Loss of energy, ΔE=21Itωi2−21(It+Ib)It2ωi2 =21(It+Ib)IbItωi2
NEET UG 2010 — Physics Mechanics
A circular disk of moment of inertia It is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed ωi. Another disk of moment of inertia Ib is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed ωf. The energy lost by the initially rotating disc due to friction is
Held on 30 Apr 2010 · Verified 9 Jul 2026.
21(It+Ib)Ib2ωi2
21(It+Ib)It2ωi2
21(It+Ib)Ib−Itωi2
21(Lt+Ib)IbItωi2
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 bullet of mass 10 g moving with 400 m/s penetrates a wall and comes to rest. The loss in kinetic energy is:
Consider a water tank shown in the figure. It has one wall at $x=L$ and can be taken to be very wide in the $z$ direction. When filled with a liquid of surface tension $S$ and density $\rho$, the liquid surface makes angle $\theta_0\left(\theta_0 \ll 1\right)$ with the $x$-axis at $x=L$. If $y(x)$ is the height of the surface then the equation for $y(x)$ is:  $\left(\operatorname{take} \theta(x)=\sin \theta(x)=\tan \theta(x)=\frac{d y}{d x}, g\right.$ is the acceleration due to gravity)
A physical quantity $P$ is related to four observations $a, b, c$ and $d$ as follows : $P=a^3 b^2 / c \sqrt{d}$ The percentage errors of measurement in $a, b, c$ and $d$ are $1 \%, 3 \%, 2 \%$ and $4 \%$ respectively. The percentage error in the quantity $P$ is
A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of $60^{\circ}$ with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )
The kinetic energies of two similar cars $A$ and $B$ are 100 J and 225 J respectively. On applying breaks, car A stops after 1000 m and car B stops after 1500 m . If $\mathrm{F}_{\mathrm{A}}$ and $\mathrm{F}_{\mathrm{B}}$ are the forces applied by the breaks on cars A and B , respectively, then the ratio $\mathrm{F}_{\mathrm{A}} / \mathrm{F}_{\mathrm{B}}$ is :
Work through every NEET UG Mechanics PYQ, year by year.