SONAR stands for 'sound navigation and ranging'.

 About 1 in 10¹² to 10¹³ s.

⟶ Stress and Young's modulus.
⟶ Work and Energy.

1 \(\mathring { A } \) = 10^-10 m
1 A.u. = 1.496 x 10¹¹ m

Acceleration = (10 m/(1s)²) 
= (10 x 10^-3 / [1/60 x 60])²
= (3600)² x 10^-2 km/h²
= 1.29 x 10⁵ km/h².

Weight of diamond = 5.42 g = 0.00542 Kg
Total weight = 1.2 + 0.00542
= 1.20542 kg = 1.2 kg

One Barn is a small unit of area used to measure area of nuclear cross-section
∴ 1 barn = 10^-28 m².

True, because human heart beats at a regular rate.

1 amu =1.66 x 10^-27 kg
∴ 1 kg = (1/1.66 x 10^-27) amu = 0.6 x 10²⁷ amu

The size of nucleus is measured in fermi
1 fermi = 10^-15 m

\(T=2\pi \sqrt { \frac { l }{ g } } or{ T }^{ 2 }=4{ \pi }^{ 2 }\frac { l }{ g } \)
\(\\ g=4{ \pi }^{ 2 }\frac { 1 }{ { T }^{ 2 } } \)
\(\\ \frac { \triangle g }{ g } =\frac { \triangle l }{ l } +2\frac { \triangle T }{ T } \)
% error in g = 1% + 2 x 2% = 5%.

\(101.6\times { 10 }^{ 6 }\) 

Q = CV \(\\ =2.0\times 20=40\mu C=40\times { 10 }^{ -6 }C\) 
\(\frac { \triangle Q }{ Q } =\frac { \triangle C }{ C } +\frac { \triangle V }{ V } =\frac { 0.1 }{ 2.0 } +\frac { 0.5 }{ 20 } =\frac { 3 }{ 40 } \) 
\(\\ \triangle Q=\frac { 3 }{ 40 } \times Q=\frac { 3 }{ 40 } \times 40\times { 10 }^{ -6 }=3\mu C\)
Hence, \(Q=(40\pm 3.0)\times { 10 }^{ -6 }C\). 

Given, Angular diameter, \(\theta =1920"\)
Distance of the sun from earth, \(r=1.496\times { 10 }^{ 11 }m\)
The diameter of the sun, d = ?
\(\theta =1920"=1920\times 4.85\times { 10 }^{ -6 }rad\)
= \(9.31\times { 10 }^{ -3 }\) rad
We know that, \(d=\theta r\)
\(d=9.31\times { 10 }^{ -3 }\times 1.496\times { 10 }^{ 11 }=1.39\times { 10 }^{ 9 }m\)
Thus, the diameter of the sun is \(1.39\times { 10 }^{ 9 }m\) 

Heat produced H is given by
\(H=\frac { { l }^{ 2 }Rt }{ J } \\ \frac { \triangle H }{ H } =2\frac { \triangle l }{ l } +\frac { \triangle R }{ R } +\frac { \triangle t }{ t } -\frac { \triangle J }{ J } \)
For maximum percentage error,
\(\\ \frac { \triangle H }{ H } \times 100=2\frac { \triangle l }{ l } \times 100+\frac { \triangle R }{ R } \times 100+\frac { \triangle t }{ t } \times 100+\frac { \triangle J }{ J } \times 100\)
= 2 x 2% + 1% + 1% + 0% = 6%

We first note that the dimension of I are [ML²]. Also the magnetic moment has the units
Am² so that its dimensions can be written as [AL 2] where A stands for the dimensions of the electric current. Finally the magnetic field vector B has the units newton (per ampere metre) so that its dimensions can be written as
\([B]=\frac { [ML{ T }^{ -2 }] }{ [A][L] } =[M{ T }^{ -2 }{ A }^{ -1 }]\)
We now assume that
T= k l^a M^b B^C
Substituting dimensions of all the quantities involved, we have
[T] = [ML²]^a[AL²]^b[MT^-2A^-1]^c
= M^a+c L^2a+2b T^-2c A^b-c
Equating powers of M, L, T and A on both sides, we have a + c = 0, 2 (a + b) = 0, - 2c = 1, b - c = 0.From the first three equations, we get c = \(\frac{-1}{2}\), a = \(\frac{-1}{2}\) and b =\(\frac{-1}{2}\)
These values are consistent with the fourth equation. Thus
a = \(\frac{1}{2}\) b = \(\frac{-1}{2}\)and c = \(\frac{-1}{2}\)
\(\therefore\)T = k I^1/2 M^-1/2 B^-1/2 =\(K\sqrt { \frac { 1 }{ MB } } \)
Experiments show that k = 2n. Therefore
T = \(2\pi \sqrt { \frac { 1 }{ MB } } \)

Given, radius of sodium atom, 
r = 2.5 \(\mathring{A}\) = \(2.5\times { 10 }^{ -10 }m\)   [\(\because \)\(\mathring{A}\)= 10^-10 m]
Volume of sodium atom = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }\)  
 \(=\frac { 4 }{ 3 } \times 3.14\times { \left( 2.5\times { 10 }^{ -10 } \right) }^{ 30 }=65.42\times { 10 }^{ -30 }{ m }^{ 3 }\)  
Number of atom in one mole of sodium
= Avogadro's number (N)      
N =\(6.023\times { 10 }^{ 23 }\) 
\(\therefore \) Atomic volume of sodium = Volume of one atom of sodium \(\times \) Number of atoms
 \(=65.42\times { 10 }^{ -30 }\times 6.023\times { 10 }^{ 23 }\\ =3.94\times { 10 }^{ -5 }{ m }^{ 3 }\) 
Mass of a mole of sodium = 23 g = \(23\times { 10 }^{ -3 }kg\) 
\(\therefore \ Average \ mass \ density \ of \ sodium=\frac { Mass }{ Volume } \)           
\(\rho =\frac { 23\times { 10 }^{ -3 } }{ 3.94\times { 10 }^{ -5 } } \simeq 5.84\times { 10 }^{ 2 }kg/{ m }^{ 2 }\simeq { 584 }kg/m^{ 3 }\) 
Density of sodium in crystalline phase
  \(=970kg/{ m }^{ 3 }=9.7\times { 10 }^{ 2 }kg/{ m }^{ 3 }\)       
The two densities are of the same order of magnitude because in solid state, atoms are tightly packed.