A disc is rolling without slipping on a surface. The radius of the disc is $$R$$. At $$t=0$$, the top most point on the disc is $$\mathrm{A}$$ as shown in figure. When the disc completes half of its rotation, the displacement of point A from its initial position is

$$_{92}^{238}A \to _{90}^{234}B + _2^4D + Q$$

In the given nuclear reaction, the approximate amount of energy released will be:

[Given, mass of $${ }_{92}^{238} \mathrm{~A}=238.05079 \times 931.5 ~\mathrm{MeV} / \mathrm{c}^{2},$$

mass of $${ }_{90}^{234} B=234 \cdot 04363 \times 931 \cdot 5 ~\mathrm{MeV} / \mathrm{c}^{2},$$

mass of $$\left.{ }_{2}^{4} D=4 \cdot 00260 \times 931 \cdot 5 ~\mathrm{MeV} / \mathrm{c}^{2}\right]$$

Under isothermal condition, the pressure of a gas is given by $$\mathrm{P}=a \mathrm{~V}^{-3}$$, where $$a$$ is a constant and $$\mathrm{V}$$ is the volume of the gas. The bulk modulus at constant temperature is equal to

Two trains 'A' and 'B' of length '$$l$$' and '$$4 l$$' are travelling into a tunnel of length '$$\mathrm{L}$$' in parallel tracks from opposite directions with velocities $$108 \mathrm{~km} / \mathrm{h}$$ and $$72 \mathrm{~km} / \mathrm{h}$$, respectively. If train 'A' takes $$35 \mathrm{~s}$$ less time than train 'B' to cross the tunnel then. length '$$L$$' of tunnel is :

(Given $$\mathrm{L}=60 l$$ )