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Byju's Answer

Standard XII

Physics

Problem Solving

In the system...

Question

In the system shown $m_{1} > m_{2} .$ System is held at rest by thread BC. Just after lower thread is Burnt,

Acceleration of

m_{2}

is upward

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Magnitude of acceleration of both blocks will be

(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) g

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Acceleration of

m_{1}

will be equal to zero

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Magnitudes of acceleration of two blocks will be non-zero and unequal

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Solution

The correct options are
A Acceleration of $m_{2}$ is upward
C Acceleration of $m_{1}$ will be equal to zero
Before burn,

$T = m_{1} g$ & $T = T_{1} + m_{2} g$
After burning the wire,
Tension in the spring does not change instantaneously.

so, for $m_{1}$
Fnet= $T - m_{1} g$ = 0
$∴ a_{m 1} = 0$
For $m_{2}$
$T_{1}$ becomes zero and $m_{2}$ will have a non-zero acceleration
$∴ T^{^{'}} - m_{2} g = m_{2} a$
$\Rightarrow a = \frac{T^{^{'}}}{m_{2}} - g$

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Similar questions

Q. In the system shown in the figure

m_{1} > m_{2}

. System is held at rest by thread

B C

. Just after the thread

B C

is cut:

(A) Acceleration of

m_{2}

will be upwards
(B) Magnitude of acceleration of both blocks will be equal to

(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) g

m_{1}

will be equal to zero
(D) Magnitude of acceleration of two blocks will be non-zero and unequal.
From the above four statements we can conclude

Q. In the system shown in the figure

m_{1} > m_{2}

. System is held at rest by thread

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. Just after the thread

B C

is cut:

(A) Acceleration of

m_{2}

will be upwards
(B) Magnitude of acceleration of both blocks will be equal to

(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) g

m_{1}

will be equal to zero
(D) Magnitude of acceleration of two blocks will be non-zero and unequal.
From the above four statements we can conclude

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Q. A block

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In the system shown m1>m2. System is held at rest by thread BC. Just after lower thread is Burnt,

In the system shown $m_{1} > m_{2} .$ System is held at rest by thread BC. Just after lower thread is Burnt,