Class 11th
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New answer posted
3 months agoContributor-Level 10
Loss in elastic potential energy = Gain in KE
½ (YA/L)x² = ½mv²
0.5 * (0.5*10? * 10? / 0.1) * (0.04)² = 20*10? ³ v²
0.5 * (5*10²) * 1.6*10? ³ = 20*10? ³ v²
0.4 = 20*10? ³ v²
v² = 20 => v = √20 ≈ 4.47 m/s
(Re-checking calculations)
0.5 * ( (0.5e9 * 1e-6) / 0.1) * (0.04)^2 = 0.5 * (5e2) * 1.6e-3 = 4.
0.5 * 20e-3 * v^2 = 10e-3 v^2
4 = 10e-3 v^2
v^2 = 400 => v = 20 m/s
New answer posted
3 months ago
Contributor-Level 9
Difference in Reading = Positive Zero Error - Negative Zero Error
= (+5) - [- (100-92)]
= 5 - [-8]
= 13
New answer posted
3 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
3 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
3 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
3 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)
New answer posted
3 months agoContributor-Level 9
(1/2)mu² = (1/2)mv² + mgl (1-cosθ)
⇒ v² = u² - 2gl (1-cosθ)
⇒ v² = 3² - 2*10*0.5* (1 - 1/2)
⇒ v² = 9 - 5 = 4
v = 2m/s
New answer posted
3 months ago
Contributor-Level 9
ω = √k_eq/μ [μ = (m? )/ (m? +m? ) (Reduced mass)]
k_eq = (k * 4k)/ (k+4k) = 4k/5
ω = √ (4k/5) / (m? / (m? +m? )
= √ (4*20/5) / (0.2*0.8)/ (0.2+0.8)
= √ (16/0.16)
= 10 rad/s
New answer posted
3 months agoContributor-Level 10
Least count = Pitch / (Total divisions on circular scale)
In 5 revolution, distance travelled = 5mm
? In 1 revolution, distance travelled = 1mm
So least count = 1/50 = 0.02mm = 0.002 cm
So, Assertion is not correct but reason is correct
New answer posted
3 months agoContributor-Level 9
r? = 10αt²î + 5β (t-5)?
v? = dr? /dt = 20αtî + 5β?
As L? = m (r? * v? )
So, at t=0, L=0
given L is same at t=t as at t=0
⇒ r? * v? = 0
⇒ (10αt²î + 5β (t-5)? ) * (20αtî + 5β? ) = 0
⇒ 50αβt² (î*? ) + 100αβt (t-5) (? *î) = 0
⇒ 50αβt² k? - 100αβt (t-5) k? = 0
⇒ 50t² - 100t (t-5) = 0
⇒ 50t² - 100t² + 500t = 0
⇒ -50t² + 500t = 0
⇒ 50t (10 - t) = 0
⇒ t = 10 second
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