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In the system shown in the figure m(1) g...

In the system shown in the figure `m_(1) gt m_(2)`. System is held at rest by thread BC. Just after the thread BC is burnt :

A

initial acceleration of `m_(2)` will be upwards

B

magnitude of initial acceleration of both blocks will be equal to `((m_(1)-m_(2))/(m_(1)+m_(2)))g`

C

initial acceleration of `m_(1)` will be equal to zero

D

magnitude of initial acceleration of two blocks will be non-zero and unequal.

Text Solution

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The correct Answer is:
a,c
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Knowledge Check

  • In the system shown in figure m_(1)gtm_(2) . System is held at rest by thread BC. Just after the thread BC is burnt.

    A
    acceleration of `m_(1)` will be equal to zero
    B
    acceleration of `m_(2)` will be downwards
    C
    magnitude of acceleration of two blocks will be non-zero and unequal
    D
    magnitude of acceleration of both the blocks will be `((m_(1)-m_(2))/(m_(1)+m_(2)))g`

  • In the system shown in figure m_(1) gt m_(2) . System is held at rest by thread BP. Just after the thread BP is burnt :

    A
    Magnitude of acceleration of both blocks will be equal to `((m_(1)-m_(2))/(m_(1)+2m_(2)))g`
    B
    Acceleration of `m_(1)` will be equal to zero
    C
    Acceleration of `m_(2)` will be upwards
    D
    Magnitudes of acceleration of two blocks will be non-zero and unequal

  • In the system in the figure m_(1)gtm_(2) system is held at rest by thread BC . Just after the thread BC is burnt:

    A
    acceleration of `m_(2)` will be upwards
    B
    magnitude of acceleration of both blocks will be equal to `((m_(1)+m_(2))/(m_(1)+m_(2)))g`
    C
    acceleration of `m_(1)` will be equal to zero
    D
    magnitude of acceleration of two blocks will be non-zro and unequal.

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