The minimum value of friction coefficient to prevent slipping is

\frac{5}{7}

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\frac{5}{12 \sqrt{3}}

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\frac{5 \sqrt{3}}{7}

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\frac{7}{5 \sqrt{3}}

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Solution

The correct option is B $\frac{5}{12 \sqrt{3}}$
$M_{T} a = M_{T} g s i n θ - f$
As frictional force $f = μ M_{T} g c o s θ$
Thus $M_{T} a = M_{T} g s i n 30 - μ M_{T} g c o s 30$
$M_{T} \frac{7 g}{24} = M_{T} g s i n 30 - μ M_{T} g c o s 30$ as $a = \frac{7 g}{24}$ (solved earlier)
$\frac{7}{24} = \frac{1}{2} - μ \frac{\sqrt{3}}{2}$
$⟹ μ = \frac{5}{12 \sqrt{3}}$

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The minimum value of friction coefficient to prevent slipping is

Q. The minimum eigen value of the following matrix is

⎡ ⎢ ⎣ \begin{matrix} 3 & 5 & 2 5 & 12 & 7 2 & 7 & 5 \end{matrix} ⎤ ⎥ ⎦

Q. Solve:

{\frac{7}{5} \times (\frac{- 3}{12})} + {\frac{7}{5} \times \frac{5}{12}}

\frac{7 + 3 \sqrt{5}}{3 + \sqrt{5}} - \frac{7 - 3 \sqrt{5}}{3 - \sqrt{5}}

find the value

a + b \sqrt{5}

Verify the following :

(i) $\frac{3}{7} \times (\frac{5}{6} + \frac{12}{13}) = (\frac{3}{7} \times \frac{5}{6}) + (\frac{3}{7} \times \frac{12}{13})$
(ii) $\frac{- 15}{4} \times (\frac{3}{7} + \frac{- 12}{5}) = (\frac{- 15}{4} \times \frac{3}{7}) + (\frac{- 15}{4} \times \frac{- 12}{5})$
(iii) $(\frac{- 8}{3} + \frac{- 13}{12}) \times \frac{5}{6} = (\frac{- 8}{3} \times \frac{5}{6}) + (\frac{- 13}{12} \times \frac{5}{6})$
(iv) $\frac{- 16}{7} \times (\frac{- 8}{9} + \frac{- 7}{6}) = (\frac{- 16}{7} \times \frac{- 8}{9}) + (\frac{- 16}{7} \times \frac{- 7}{6})$

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The minimum value of friction coefficient to prevent slipping is

The correct option is B 512√3MTa=MTgsinθ−fAs frictional force f=μMTgcosθThus MTa=MTgsin30−μMTgcos30 MT7g24=MTgsin30−μMTgcos30 as a=7g24 (solved earlier) 724=12−μ√32⟹μ=512√3