Physics Guru

A charge q is placed at a distance of $\frac {a}{\sqrt {24}}$ above the centre of a horizontal, equilateral triangular surface of edge a. Find the flux of the electric field through the equilateral triangular surface. Solution Let us evaluate if we can have regular tetrahedron as the closed Gaussian surface. The altitude of tetrahedron having side a is given by $H = a \sqrt {\frac {2}{3}}$ Distance of centroid from any face = $\frac {H}{4} = \frac {a}{\sqrt {24}}$ So, we can imagine a regular tetrahedron as the closed Gaussian surface with q placed at the centroid. Total flux $= \frac {q}{\epsilon_0} = 4 \times $ flux though one surface So, flux through one triangular surface $= \frac {q}{4\epsilon_0}$

Find a if ${S_n} = {1^4} + {2^4} + .......... + {n^4} = a{n^5} + b{n^4} + c{n^3} + d{n^2} + en + f$, where n is a natural number and a, b, c, d, e, f are constants. Solution We have, ${S_{n - 1}} = {1^4} + {2^4} + .......... + {(n - 1)^4}$ $\therefore {S_n} - {S_{n - 1}} = {n^4}$ $\therefore a\{ {n^5} - {(n - 1)^5}\}  + b\{ {n^4} - {(n - 1)^4}\}  + c\{ {n^3} - {(n - 1)^3}\}  + ..... = {n^4}$ In LHS, $n^4$ can only come from $a\{ {n^5} - {(n - 1)^5}\} $ as other terms have lower power on n. Using binomial expansion of ${(n - 1)^5}$ and equating the coefficient of $n^4$, $a.5 = 1$ $ \Rightarrow a = \frac{1}{5}$

There are two identical right-triangular ramps. Their vertical faces are parallel and facing each other, separated by a horizontal distance of 5 m, and their upper ends are at the same height. The inclined surface of each ramp makes an angle of 15° with the horizontal. A small toy car approaches the first ramp along a horizontal road and moves onto the inclined surface with constant speed. What is the minimum speed with which the car should enter the ramp so that it clears the gap and lands safely on the inclined surface of the second ramp? Neglect air resistance. Take $𝑔=10 m/s^2$. Solution The car leaves the first ramp with speed u at $15^\circ$. For the minimum speed, its projectile range must be 5 m. $R=\frac {u^2sin 2\theta}{g}$ $\therefore 5=\frac {u^2sin30}{10}$ $\Rightarrow u = 10m/s$

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 = $\frac{{\vec r}}{1} = 24\hat i + 12\hat k$ $m{s^{ - 1}}$