masses $m_{1}, m_{2}$ and $m_{3}$ are connected by massless strings, as shown, on a frictionless table. They are pulled with a force $T_{3} = 40 N$ . If $m_{1} = 10 kg, m_{2} = 6 kg$ and $m_{3} = 4$ $kg$ , the tension $T_{2}$ will be

1

20 N

2

40 N

3

10 N

4

32 N

Explanation:

D Given $m_{1} = 10 kg, m_{2} = 6 kg, m_{3} = 4 kg, T_{3} =$ $40 N$

Total mass of system $= m_{1} + m_{2} + m_{3} = 20 kg$
$\begin{array}{r} ∴ a = \frac{F}{m} \\ a = \frac{40}{20} = 2 m / s^{2} \end{array}$
Because string $T_{2}$ pulls $m_{1}$ and $m_{2}$ block.
$T_{2} = (m_{1} + m_{2}) a$
$T_{2} = (10 + 6) \times 2$
$T_{2} = 32 N$

BITSAT-2011

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