(i) Accleration = \(\frac{velocity}{time} \) \(\therefore\) [Acceleration] = \(\frac{Velocity}{Time}\) = \(\frac { L{ T }^{ -1 } }{ T } =L{ T }^{ -2 }\)
(ii) Angle = \(\frac{Distance}{Distance}\)  \(\therefore\) [angle is dimensionless.
(iii) Density=\(\frac{Mass}{Volume}\) \(\therefore\) Density = \(\frac{Mass}{Volume}\) = \(\frac { M }{ { L }^{ 3 } } =M{ L }^{ 3 }\)
(iv) Kinetic energy = \(\frac{1}{2}\) Mass x Velocity²
 [Kinetic energy] = [Mass] x [Velocity]² = ML²T^-2
(v) Constant of gravitation occurs in Newton's law of gravitation
\(F=G\frac { { m }_{ 1 }{ m }_{ 2 } }{ { d }^{ 2 } }\)
\( \\ \therefore [G]=\frac { [F][{ d] }^{ 2 } }{ [{ m }_{ 1 }][{ m }_{ 2 }] } =\frac { ML{ T }^{ -2 } }{ MM } ={ M }^{ -1 }{ L }^{ 3 }{ T }^{ -2 }\)
(vi) Permeability occurs in Ampere's law of force
\(\triangle F=\mu \frac { ({ i }_{ 1 }\triangle { l }_{ 1 })({ i }_{ 2 }\triangle { l }_{ 2 })sin\theta }{ { r }^{ 2 } }\)
\( \\ [\mu ]=\frac { [\triangle F][{ r }^{ 2 }] }{ [{ i }_{ 1 }\triangle { l }_{ 1 }][{ i }_{ 2 }\triangle { l }_{ 2 }] } =\frac { ML{ T }^{ -2 } }{ AL.AL } =ML{ T }^{ -2 }{ A }^{ -2 }\)

(i) [R.H.S] = \(\sqrt { \frac { [P] }{ D] } } \)
\(=\sqrt { \frac { M{ L }^{ -1 }{ T }^{ -2 } }{ M{ L }^{ -3 } } } =L{ T }^{ -1 }\)
[L.H.s] = [v]=LT^-1
[R.H.S] = [L.H.S]
Hence, the relation is correct.
(ii) \([R.H.S]=\frac { 1 }{ [l] } \sqrt { \frac { [F] }{ M] } } \)
\(=\frac { 1 }{ L } \sqrt { \frac { ML{ T }^{ -2 } }{ M{ L }^{ -1 } } } =\frac { 1 }{ L } L{ T }^{ -1 }={ T }^{ -1 }\)
[L.H.S] = \(\frac{1}{time}=\frac{1}{T}\) = T^-1

Mass = Volume x density
Volume of earth = \(\frac { 4 }{ 3 } \pi { R }^{ 3 }\)
\(=\frac { 4 }{ 3 } \times 3.142\times (6.37\times { 10 }^{ 6 }){ m }^{ 3 }\)
Mass of earth\(=\frac { 4 }{ 3 } \times 3.142\times (6.37\times { 10 }^{ 6 }){ m }^{ 3 }\)
= 5974.01 x 10²¹ kg = 5.97401 x 10²⁴ kg
The radius has three significant figures and the density has four. Therefore, the final result should be rounded upto three significant figures. Hence, mass of the earth = 5.97 x 10²⁴ kg.

All the terms should have the same dimension
\([a]=\left[ \frac { x }{ { T }^{ 2 } } \right] =[L{ T }^{ -2 }]\)
\(\\ [b]=\left[ \frac { x }{ t } \right] =[L{ T }^{ -1 }]\)
\(\\ [c]=[x]=[L]\)

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

Dimensions of K.E. \(=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\)
Dimensions of \(\frac{3}{16} m v^2=[\mathrm{M}]\left[\mathrm{LT}^{-1}\right]^2=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\)
Dimensions of \(\frac{1}{2} m v^2+m a=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]+\left[\mathrm{MLT}^{-2}\right]\)
So, (ii) formula is incorrect as its dimensions are different from the dimensions of kinetic energy.

