Past NEET Papers

Hint

In steady-state, the amount of heat flowing from one face to the other face in time \(d t\) is given by
\(
\begin{aligned}
&\mathrm{H}=\frac{\mathrm{kA}\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right) \mathrm{dt}}{\ell} \\
&\Rightarrow \frac{\mathrm{dH}}{\mathrm{dt}}=\frac{\mathrm{kA}}{\ell} \Delta \mathrm{T}(k=\text { coefficient of thermal conductivity})
\end{aligned}
\)
\(
\therefore \mathrm{k}=\frac{\ell \mathrm{dH}}{\mathrm{A} \text { dt } \Delta \mathrm{T}}
\)
Unit of \(k=\mathrm{Wm}^{-1} \mathrm{~K}^{-1}\)

Hint

In CGS system, density \(d=4 \frac{\mathrm{g}}{\mathrm{cm}^{3}}\)
unit of length \(=1 \mathrm{~cm}\)
unit of mass \(=1 \mathrm{~g}\)
The unit of mass is \(100 \mathrm{~g}\) and unit of length is 10 \(\mathrm{cm}\), so substitute these values
\(
\begin{aligned}
\text { Density } &=\frac{4\left(\frac{100 \mathrm{~g}}{100}\right)}{\left(\frac{10}{10} \mathrm{~cm}\right)^{3}}=\frac{\left(\frac{4}{100}\right)}{\left(\frac{1}{10}\right)^{3}} \frac{(100 \mathrm{~g})}{(10 \mathrm{~cm})^{3}} \\
=\frac{4}{100} \times(10)^{3} \cdot \frac{100 \mathrm{~g}}{(10 \mathrm{~cm})^{3}}=40 \text { unit }
\end{aligned}
\)

Hint

\(\varepsilon_{o}=\frac{q^{2}}{\left(r^{2}\right) 4 \pi F}\)
\(\Rightarrow\) unit of \(\varepsilon_{0}\) is (coulomb) \({ }^{2} /\) newton-metre \({ }^{2}\)

Hint

According to Stefan’s law, \(\mathrm{E}=\sigma \mathrm{AT}^{4}\) where, \(E\) is energy dissipated per second, \(\mathrm{A}=\) surface area
\(\mathrm{T}=\) absolute temperature
\(
\sigma=\frac{E}{A T^{4}}=\frac{W}{\mathrm{~m}^{2} K^{4}}
\)

Hint

As we know force \(=\) Mass \(\times\) Acceleration \(=\) Mass \(\times\) length \(\times\) time\(^{-2}=(10 \mathrm{~g})(10 \mathrm{~cm})(0.1 \mathrm{~s})^{-2}\) \(=\left(10^{-2} \mathrm{~kg}\right)\left(10^{-1} \mathrm{~m}\right)\left(10^{-1} \mathrm{~s}\right)^{-2}=10^{-1} \mathrm{~N}\).

Hint

Stress \(=\frac{\text { Force }}{\text { Area }}\)
Dimensions of force \(=\left[\mathrm{MLT}^{-2}\right]\)
Dimensions of area \(=\left[\mathrm{L}^{2}\right]\)
\(\therefore\) Stress \(=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^{2}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\)

Hint

As \(x=a t+b t^{2}\)
According to the concept of dimensional analysis and principle of homogeneity \(\therefore \quad\) unit of \(x=\) unit of \(b t^{2}\)
\(\therefore \quad\) unit of \(b=\frac{\text { unit of } x}{\text { unit of } t^{2}}=\mathrm{km} / \mathrm{s}^{2}\)

