Class 11th
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New answer posted
2 months agoContributor-Level 10
f? = f? (330/ (330-V) f? = f? (330/ (330+V)
(f? - f? )/f? * 100 = 2% ⇒ (330/ (330-V) - (330/ (330+V) = 0.02
⇒ 330 [ (2V)/ (330)²-V²)] = 0.02 ⇒ 330V = 0.01 * (330)² ⇒ V = 3.3 m/s (approx.)
New answer posted
2 months agoContributor-Level 10
(K (4A)/l) (100 – θ) = (KA/l) (θ – 50)
⇒ 400 – 4θ = θ – 50 ⇒ 5θ = 450 ⇒ θ = 90°C
New answer posted
2 months agoContributor-Level 10
v? = 3tî v? = 24cos 60°î + 24sin 60°? = 12î + 12√3?
v? = v? – v? = (12 – 3t)î + 12√3?
It is minimum when 12 - 3t = 0 ⇒ t = 4sec
New answer posted
2 months agoContributor-Level 10
Zero error = 0.0 + 4 * 0.01 = 0.04 cm
Reading = 2.5a + 7 * 0.01 = 2.57 cm
Correct reading = 2.57 cm - 0.04 cm = 2.53 cm
New answer posted
2 months agoContributor-Level 10
at t = t? , x? = 0 ⇒ 0 = asin (ωt? + π/6)
⇒ ωt? + π/6 = π
at t = t? , x? = 0 ⇒ 0 = asin (ωt? + π/4)
⇒ ωt? + π/4 = π
From (i) and (ii)
ω (t? - t? ) = π/4 - π/6 ⇒ t? - t? = π/ (12ω)
New answer posted
2 months agoContributor-Level 10
ΔW = area under P — t graph
= (1/2) (4 + 6) * 7 = 35 J
Work done = change in KE ⇒ 35 = (1/2) * 2 * v² - (1/2) * 2 * (1)² ⇒ v = 6 m/s
New answer posted
2 months agoContributor-Level 10
U = kr² ⇒ F = -dU/dr = -2kr ; 2kr = mv²/r ⇒ v = √ (2k/m) or T = 2πr/v = 2π√ (m/2k)
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