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New answer posted
10 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
10 months agoContributor-Level 9
I = I? e? /?
= (20/10000) e^- (1*10? / 10*10? ³)
= 2 * 10? ³ e? ¹
The provided solution calculates as:
= 2 * 10? ³ e? ¹
= 2e? ¹ mA
= 2 * 0.37mA
= 0.74 mA = 74/100 mA
(Note: There seems to be a calculation discrepancy in the source image, the steps shown lead to I = 2e? ¹ mA)
New answer posted
10 months agoContributor-Level 10
β = λD/d or β ∝ λ
Also, we know that λ_blue < _orange
So, β_orange > β_blue
So, dist b/w consecutive fringes will decrease.
New answer posted
10 months agoContributor-Level 9
F_net = -kq²/ (l+x)² + kq²/ (l-x)²
= kq² [ (l+x)² - (l-x)² / (l²-x²)² ]
= kq² [ 4lx / (l²-x²)² ]
For x << l,
ma ≈ -4kq²x / l³
a = - (4kq²/ml³)x
ω² = 4kq²/ml³
ω = √ (4kq²/ml³)
= √ (4 * 9*10? * 10 / (1*10? * 1)
= 6 * 10? rad/s
= 6000 * 10? rad/s

New answer posted
10 months agoContributor-Level 10
As charge remains same. So,
Initial total charge Q = (2C)V + CV = 3CV
When dielectric is inserted in C, its new capacitance is KC.
The capacitors 2C and KC are in parallel.
Equivalent capacitance C_eq = 2C + KC = (K+2)C
New potential Vc = Q/C_eq = 3CV/ (K+2)C) = 3V/ (K+2)
New answer posted
10 months agoContributor-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
10 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
10 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
New answer posted
10 months agoContributor-Level 10
A→B & D→C (isothermal process)
So, TA = TB & TD = TC. Now B→C & D→A (adiabatic process)
|WBC| = nR/ (γ-1) (TB - TC)
|WAD| = nR/ (γ-1) (TA - TD) = nR/ (γ-1) (TB - TC)
∴ |WBC| = |WAD|
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