576 - Physics Questions Answers

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

$$\begin{gathered} let u and v are the speeds of boy and escalator \\ according to given condition \\ \left[1\right(u+v)T-2(u-v\left)T\right]n_1=\left[2\right(u+v)T-1(u-v\left)T\right]n_2 \\ \Rightarrow \left[1\right(u+v)-2(u-v\left)\right]T{n}_1=\left[2\right(u+v)-1(u-v\left)\right]T{n}_2 \\ Given that T{n}_1=250 and T{n}_2=50 \\ \Rightarrow \left[1\right(u+v)-2(u-v\left)\right]250=\left[2\right(u+v)-1(u-v\left)\right]50 \\ now solve it for v, u is given \end{gathered}$$

$$\begin{gathered} \frac{k{q}_1{q}_2}{{l_3}^2}=\frac{k{q}_2{q}_3}{{l_1}^2}=\frac{k{q}_3{q}_1}{{l_2}^2} \\ {l}_1+l_2+l_3=L \\ from these 2 equations \\ \frac{L \frac{1}{\sqrt{q_1}}}{\frac{1}{\sqrt{q_1}}+\frac{1}{\sqrt{q_2}}+\frac{1}{\sqrt{q_3}}},\frac{L \frac{1}{\sqrt{q_2}}}{\frac{1}{\sqrt{q_1}}+\frac{1}{\sqrt{q_2}}+\frac{1}{\sqrt{q_3}}},\frac{L \frac{1}{\sqrt{q_3}}}{\frac{1}{\sqrt{q_1}}+\frac{1}{\sqrt{q_2}}+\frac{1}{\sqrt{q_3}}} \end{gathered}$$

$$\begin{gathered} X_C=50\Omega and X_L=100\Omega are in parallel. \\ Resul\tan t is 100\Omega capaci\tan ce. \\ Now 100\Omega capaci\tan ce and 100\Omega induc\tan ce give reac\tan ce 0\Omega \\ and resis\tan ce is 100\Omega . So only option D is correct. \end{gathered}$$