MHT CET · Physics · Laws of Motion
On dry road, the maximum speed of a vehicle along a circular path is ' \(\mathrm{V}\) '. When the road becomes wet, maximum speed becomes \(\frac{V}{2}\). If coefficient of friction of dry road is ' \(\mu\) ' then that of wet road is
- A \(\frac{2 \mu}{3}\)
- B \(\frac{\mu}{4}\)
- C \(\frac{\mu}{3}\)
- D \(\frac{3 \mu}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mu}{4}\)
Step-by-step Solution
Detailed explanation
The equation for maximum velocity is \(\mathrm{V}=\sqrt{\mu \mathrm{rg}}\)
When the road becomes wet the equation becomes
\(\frac{\mathrm{V}}{2}=\sqrt{\mu^{\prime} \mathrm{rg}}\)
Dividing equation (i) with equation (ii),
\(\frac{\mathrm{V}}{\mathrm{V}} \)
\(=\frac{\sqrt{\mu \mathrm{rg}}}{\sqrt{\mu^{\prime} \mathrm{rg}}} \)
\(\therefore 2 =\frac{\sqrt{\mu}}{\sqrt{\mu^{\prime}}} \)
\(\therefore \mu^{\prime} =\frac{\mu}{4}\)
When the road becomes wet the equation becomes
\(\frac{\mathrm{V}}{2}=\sqrt{\mu^{\prime} \mathrm{rg}}\)
Dividing equation (i) with equation (ii),
\(\frac{\mathrm{V}}{\mathrm{V}} \)
\(=\frac{\sqrt{\mu \mathrm{rg}}}{\sqrt{\mu^{\prime} \mathrm{rg}}} \)
\(\therefore 2 =\frac{\sqrt{\mu}}{\sqrt{\mu^{\prime}}} \)
\(\therefore \mu^{\prime} =\frac{\mu}{4}\)
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