Hint

As force \(F=\frac{e^{2}}{4 \pi \varepsilon_{0} r^{2}} \Rightarrow \frac{e^{2}}{4 \pi \varepsilon_{0}}=r^{2} \cdot F\)
Putting dimensions of \(r\) and \(F\), we get,
\(\Rightarrow\left[\frac{e^{2}}{4 \pi \varepsilon_{0}}\right]=\left[M L^{3} T^{-2}\right]\)
\(\begin{aligned}
&\text { Also, force, } F=\frac{G m^{2}}{r^{2}} \\
&\Rightarrow \quad[G]=\frac{\left[M L T^{-2}\right]\left[L^{2}\right]}{\left[M^{2}\right]} \\
&\Rightarrow \quad\left[\frac{1}{c^{2}}\right]=\frac{1}{\left[L^{2} T^{-2}\right]}=\left[L^{-2} T^{2}\right] \\
&\text { and } \quad \\
&\text { Now, checking optionwise} \\
&=\frac{1}{c^{2}}\left(\frac{G e^{2}}{4 \pi \varepsilon_{0}}\right)^{1 / 2}=\left[L^{-2} T^{2}\right]\left[L^{6} T^{-4}\right]^{1 / 2}=[L]
\end{aligned}\)

Hint

We know that
Surface tension \((S)=\frac{\text { Force }[F]}{\text { Length }[L]}\)
So, \(\quad[S]=\frac{\left[M L T^{-2}\right]}{[L]}=\left[M L^{0} T^{-2}\right]\)
Energy \((E)=\) Force \(\times\) displacement
\(\Rightarrow \quad[E]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]\)
Velocity \((v)=\frac{\text { displacement }}{\text { time }}\)
\(\Rightarrow \quad[\mathrm{V}]=\left[\mathrm{LT}^{-1}\right]\)
As, \(\quad S \propto E^{a} v^{b} T^{c}\)
where, \(a, b, c\) are constants.
From the principle of homogeneity,
\(
\begin{gathered}
{[\mathrm{LHS}]=[\mathrm{RHS}]} \\
\Rightarrow \quad\left[M L^{0} \mathrm{~T}^{-2}\right]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]^{a}\left[\mathrm{LT}^{-1}\right]^{b}\left[\mathrm{~T}^{c}\right. \\
\Rightarrow \quad\left[M L^{0} \mathrm{~T}^{-2}\right]=\left[\mathrm{M}^{a} \mathrm{~L}^{2 a+b} \mathrm{~T}^{-2 a-b+c}\right]
\end{gathered}
\)
Equating the power on both sides, we get
\(
\begin{array}{ll}
& a=1,2 a+b=0, b=-2 \\
\Rightarrow & -2 a-b+c=-2 \\
\Rightarrow & c=(2 a+b)-2=0-2=-2 \\
\text { So } & {[S]=\left[E v^{-2} T^{-2}\right]=\left[E v^{-2} T^{-2}\right]}
\end{array}
\)

Hint

Key Concept According to the principle of homogeneity of dimension states that a physical quantity equation will be dimensionally correct if the dimensions of all the terms occurring on both sides of the equations are the same. Given the critical velocity of the liquid flowing through a tube are expressed as
\(
v_{c} \propto \eta^{n} \rho^{y} r^{z}
\)
Coefficient of viscosity of liquid,
\(
\eta=\left[M L^{-1} \mathrm{~T}^{-1}\right]
\)
Density of liquid, \(\rho=\left[\mathrm{ML}^{-3}\right]\)
Radius of a tube \(r=[L]\)
Critical velocity of liquid \(v_{c}=\left[M L^{0} \mathrm{~T}^{-1}\right]\)
\(\Rightarrow\left[M^{0} L^{1} T^{-1}\right]=\left[M L^{-1} T^{-1}\right]^{x}\left[M L^{-3}\right]^{y}[L]^{z}\)
\(\left[M^{0} L^{1} T^{-1}\right]=\left[M^{x+y} L^{-x-3 y+z} T^{-x}\right]\)
Comparing exponents of \(M, L\) and \(L\), we get
\(
\begin{gathered}
x+y=0,-x-3 y+z=1,-x=-1 \\
\Rightarrow \quad z=-1, x=1, y=-1
\end{gathered}
\)

