Physics Guru

An astronaut sitting in a chair in a satellite feels weightless. However, Moon is also a satellite and the astronaut does not feel weightless there. Are laws of Physics different for man-made satellite and the God made satellite?

A car is moving with uniform velocity. Is it possible to change velocity of the car without pressing accelerator pedal harder?

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)

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR} $ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1 $ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$

Let the acute angle bisector of the two planes $x-2y-2z+1=0$ and $2x-3y-6z+1=0$ be the plane P. Then which of the following points lies on P? (A) $(0,2,-4)$ (B) $(4,0,-2)$ (C) $(-2,0,-\frac {1}{2})$ (D) $(3,1,-\frac {1}{2})$ Solution Bisectors are given by, $\frac{{x - 2y - 2z + 1}}{3} =  \pm \frac{{2x - 3y - 6z + 1}}{7}$ $ \Rightarrow 7x - 14y - 14z + 7 =  \pm (6x - 9y - 18z + 3)$ Hence, $x-5y+4z+4=0$ & $13x-23y-32z+10=0$ Let $\theta $ be the angle between $x-2y-2z+1=0$ & $x-5y+4z+4=0$ $\cos \theta  = \frac{{|1 + 10 - 8|}}{{3 \times \sqrt {42} }} = \frac{1}{{\sqrt {42} }} < \frac{1}{{\sqrt 2 }}$ $\theta > 45^\circ $ So, $x-5y+4z+4=0$ is the obtuse angle bisector. $\therefore $ Acute angle bisector $P \equiv 13x - 23y - 32z + 10 = 0$ The point $(-2,0,-\frac {1}{2})$ satisfies P. Answer: (C)

80 g of copper sulphate $CuSO_4 .5H_2 O $ is dissolved in deionised water to make 5 L of solution. The concentration of the copper sulphate solution is $x \times 10^{-3} mol L^{-1} $. The value of x is _ _ _ _ . (Nearest integer) [Atomic masses - Cu:63.54 u, S:32 u, O: 16 u, H: 1 u] Solution $Molarity = \frac {n}{V} = \frac {w}{M_0 L} $ $\therefore Molarity = \frac {80}{249.54 \times 5 } \approx 64 \times 10^{-3} mol L^{-1} $ $\therefore x = 64 $

For the reaction $2NO_2 (g) \rightleftharpoons N_2O_4 (g) $, when $\Delta S = -176.0 JK^{-1} $ and $\Delta H = -57.8 kJ mol^{-1} $, the magnitude of $\Delta G $ at 298 K for the reaction is _ _ _ _ $kJ mol^{-1} $. (Nearest integer) Solution We have, $\Delta G = \Delta H - T \Delta S $ $\Delta S = -176.0 JK^{-1} = -0.176 kJ K^{-1} $ $\Delta G =  -57.8 - 298 \times (-0.176) \approx -57.8+52.5 = -5.3 $ $\therefore |\Delta G | = 5.3  kJ mol^{-1} $ Ans: 5