Physics Guru

A negative point charge -q is brough near an isolated metal cube. (A) The cube becomes positively charged (B) The cube becomes negatively charged (C) The interior of the cube remains charge free and the surface gets nonuniform charge distribution (D) Protons inside the metal cube move towards -q making the interior of the cube negatively charged and the surface close to -q positively charged Answer The charge does not stay inside the metal cube. The charge has to reside on the surface. Moreover, the free electrons move farther away from -q so there is nonuniform surface charge distribution. Hence, (C)

At t = 0, three particles A, B and C are situated at the vertices of an equilateral triangle ABC of side 1.5d. Each particle moves with constant speed v such that A always has its velocity along AB, B along BC and C along CA. At what time will the particles meet? Solution Velocity of A along AB = $v$ Velocity of B along BA = $v cos 60^\circ = 0.5v$ Relative velocity of B towards A = $v+0.5v=1.5v$ Displacement covered by B to reach A with respect to A = 1.5d time = $\frac {d_{rel}}{v_{rel}}=\frac {1.5d}{1.5v}=\frac {d}{v}$

Why are small steps advisable while walking on wet path? Answer Let F be the force that the person exerts on the ground. The ground exerts equal and opposite force F on the person. Let $\theta$ be the angle that the force F exerted on the person by the ground makes with the vertical. The force F that ground exerts on the person can be resolved into two components: $F cos \theta = N$ (normal reaction) $F sin \theta = Friction \leq \mu. N$ So, $F sin \theta \leq \mu. F cos \theta $ So, $tan \theta \leq \mu $ or $(tan \theta)_{max} = \mu $ The coefficient of friction $\mu $ decreases when the surface is wet. So, the maximum permissible $\theta $ also decreases. Hence, the legs should be kept more nearly vertical, which is achieved by taking smaller steps.

A projectile is projected from the origin with a speed of $20m{s^{ - 1}}$ at an angle of ${63^\circ }$ above the horizontal. After a delay of 1s, a drone is launched from the same point with constant velocity. The motion of both the projectile and the drone takes place in the vertical xz-plane. If the drone strikes the projectile 2s after the projectile is launched, find the velocity of the drone. Ignore air resistance. [$g = 10m{s^{ - 2}}$, $\sin {63^\circ } = \frac{4}{5}$ and $\cos {63^\circ } = \frac{3}{5}$] Solution For projectile motion, $x = 20\cos {63^\circ }.2 = 24m$ $z = 20\sin {63^\circ }.2 - \frac{1}{2}{.10.2^2} = 32 - 20 = 12m$ Position vector of the projectile when the drone strikes it,  $\vec r = 24\hat i + 12\hat k$ m Velocity of drone = $\frac{{\vec r}}{1} = 24\hat i + 12\hat k$ $m{s^{ - 1}}$

Find the minimum value of the function $f(x, y)$ for $x, y > 0:$ $f(x,y)=\sqrt{x^{2}+1}+\sqrt{\frac{y^{2}}{x^{2}}+4}+\sqrt{\frac{9}{y^{2}}+9}$ Solution Let, $a = x,b = \frac{y}{x},c = \frac{3}{y}$ Then $a,b,c > 0$ and $abc = x.\frac{y}{x}.\frac{3}{y} = 3$ So $f = \sqrt {{a^2} + 1}  + \sqrt {{b^2} + 4}  + \sqrt {{c^2} + 9} $ Now, Minkowski inequality (in its 2-dimensional form) for ${a_i},{b_i} \ge 0$, $\sum\limits_{i = 1}^n {\sqrt {a_i^2 + b_i^2} }  \ge \sqrt {{{\left( {\sum\limits_{i = 1}^n {{a_i}} } \right)}^2} + {{\left( {\sum\limits_{i = 1}^n {{b_i}} } \right)}^2}} $ $\therefore f = \sqrt {{a^2} + {1^2}}  + \sqrt {{b^2} + {2^2}}  + \sqrt {{c^2} + {3^2}}  \ge \sqrt {{{(a + b + c)}^2} + {{(1 + 2 + 3)}^2}} $ $\therefore f \ge \sqrt {{{(a + b + c)}^2} + 36} $ Now, $A.M. \ge G.M$ or $\frac{{a + b + c}}{3} \ge {(abc)^{1/3}}$ $ \Rightarrow a + b + c \ge {3^{4/3}}$ $\therefore f \ge \sqrt {{3^{8/3}} + 36} $ $ \Rightarrow f \ge \sqrt {{{9.9}^{1/3}} + 9.4} ...

What should be the minimum value of amplitude so that the particle speed can match or exceed the wave speed in sinusoidal progressive wave $y = A sin (ωt − \frac {2\pi}{\lambda}x)$? (A) Particle speed can never exceed wave speed (B) $A_{min} = \lambda$ (C) $A_{min} = \frac {\lambda}{2\pi}$ (D) Particles are not present as the wave travels in vacuum Solution Particle speed = $|\frac {\partial y}{\partial t}| = Aω |cos (ωt − \frac {2\pi}{\lambda}x)|$ So, particle speed is less than or equal to Aω. Wave speed = $\nu.\lambda = \frac {ω}{2\pi}.\lambda$ If particle speed were to match or exceed wave speed, $Aω |cos (ωt − \frac {2\pi}{\lambda} x)| \geq \frac {ω}{2\pi}.\lambda$ $\therefore A \geq \frac{{\frac{\lambda }{{2\pi }}}}{{\left| {cos\left( {\omega t - \frac{{2\pi }}{\lambda }x} \right)} \right|}}$ For A to be min. cosine should be maximum = 1 $A_{min} \geq \frac {\lambda}{2\pi}$ When $A_{min} = \frac {\lambda}{2\pi}$, particle speed can match wave speed periodically and when $A_{min} ...

A stationary man standing on a circular platform that is free to rotate is holding an umbrella starts rotating it anticlockwise. The platform starts rotating clockwise due to conservation of angular momentum. Man also rotates clockwise along with the platform.