Hint

We know that
\(F=m a\)
\(\Rightarrow F=\frac{m v}{t} \Rightarrow m=\frac{F t}{v}\)
\([M]=\frac{[F][T]}{[v]}=\left[\mathrm{Fv}^{-1} \mathrm{~T}\right]\)

Hint

Work \(=\) Force \(\times\) displacement
\(
\begin{aligned}
&=\left[\mathrm{MLT}^{-2}\right][\mathrm{L}] \\
&=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right] \\
&\text { Torque }=\text { Force } \times \text { force arm } \\
&\quad=\text { mass } \times \text { acceleration } \times \text { length } \\
&\quad=[\mathrm{M}] \times\left[\mathrm{LT}^{-2}\right] \times[\mathrm{L}]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]
\end{aligned}
\)

Hint

\(\left(\mu_{0} \varepsilon_{0}\right)^{-1 / 2}=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}} \Rightarrow c=\) speed of light where \(\varepsilon_{0}=\) permittivity of free space \(\mu_{0}=\) permeability of free space So dimensions of \(\left(\mu_{0} \varepsilon_{0}\right)^{-1 / 2}\) will be \(\left[\mathrm{LT}^{-1}\right]\)

Hint

As we know that,
Dimension of \(\varepsilon_{0}=\left[M^{-1} L^{-3} T^{4} A^{2}\right]\)
Dimension of \(E=\left[M L T^{-3} A^{-1}\right]\)
So, dimension of
\(
\begin{aligned}
\frac{1}{2} \varepsilon_{0} E^{2} &=\left[M^{-1} L^{-3} T^{4} A^{2}\right] \times\left[M L T^{-3} A^{-1}\right]^{2} \\
&=\left[M L^{-1} T^{-2}\right]
\end{aligned}
\)

Hint

\(\begin{aligned}
&\text { (d) Pressure }=\frac{\text { Force }}{\text { Area }} \\
&\Rightarrow \frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^{2}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] \\
&\Rightarrow a=1, b=-1, c=-2 .
\end{aligned}\)

Hint

Energy density \(=\frac{\text { Energy }}{\text { Volume }}\) \(\Rightarrow \frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^{3}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\)
Refractive index and dielectric constant have no dimensions.
Young’s Modulus \(=\frac{F}{A} \times \frac{l}{\Delta l}\)
\(\Rightarrow \frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^{2}[\right.} \cdot \frac{[\mathrm{L}]}{[\mathrm{L}]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\)
Magnetic field, \(B=\frac{F}{i \ell}=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{A}][\mathrm{L}]}\) \(=\left[\mathrm{MT}^{-2} \mathrm{~A}^{-1}\right]\).

Hint

As we know that
\(
\mathrm{R}=\frac{[\mathrm{V}]}{[\mathrm{I}]}=\frac{w}{q \cdot i}=\frac{w}{i \cdot t \cdot i}
\)
Dimension of Resistance
\(
=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{I}^{-1}\right]}{[\mathrm{T}][\mathrm{I}]}=\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{I}^{-2}\right]
\)

Hint

Dimension of a.t = dimension of velocity
\(
a . t=\left[\mathrm{LT}^{-1}\right] \Rightarrow\left[a=\mathrm{LT}^{-2}\right]
\)
Dimension of \(c=\) dimension of \(t\)
(two physical quantity of same dimension can only be added)
So, dimension of \(c=[T]\)
Dimension of \(\frac{b}{t+c}=\) Dimension of velocity
\(
\begin{aligned}
&\frac{b}{\mathrm{~T}+\mathrm{T}}=\left[\mathrm{LT}^{-1}\right]\left[\mathrm{LT}^{-1}\right] \Rightarrow\left[b \cdot \mathrm{T}^{-1}\right]=\left[\mathrm{LT}^{-1}\right] \\
&\Rightarrow b=[\mathrm{L}]
\end{aligned}
\)
So, answer is \(\left[\mathrm{LT}^{-2}\right],[\mathrm{L}]\) and [T]

