Class 11th

Radius of gyration of a solid surface,

${\mathrm{K}}_{\mathrm{S}}=\sqrt[]{\frac{2}{5}\mathrm{R}}$

Radius of gyration of a hollow surface,

${\mathrm{K}}_{\mathrm{H}}=\sqrt[]{\frac{2}{3}}\mathrm{R}$

$\Rightarrow \frac{{\mathrm{K}}_{\mathrm{S}}}{{\mathrm{K}}_{\mathrm{H}}}=\sqrt[]{\frac{3}{5}}=\frac{\sqrt[]{3}}{\sqrt[]{5}}$

The angular acceleration direction is given along angular velocity or opposite to angular velocity depending upon whether angular velocity magnitude is increasing or decreasing and this direction remains along the axis of circular motion.

Potential energy stored in the spring $=\frac{1}{2}{\mathrm{k}\mathrm{x}}^2$

$\begin{array}{rr}& \text{Now}\frac{1}{2}\mathrm{k}(2{)}^2=\mathrm{U}\\ & \&\frac{1}{2}\mathrm{k}(8{)}^2=U^{\mathrm{\text{'}}}\mathrm{}\text{(say)}\\ & \Rightarrow U^{\mathrm{\text{'}}}=\frac{64}{4}U=16U\end{array}$

${\mathrm{h}}_{\text{max}}=\frac{{\mathrm{u}}^2{\mathrm{s}\mathrm{i}\mathrm{n}}^2??}{2\mathrm{g}}=\frac{280*280}{2*9.8}*\frac{1}{4}$

$=1000\mathrm{m}$

Average speed $=\left(\frac{4{\mathrm{v}}^2}{3\mathrm{v}}\right)$ $\left(\frac{{}^{}}{}\right)$

$=\frac{4\mathrm{v}}{3}$

(d) Initial velocity  $=-v\hat{j}$ $\stackrel{}{}$

Final velocity $=v\hat{i}$ $\stackrel{}{}$

Change in velocity  $=v\hat{i}-(-v\hat{j})$

$\stackrel{}{}\stackrel{}{}$$\stackrel{}{}\stackrel{}{}$ $=v(\hat{i}+\hat{j})$ Momentum gain is along $\stackrel{\mathrm{ˆ}}{\mathrm{i}}+\stackrel{\mathrm{ˆ}}{\mathrm{j}}$ $\stackrel{}{}\stackrel{}{}$

$$ $\Rightarrow$  Force experienced is along $\hat{i}+\hat{j}$ $\stackrel{}{}\stackrel{}{}$

$$  $\Rightarrow$ Force experienced is in North-East direction.

$m=\rho \pi r^{2}1$

$\begin{array}{rr}& \rho =\frac{\mathrm{m}}{\pi {\mathrm{r}}^{2}1}\\ & \frac{\mathrm{\Delta }\rho }{\rho }=\frac{\mathrm{\Delta }\mathrm{m}}{\mathrm{m}}+\frac{2\mathrm{\Delta }\mathrm{r}}{\mathrm{r}}+\frac{\mathrm{\Delta }\mathrm{l}}{1}\\ & \frac{\mathrm{\Delta }\rho }{\rho }*100=\frac{0.002}{0.4}*100+\frac{2*0.001}{0.3}*100+\frac{0.02}{5}*100\\ & =\frac{0.2}{0.4}+\frac{0.2}{0.3}+\frac{2}{5}\\ & =0.5+0.67+0.4\\ & =1.57\\ & =1.6\mathrm{\%}\end{array}$

As the factors controlling temperature and voltage supply are beyond prediction and control so the error occurred due to unpredictable fluctuations of temperature and voltage would be random errors.

Kindly consider the following Image 

lim (x→0) [a e? - b cos (x) + c e? ] / (x sin (x) = 2
Using Taylor expansions around x=0:
lim (x→0) [a (1+x+x²/2!+.) - b (1-x²/2!+.) + c (1-x+x²/2!+.)] / (x * x) = 2
lim (x→0) [ (a-b+c) + x (a-c) + x² (a/2+b/2+c/2) + O (x³)] / x² = 2
For the limit to exist, the coefficients of lower powers of x in the numerator must be zero.
a - b + c = 0
a - c = 0 ⇒ a = c
Substituting a=c into the first equation: 2a - b = 0 ⇒ b = 2a.
The limit becomes: lim (x→0) [x² (a/2 + b/2 + c/2)] / x² = (a+b+c)/2
(a + b + c) / 2 = 2 ⇒ a + b + c = 4.

|z+i|/|z-3i| = 1 ⇒ |z+i| = |z-3i|. This means z is on the perpendicular bisector of the segment from -i to 3i. The midpoint is i, so z = x+i.
w = z? - 2z + 2. Let z = x + iy.
w = (x² + y²) - 2 (x + iy) + 2 = (x² - 2x + 2 + y²) - 2iy.
Re (w) = x² - 2x + 2 + y² = (x - 1)² + 1 + y².
From the first condition, y=1. Re (w) = (x - 1)² + 1 + 1 = (x - 1)² + 2.
Re (w) is minimum for x = 1.
The common z is z = 1 + i.
w = (1+i) (1-i) - 2 (1+i) + 2 = 2 - 2 - 2i + 2 = 2 - 2i.
w² = (2 - 2i)² = 4 (1 - 2i - 1) = -8i.
w? = (-8i)² = -64 ∈ R.
∴ least n ∈ N for which w? ∈ R is n=4.