Physics Guru

 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

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

When brakes are applied to stop the car, how is the car stopped as a system cannot apply force on itself.

A boy standing at the rear end of a long railroad car throws a ball vertically upwards. The car is moving on the straight horizontal road with a deceleration of $1 m/s^2$ and the projection velocity in the vertical direction is 10 m/s. How far ahead of the boy will the ball fall on the car? Assume that the car keeps on moving throughout the motion of the ball. [$g=10 m/s^2$]

A car driver going at some speed v engaged in phone call all of a sudden finds a wide wall at a distance d. Are hard brakes applied in panic or car suddenly turned in a circle of radius d more likely to avoid collision with the wall? Assume that the driving balance is not lost.

 Select the correct option: (A) $1.005^{200} < 2$ (B) $1.005^{200} = 2$ (C) $1.005^{200} \leq 2$ (D) $1.005^{200} > 2$

The ranges and heights for two projectiles projected with the same initial velocity at angles $42^\circ $ and $48^\circ $ with the horizontal are $R_1\,,R_2$ and $H_1\,,H_2$ respectively. Choose the correct option: (A) $R_1 < R_2 $ and $H_1 < H_2 $ (B) $R_1 > R_2 $ and $H_1 = H_2 $ (C) $R_1 = R_2 $ and $H_1 = H_2 $ (D) $R_1 = R_2 $ and $H_1 < H_2 $ Solution $R=\frac {u^2 sin 2\theta }{g} $, $H=\frac {u^2 sin^2 \theta }{2g}$  $\frac {R_1}{R_2} = \frac {sin 2 \times 42^\circ}{sin 2\times 48^\circ} = \frac {cos 6^\circ}{cos 6^\circ} = 1 $ $\frac {H_1}{H_2}= \frac {sin^2 42^\circ }{sin^2 48^\circ } = tan^2 42^\circ < 1 $ Answer: (D)

If the sum of the coefficients in the expansion of $(x+y)^n$ is 4096, then the greatest coefficient in the expansion is _ _ _ _ . Solution $C_0 + C_1 + C_2 + C_3 + ......................... + C_n =4096 $ $\therefore 2^n = 4096 =2^{12} $ $\Rightarrow n = 12 $ Greatest coefficient = ${}^{12}{C_6} = 924$

The distance of line $3y-2z-1=0=3x-z+4$ from the point $(2,-1,6)$ is: (A) $2\sqrt 6$ (B) $\sqrt {26}$ (C) $2\sqrt 5 $ (D) $4\sqrt 2 $ Solution We have, $3x-z+4=0$ or $z=3x+4$ & $3y-2z-1=0$ or $3y-2(3x+4)-1=0$ or $y=2x+3$ Any point $(x,y,z)$ on the line $ \equiv (t,2t + 3,3t + 4)$ Let $d$ be the distance between $(2,-1,6)$ & $ (t,2t + 3,3t + 4)$ Then, $d^2=(t-2)^2+(2t+3+1)^2+(3t+4-6)^2=14t^2+24$ Minimum $d$ = Required answer = $\sqrt {24}=2\sqrt 6$ when t = 0. Answer: (A)

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R $. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $ \Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$

$cos^{-1}(cos(-5))+sin^{-1}(sin(6))-tan^{-1}(tan(12)) $ is equal to:  (The inverse trigonometric functions take the principal values) (A) $3\pi - 11$ (B) $3\pi + 1 $ (C) $4\pi - 11$ (D) $4\pi - 9$ Solution The given expression can be rewritten as, $cos^{-1}cos5+sin^{-1}sin 6-tan^{-1} tan12 $ Considering the principal values the given expression can be further rewritten as, $cos^{-1}cos(2\pi - 5)+sin^{-1}sin (-(2\pi- 6))-tan^{-1} tan(-(4\pi -12)) $ Or $(2\pi - 5)+ (-(2\pi- 6))-(-(4\pi-12)) = 4\pi -11 $ Answer: (C)