Hint

Dimension formula for the planck’s constant, \(h=\left[\mathrm{ML}^{2} \mathrm{~T}^{1}\right]\)
Dimension formula for the moment of inertia, \(\mathrm{I}=\left[\mathrm{ML}^{2}\right]\)
So, the ratio between the plank’s constant and moment of inertia is
\(
\begin{aligned}
&\Rightarrow \frac{h}{I}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]}{\left[\mathrm{ML}^{2}\right]} \Rightarrow\left[\mathrm{T}^{-1}\right] \\
&\Rightarrow \frac{\mathrm{h}}{\mathrm{I}}=\left[\mathrm{T}^{-1}\right] \Rightarrow \text { dimension of frequency }
\end{aligned}
\)

Hint

\(\begin{aligned}
&\text { (b) } F=\frac{G M_{1} m_{2}}{r^{2}} \Rightarrow G=\frac{F r^{2}}{M_{1} m_{2}} \\
&\therefore \text { dimension of } \mathrm{G} \text { is } \frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^{2}\right]}{[\mathrm{M}][\mathrm{M}]} \\
&\quad=\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]
\end{aligned}\)

Hint

\(\begin{aligned}
E &=h v \\
\Rightarrow & h=\text { Planck’s constant }=\frac{\text { Energy }(E)}{\text { frequency }(v)} \\
\therefore & {[h]=\frac{E}{v}=\frac{\left[M L^{2} T^{-2}\right]}{\left[T^{-1}\right]}=\left[M L^{2} T^{-1}\right] }
\end{aligned}\)

Angular momentum
= Moment of inertia \(\times\) angular velocity
or \(L=1 \times \omega=m r^{2} \omega \quad\left[\because l=m r^{2}\right]\)
\(\therefore \quad[L]=[M]\left[L^{2}\right]\left[\mathrm{T}^{-1}\right]=\left[M L^{2} \mathrm{~T}^{-1}\right]\)

Hint

Force has dimension [MLT \({ }^{-2}\) ] while impulse has dimension \(\left[\mathrm{MLT}^{-1}\right]\), both have different dimensions.

Hint

Dimension of magnetic flux
= Dimension of magnetic field \(\times\) Dimension of area
\(
\left[\mathrm{ML}^{0} \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]\left[\mathrm{L}^{2}\right]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]
\)

Hint

The viscous force on a sphere of radius r is

\(\mathrm{F}=6 \pi \eta a v\)
\(
\eta=\frac{F}{6 \pi a v}=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{L}]\left[\mathrm{LT}^{-1}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]
\)

Hint

According to the principle of homogeneity quantity with same dimension can be added or subtracted.
Hence, Dimension of \(\mathrm{P}=\) Dimension of \(\frac{a}{V^{2}}\)
\(
\begin{aligned}
&\Rightarrow \text { Dimension of } \frac{\text { Force }}{\text { Area }}=\text { Dimension of } \frac{a}{V^{2}} \\
&\Rightarrow\left[\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^{2}}\right]=\frac{\mathrm{a}}{\left[\mathrm{L}^{3}\right]^{2}} \Rightarrow a=\left[\mathrm{ML}^{5} \mathrm{~T}^{-2}\right]
\end{aligned}
\)

Hint

Given, the main scale reading, MSR \(=0\)
The circular scale reading, \(\mathrm{CSR}=52\) divisions
Now, we shall determine the least count of the screw gauge,
\(
\mathrm{LC}=\frac{p}{n}
\)
Here, \(p\) is the pitch of the screw, \(n\) is the number of circular divisions in one complete revolution.
\(
\begin{aligned}
&\qquad L C=\frac{1}{100} \mathrm{~mm} \\
&\Rightarrow L C=0.01 \mathrm{~mm} \\
&\Rightarrow L C=0.001 \mathrm{~cm} \\
&\text { Thus, the least count of the screw gauge } \\
&\text { is } 0.001 \mathrm{~cm} \text {. } \\
&\text { Therefore, diameter of the wire of screw } \\
&\text { gauge, } \\
&\qquad D=M S R+(C S R \times L C) \\
&\Rightarrow D=0+(52 \times 0.001) \\
&\Rightarrow D=0.052 \mathrm{~cm}
\end{aligned}
\)

