409 - Physics Questions Answers

No meaning can be incurred by this language. Rectify it for getting answer

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

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

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