575 - Physics Questions Answers

$$\begin{gathered} Let the acceleration of hemisphere is a_h \\ N \sin \theta = M{a}_h \\ \Rightarrow N/m{a}_h = M/m\sin \theta \\ Here pseudo force is considered due to acceleration of hemisphere \end{gathered}$$

$$\begin{gathered} Potential difference from centre to surface = 3V/2 - V = V/2 \\ here V = kq/r (q is charge of sphere) \\ m{v_r}^2/2 = VQ/2 (Q is charge of particle) \\ {v}_r is radial velocity \\ Besides \tan gential velocity due to rotation of sphere = r\omega =r2\pi n \\ now solve \end{gathered}$$

$$\begin{gathered} By volume comparision radius of combined drop = 2^{1/3}r \\ net charge in the new drop = 2q \\ now put these values in the formula of potential \end{gathered}$$

$$\begin{gathered} In first case F = 2T = 150 \\ so T = 75 newton, \\ Now in second case system is free to move \\ acceleration of the system a = F/20 \\ for 10 kg rod F-2T = 10a = 10(F/20) = F/2 \\ \Rightarrow F-(F/2)=2T \Rightarrow F=4T = 300 newton \end{gathered}$$

$$\begin{gathered} Moment of inertia of a cube about a line perpendicular to \\ its one face and pas\sin g through its centre is m{a}^2/6 \\ so about diagonal inertia = l^2{I}_x+m^2{I}_y+n^2{I}_z \\ = (l^2+m^2+n^2)I_x (here I_x=I_y=I_z) \\ = I_x = m{a}^2/6 \\ l,m,n are direction \cos ines of three lines pas\sin g througe \\ centre of cube and perpendicular to each other \end{gathered}$$

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$$\begin{gathered} Since thickness is same so force per unit length should be same \\ Force per unit length = F/2\pi R \\ Charge density in the sphere is proportional to E \\ so F \alpha qE \alpha \sigma AE \alpha E^2{R}^2 \\ Thus force per unit length \alpha E^2{R}^2/2\pi R \alpha E^{2}R \\ Now compare the two cases \end{gathered}$$

$$\begin{gathered} At a general point \\ -2T\sin \theta = ma (\theta is measured from horizontal) \\ \Rightarrow a = -2Mg\theta /m = -2Mgy/ml \\ compare with a = -{\omega }^{2}y \\ \omega =\sqrt{\frac{2Mg}{ml}} \left(1\right) \\ Now for max displacement of small m \\ mg(h+y) = 2Mgx and x = y^2/2l \\ on solving this quadratic eq. for y/l we get \\ \frac{y}{l}-\frac{m}{2M}=\sqrt{\frac{m}{M}\frac{h}{l}}\Rightarrow \frac{y}{l}=\sqrt{\frac{m}{M}\frac{h}{l}} (given that \frac{m}{M} <<\frac{h}{l}) \\ so x = \frac{m}{2M}h \left(2\right) \\ so v_{max} = x\omega \\ now solve \end{gathered}$$