physics ncert solutions class 11th
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a month agoNew answer posted
a month agoContributor-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
2 months agoContributor-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
2 months agoContributor-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
2 months agoContributor-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
2 months agoContributor-Level 10
λ = kT / (√2πd²P)
= (1.38*10? ²³ * 300) / (√2 * 3.14 * (0.3*10? )² * 1.01*10? )
≈ 102 nm
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