Physics Guru

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

Which phenomenon of light is responsible for rain blocking the vision? (A) Total internal reflection (B) Dispersion (C) Interference (D) Scattering

A commercial plane experiences a complete loss of all engine power at an altitude of 35,000 feet (1 km = 3280.84 feet) and an airspeed of 450 knots (1 knot = 1.852 km/h). The Ram Air Turbine (RAT) is immediately deployed and the pilots gets ready for emergency landing. The RAT is designed to generate 20 kVA of electrical power with power generation efficiency of 85%. Power factor is 0.8 for a typical aircraft electrical system. The aircraft's essential systems require a continuous power supply of 13.5 kW. The aircraft has an average glide ratio of 15:1 (meaning for every 15 units of horizontal distance covered, it loses 1 unit of altitude). For how many minutes can the RAT continuously power the essential systems if the aircraft maintains a constant glide until it reaches the ground level for an emergency landing? (Ignore any potential battery backup before touchdown and also ignore RAT after touchdown for this calculation).