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New answer posted
10 months agoContributor-Level 9
ε = |dΦ/dt| = A|dB/dt|
Given B (t), dB/dt = (4/π) * 10? ³ * (-1/100)
ε = (π * 1²) * | (4/π) * 10? ³ * (-1/100)|
= 4 * 10? V
To find when B=0:
B = 0 ⇒ 1 - t/100 = 0
⇒ t = 100 second
Energy Dissipated, E = P * t = (ε²/R) * t
E = (4*10? )² / (2*10? ) * 100 = 80mJ
New answer posted
10 months agoContributor-Level 10
As we know, β = Ic/Ib = 24 (Given)
R = 1000Ω, ΔV = 0.6V, Ic = ΔV/R = 0.6/1000 = 6*10? A
So, Ib = Ic/β = (6*10? )/24 = 25*10? A
or Ib = 25µA
New answer posted
10 months agoContributor-Level 9
We know, μ? = 1 + χ?
B? ∝ μ?
B ∝ μ
(ΔB/B? ) = (B-B? )/B? = (μ-μ? )/μ? = μ/μ? - 1 = μ? - 1 = χ?
% (ΔB/B) = χ? * 100 = 2.2 * 10? * 100 = 22/10?
(Note: The value for χ? in the source image appears to have a typo as 2.2 * 10? , it has been corrected to 2.2 * 10? to match the final answer.)
New answer posted
10 months ago
Contributor-Level 9
Difference in Reading = Positive Zero Error - Negative Zero Error
= (+5) - [- (100-92)]
= 5 - [-8]
= 13
New answer posted
10 months agoContributor-Level 10
-dT/dt = K [T - Ts]
- (61-59)/4 = K [ (61+59)/2 - 30]
-0.5 = K [60 - 30] = 30K
So, K = -1/60 min? ¹
Again - (51-49)/t = K [ (51+49)/2 - 30]
-2/t = (-1/60) [50-30] = -20/60 = -1/3
t = 6 min
New answer posted
10 months agoContributor-Level 10
L = 0.07H and R = 12Ω
Vs = 220V, 50Hz
XL = 2πfL = 2 (22/7) (50) (0.07) = 22Ω
Z = √ (R² + XL²) = √ (12² + 22²) = √ (144+484) = √628 ≈ 25Ω
i = V/Z = 220/25 = 8.8A
tan φ = XL/R = 22/12 = 11/6
φ = tan? ¹ (11/6)
New answer posted
10 months agoContributor-Level 9
Given: Power in R_B = 200 W, R_B = 50 Ω, V_source = 200 V
I² R_B = 200
I² (50) = 200
I² = 4 ⇒ I = 2A
Also, I = V_source / (R + R_B)
2 = 200 / (R + 50)
2 (R + 50) = 200
R + 50 = 100
R = 50Ω
New answer posted
10 months agoContributor-Level 9
From conservation of momentum:
2*4 = 2v? + mv?
Given v? = 1 m/s (interpreted from intermediate steps)
8 = 2 (1) + mv?
mv? = 6 . (i)
From coefficient of restitution (e=1 for elastic collision):
e = (v? - v? )/ (u? - u? )
1 = (v? - v? )/ (4 - 0)
-1 = (v? - v? )/ (0 - 4) (as written in the image)
⇒ 4 = v? - 1
⇒ v? = 5 . (ii)
Put (2) in (1), m (5) = 6
m = 1.2kg
New answer posted
10 months agoContributor-Level 10
E = ½mω²a² . (i)
When KE = 3E/4, PE = E - KE = E/4
E/4 = ½mω²y² . (ii)
Divide eq? (i) by eq? (ii) we get
4 = a²/y² => y = a/2
New answer posted
10 months agoContributor-Level 10
h = u²/2g, u = √2gh
Now, S = h/3
S = ut + ½at²
h/3 = √2ght - ½gt²
t² - 2√ (2h/g)t + 2h/3g = 0
Using quadratic formula for t:
t = ( 2√ (2h/g) ± √ (8h/g) - 4 (2h/3g) / 2
t = √ (2h/g) ± √ (2h/g - 2h/3g) = √ (2h/g) ± √ (4h/3g)
t? /t? = (√ (2h/g) - √ (4h/3g) / (√ (2h/g) + √ (4h/3g)
t? /t? = (√2 - √ (4/3) / (√2 + √ (4/3) = (√6 - 2)/ (√6 + 2)
(Note: There is a calculation error in the provided solution. Re-evaluating the physics.)
h/3 = (√2gh)t - ½gt²
(g/2)t² - (√2gh)t + h/3 = 0
t = (√2gh ± √ (2gh - 4 (g/2) (h/3)/g = (√2gh ± √ (4gh/3)/g
t? /t? = (√2gh - 2√gh/√3)/ (√2gh + 2√gh/√3)
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