NEET Laws of Motion MCQs & PYQs

Practice NEET Laws of Motion Questions

Question 11:

easy

Consider a car moving along a straight horizontal road with a speed of \(72\text{ km/h}\). If the coefficient of static friction between the tyres and the road is 0.5, the shortest distance, in which the car can be stopped is (taking \(g = 10\text{ m/s}^2)\):

(1992)

Convert speed to m/s: \(v = 72 \times 5/18 = 20\text{ m/s}\). Deceleration due to friction is \(a = \mu_s g = 0.5 \times 10 = 5\text{ m/s}^2\). Using \(v^2 = u^2 + 2as\), \(0 = (20)^2 - 2(5)s\). So, \(10s = 400\) and \(s = 40\text{ m}\).

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Question 12:

moderate

A heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is 0.25, then the maximum fraction of the length of the chain that can hang over one edge of the table is:

(1991)

For equilibrium, the weight of the hanging part must equal the maximum static friction on the table. If (x) is the fraction hanging, then \(xMg = mu(1-x)Mg\). So, \(x = \mu(1-x)\). This gives \(x = \frac{\mu}{1+\mu}\). With \(\mu = 0.25\), \(x = \frac{0.25}{1+0.25} = \frac{0.25}{1.25} = 0.20\) or (20%).

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