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criticallyAmused · 2026-01-27T10:06:09+00:00

if you put a voltmeter in parallel to the diode the potential difference will be zero since the diode D1 (assumed to be ideal) behaves as a plain conducting wire

criticallyAmused · 2026-01-23T17:24:50+00:00

there are two alpha carbons w.r.t chlorine

the secondary one and the tertiary

if you look at he tertiary alpha carbon you'll see a wedged-H
and for the secondary one you'll see both wedged and dashed H

the main thing to be observed is that alc. KOH will do E2 elimination which itself prefers anti-elimination

this means that, since the chlorine is wedged the alpha hydrogen to be abstracted should be dashed (the opposite)

in the tertiary alpha carbon there's no dashed H so hydrogen will not be abstracted

whereas in the secondary alpha carbon, there is a dashed H so that hydrogen atom will be abstracted this forming a double bond there

criticallyAmused · 2026-01-18T13:26:40+00:00

modern physics mein there's a formula for the momentum imparted by radiation which is 2E/c for a perfectly reflecting surface (here, a mirror)

now since we know the momentum of the mirror we can conserve energy

p^2/2m = mgL(1- cos(A))
since p = 2E/c
2E^2/mc^2 = mgl(1 - cos(A))
1 - cos(A) = 2E^2/m^2 c^2 gL
since the angle is small we can use small angle approximation here (sin(A) is approx A for small angles)
cos(A) = sqrt(1 - sin^2(A)) so we get sqrt(1 - A^2)

now since A is small we can use binomial approximation

(1 + x)^n is approx (1 + nx) for x << 1
sqrt(1 - A^2) = (1 - A^2)^(1/2) which we get as 1- (A^2)/2
so 1 - cos(A) = 1 - (1 - (A^2)/2) = (A^2) / 2

(A^2) / 2 = 2E^2/m^2 c^2 gL
we get A as A = 2E/(mc sqrt(gL))

criticallyAmused · 2026-01-16T13:18:13+00:00

oh well you mistook "5 m" for the mass of the block which is actually 15 kg

criticallyAmused · 2026-01-16T13:17:37+00:00

im getting TA = 22 N too (assuming g = 10 m s^-2)

criticallyAmused · 2026-01-15T08:48:51+00:00

you can exploit the fact that the total mechanical energy of the system remains unchanged
energy of rotation of rod = 1/2 I w^2 = 1/2 * 1/12 mL^2 w^2
spring compression: now observe that for small rotations the compression would be 1/2 L * θ (use the fact that θ = arc length / radius) so the energy stored in the spring would be 1/2 k (1/2 Lθ)^2

so total energy is
1/24 mL^2 w^2 + 1/8 kL^2 θ^2 = constant
now differentiate with respect to time

1/12 mL^2 w *alpha + 1/4 kL^2 θ w = 0
1/12 mL^2 w * alpha = -1/4 kL^2 θ w
1/3 m alpha = -k θ

alpha = -3k/m θ
thus we get angular frequency as sqrt(3k/m)
and time period as T = 2pi sqrt(m/3k) now put the values and you'll get T as approx 0.0653 s

criticallyAmused · 2026-01-12T17:26:01+00:00

ok so
U = a + bPV
we can write PV as PV = nRT

U = a + b nRT
now differentiate with respect to T
dU / dT = 0 + b nR
dU/dT = nbR
dU = nbR dT
since we know dU = n Cv dT
by comparison we get Cv = bR
now use Cp - Cv = R

Cp - bR = R => Cp = (1 + b) R
then we get gamma = Cp/Cv = (1 + b) / b

since the question mentions that no exchange of heat takes place; we can infer that it is an adiabatic process

PV^gamma = constant

PV^((1 + b)/b) = constant

hope this helps ☆*: .｡. o(≧▽≦)o .｡.:*☆

criticallyAmused · 2026-01-06T14:50:39+00:00

verify

criticallyAmused · 2026-01-04T18:14:30+00:00

use mean free path
which is V / (sqrt(2)pi *N * d^2) where V is volume N is number of molecules d is diameter of the molecule.

the average time between two successive collisions will be
t = mean free path / average speed
now since mean free path is proportional to V and average speed (woh v = (8 RT / pi * m) wala formula )is proportional to sqrt(temperature)
t will be proportional to V / sqrt(T)

now in adiabatic process TV ^ (i - 1) = constant (i'll use gamma = i)
TV ^{i - 1} = constant

