JEE Advance - Physics (2024 - Paper 1 Online - No. 12)
Explanation
$$\begin{aligned} & A v=\mathrm{av}_1 \\ & A\left(-\frac{d y}{d t}\right)=a \sqrt{2 g y} ; d t=\frac{A}{a \sqrt{2 g}} \cdot \frac{-d y}{\sqrt{y}} \\ & \int_\limits 0^{t_1} d t=\frac{A}{a \sqrt{2 g}} \int_\limits h^0-\frac{d y}{\sqrt{y}} \\ & t_1=\frac{A}{a \sqrt{2 g}} 2 \sqrt{h} ; t_1=\frac{A}{a} \sqrt{\frac{2 h}{g}} \end{aligned}$$
$$A v^{\prime}=a v_2$$
$$ \begin{aligned} & A\left(-\frac{d y}{d t}\right)=a \sqrt{2 g(H+y)} \\ & d t=-\frac{A}{a \sqrt{2 g}} \frac{d y}{\sqrt{H+y}} \\ & \int_\limits0^{t_2} d t=-\frac{A}{a \sqrt{2 g}} \int_\limits{\mathrm{H}}^0 \frac{d y}{\sqrt{H+y}} \\ & t_2=\frac{A}{a \sqrt{2 g}}(2)(\sqrt{H+h}-\sqrt{H}) \& H=\frac{16 h}{9} \\ & =\frac{A}{a} \sqrt{\frac{2 h}{g}}\left(\frac{5}{3}-\frac{4}{3}\right) \\ & t_2=\frac{A}{a} \sqrt{\frac{2 h}{g}}\left(\frac{1}{3}\right) \\ & \text { ratio } \frac{t_1}{t_2}=3 \end{aligned}$$
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