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Arrangement is shown in given figure. If coefficient of friction between 2kg block and table is 0.2 . What would be the maximum mass (in kg) value of block B. So that the two blocks do not move.(10=m/s^2).

Option: 1
0.4

Option: 2
0.2

Option: 3
4

Option: 4
2

Answers (1)

best_answer

As we learnt

Minimum mass hung to start Motion -

When m₁ is placed on table (rough)

For liming condition

$T=F_{l}$

$m_{2}g=\mu R$

$m_{2}g=\mu m_{1}g$

$m_{2}=\mu m_{1}$

Also $\mu=\frac{m_{2}}{m_{1}}$

- wherein

T = Tension in string

$F_{l}=$ Limiting friction

$\mu=$ Coefficient of friction

For block A not to move

? $T=\mu m_{A}g$

For block B

$T= m_{B}g$

for equilibrium

$T =\mu m_{A}g=m_{B}g$

$m_{B}=\mu m_{A}=0.2*2=0.4kg$

Posted by

Irshad Anwar

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