Class 12th

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New answer posted

a year ago

0 Follower 27 Views

V
Vishal Baghel

Contributor-Level 10

Tsinθ = qE
Tcosθ = mg
tanθ = qE/mg
E = V/d_eff. V = V? - (-V? ) = V? +V?
C? =kε? A/t, C? =ε? A/ (d-t).
Capacitors are in series. Q=C_eqV. E inside dielectric = σ/ (kε? ) = Q/ (Akε? ).
E in air = σ/ε? = Q/Aε?
Let's follow the solution.
Q = (C? / (C? +C? ) (V? +V? ).
E = E_air = Q/ (Aε? )
tanθ = qQ/ (Aε? mg) = q/ (Aε? mg) * (C? / (C? +C? ) (V? +V? )
Substitute C? and C?
This is getting complicated. The solution directly uses an expression for E. E = Q/C? Aε?

New answer posted

a year ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

a year ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

ω = 2πf = 1/√ (LC)
L = 1/ (4π²f²C) = 1/ (4π² * 60² * 0.1*10? ) = 70.3 mH

New answer posted

a year ago

0 Follower 87 Views

V
Vishal Baghel

Contributor-Level 10

E? (due to A and G) = 2 * (kq/l²)cos (45) = √2 kq/l² (downwards)
E? (due to B and F) = 2 * (kq/l²)cos (45) = √2 kq/l² (towards left)
E? (due to C and H) = 0
E? (due to D and E) = 0
Resultant E = √ (E? ²+E? ²) = √ (2 (kq/l²)²+2 (kq/l²)²) = 2kq/l²
(Solution in the image seems to be different.)

New answer posted

a year ago

0 Follower 11 Views

V
Vishal Baghel

Contributor-Level 10

Energy of electron = 3 eV
It forms H atom in n=3 state. Energy released E = 3 - (-13.6/9) = 3 + 1.51 = 4.51 eV.
Photon Energy = 4.51 eV
Threshold energy = hc/λ = 12400eVÅ / 4000Å = 3.1 eV.
kE_max = 4.51 - 3.1 = 1.41 eV

 

New answer posted

a year ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Density of nucleus is constant.

New answer posted

a year ago

0 Follower 25 Views

V
Vishal Baghel

Contributor-Level 10

R = R? [1 + α? T]
⇒ 16 = R? [1 + α (15 - T? )]
20 = R? [1 + α (100 - T? )]
Assuming T? = 0°C:
16 = R? (1 + 15α)
20 = R? (1 + 100α)
Dividing the equations:
16/20 = (1+15α)/ (1+100α)
16 + 1600α = 20 + 300α
1300α = 4
α = 4/1300 ≈ 0.003 °C? ¹

New answer posted

a year ago

0 Follower 32 Views

V
Vishal Baghel

Contributor-Level 10

As b > a
The magnetic field inside the wire (rR) is B = µ? I/ (2πr).
For wire with radius a, B increases linearly to r=a, then decreases. For wire with radius b, B increases linearly to r=b, then decreases. Since a⇒ B? > B?
B? = µ? I / 2πa
B? = µ? I / 2πb
(Note: The question is likely asking for the graph representation, which is option A based on the formulas.)

New answer posted

a year ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

dy/dx = 2y/ (xlnx).
dy/y = 2dx/ (xlnx).
ln|y| = 2ln|lnx| + C.
ln|y| = ln (lnx)²) + C.
y = A (lnx)².
(ln2)² = A (ln2)². ⇒ A=1.
y = f (x) = (lnx)².
f (e) = (lne)² = 1² = 1.

New answer posted

a year ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

P (at least one head) = 1 - P (no heads) = 1 - (1/2)? ≥ 0.9.
0.1 ≥ (1/2)?
10 ≤ 2?
n=3, 2³=8. n=4, 2? =16.
Minimum value of n is 4.

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