Chapter 5
Master Resource Book in JEE Main Physics · 62 exercises
Problem 39
A heavy small-sized sphere is suspended by a string of length \(l\). The sphere rotates uniformly in a horizontal circle with the string making an angle \(\theta\) with the vertical. Then, the time-period of this \(\begin{array}{ll}\text { conical pendulum is } & \text { [BVP Engg. 2008] }\end{array}\) (a) \(t=2 \pi \sqrt{\frac{l}{g}}\) (b) \(t=2 \pi \sqrt{\frac{l \sin \theta}{g}}\) (c) \(t=2 \pi \sqrt{\frac{l \cos \theta}{g}}\) (d) \(t=2 \pi \sqrt{\frac{l}{g \cos \theta}}\)
6 step solution
Problem 39
What is the smallest radius of a circle at which a cyclist can travel if its speed is \(36 \mathrm{kmh}^{-1}\), angle of inclination is \(45^{\circ}\) and \(g=10 \mathrm{~ms}^{-2}\) ? (a) \(20 \mathrm{~m}\) (b) \(10 \mathrm{~m}\) (c) \(30 \mathrm{~m}\) (d) \(40 \mathrm{~m}\)
4 step solution
Problem 40
A stone of mass \(2 \mathrm{~kg}\) is tied to a string of length \(0.5 \mathrm{~cm}\). If the breaking tension of the string is \(900 \mathrm{~N}\), then the maximum angular velocity, that stone can have in uniform circular motion is [Kerala CET 2010] (a) \(30 \mathrm{rad} / \mathrm{s}\) (b) \(20 \mathrm{rad} / \mathrm{s}\) (c) \(10 \mathrm{rad} / \mathrm{s}\) (d) \(25 \mathrm{rad} / \mathrm{s}\)
6 step solution
Problem 40
The angle which the bicycle and its rider must make with the vertical when going round a curve of \(7 \mathrm{~m}\) radius at \(5 \mathrm{~ms}^{-1}\) is (a) \(20^{\circ}\) (b) \(15^{\circ}\) (c) \(10^{\circ}\) (d) \(5^{\circ}\)
6 step solution
Problem 41
Centripetal acceleration is (a) a constant vector (b) a constant scalar (c) a magnitude changing vector (d) not a constant vector
4 step solution
Problem 41
A car rounds an unbanked curve of radius \(92 \mathrm{~m}\) without skidding at a speed of \(26 \mathrm{~ms}^{-1} .\) The smallest possible coefficient of static friction between the tyres and the road is (a) \(0.75\) (b) \(0.60\) (c) \(0.45\) (d) \(0.30\)
6 step solution
Problem 42
A particle moves in a circle of radius \(5 \mathrm{~cm}\) with constant speed and time period \(0.2 \pi \mathrm{s}\). The acceleration of the particle is (a) \(5 \mathrm{~m} / \mathrm{s}^{2}\) (b) \(15 \mathrm{~m} / \mathrm{s}^{2}\) (c) \(25 \mathrm{~m} / \mathrm{s}^{2}\) (d) \(36 \mathrm{~m} / \mathrm{s}^{2}\)
5 step solution
Problem 42
A particle moves in circular path of radius \(R\). If centripetal force \(F\) is kept constant but the angular velocity is double, the new radius of the path will be (a) \(2 R\) (b) \(R / 2\) (c) \(R / 4\) (d) \(4 R\)
6 step solution
Problem 43
A car of mass \(1000 \mathrm{~kg}\) negotitates a banked curve of radius \(90 \mathrm{~m}\) on a frictionless road. If the banking angle is \(45^{\circ}\), the speed of car is (a) \(20 \mathrm{~ms}^{-1}\) (b) \(30 \mathrm{~ms}^{-1}\) (c) \(5 \mathrm{~ms}^{-1}\) (d) \(10 \mathrm{~ms}^{-1}\)
5 step solution
Problem 43
A curved road of \(50 \mathrm{~m}\) radius is banked at correct angle for a given speed. If the speed is to be doubled keeping the same banking angle, the radius of curvature of the road should be changed to (a) \(25 \mathrm{~m}\) (b) \(100 \mathrm{~m}\) (c) \(150 \mathrm{~m}\) (d) \(200 \mathrm{~m}\)
5 step solution
Problem 44
A motorcycle moving with a velocity of \(72 \mathrm{kmh}^{-1}\) on a flat road takes a turn on the road at a point where the radius of curvature of the road is \(20 \mathrm{~m}\). The acceleration due to gravity is \(10 \mathrm{~ms}^{-2}\). In order to avoid skidding, he must not bent with respect to the vertical plane by an angle greater than [BVP Engg. 2006] (a) \(\theta=\tan ^{-1}(2)\) (b) \(\theta=\tan ^{-1}(6)\) (c) \(\theta=\tan ^{-1}(4)\) (d) \(\theta=\tan ^{-1}(25.92)\)
4 step solution
Problem 45
A particle is moving in a circle of radius \(R\) in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at \(t=0\) is \(v_{0}\), the time taken to complete the first revolution is [Kerala CET 2005] (a) \(\frac{R}{v_{0}}\) (b) \(\frac{R}{v_{0}}\left(1-e^{-2 \pi}\right)\) (c) \(\frac{R}{v_{0}} e^{-2 \pi}\) (d) \(\frac{2 \pi R}{v_{0}}\)
8 step solution