Hint

Given, fundamental physical quantities are force \([F]\), acceleration \([a]\) and time \([T]\).
Now, we shall determine the dimensions of the energy.
Energy depends on force, acceleration and time as,
\(
\begin{aligned}
{[E] } &=[F]^{a}[a]^{b}[T]^{c} \\
\Rightarrow\left[M L^{2} T^{-2}\right] &=\left[M L T^{-2}\right]^{a}\left[L T^{-2}\right]^{b}[T]^{c} \\
\Rightarrow\left[M L^{2} T^{-2}\right] &=[M]^{a}[L]^{a+b}[T]^{-2 a-2 b+c}
\end{aligned}
\)
Comparing the powers of \(M, L\) and \(T\) on both sides, we get
\(
\begin{aligned}
&a=1, a+b=2 \\
&\text { and }-2 a-2 b+c=-2 \\
&\Rightarrow 1+b=2 \Rightarrow b=1 \\
&\Rightarrow-2(1)-2(1)+c=-2 \Rightarrow c=2
\end{aligned}
\)
The dimensions of the energy are \(\left[F^{1}\right][a][T]^{2}\).

Hint

The dimensions of energy
\(
\begin{gathered}
{[E]=[F] \cdot[d]} \\
\Rightarrow[E]=\left[M L T^{-2}\right][L] \Rightarrow[E]=\left[M L^{2} T^{-2}\right]
\end{gathered}
\)
As we know that, the expression of gravitational force,
\(
\begin{gathered}
F=\frac{G M_{1} M_{2}}{r^{2}} \Rightarrow G=\frac{F r^{2}}{M_{1} M_{2}} \\
\therefore[G]=\frac{[F]\left[r^{2}\right]}{\left[M_{1}\right]\left[M_{2}\right]} \Rightarrow[G]=\frac{\left[M L T^{-2}\right][L]^{2}}{[M][M]} \\
\Rightarrow[G]=\left[M^{-1} L^{3} T^{-2}\right]
\end{gathered}
\)
The dimensions of
\(
\frac{E}{G}=\frac{\left[M L^{2} T^{-2}\right]}{\left[M^{-1} L^{3} T^{-2}\right]} \Rightarrow\left[\frac{E}{G}\right]=\left[M^{2} L^{-1} T^{0}\right]
\)

Hint

\(\frac{L}{R}\) is the time constant of \(R-L\) circuit so, the dimensions of \(\frac{L}{R}\) is same as that of time.
Alternative
\(\frac{\text { Dimensions of } L}{\text { Dimensions of } R}=\frac{\left[M L^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]}{\left[M L^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]}=[\mathrm{T}]\)

Hint

Therefore, gravitational constant \((G)\) is a dimensional constant. Value of \(\mathrm{G}=6.67 \times 10^{-}\) \({ }^{11} \mathrm{~m}^{2} / \mathrm{kg} / \mathrm{sec}^{2}\)
dimension of \(G=\left[M^{-1} L^{3} \mathrm{~T}^{-2}\right]\)

Hint

In \(p=p_{0} \exp \left(-\alpha t^{2}\right)\), where \(\alpha t^{2}\) where is dimensionless
\(
\therefore \alpha=\frac{1}{t^{2}}=\frac{1}{\left[T^{2}\right]}=\left[T^{-2}\right]
\)

