As relative density is defined as the ratio of the density of given substance and the density of standard distance (water), it is a dimensionless quantity.

The dimensions of LHS are \([\mathrm{M}]\left[\mathrm{L} \mathrm{T}^{-1}\right]^{2}=[\mathrm{M}]\left[\mathrm{L}^{2} \mathrm{~T}^{-2}\right]\)
\(=\left[\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-2}\right]\)
The dimensions of RHS are \([\mathrm{M}]\left[\mathrm{L} \mathrm{T}^{-2}\right] \quad[\mathrm{L}]=[\mathrm{M}]\left[\mathrm{L}^{2} \mathrm{~T}^{-2}\right]\)
\(=\left[\mathbf{M} \mathrm{L}^{2} \mathrm{~T}^{-2}\right]\)
The dimensions of LHS and RHS are the same and hence the equation is dimensionally correct.

10⁷ CGS units.

Here, x = [L]
t = [T]; x = ct²
[L] = c × [T²]
\(\Rightarrow \frac{[L]}{\left[T^2\right]}=c \Rightarrow c=\left[L T^{-2}\right]\)

No, it is not possible.

Molar gas constant [R]=[ML²T⁻²K⁻¹mol⁻¹]

In the given relation dimensions of LHS terms v are [LT^-1]. Dimensional formula for E and p are [ML^-1T^-2] and [ML^-3].
Dimensions of RHS \(=\sqrt { \frac { { ML }^{ -1 }{ T }^{ -2 } }{ { ML }^{ -3 } } } =\sqrt { { L }^{ 2 }{ T }^{ -2 } } \)
= [LT^-1]
As dimensions of LHS and RHS of the equation are same, Hence the equation is dimensionally correct.

It least count of physical balance is 0.1 g, the mass measured will be correctly represented as m=(156.3±0.1)g

No, the statement is true

No, the inertial and gravitational mass of a body are equivalent.

Given, Total time (t) = 100 yr
= 100 x 365\(\frac { 1 }{ 4 } \) days
= 100 x 365 \(\frac { 1 }{ 4 } \) x 24 h
= 100 x 365 \(\frac { 1 }{ 4 } \) x 24 x 60 x 60 s
Difference in time (\(\triangle \) t) = 0.02 s
Error in 1s = \(\frac { 0.02 }{ 100\times365\frac { 1 }{ 4 } \times24\times60\times60 } \)
\(\\ =\frac { 2\times{ 10 }^{ -2 }\times4 }{ 1461\times24\times36\times{ 10 }^{ 4 } }\)
\( \\ =6.34\times{ 10 }^{ -12 }s\)
\(\\ \approx { 10 }^{ -12 }s\)
Therefore, the accuracy of the standard cesium clock in measuring a time interval of 1 s is 10^-12 s.

Given, Magnification of microscope = 100
Observed width of the hair = 3.5 mm
Estimates on the thickness of hair is given by,
Magnification = \(\frac { observed \ width }{ Real \ width } \)
or Real width =   \(\frac { Observed\ width }{ Magnification } =\frac { 3.5 }{ 100 } \)
= 0.035 mm

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.

Two

No, since length is fundamental quantity and velocity is the derived quantity.

(i) Sincerity, Curiosity, dedicated and helping nature
(ii) Unit system are :
(a) FPS system
(b) MKS system
(c) CGS system

\(f\propto { m }^{ a }{ v }^{ b }{ r }^{ c }\)
\(\\ [f]=k[{ m] }^{ a }[{ v] }^{ b }[{ r] }^{ c }\)
where k is a constant
or \([ML{ T }^{ -2 }]=[{ M] }^{ a }[L{ T }^{ -1 }][{ L] }^{ c }\)
Compare the powers of M, Land T, we have
a = 1, b + C = 1 and - b = - 2
Solving above equations, we get
a = 1, b = 2 and c = - 1
\(\therefore\) f = kM¹ v² r^-1
\(f=\frac { m{ v }^{ 2 } }{ r } \)
Here k=1
\(f=\frac { m{ v }^{ 2 } }{ r } \)