The correct option is A T=\(k \sqrt{ρr^3/S}\)
Dimensional formula of,
Time period, T
[T]=[M⁰L⁰T¹]
Surface tension, S
[S]=[ML⁰T⁻²]
Radius, r
[r]=[M⁰LT⁰]
Density, ρ
[ρ]=[M¹L⁻³T⁰]
Let us suppose the relation is
T=kρ^ar^bS^c
⇒[T]=[kρ^ar^bS^c]
⇒[M⁰L⁰T¹]=[M¹L⁻³T⁰]^a × [M⁰LT⁰]^b × [ML⁰T⁻²]^c
⇒[M⁰L⁰T¹]=[M^(a+c)L^(−3a+b)T^(−2c)]
On comparing, we get,
a+c=0(1)
−3a+b=0(2)
−2c=1(3)
On solving these equations, we get,
a=1/2,b=3/2 and c=−1/2
Thus,
T=kρ^ar^bS^c ⇒\(T=\frac { { pr }^{ 3 } }{ s } \)

Let, \(v \propto \rho^a\)
\(k\propto E^b\)
Combining the equations (i) and (ii), we get
\(v=k \rho^a E^b\)
where k is a dimensionless constant of proportionally.
Writing the dimensional formula of \(\mathrm{v}, \rho\) and E in the equation (iii), we get
\(\left[L T^{-1}\right]=\left[M L^{-3}\right]^a\left[M L^{-1} T^{-2}\right]^b\)
or \(\left[M^0 L T^{-1}\right]=\left[M^{a+b} L^{-3 a-b} T^{-2 b}\right]\)
Comparing the dimensions of M,L and T on the two sides, we get
a+b=0
-3 a-b=1
and -2 b=-1 or b=1 / 2
Substituting for b in the equation (iv), we get
a+1 / 2=0 or a=-1 / 2
In the equation (iii), putting the values of a and b, we get
\(v=k \rho^{-1 / 2} E^{1 / 2} \text { or } v=k \sqrt{\frac{E}{\rho}}\)
The constant of proportionally k is found to be 1. Therefore the above relation becomes
\(v=k\sqrt { \frac { E }{ \rho } } \)

\(75.5\times 125.2\times 0.51=4820.826=4800\)
[rounded off upto two significant figures]

Given, mass of the box (m) = 2.3 Kg
mass  of first gold piece (m₁ ) = 20.15 g = 0.02015 Kg
Mass of second gold piece (m₂) = 20.17 g = 0.02017 Kg
Difference in masses of gold pieces ( \(\triangle m\) ) = m₂ - m₁
= 20.17 - 20.15 = 0.02 g
(The masses of two gold pieces has two decimal places, therefore, it is correct up to two places of decimal)

Given, Length (l) = 4.234 m
Breadth (b) = 1.005 m
Thickness (t) = 2.01 cm = 0.0201 m
Area of sheet (A) \(=2(l\times b+b\times t+t\times l)\) 
\(=2\left[ \left( 4.234\times 1.005 \right) +\left( 1.005\times 0.0201 \right) +\left( 0.0201\times 4.234 \right) \right] \)  
\(\\ =2\times 4.3604739\)
\(=8.7209578{ m }^{ 2 }\)
As, thickness has least number of significant figures 3, therefore, rounding off area up to the three significant figures, we get
Area of sheet (A) = 8.72m^-2
Volume of sheet (V) = \(l\times b\times h\) 
= \(4.234\times 1.005\times 0.0201\) 
= 0.0855289
Rounding off up to three significant figures, we get Volume of the sheet = 0.0855m³.

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, l = \((10.5\pm 0.2)\)cm, b = \((5.2\pm 0.1)\)cm
Perimeter of a rectangle, p = 2(l + b)
= 2(10.5 + 5.2) = 31.4 cm
 \(\triangle p=\pm 2(\triangle l+\triangle b)\) 
\(=\pm 2(0.2+0.1)=\pm 0.6\)
Perimeter of a rectangle \(=(31.4\pm 0.6)\)