Class 11th

|Z|max + |Z|min
= 6 + 0
= 6

In isothermal process, temperature is constant.

In isochoric process, volume is constant.

In adiabatic process, there is no exchange of heat.

In isobaric process, pressure is constant.

$Q_1=\underset{water}{20°C}\to 100°steam$

$Q_1=mc\Delta T+mL$

$=m\left[c\Delta T+L\right]$

$=31000=31*10^{3}cal$

$Q_2=\frac{293}{373}*31*10^3$

 

f (x) = x√ (1 – x²)
∴ 1 - x² ≥ 0
⇒ x ∈ [-1, 1]
Put x = cos θ
⇒ f (θ) = cos θ|sin θ|
If sin θ > 0, f (θ) = sin θ cos θ
= (sin 2θ)/2, f (θ) ∈ [-1/2, 1/2]
if sin θ < 0, f () = sin cos
= - (sin 2θ)/2, f (θ) ∈ [-1/2, 1/2]

5 red, 3 yellow
There are 3 points in sample space either both are red or both are yellow or one is red other is yellow.
⇒ P (E) = (3.5+5.3) / (3.5+5.3+5.4+3.2) = 15/28

For part AM, slope of v – t graph is constant but negative. For part MB, slope of v – t graph is constant but positive.

$\tau =\eta \frac{v}{h}$  

              Given

              $\tau =10^{-3}N/m^2$  

              (shear stress)

              h =?

              v = 36 km/hr = 10 m/sec

              $10^{-3}=\frac{10^{-2}*10}{h}$

h = 100 m

$\gamma =3\alpha$

$\Delta V=\gamma V\Delta T=3\alpha .a^3.\Delta T$

x = t² - t + 1 . (i)
y = t² + t + 1 . (ii)
(i) + (ii)
x + y = 2 (t² + 1) . (iii)
(i) – (ii)
x - y = -2t . (iv)
from (iii) and (iv)
x² – 2xy + y² – 2x – 2y + 4 = 0
Here H² = ab and Δ≠ 0
This is parabola, so e = 1

$F_{1}cos45°+F_{2}cos45°+F_3=\frac{m{v}^2}{r}$

$\Rightarrow \frac{G{m}^2}{{\left(\sqrt[]{2}r\right)}^2}.\frac{1}{\sqrt[]{2}}+\frac{G{m}^2}{{\left(\sqrt[]{2}r\right)}^2}.\frac{1}{\sqrt[]{2}}+\frac{G{m}^2}{{\left(2r\right)}^2}=\frac{m{v}^2}{r}$  

  $\Rightarrow v=\sqrt[]{\frac{Gm}{4r}\left(2\sqrt[]{2}+1\right)}$            

Putting m = 1 kg and r = 1 m,

$v=\frac{1}{2}\sqrt[]{G\left(1+2\sqrt[]{2}\right)}$