Given, x = 4.29 light years = 4.29 x 9.46 x 10¹⁵ m
= \(\frac { 4.29\times9.46\times{ 10 }^{ 15 } }{ 3.08\times{ 10 }^{ 16 } } par\ sec\ =1.317par\ sec\)
\(\\ \theta =\frac { l }{ r } =\frac { 2AU }{ x } \)
Radius of the Earth orbit = 1 AU
= 1.496 x 10¹¹m
\(=\frac { 2\times1.496\times{ 10 }^{ 11 } }{ 4.29\times9.46\times{ 10 }^{ 15 } } rad\)
\(\\ =7.39\times{ 10 }^{ -6 }rad\)
\(\\ =.39\times{ 10 }^{ -6 }\times\frac { 180\times60\times60 }{ \pi } s=1.52s\) 

Mean absolute error\(=\frac { \sum { \left| \triangle { T }_{ i } \right| } }{ n }\)
\( \\ \triangle \overline { T } =\frac { 0.01+0.06+0.20+0.09+0.18 }{ 5 } \)
\(\\ =\frac { 0.54 }{ 5 } =0.108s=0.11s\) 

As,      \(x=\frac { { a }^{ 3 }{ b }^{ 3 } }{ c\sqrt { d } } \) 
 \(\therefore \quad\frac { \triangle x }{ x } =\pm \left[ 2\frac { \triangle a }{ a } +3\frac { \triangle b }{ b } +\frac { \triangle c }{ c } +\frac { 1 }{ 2 } \frac { \triangle d }{ d } \right] \)     
Percentage error in x is given by 
 \(\frac { \triangle x }{ x } \times 100\) = \(\pm\) \([2\times 2\)% + \(3\times 1\)% +3% + \(\frac { 1 }{ 2 } \times 4\)%\(]\) 
 \(=\pm 12\)%

Let the distance covered is S,
Then S = K u^a  f^b t^c ; where k is a constant. Writing dimensions on both the sides, we have
[L] = [LT^-1]^a[LT^-2]^b[T]^c
or [L] = [L^a+bT^-a-2b+c]
Comparing powers on both sides, we get
1 = a + b
and 0 = - a - 2b + C
We have only two equations with three unknowns, therefore, we split the problem into two parts.
(a) Let the body have no acceleration,
then S = k₁u^at^b
or [L] = [LT^-1]^a[T]^b
= [L^aT^-a+b] 
or a = 1
-a + b = 0 or b = 1
S = k₁ut  ------------------(i)
(b) Suppose the body has no initial velocity
then S=k₂f^at^b
[L] = k₂[LT^-2]^a[T]^b
or [L] =[L^aT^-2a+b] or a = 1
- 2a + bOor b = 2a = 2
S = k₂ft² ------------------(ii)
If the body has both the initial velocity and acceleration comparing (i) and (ii), we get,
S = k₁ ut + k₂ ft²
This is the required equation
If we put k₁= 1, k₂ =\(\frac{1}{2}\), we get
S = ut +\(\frac{1}{2}\)ft²

Given d=50 m,  \({ \theta }_{ 1 }=60°,\quad { \theta }_{ 2 }=30°\)
h = \(\frac { d }{ \cot { { \theta }_{ 2 } } -\cot { { \theta }_{ 1 } } } =\frac { 50 }{ \cot { 30° } -\cot { 60° } } \)
or h = \(\frac { 50 }{ \sqrt { 3 } -1/\sqrt { 3 } } =\frac { 50\sqrt { 3 } }{ 3-1 } =25\sqrt { 3 } \)
= \(25\times 1.732\)
= 43.3 m

The dimensional formula of energy = [ML²T^-2]
Let M₁ ,L_1, T₁ and M₂ ,L₂ ,T₂ are the units of mass, length and time in given two systems
\(M_{ 1 }=1 \ kg,\quad M_{ 2 }=\alpha \ kg\)
\(\\ L_{ 1 }=1 \ m,\quad L_{ 2 }=\beta \ m\)
\(\\ T_{ 1 }=1 \ s,\quad T_{ 2 }=\gamma \ s\)
For any physical quantity, the product of its magnitude and unit is always constant
 \({ n }_{ 1 }{ u }_{ 1 }={ n }_{ 2 }{ u }_{ 2 }\)
\(\\ or\ { n }_{ 2 }={ n }_{ 1 }\frac { u_{ 1 } }{ u_{ 2 } } =4.2\times [\frac { [{ M }_{ 1 }{ L }_{ 1 }^{ 2 }{ T }_{ 1 }] }{ [{ M }_{ 2 }{ L }_{ 2 }^{ 2 }{ T }_{ 2 }^{ -2 }] }\)
\( \\ =4.2\left[ \frac { { M }_{ 1 } }{ { M }_{ 2 } } \right] \times \left[ \frac { { L }_{ 1 } }{ { L }_{ 2 } } \right] ^{ 2 }\times \left[ \frac { { T }_{ 1 } }{ { T }_{ 2 } } \right] ^{ -2 }\)
\(\\ =4.2\left[ \frac { 1 }{ \alpha } \right] kg\times \left[ \frac { 1 }{ \beta } m \right] ^{ 2 }\left[ \frac { 1 }{ \gamma } s \right] ^{ -2 }\)
\(\\ n_{ 2 }=4.2{ \alpha }^{ -1 }{ \beta }^{ -2 }{ \gamma }^{ 2 } \text{ new unit}\)
\(\therefore \ 1 \ cal=4.2{ \alpha }^{ -1 }{ \beta }^{ -2 }{ \gamma }^{ 2 } \text{ new unit}\)

