Two blocks   A and   B of masses m A = 1   k g and m B = 3   k g are kept on the table…

Two blocks $\mathrm{A}$ and $\mathrm{B}$ of masses ${\mathrm{m}}_{\mathrm{A}}=1\mathrm{ }\mathrm{k}\mathrm{g}$ and ${\mathrm{m}}_{\mathrm{B}}=3\mathrm{ }\mathrm{k}\mathrm{g}$ are kept on the table as shown in figure. The coefficients of friction between $\mathrm{A}$ and $\mathrm{B}$ is $0.2$ and between $\mathrm{B}$ and the surface of the table is also $0.2.$ The maximum force $\mathrm{F}$ that can be applied on $\mathrm{B}$ horizontally, so that the block $\mathrm{A}$ does not slide over the block $\mathrm{B}$ is : [Take $\mathrm{ }\mathrm{g}=10\mathrm{ }\mathrm{m}/{\mathrm{s}}^2$ ]

1. $16\mathrm{ }\mathrm{N}$
2. $12\mathrm{ }\mathrm{N}$
3. $40\mathrm{ }\mathrm{N}$
4. $8\mathrm{ }\mathrm{N}$

Solution

Maximum possible acceleration of $1\mathrm{ }\mathrm{k}\mathrm{g}$ block $a=\frac{\mathrm{\mu }\mathrm{m}\mathrm{g}}{\mathrm{m}}=\mathrm{\mu }\mathrm{g}=2\mathrm{ }\mathrm{m}/{\mathrm{s}}^2$

For $1\mathrm{ }\mathrm{g}$ block not to slide the maximum acceleration of both blocks should be

${\mathrm{F}}_{\mathrm{m}\mathrm{a}\mathrm{x}}-\mathrm{\mu }\mathrm{m}\mathrm{g}=\mathrm{m}{a}_{\mathrm{m}\mathrm{a}\mathrm{x}}$
${\mathrm{F}}_{\mathrm{m}\mathrm{a}\mathrm{x}}-0.2\times 4\times 10=4\times 2$
${\mathrm{F}}_{\mathrm{m}\mathrm{a}\mathrm{x}}-\mathrm{ }8=8$
${\mathrm{F}}_{\mathrm{m}\mathrm{a}\mathrm{x}}=16\mathrm{ }\mathrm{N}$

Asked in: JEE Main 2019 (10 Apr Shift 2)

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