Given:
Initial speed =360054×1000=15ms−1
∣at∣=∣ac∣
⇒vdxdv=Rv2⇒∫15vvdv=R1∫0xdx
⇒v=15eRx
⇒dtdx=15eRx
Distance covered in first quarter of revolution =42πR=2πR.
Therefore,
∫02πRe−Rxdx=15∫0t0dt
⇒t0=40(1−e−2π)
Hence, t=40.
JEE Main 2023 — Physics Mechanics
A car is moving on a circular path of radius 600m such that the magnitudes of the tangential acceleration and centripetal acceleration are equal. The time taken by the car to complete first quarter of revolution, if it is moving with an initial speed of 54kmh−1 is t(1–e–2π)s. The value of t is ______.
Held on 29 Jan 2023 · Verified 6 Jul 2026.
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
A small bob $A$ of mass $m$ is attached to a massless rigid rod of length 1 m pivoted at point $P$ and kept at an angle of $60^{\circ}$ with vertical as shown in figure. At distance of 1 m below point $P$, an identical bob $B$ is kept at rest on a smooth horizontal surface that extends to a circular track of radius $R$ as shown in figure. If bob $B$ just manages to complete the circular path of radius $R$ upto a point $Q$ after being hit elastically by $\operatorname{bob} A$, then radius $R$ is $\_\_\_\_$ m. 
In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are $100$ divisions in circular scale and pitch of screw gauge is $0.1$ mm. When diameter of a sphere is measured, the reading of main scale is $5$ mm and $50^{\text{th}}$ division of circular scale coincides with the reference line of main scale. The diameter of sphere is _______ mm.
Two masses $m$ and 2 m are connected by a light string going over a pulley (disc) of mass 30 m with radius $r=0.1 \mathrm{~m}$. The pulley is mounted in a vertical plane and it is free to rotate about its axis. The 2 m mass is released from rest and its speed when it has descended through a height of 3.6 m is $\_\_\_\_$ $\mathrm{m} / \mathrm{s}$. (Assume string does not slip and $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$)
Water flows through a horizontal tube as shown in the figure. The difference in height between the water columns in vertical tubes is 5 cm and the area of cross-sections at $A$ and $B$ are $6 \mathrm{~cm}^{2}$ and $3 \mathrm{~cm}^{2}$ respectively. The rate of flow will be $\_\_\_\_$ $\mathrm{cm}^{3} / \mathrm{s}$. (take $g=10 \mathrm{~m} / \mathrm{s}^{2}$) 
Two circular discs of radius each 10 cm are joined at their centres by a rod of length 30 cm and mass 600 gm as shown in figure. If the mass of each disc is 600 gm and applied torque between two discs is $43 \times 10^{5}$ dyne. cm, the angular acceleration of the discs about the given axis $A B$ is $\_\_\_\_$ $\mathrm{rad} / \mathrm{s}^{2}$. 
Work through every JEE Main Mechanics PYQ, year by year.