Q175 / 225 · JEE Main 2003 · Physics

A marble block of mass $2 \mathrm{~kg}$ lying on ice when given a velocity of $6 \mathrm{~m} / \mathrm{s}$ is stopped by friction in $10 \mathrm{~s}$. Then the coefficient of friction is

  • A
    $0.02$
  • B
    $0.03$
  • C
    $0.04$
  • Correct answer
    D
    None of these

Solution

$a=\mu g=\frac{6}{10}$ [ using $\mathrm{v}=\mathrm{u}+\mathrm{at}$ ] $\Rightarrow \mu=\frac{6}{10 \times g}=\frac{6}{10 \times 10}=0.06$

Hence, the correct answer is option: 'D'.

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