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 $=\frac{1}{2}{\mathrm{k}\mathrm{x}}^2$

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

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

=1000 m

Initial velocity $=-v\hat{j}$

Final velocity $=-v\hat{j}$  

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

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

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

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

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

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

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

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

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.

The valence bond theory explains the covalent bond formation for two half-filled orbitals. The bond strength depends on several factors, including the extent of overlap of the two atomic orbitals. The bond strength is directly proportional to the extent of overlap.

In simple words, the greater the amount of overlap between two orbitals, the stronger the covalent bond will be.

JEE asks many questions based on the comparison of bond strength for two different pairs of atomic orbitals forming a covalent bond. For example, why H–F bond stronger than the F–F bond?