\(Velocity \ gradient=\frac { Velocity }{ Distance } =\frac { \left[ LT^{ -1 } \right] }{ \left[ L \right] } =\left[ { T }^{ -1 } \right] \)

22%

Relative error = \(\frac { \triangle \bar { d } }{ d } =\frac { 0.01 }{ 0.39 } =0.0256\)

Three

Mean value of refractive index
\({ \mu }_{ m }=\frac { { \mu }_{ 1 }+{ \mu }_{ 2 }+{ \mu }_{ 3 }+{ \mu }_{ 4 }+{ \mu }_{ 5 }+{ \mu }_{ 6 }+{ \mu }_{ 7 }+{ \mu }_{ 8 } }{ 8 } \)
\(\\ =\frac { 1.29+1.33+1.34+1.35+1.32+1.36+1.33+1.30 }{ 8 }\)
\( \\ =\frac { 10.62 }{ 8 } =1.33\)

Radius of the lunar orbit around the earth = Distance between the moon and the earth Time taken by the laser beam from earth to moon and then back fo the earth = 2.56 s.
\(\therefore\) Time taken by the laser beam to go from earth to the moon is t = \(\frac{2.56}{2}\) = 1.28 s
Speed of the laser beam (i.e., light),
c = 3 x 10⁸ ms^-1
\(\therefore\) Distance between moon and earth,
S = c t = 3 x 10⁸ x 1.28
= 3.84 x 10⁸ m = 3.84 x 10⁵ km.

Let X be the desired number, then
\(X=k{ e }^{ u }{ \varepsilon }_{ 0 }^{ x }{ m }_{ e }^{ -2 }{ m }_{ p }^{ -1 }{ c }^{ y }{ G }^{ z }\)
Here k is a dimensionless constant and x, y, z and u are unknowns, whose value is to be obtained from the principle of homogeneity of dimensions. ow
[X] = M⁰ L⁰T¹ Q⁰
[e] = M⁰L⁰T⁰Q
[£0] = M^-1L^-3T²Q
[C] = M⁰L³T^-1
[G] = M^-1L³T^-2
Substituting dimensions of parameters involved in equation (i), we get
M⁰ L⁰ T¹ Q^O = Q^u [M^-1L^-3T²Q²]M^-2M^-1(LT^-1)^y(M^-1L³T^-2)^z
= M^-x-3-z L^-3x+y+3z T^2x-y+2x Q^u+2x
From the principle of homogeneity of dimensions
-x-3-z= 0
- 3x + Y + 3z =0
2x-y-2z =1
u + 2x=0
Solving eqns. (ii), (iii), (iv) and (v), we get
u + 4, x = - 2, Y = - 3, Z = - 1
\(x=k{ e }^{ 4 }{ \varepsilon }_{ 0 }^{ -2 }{ m }_{ e }^{ -2 }{ m }_{ p }^{ -1 }{ c }^{ -3 }{ G }^{ -1 }\ \)
\(x=\frac { k{ e }^{ 4 } }{ { \varepsilon }_{ 0 }^{ 2 }{ m }_{ e }^{ 2 }{ m }_{ p }^{ }{ c }^{ 3 }{ G } } \)
Experiments show that
k = \(\frac { 1 }{ 16{ \pi }^{ 2 } } \)
x\(=\frac { { e }^{ 4 } }{ { 16{ \pi }^{ 2 }\varepsilon }_{ 0 }^{ 2 }{ m }_{ e }^{ 2 }{ m }_{ p }^{ }{ c }^{ 3 }{ G } } \)
Substituting values of all known parameters we find that the value of x is nearly 15 billion years, which is approximately equal to the present estimate of the age of the universe.

It is given that viscous force F depends on (i) radius r of steel ball, (ii) coefficient of viscosity \(\eta \) of viscous liquid (iii) Speed v of the ball
i.e., F = kr^a\(\eta \)^bv^c, where k is dimensionless constant Dimensional formula of force
F = [MLT^-2], r=[L]
\(\eta \) = [M¹L^-1T^-1] and v=[LT^-1], we have
[MLT^-2] = [L]^a[M¹L^-1T^-1]^b[LT^-1]^c
= [M^aL^a-b+cT^-b-c]
Comparing powers of M,L and T on either side of equation, we get
a =1
a - b + c = 1
-b - c = -2
On solving, these above equations, we get
a = 1, b = 1 and c = 1
Hence, the relation becomes
F = kr\(\eta \)v

Here, \({ R }_{ 1 }=\left( 100\pm 3 \right) \Omega \), \({ R }_{ 2 }=\left( 200\pm 4 \right) \Omega \) 
Series combination
R = R₁ + R₂.​​​​​​ = 100 + 200 = 300\(\Omega \) 
\(\triangle { R }=\pm \left( \triangle { R }_{ 1 }+\triangle { R }_{ 2 } \right) =\pm \left( 3+4 \right) =\pm 7\Omega \) 
 \(\therefore \quad R=\left( 300\pm 7 \right) \Omega \).

Given, molar volume of one mole of hydrogen
= 22.4 L = \(22.4\times { 10 }^{ -3 }{ m }^{ 3 }\) 
Diameter of hydrogen molecules (d) = 1\(\mathring{A}\) = 10^-10m
\(\therefore \) Radius of hydrogen molecule (r) = \(\frac { d }{ 2 } =\frac { { 10 }^{ -10 } }{ 2 } \) 
 \(=0.5\times { 10 }^{ -10 }m\)  
Volume of one molecule of hydrogen = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }\) 
 \(=\frac { 4 }{ 3 } \times 3.14\times (0.5\times { 10 }^{ -10 })\)  
\(=5.234\times { 10 }^{ -31 }{ m }^{ 3 }\)
Number of molecules in one mole hydrogen = Avogadro's number (N) = \(6.023\times { 10 }^{ 23 }\) 
\(\therefore \) Atomic volume of one mole of hydrogen = Number of molecules in one mole of hydrogen \(\times \) Volume of one molecule of hydrogen
\(=6.023\times { 10 }^{ 23 }\times 5.234\times { 10 }^{ -31 }\)   
\(\\ =3.152\times { 10 }^{ -7 }{ m }^{ 3 }\)
\(\therefore \ \frac { Molar \ Volume }{ Atomic \ Volume } =\frac { 22.4\times { 10 }^{ -3 } }{ 3.152\times { 10 }^{ -7 } } = 7.1\times { 10 }^{ 4 }\) 
This ratio is very large, which shows that the intermolecular separation in a gas is much larger than the size of a molecule.