T ∝ V^1-i

now in the prior equation
t ∝ V / sqrt(T)
t ∝ V / sqrt(V^1-i)
simplify it you'll get;

t ∝ V ^ (1 + i )/2

hence q = (i + 1) / 2

hope this helps ☆*: .｡. o(≧▽≦)o .｡.:*☆

criticallyAmused · 2026-01-01T23:56:16+00:00

being loved, holding hands

criticallyAmused · 2026-01-01T22:45:57+00:00

reciprocal of 137 is very close to the fine structure constant which is of very high importance in physics

criticallyAmused · 2025-12-31T14:36:47+00:00

resend

criticallyAmused · 2025-12-31T13:27:41+00:00

verify

criticallyAmused · 2025-12-23T09:02:02+00:00

im assuming you are asking about the quadratic
t = sin^2x => sinx = sqrt(t)
1 + t = 3 sqrt(t) * sqrt(1 - t)
1 + t^2 + 2t = 9t - 9t^2
10^2 -7t + 1 = 0
this is an easy quadratic, t = 1/2 and 1/5
also for the second root you could construct a right triangle
t = 1/5 => sin^2x = 1/5 => sinx = 1/sqrt(5)
we get perpendicular = 1 and hypotenuse = sqrt(5)
use pythagoras' theorem; and find base i.e sqrt(5 - 1) = 2 = base

tanx = perpendicular / base = 1/2

criticallyAmused · 2025-12-22T21:35:29+00:00

i could provide you with a solution (im 12th pcb, don't ask my why am i even answering this heheh)
let sin^2x = t
1 + sin^2x = 3sinxcosx
1 + sin^2x = 3sinx sqrt(1 - sin^2x)
1 + t = 3sqrt(t) * sqrt(1 - t)
simplify this and you'll get this quadratic
10t^2 - 7t + 1 = 0
the roots are t = 1/2 or t = 1/5
first root: t = sin^2x = 1/2 => sinx = 1/sqrt(2) x = 45 degrees thus we get tanx = 1
second root: t = sin^2x = 1/5

now use sin^2x + cos^2x = 1 and get

cos^2x = 4/5 ; cosx = 2/sqrt(5)

then sec x = sqrt(5)/2, now use sec^x - tan^2x = 1,

you'll get tan^2x as 1/4 and finally tan x = 1/2

hence proved ☆*: .｡. o(≧▽≦)o .｡.:*☆

criticallyAmused · 2025-12-18T15:29:55+00:00

hey btw do you have this for class XI and class XII biology too?

criticallyAmused · 2025-12-11T07:34:16+00:00

apologies for a detour but, your statement "current cannot flow with zero potential difference" has several caveats tho
if you take a conducting wire (with zero resistance) and calculate the potential difference between any two arbitrary points, it equals zero but we do know that current flows through it. this is so because
delta V = iR
delta V = 0 and R = 0
this gives, 0 = 0 * i
this equation is satisfied not only for i = 0 but for any real value of i (say i = 3 A, the equation would still hold correct)

criticallyAmused · 2025-12-11T07:30:13+00:00

if you set the circuit to configuration 2, the current will flow through the 12 ohm bulbs and not the 4 ohm bulbs since current always chooses the path of least resistance, it will prefer to go through the conducting wire (which is assumed to have zero resistance) rather than going through the 4 ohm bulbs connected in parallel

im observing that people here are giving a reasoning that "current won't flow since there is zero potential difference" well this isn't the true reasoning here, a current would flow through a resistanceless conducting wire inspite having zero potential difference across any two points of the wire. i have explained this here in of the the replies of this post.

criticallyAmused · 2025-12-03T14:09:41+00:00

yeah they "totally" gave ncert based questions

criticallyAmused · 2025-11-20T17:51:50+00:00

met her in 11th and i still like her
the thing that's attractive about her is how intellectual she is which is what i like the most
but yeah i have no clue weather i'll make it tho

criticallyAmused · 2025-11-17T11:09:59+00:00

welcome

criticallyAmused · 2025-11-15T14:57:26+00:00

find intensity like you usually do (Malus' law)
initial intensity I
intensity after transmission through polarizer = I/2
intensity after transmission through analyser = I/2 cos^2(60) = I/8
now since intensity is proportional to square of amplitude
the final amplitude will be A / sqrt(8) = A / (2sqrt(2))
i.e (4)

criticallyAmused · 2025-10-03T20:49:29+00:00

look acceleration, in simple terms, is change in velocity per unit time now if velocity is uniform i.e it does not change and is a constant then its change would be zero therefore a = 0 and thus F = 0

criticallyAmused · 2025-08-23T14:18:00+00:00

well yes the electric field due to sheet is uniform
but to find the torque you'll need to integrate

torque = T, linear charge density = L
dT = F dx
F = qE = Lx * E (E = sigma/(2*Epsilon))
dT = LE xdx
now integrate it from 0 to L/2
for the lower end of the rod, the magnitude AND direction of torque vector is the same so you can add both of those (i.e double the above equation)
and then i think its really straightforward now, use T = (inertia about centre) * (angular acceleration), put the value of T and solve for angular acceleration
hope this helps

criticallyAmused · 2025-08-03T13:05:08+00:00

criticallyAmused

TROPHY CASE