Suppose period of oscillation T depends on radius of star R, mean density of fluid p and universal gravitational constant (G) as
T = kR^ap^bG^c, where k is a dimensionless constant
[M⁰L⁰T¹] = [L]^a[ML^-3]^b[M^-1L³T^-2]^c
= M^b-cL^a-3b+3c T^-2c
Comparing powers of M,L and T, we have
b - c = 0
a - 3b + 3c = 0 and - 2c-=1
On simplifying these equations, we get
c = -1/2, b = -1/2, a = 0
Thus, we have T = kp^-1/2G^-1/2 =\(\frac { k }{ \sqrt { \rho G } } \)     

Here, \({ R }_{ 1 }=\left( 100\pm 3 \right) \Omega \), \({ R }_{ 2 }=\left( 200\pm 4 \right) \Omega \) 
Parallel combination
 \(\frac { 1 }{ { R }^{ ' } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } \) \(=\frac { 1 }{ 100 } +\frac { 1 }{ 200 } =\frac { 3 }{ 200 } \) 
 \({ R }^{ ' }=\frac { 200 }{ 3 } =66.7\Omega \) 
 \(\frac { { \triangle R }^{ ' } }{ { R }^{ { '2 } } } =\frac { \triangle { R }_{ 1 } }{ { R }_{ 1 }^{ 2 } } +\frac { \triangle { { R }_{ 2 } } }{ { R }_{ 2 }^{ 2 } } \)  
\({ \triangle R }^{ ' }=\triangle { R }_{ 1 }{ \left( \frac { { R }^{ ' } }{ { R }_{ 1 } } \right) }^{ 2 }+\triangle { { R }_{ 2 } }{ \left( \frac { { R }^{ ' } }{ { R }_{ 2 } } \right) }^{ 2 }\)
\(=3{ \left( \frac { 200 }{ 3\times 100 } \right) }^{ 2 }+4{ \left( \frac { 200 }{ 3\times 200 } \right) }^{ 2 }=1.8\Omega \)
\(\\ Hence,\quad { R }^{ ' }=\left( 66.7\pm 1.8 \right) \Omega \)

Given, Mass of the Sun =\(2.0\times { 10 }^{ 30 }kg\) 
 Radius of the Sun =\(7.0\times { 10 }^{ 8 }m\)
Density of the Sun = \(\frac { Mass \ of \ the \ Sun \ (M) }{ Volume \ of \ the\ Sun \ (S) } \)  
\(\left[ \therefore \text{Density}=\frac { Mass }{ Volume } \right] \)  
  \(\rho =\frac { M }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } =\frac { 3 }{ 4 } \frac { M }{ \pi { R }^{ 3 } }\)     
\( \\ =\frac { 3\times 2.0\times { 10 }^{ 30 } }{ 4\times 3.14\times { (7.0\times { 10 }^{ 8 }) }^{ 3 } } \)
\(\\ =\frac { 3\times { 10 }^{ 30 } }{ 6.28\times 343\times { 10 }^{ 24 } }\)
\( \\ =1.392\times { 10 }^{ 3 }\)
\( \rho \approx 1.4\times { 10 }^{ 3 }kg/{ m }^{ 3 }\)
This density is the order of density of solids and liquids and not of gases.
The temperature of inner core of the sun is 10⁷ K while the temperature of the outer layers is nearly 6000 K. At so high temperature, no matter can be exist in its solid or liquid state.
Every matter is highly ionised and present as a mixture of nucleus, free electr4onbs and ions which is called plasma. The density of plasma is so high due to inward gravitational attraction on outer layers due o inner layers of the sun.