Physics Guru

Let $x>0, y>0, z>0$ satisfy $x+y+z=3$. Prove that, $\frac{{{x^2}}}{{1 + x}} + \frac{{{y^2}}}{{1 + y}} + \frac{{{z^2}}}{{1 + z}} \ge \frac{3}{2} - \frac{{{{(x - y)}^2} + {{(y - z)}^2} + {{(z - x)}^2}}}{{12}}$ Solution Using Titu Andreescu's lemma, $\frac{{a_1^2}}{{{b_1}}} + \frac{{a_2^2}}{{{b_2}}} + ............. + \frac{{a_n^2}}{{{b_n}}} \ge \frac{{{{({a_1} + {a_2} + ........ + {a_n})}^2}}}{{{b_1} + {b_2} + ....... + {b_n}}}$ for ${b_i} > 0$ We have, $\frac{{{x^2}}}{{1 + x}} + \frac{{{y^2}}}{{1 + y}} + \frac{{{z^2}}}{{1 + z}} \ge \frac{{{{(x + y + z)}^2}}}{{(1 + x) + (1 + y) + (1 + z)}}$ $ \Rightarrow \frac{{{x^2}}}{{1 + x}} + \frac{{{y^2}}}{{1 + y}} + \frac{{{z^2}}}{{1 + z}} \ge \frac{{{{(x + y + z)}^2}}}{{3 + (x + y + z)}}$ $ \Rightarrow \frac{{{x^2}}}{{1 + x}} + \frac{{{y^2}}}{{1 + y}} + \frac{{{z^2}}}{{1 + z}} \ge \frac{3}{2} \because x + y + z = 3$ The expression $\frac{{{{(x - y)}^2} + {{(y - z)}^2} + {{(z - x)}^2}}}{{12}}$ is non-negative. So, if $\frac{{{x^2}}}{...

A man walking at speed v and an old man walking at speed 0.5v enter a circular park through the same gate and walk in opposite directions. If the man completes two rounds and exits through the same gate, how many times will they meet each other? Solution Let the circumference of the circular park be C. Distance covered by the man = 2C Distance covered by the old man = C Relative distance covered = 3C One meeting takes place after one relative round or when C relative distance is covered. So, three meetings would take place after 3C relative distance is covered. This means that just when the man is about to exit from the gate after two rounds, the old man also reaches the gate after one round. The three meetings will be angularly equispaced. In other words there will be a meeting after every 120 degree angle.

Solve the differential equation, $(x - y)\frac{d}{{dx}}\sqrt {\frac{y}{x}}  = x + y + 2\sqrt {xy} ;x > 0,y > 0$ Solution We have, $\left( {\sqrt x  + \sqrt y } \right)\left( {\sqrt x  - \sqrt y } \right)\frac{d}{{dx}}\sqrt {\frac{y}{x}}  = {\left( {\sqrt x  + \sqrt y } \right)^2}$ $ \Rightarrow \frac{d}{{dx}}\sqrt {\frac{y}{x}}  = \frac{{\sqrt x  + \sqrt y }}{{\sqrt x  - \sqrt y }}$ $ \Rightarrow \frac{d}{{dx}}\sqrt {\frac{y}{x}}  = \frac{{1 + \sqrt {\frac{y}{x}} }}{{1 - \sqrt {\frac{y}{x}} }}$ $ \Rightarrow \frac{{dt}}{{dx}} = \frac{{1 + t}}{{1 - t}};t = \sqrt {\frac{y}{x}} $ $ \Rightarrow \frac{{1 - t}}{{1 + t}}dt = dx$ $ \Rightarrow \int {\frac{{2 - (t + 1)}}{{1 + t}}} dt = \int {dx} $ $ \Rightarrow \int {\left( {\frac{2}{{1 + t}} - 1} \right)} dt = \int {dx} $ $ \Rightarrow 2\ln |1 + t| - t = x + C$ $ \Rightarrow 2\ln \left| {1 + \sqrt {\frac{y}{x}} } \right| - \sqrt {\frac{y}{x}}  = x + C$

A particle moves in the xy-plane along the parabolic trajectory $y=x^2$. As the particle approaches the origin, evaluate $L=\mathop {\lim }\limits_{x \to 0} \frac{{{x^2}y({e^{xy}} - 1 - xy)}}{{{{({x^2} + {y^2})}^2}}}$. Solution Substituting $y=x^2$, we have $L=\mathop {\lim }\limits_{x \to 0} \frac{{{x^4}({e^{x^3}} - 1 - x^3)}}{{{{({x^2} + {x^4})}^2}}}$ Or, $L=\mathop {\lim }\limits_{x \to 0} \frac{{{x^4}({e^{x^3}} - 1 - x^3)}}{x^4{{{(1 + {x^2})}^2}}}=\mathop {\lim }\limits_{x \to 0} \frac{{{e^{x^3}} - 1 - x^3}}{{{{(1 + {x^2})}^2}}}$ It may seem like exponential series expansion formula should be used, but it is not required. Direct substitution of x = 0 yields L = 0.

