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Byju's Answer

Standard XII

Physics

Tension in a String

Two blocks ...

Question

Two blocks $m_{1}$ and $m_{2}$ lying on a horizontal surface are connected to one another by a massless inextensible string. A constant force $F$ is applied as shown in figure and block $m_{2}$ just starts moving. Then $F$ is ( $μ$ is the coefficient of friciton):

μ (m_{1}) g

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μ (m_{1} + m_{2}) g

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μ (m_{1} - m_{2}) g

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μ (m_{1} m_{2}) g

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Solution

The correct option is B $μ (m_{1} + m_{2}) g$
As the string is inextensible so there is no tension. The block $M_{2}$ will start to move if the force $F$ balances the frictional force of the total system. i.e,
$F = μ N = μ (M_{1} + M_{2}) g$

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Two blocks m1 and m2 lying on a horizontal surface are connected to one another by a massless inextensible string. A constant force F is applied as shown in figure and block m2 just starts moving. Then F is (μ is the coefficient of friciton):

The correct option is B μ(m1+m2)gAs the string is inextensible so there is no tension. The block M2 will start to move if the force F balances the frictional force of the total system. i.e,F=μN=μ(M1+M2)g

Two blocks $m_{1}$ and $m_{2}$ lying on a horizontal surface are connected to one another by a massless inextensible string. A constant force $F$ is applied as shown in figure and block $m_{2}$ just starts moving. Then $F$ is ( $μ$ is the coefficient of friciton):

The correct option is B $μ (m_{1} + m_{2}) g$
As the string is inextensible so there is no tension. The block $M_{2}$ will start to move if the force $F$ balances the frictional force of the total system. i.e,
$F = μ N = μ (M_{1} + M_{2}) g$