physics ncert solutions class 11th

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New answer posted

5 months ago

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A
alok kumar singh

Contributor-Level 10

g_h = g (1-2h/R). g_d = g (1-d/R). Given h=d.
g_h = g (R/ (R+h)² ≈ g (1-2h/R). This differs from the image.
The image has g/ (1+h/R)² = g (1-h/R). This leads to h² + hR - R²=0. h = R (√5-1)/2.

New answer posted

5 months ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

F = -dU/dr = - [-12A/r¹³ + 6B/r? ]
F=0 ⇒ r= (2A/B)¹/?
U (at r= (2B/A)¹/? ) = -A²/4B

New answer posted

5 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

V_rms = √ (3RT/M)
V_N? = V_H?
√ (3RT_N? / M_N? ) = √ (3RT_H? / M_H? )
573/28 = T_H? /2
⇒ T_H? = 40.928

New answer posted

5 months ago

0 Follower 13 Views

V
Vishal Baghel

Contributor-Level 10

Initially S? L = 2m
S? L = √2² + (3/2)²
S? L = 5/2 = 2.5 m
? x = S? L - S? L = 0.5 m
So since λ = 1 m. ∴? x = λ/2
So white listener moves away from S? Then? x (= S? L − S? L) increases and hence, at? x = λ first maxima will appear.? x = λ = S? L − S? L.
1 = d - 2 ⇒ d = 3 m.

New answer posted

5 months ago

0 Follower 11 Views

A
alok kumar singh

Contributor-Level 10

90 ?

Sol. | P ? + Q ? | = | P ? |

P 2 + Q 2 + 2 P Q c o s ? θ = P 2

Q + 2 P c o s ? θ = 0

c o s ? θ = - Q 2 P

t a n ? α = 2 P s i n ? θ 2 P c o s ? θ + Q =

? [ 2 P c o s ? θ + Q = 0 ]

α = 90 ?

 

New answer posted

5 months ago

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R
Raj Pandey

Contributor-Level 9

τ 1 T

New answer posted

5 months ago

0 Follower 9 Views

A
alok kumar singh

Contributor-Level 10

λ = kT / (√2πd²P)
= (1.38*10? ²³ * 300) / (√2 * 3.14 * (0.3*10? )² * 1.01*10? )
≈ 102 nm

New answer posted

5 months ago

0 Follower 101 Views

V
Vishal Baghel

Contributor-Level 10

1 litre, T = 300K, P = 2 atm, KE = 2*10? J/molecule, no of molecule =?
No. of molecules = (no of moles) * NA = nNA
Also, n = PV/RT = PV/ (NAkT)
KE = (3/2)kT = 2*10? J [Given]
kT = (4/3)*10?
P = 2 atm = 2 * 1.013 * 10? N/m²
vol = 1 lit = 10? ³ m³
No. of molecules = PV/kT = (2*1.013*10? * 10? ³)/ (4/3)*10? ) ≈ 1.5 * 10¹¹

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