Physics Guru

Flying Bird ~ Q & A

noreply@blogger.com (Manish Verma) — Wed, 07 Jan 2026 16:25:00 +0000

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.

Atomic Electrons Quiz

noreply@blogger.com (Manish Verma) — Tue, 11 Nov 2025 18:06:00 +0000

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.

\[\frac{nnnnnnnnn}{{n + n + n + n + n + n + n + n + n}} = ?\]

noreply@blogger.com (Manish Verma) — Tue, 28 Oct 2025 16:42:00 +0000

\[\frac{{{\rm{9 \: digit \: number \: nnnnnnnnn}}}}{{n + n + n + n + n + n + n + n + n}} = \]

(A) not an integer
(B) 12345678
(C) depends on n
(D) 12345679

Hexadecimal Numbers ~ Permutations

noreply@blogger.com (Manish Verma) — Sun, 26 Oct 2025 09:23:00 +0000

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?

Will Planes Cross?

noreply@blogger.com (Manish Verma) — Thu, 07 Aug 2025 09:37:00 +0000

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

Why Rains Block Vision?

noreply@blogger.com (Manish Verma) — Wed, 30 Jul 2025 15:52:00 +0000

Which phenomenon of light is responsible for rain blocking the vision?

(A) Total internal reflection
(B) Dispersion
(C) Interference
(D) Scattering

Plane Engine Failure Quiz

noreply@blogger.com (Manish Verma) — Thu, 24 Jul 2025 08:40:00 +0000

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).

Neutron Star

noreply@blogger.com (Manish Verma) — Mon, 14 Jul 2025 16:04:00 +0000

A spherical neutron star has density equal to that of the nucleus. The radius of the neutron star having mass $5.0 \times 10^{30}$ kg is

(A) $\sqrt 2$ km
(B) $\sqrt 3$ km
(C) $\sqrt 2 \times 10^4 $ m
(D) $\sqrt 3 \times 10^4 $ m

Plane on Runway

noreply@blogger.com (Manish Verma) — Fri, 20 Jun 2025 13:18:00 +0000

Consider an aeroplane covering the entire runway 3.5 km long before take-off. The initial speed of the plane is zero and the final speed just before take-off is 174 knots (1 knot = 1.852 kmph). Assuming the acceleration to be constant, how much time is taken to cover the runway?

Earthen Pot in Summer

noreply@blogger.com (Manish Verma) — Wed, 11 Jun 2025 16:31:00 +0000

It is a common practice in India to keep drinking water in earthen pot in summer so as to cool it. What makes it work?

Does car brake apply decelerating force?

noreply@blogger.com (Manish Verma) — Fri, 06 Jun 2025 16:30:00 +0000

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

Man-Made Vs God-Made

noreply@blogger.com (Manish Verma) — Sun, 25 May 2025 11:21:00 +0000

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?

Velocity Change without Excess Petrol

noreply@blogger.com (Manish Verma) — Thu, 15 May 2025 13:02:00 +0000

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

Motion on Long Railroad Car

noreply@blogger.com (Manish Verma) — Tue, 17 Sep 2024 04:09:00 +0000

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. [$g=10 m/s^2$]

Car ~ Turning OR Braking

noreply@blogger.com (Manish Verma) — Wed, 03 May 2023 04:25:00 +0000

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.

$1.005^{200}$ Vs 2

noreply@blogger.com (Manish Verma) — Mon, 14 Nov 2022 04:00:00 +0000

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 ....

noreply@blogger.com (Manish Verma) — Fri, 24 Jun 2022 02:06:00 +0000

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 $

(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)

Sum of the coefficients in the expansion of $(x+y)^n$ ....

noreply@blogger.com (Manish Verma) — Thu, 23 Jun 2022 01:45:00 +0000

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 $(2,-1,6)=?$

noreply@blogger.com (Manish Verma) — Wed, 22 Jun 2022 02:15:00 +0000

The distance of line $3y-2z-1=0=3x-z+4$ from the point $(2,-1,6)$ is:

(A) $2\sqrt 6$

(B) $\sqrt {26}$

(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)

$f(x)=x^6+2x^4+x^3+2x+3 $$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$$n=?$

noreply@blogger.com (Manish Verma) — Tue, 21 Jun 2022 02:17:00 +0000

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}sin6$-$tan^{-1}tan12 =? $

noreply@blogger.com (Manish Verma) — Mon, 20 Jun 2022 02:20:00 +0000

$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 $

(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) ....

noreply@blogger.com (Manish Verma) — Sun, 19 Jun 2022 02:11:00 +0000

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, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$

Planes $x-2y-2z+1=0$ & $2x-3y-6z+1=0$ ....

noreply@blogger.com (Manish Verma) — Sat, 18 Jun 2022 01:38:00 +0000

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) $(4,0,-2)$

(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 $ ....

noreply@blogger.com (Manish Verma) — Fri, 17 Jun 2022 02:17:00 +0000

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) $ ....

noreply@blogger.com (Manish Verma) — Thu, 16 Jun 2022 02:11:00 +0000

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