MHT CET · Physics · Laws of Motion
A body weighing \(20 \mathrm{~kg}\) just slides down a rough inclined plane that rises 5 in 12 . The coefficient of friction is
- A \(0.46\)
- B \(4.6\)
- C \(0.52\)
- D \(0.12\)
Answer & Solution
Correct Answer
(A) \(0.46\)
Step-by-step Solution
Detailed explanation
The given situation can be shown as
As the plane rises 5 in 12
\(\therefore \quad \sin \theta=\frac{5}{12}\)
and \(\cos \theta=\sqrt{1-\sin ^{2} \theta}=\sqrt{1+\left(\frac{5}{12}\right)^{2}}\)
\(=\frac{\sqrt{119}}{12}\)
So, the coefficient of friction,
\(\mu=\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{5}{12} \times \frac{12}{\sqrt{119}}=\frac{5}{\sqrt{119}}=0.46\)
As the plane rises 5 in 12
\(\therefore \quad \sin \theta=\frac{5}{12}\)
and \(\cos \theta=\sqrt{1-\sin ^{2} \theta}=\sqrt{1+\left(\frac{5}{12}\right)^{2}}\)
\(=\frac{\sqrt{119}}{12}\)
So, the coefficient of friction,
\(\mu=\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{5}{12} \times \frac{12}{\sqrt{119}}=\frac{5}{\sqrt{119}}=0.46\)
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