Hint

Pressure
\(
(p)=\frac{\text { Force }}{\text { Area }}=\frac{\left[M L T^{-2}\right]}{\left[L^{2}\right]}=\left[M L^{-1} T^{-2}\right]
\)
Velocity, \(\quad v=\left[\mathrm{LT}^{-1}\right]\)
From principle of homogeneity, the dimensions of \(r^{2}\) and \(x^{2}\) are same.
So, the dimensions of viscosity,
\(
\eta=\frac{\left[M L^{-1} T^{-2}\right]\left[L^{2}\right]}{\left[L T^{-1}\right][L]}=\left[M L^{-1} T^{-1}\right]
\)

Hint

Let the expression, \(\alpha=P^{x} S^{y} c^{z} \ldots\) (i) and given that dimension of \(\alpha=\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right] \ldots\) (ii) \(=\) dimensionless

Dimension fo radiation pressure \(P=\frac{\text { Force }}{\text { Area }}\)
\(
=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^{2}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]
\)
Dimension of radiation energy/unit area unit time
\(
S=\frac{\text { Energy }}{\text { Area } \times \text { Time }}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^{2}\right][\mathrm{T}]}=\left[\mathrm{MT}^{-3}\right]
\)
Dimension of speed of light, \(c=\left[\mathrm{LT}^{-1}\right]\)
By equation (i) we get,
So, the dimension of \(\alpha=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]^{x}\left[\mathrm{MT}^{-3}\right\}^{y}\)
\(\left[\mathrm{LT}^{-1}\right]^{z}\)
According to equation (ii),
\(\Rightarrow\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]^{x}\left[\mathrm{MT}^{-3}\right]^{y}\left[\mathrm{LT}^{-1}\right]^{z}\)
\(\Rightarrow\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]=\left[\mathrm{M}^{x+y} \mathrm{~L}^{-x+z} \mathrm{~T}^{-2 x-3 y-z}\right]\)
Applying the principle of homegenity of dimension we get,
\(x+y=0\)
\(-x+z=0\)
\(-2 x-3 y-z=0 \quad\)…(v)
After solving above three equation we get, \(x=1 ; y=-1 ; z=1\)

Hint

From Biot-Savart law
\(
\begin{aligned}
d B &=\frac{\mu_{0}}{4 \pi} \frac{I d l \sin \theta}{r^{2}} \\
|d| &=\text { current element } \\
r &=\text { displacement vector } \\
\mu_{0} &=\frac{4 \pi r^{2}(d B)}{|d| \sin \theta}=\frac{\left[L^{2}\right]\left[M T^{-2} A^{-1}\right]}{[A][L]} \\
&=\left[M L T^{-2} A^{-2}\right]
\end{aligned}
\)

Hint

\(\begin{aligned}
&\text { As F }=-\eta A \frac{d v}{d z} \Rightarrow \eta=-\frac{F}{A\left(\frac{d v}{d z}\right)} \\
&\text { As } \quad F=\left[M L T^{-2}\right], A=\left[L^{2}\right] \\
&d v=\left[L T^{-1}\right], d z=[L] \\
&\therefore \quad \eta=\frac{\left[M L T^{-2}\right][L]}{\left[L^{2}\right]\left[L T^{-1}\right]}=\left[M L^{-1} T^{-1}\right]
\end{aligned}\)

Hint

\(f=c m^{x} k^{y}\);
Spring constant \(k=\) force/length.
\(
\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1}\right]=\left[\mathrm{M}^{x}\right]\left[\mathrm{MT}^{-2} ]^{y}=\left[\mathrm{M}^{x+y} \mathrm{~T}^{-2 y}\right]\right.
\)
\(\Rightarrow x+y=0,-2 y=-1\) or \(y=\frac{1}{2}\)
Therefore, \(x=-\frac{1}{2}\)

Hint

\(
\begin{aligned}
\text { Pressure } &=\frac{\text { Force }}{\text { Area }}=\frac{F}{A}=\frac{\left[M L T^{-2}\right]}{\left[L^{2}\right]} \\
&=\left[M L^{-1} T^{-2}\right]
\end{aligned}
\)

