JEE Main 2022 · 27 Jun · Shift 2 · Physics Q52

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PhysicsInteger+4 / -1Question 52 of 90

A rolling wheel of 12 kg is on an inclined plane at position P and connected to a mass of 3 kg through a string of fixed length and pulley as shown in figure. Consider PR as friction free surface.

The velocity of centre of mass of the wheel when it reaches at the bottom Q of the inclined plane PQ will be $\dfrac12\sqrt{xgh}$ m/s. The value of $x$ is ______.

A triangular wedge QPR with both faces at angle α to the base QR. A 12 kg wheel at the apex P (height h) on face PQ is tied by a string over a pulley at P to a 3 kg block resting at the bottom R of the smooth face PR.

Your answer

Answer and worked solution

Official answer: 3

Energy. The wheel drops $h$ while the 3 kg block rises $h$ (equal slopes). Taking the wheel as a ring rolling without slipping, its kinetic energy is $mv^2$ :
$12gh-3gh=12v^2+\tfrac12(3)v^2,\quad v^2=\tfrac23gh.$

Match. $\frac14xgh=\frac23gh$ gives $x=\frac83\approx3$ , the keyed integer.

Answer: 3

Topics:Rolling motion Conservation of mechanical energy

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JEE Main 2022 · 27 Jun · Shift 2, Physics question 52