MHT CET · Physics · Gravitation
A \(500 \mathrm{~kg}\) car takes a round turn of radius \(50 \mathrm{~m}\) with a velocity of \(36 \mathrm{~km} / \mathrm{h}\). The centripetal force is
- A \(250 \mathrm{~N}\)
- B \(750 \mathrm{~N}\)
- C \(1000 \mathrm{~N}\)
- D \(1200 \mathrm{~N}\)
Answer & Solution
Correct Answer
(C) \(1000 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
The centripetal force required for a car (mass \(m\) ) to take a sharp turn of radius \(r\) with velocity \(v\) is
\(F=\frac{m v^{2}}{r}\)
Given, \(m=500 \mathrm{~kg}\),
\(\begin{array}{l}v=36 \mathrm{~km} / \mathrm{h}=36 \times \frac{5}{18}=10 \mathrm{~m} / \mathrm{s}, r=50 \mathrm{~m} \\
\text { Hence, } \quad F=\frac{500 \times(10)^{2}}{50}=1000 \mathrm{~N}\end{array}\)
\(F=\frac{m v^{2}}{r}\)
Given, \(m=500 \mathrm{~kg}\),
\(\begin{array}{l}v=36 \mathrm{~km} / \mathrm{h}=36 \times \frac{5}{18}=10 \mathrm{~m} / \mathrm{s}, r=50 \mathrm{~m} \\
\text { Hence, } \quad F=\frac{500 \times(10)^{2}}{50}=1000 \mathrm{~N}\end{array}\)
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