Class 12th
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New answer posted
a year ago
Contributor-Level 10
Sol. eVslop = hc/λ - φ
Slope of curve
tan θ = hc/e = constant
as intensity of incident radiation is increased, there will be no effect on graph
New answer posted
a year agoContributor-Level 10
Q? = CV?
U? = ½CV? ²
q? = (q? C)/ (C + C/2) = 2q? /3
q? = (q? (C/2)/ (C + C/2) = q? /3; Uf = q? ²/2C + q? ²/2 (C/2)
= (4q? ²/9x2C) + (q? ²/9C)
= (6q? ²/18C) = q? ²/3C = CV? ²/3
Energy loss in the process = U? – Uf
= ½CV? ² - CV? ²/3
= CV? ²/6
New answer posted
a year agoContributor-Level 10
In Rutherford's model, electrostatic force provides the centripetal force:
∈
Kinetic Energy (K):
Potential Energy (U):
Total Energy (E):
Note that, E is the Epsilon symbol.
New answer posted
a year agoContributor-Level 10
Since, lim (x→0) f (x)/x exist ⇒ f (0) = 0
Now, f' (x) = lim (h→0) (f (x+h)-f (x)/h = lim (h→0) (f (h)+xh²+x²h)/h (take y = h)
= lim (h→0) f (h)/h + lim (h→0) (xh) + x²
⇒ f' (x) = 1 + 0 + x²
⇒ f' (3) = 10
New answer posted
a year agoContributor-Level 10
D = |1 -2 3; 2 1; 1 -7 a| = 0 ⇒ a = 8
also, D? = |9 -2 3; b 1; 24 -7 8| = 0 ⇒ b = 3
hence, a-b = 8-3=5
New answer posted
a year agoContributor-Level 10
x dy/dx - y = x² (xcosx + sinx), x > 0
dy/dx - y/x = x (xcosx+sinx) ⇒ dy/dx + Py = Q
So, I.F. = e^ (∫-1/x dx) = 1/|x| = 1/x (x > 0)
Thus, y/x = ∫ 1/x (x (xcosx+sinx)dx
⇒ y/x = xsinx + C
? y (π) = π ⇒ C = 1
So, y = x²sinx + x ⇒ (y)? /? = π²/4 + π/2
Also, dy/dx = x²cosx + 2xsinx + 1
⇒ d²y/dx² = -x²sinx + 4xcosx + 2sinx
⇒ [d²y/dx²] at π/2 = -π²/4 + 2
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