The sketch between cause (magnitude of acceleration) and effect (magnitude of displacement) is a straight line.

Average K.E of each degree of freedom
\(=\frac{1}{2} k T=\frac{1}{2} N T[k=N]\)
N - no. of moles
T - temperature

Moment of Inertia = MK²
K-Radius of gyration
During heating K would be increased.
Hence moment of inertia will increase

\(\text { Rigidity modulus }=\frac{1}{3} \times \text { Young's modulus }\)
\(\text { Poisson ratio }=\frac{\text { lateral strain }}{\text { longitudinal strain }}\)

T² = C (R + h)³
\(\therefore T \alpha (R_E)^\frac{3}{2}\)
R_E = 2R_E
\(\therefore T \alpha (2R_E)^\frac{3}{2}\)
Time period increases by \(2^\frac{2}{3}\)= 2\(\sqrt 2\) 
No. of days in a year = (365.4) \(\times\)2\(\sqrt 2\)
= 1032

\(m=3 \mathrm{~kg} \quad r=40 \times 10^{-2} \mathrm{~m}=0.4 \mathrm{~m}\)
\(\text { Force }=30 N\)
Moment of inertia of a hollow-cylinder I= MR²
\(=3 \times\left(40 \times 10^{-2}\right)^{2} \)
\(=3 \times 0.4 \times 0.4=0.48 \mathrm{kgm}^{2} \)
\( F R =I \alpha \)
\(30 \times 40 \times 10^{-2}=0.48 d \)
\(d=\frac{12}{0.48}=\frac{1200}{48}=25 \mathrm{rad} \mathrm{s}^{-2} \)
\(\alpha=25 \mathrm{rad} \mathrm{s}^{-2} \)

\(m_{1} =1, \quad m_{2}=2 \)
\(K . E . =m g(h-x) \)
\(\text { For both balls }(h-x)\)
\(=40 \text { i.e.) Same }\)
\(g=\text { constant }\)
\(\therefore K \cdot E_{1}=m_{1} g(h-x)=m_{1} g \times 40 \)
\(K \cdot E_{2}=m_{2} g(h-x)=m_{2} g \times 40 \)
\(\therefore \frac{K \cdot E_{1}}{K \cdot E_{2}}=\frac{m_{1} g \times 40}{m_{2} g \times 40}=\frac{m_{1}}{m_{2}} \)
\( \therefore K \cdot E_{1}: K \cdot E_{2}=1: 2 \)

Unit vector specifies only direction
\(\hat{A}=\frac{\vec{A}}{|\vec{A}|} \)
\(\therefore \hat{i}+\hat{j}=\frac{\hat{l}+\hat{j}}{|\hat{l}+\hat{j}|} \)
\(\hat{i} \text { and } \hat{j} \text { are orthogonal components of vectors. }\)
\(\therefore|\hat{i}+\hat{j}|=\sqrt{1^{2}+1^{2}}=\sqrt{1+1}=\sqrt{2}\)

\(\text { Error in radius }=2 \%\)
\(\Delta r=\frac{2}{100}=0.02\)
\(\text {Volume of the sphere }=\frac{4}{3} \pi r^{3}\)
\(V =\frac{4}{3} \pi r^{3} \)
\(\frac{d V}{V} =\frac{4}{3} \pi \times 3 r^{2} d r \)
\(=3 d r=3(2 \%)=6 \%\)