Hint

Torque \(\tau=\) Force \(\times\) distance
So dimensional formula,
\(
=\left[\mathrm{MLT}^{-2}\right][\mathrm{L}]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]
\)

Hint

\(L=\varepsilon\left(\frac{d t}{d I}\right)=\frac{W}{q}\left[\frac{d i}{d t}\right]=\frac{W}{i \cdot t}\left[\frac{d i}{d t}\right]\)
or, \([\mathrm{L}]=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right][\mathrm{T}]}{[\mathrm{AT}][\mathrm{A}]}=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]\)

Hint

For angular momentum, the dimensional formula is \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]\). For other three, it is \(\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\)

Hint

Angular momentum
\(=\) Momentum of inertia \(\times\) Angular velocity
So dimensional formula,
\(
\begin{aligned}
&=\left[\mathrm{ML}^{2}\right] \times\left[\mathrm{T}^{-1}\right] \\
&=\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]
\end{aligned}
\)

Hint

\(
\begin{aligned}
\mathrm{CR} &=\left(\frac{q}{V}\right)\left(\frac{V}{i}\right) \Rightarrow\left(\frac{i \cdot t}{i}\right) \Rightarrow t=\text { time } \\
&=[\mathrm{T}]=\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1}\right]
\end{aligned}
\)
\(\mathrm{RC}\) is the time constant of the circuit.

Hint

The difference between \(9.99 \mathrm{~m}\) and \(0.0099 \mathrm{~m}\) is

\(
=9.99-0.0099=9.9801 \mathrm{~m}
\)
Taking significant figures into account, as both the values have two significant figures after the decimal. So, their difference will also have two significant figures after the decimal, i.e. \(9.98 \mathrm{~m}\).

Hint

Least count of screw gauge \(=0.01 \mathrm{~mm}\) Least count
\(
\begin{aligned}
&=\frac{\text { Pitch }}{\text { No. of divisions on circular scale }} \\
&\Rightarrow 0.01 \mathrm{~mm}=\frac{\text { Pitch }}{50} \\
&\Rightarrow \text { Pitch }=0.5 \mathrm{~mm}
\end{aligned}
\)

Hint

Given, \(x=\frac{A^{2} B^{1 / 2}}{C^{1 / 3} D^{3}}\)
\(\%\) error, \(\frac{\Delta \mathrm{x}}{\mathrm{x}} \times 100=2 \frac{\Delta \mathrm{A}}{\mathrm{A}} \times 100+\frac{1}{2} \frac{\Delta \mathrm{B}}{\mathrm{B}} \times\)
\(
\begin{aligned}
& 100+\frac{1}{3} \frac{\Delta \mathrm{C}}{\mathrm{C}} \times 100+3 \frac{\Delta \mathrm{D}}{\mathrm{D}} \times 100 \\
=& 2 \times 1 \%+\frac{1}{2} \times 2 \%+\frac{1}{3} \times 3 \%+3 \times 4 \% \\
=& 2 \%+1 \%+1 \%+12 \%=16 \%
\end{aligned}
\)

Hint

\(\begin{aligned}
&\text { (d) } n \mathrm{VSD}=(n-1) \mathrm{MSD} \\
&1 \mathrm{VSD}=\frac{(n-1)}{n} \mathrm{MSD} \\
&\text { L.C. }=1 \mathrm{MSD}-1 \mathrm{VSD}=1 \mathrm{MSD}-\frac{(n-1)}{n} \mathrm{MSD} \\
&=\frac{1}{n} \mathrm{MSD}=\frac{1}{n} \times \frac{1}{n}=\frac{1}{n^{2}}
\end{aligned}\)

Hint

Diameter of the ball
\(
\begin{aligned}
&=\mathrm{MSR}+\operatorname{CSR} \times(\text { least count })-\text { zero error } \\
&=0.5 \mathrm{~cm}+25 \times 0.001-(-0.004) \\
&=0.5+0.025+0.004=0.529 \mathrm{~cm}
\end{aligned}
\)