The real solutions to the the septic equation \[{x^7} - 3{x^6} - 8{x^5} + 24{x^4} - 7{x^3} + 21{x^2} - 18x + 54 = 0\] are: (A) 3, 3, -3 (B) -3, -3, 3 (C) No real solutions (D) Seven overall Solution Grouping, $({x^7} - 3{x^6}) + ( - 8{x^5} + 24{x^4}) - (7{x^3} - 21{x^2}) + ( - 18x + 54) = 0$ $ \Rightarrow {x^6}(x - 3) - 8{x^4}(x - 3) - 7{x^2}(x - 3) - 18(x - 3) = 0$ $ \Rightarrow (x - 3)({x^6} - 8{x^4} - 7{x^2} - 18) = 0$ So, x = 3 is one solution. Now, ${x^6} - 8{x^4} - 7{x^2} - 18 = 0$ Let, ${x^2} = t$ $\therefore {t^3} - 8{t^2} - 7t - 18 = 0$ t = 9 satisfies the equation. $\therefore {t^2}(t - 9) + t(t - 9) + 2(t - 9) = 0$ $ \Rightarrow (t - 9)({t^2} + t + 2) = 0$ $\therefore t = 9 = {x^2}$, ${t^2} + t + 2 \ne 0$ since ${t^2} + t + 2 > 0$ $\therefore x =  \pm 3$ $\therefore x = 3,3, - 3$ Hence, (A).

Q: A ball is moving on a horizontal table with some velocity. It stops on its own after moving some distance. Which external force(s) cause(s) change in the linear momentum of the ball (ignore air)? A: The ball loses momentum due to two effects: Friction due to some slipping between the ball and the table. Deformation of the ball and/or the table during rolling, which shifts the line of action of the normal reaction slightly ahead of the vertical through the centre. This gives the contact force a small backward horizontal component (rolling resistance), which opposes the motion and reduces the ball's momentum.

The glowing effect happens because the setting sun hits the waterfall at a certain angle for a few minutes in late February. The atmosphere scatters the sunlight, allowing only the longer red-orange wavelengths to pass through. This remaining red-orange light is what creates the glowing effect.

Evaluate, $I = \int {\frac{{{x^3} + 2x + 1}}{{{x^2} + 1}}} dx$ We have, $I = \int {\frac{{{x^3} + x + x + 1}}{{{x^2} + 1}}} dx$ = $ \int {\frac{{x({x^2} + 1) + x + 1}}{{{x^2} + 1}}} dx$ $ = \int {x + \frac{{x + 1}}{{{x^2} + 1}}dx} $ $ = \int {xdx + \int {\frac{x}{{{x^2} + 1}}} } dx + \int {\frac{{dx}}{{1 + {x^2}}}} $ $ = \frac{{{x^2}}}{2} + \frac{1}{2}\int {\frac{{2x}}{{{x^2} + 1}}} dx + {\tan ^{ - 1}}x + C$ $ = \frac{{{x^2}}}{2} + \frac{1}{2}\ln ({x^2} + 1) + {\tan ^{ - 1}}x + C$

Evaluate, $I = \int\limits_0^\infty {\frac{{dx}}{{{{(1 + {x^2})}^5}}}} $ $Let,x = \tan \theta $ $\therefore dx = {\sec ^2}\theta d\theta $ $\therefore I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sec }^2}\theta }}{{{{(1 + {{\tan }^2}\theta )}^5}}}} d\theta $ $\therefore I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sec }^2}\theta }}{{{{({{\sec }^2}\theta )}^5}}}} d\theta $ $\therefore I = \int\limits_0^{\frac{\pi }{2}} {{{\cos }^8}\theta d\theta } $ $\therefore I = \frac{{7 \cdot 5 \cdot 3 \cdot 1}}{{8 \cdot 6 \cdot 4 \cdot 2}} \times \frac{\pi }{2} = \frac{{7 \cdot 5}}{{8 \cdot 2 \cdot 4 \cdot 2}} \times \frac{{22}}{{7.2}}$ $\therefore I = \frac{{55}}{{128}}$

Q: A bird while flying takes a right turn, where does it get the centripetal force from? A: The centripetal force is provided by the bird’s wings as it changes the direction of its flight. As the bird turns to the right, it must adjust the angle of its wings to produce a component of the force towards the center of the turn.

Among the following, the correct statement(s) for electrons in an atom is(are)  (A) Uncertainty principle rules out the existence of definite paths for electrons. (B) The energy of an electron in 2s orbital of an atom is lower than the energy of an electron that is infinitely far away from the nucleus. (C) According to Bohr’s model, the most negative energy value for an electron is given by n = 1, which corresponds to the most stable orbit. (D) According to Bohr’s model, the speed of electrons increases with increase in values of n.

 $\frac{{{\rm{9 \: digit \: number \: nnnnnnnnn}}}}{{n + n + n + n + n + n + n + n + n}} = $ (A) not an integer (B) 12345678 (C) depends on n (D) 12345679

How many 3-digit hexadecimal numbers can be formed if each digit can be any of the characters from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F}, with the restriction that the starting/left digit cannot be 0?

Plane A flies at a constant speed of 300 km/h. Plane B is 225 km behind plane A and flies at an initial speed of 420 km/h, but due to fuel limitations, its speed decreases linearly by 30 km/h every hour. Assume both planes fly in horizontal path at safe distance from each other.  (A) They will cross each other at t = 3 hrs. only (B) They will cross each other at t = 5 hrs. only (C) They will cross each other two times (D) They will never cross each other