Hint

Given, \(\mathrm{P}=\frac{a^{3} b^{2}}{c d}\)
Therefore, \(\frac{\Delta \mathrm{P}}{\mathrm{P}} \times 100 \%=3 \frac{\Delta a}{a} \times 100 \%\)
\(
\begin{aligned}
&+2 \frac{\Delta b}{b} \times 100 \%+\frac{\Delta c}{c} \times 100 \%+\frac{\Delta d}{d} \times 100 \% \\
&=3 \times 1 \%+2 \times 2 \%+3 \%+4 \%=14 \%
\end{aligned}
\)

Hint

Volume of a sphere, \(V=\frac{4}{3} \pi r^{3}\)
\(
\therefore \quad \frac{\Delta V}{V} \times 100=\frac{3 \times \Delta r}{r} \times 100
\)
Here \(\frac{\Delta r}{r} \times 100=2 \%\)
\(
\therefore \quad \frac{\Delta V}{V} \times 100=3 \times 2 \%=6 \%
\)

Hint

As we know, density \(=\frac{\text { mass }}{\text { volume }}\)
Maximum error in the measurement of density \(\rho=\frac{M}{L^{3}}\)
\(
\therefore \quad \frac{\Delta \rho}{\rho}=\frac{\Delta \mathrm{M}}{\mathrm{M}}+3 \frac{\Delta \mathrm{L}}{\mathrm{L}}
\)
\(\%\) error in density \(=\%\) error in Mass
\(
\begin{aligned}
&\quad+3(\% \text { error in length }) \\
&=4+3(3)=13 \%
\end{aligned}
\)

Hint

Percentage error in mass \(\left(\frac{\Delta m}{m} \times 100\right)=2\) and percentage error in speed \(\left(\frac{\Delta v}{v} \times 100\right)=3\).
Kinetic energy, \(k=\frac{1}{2} m v^{2}\).
\(\therefore\) Error in measurement of kinetic energy \(\frac{\Delta K}{K}=\frac{\Delta m}{m}+2\left(\frac{\Delta v}{v}\right)\)
\(
\begin{aligned}
&=\left(\frac{2}{100}\right)+\left(2 \times \frac{3}{100}\right)=\frac{8}{100}=8 \% \\
&\therefore \% \text { age error }=8 \% .
\end{aligned}
\)

Hint

Least count \(=1 \mathrm{MSD}-1 \mathrm{VSD}\)
\(
\begin{aligned}
&=1 \mathrm{MSD}-\left(\frac{\mathrm{N}-1}{\mathrm{~N}}\right) \mathrm{MSD} \\
&=\frac{1}{\mathrm{~N}} \mathrm{MSD}=\frac{1}{\mathrm{~N}} \times \frac{1}{10} \mathrm{~cm}=\frac{1}{10 \mathrm{~N}}
\end{aligned}
\)

Hint

Density, \(\mathrm{D}=\frac{\operatorname{Mass}(\mathrm{M})}{\operatorname{Volume}(\mathrm{V})}\)
\(
\therefore \frac{\Delta \mathrm{D}}{\mathrm{D}}=\frac{\Delta \mathrm{M}}{\mathrm{M}}+\frac{\Delta \mathrm{V}}{\mathrm{V}}=\left(\frac{0.01}{22.42}+\frac{0.1}{4.7}\right) \times 100=2 \%
\)

Hint

\(
\begin{array}{|l|l|}
\hline {\text { Waves }} & {\text { Wavelength }} \\
\hline \text { AM radio waves } & 10^2 \mathrm{~m} \\
\hline \text { Microwaves } & 10^{-2} \mathrm{~m} \\
\hline \text { Infrared radiations } & 10^{-4} \mathrm{~m} \\
\hline \text { X-rays } & 10^{-10} \mathrm{~m} \\
\hline
\end{array}
\)

\(
\begin{aligned}
& \text { (a) – (ii) } \\
& \text { (b) – (iii) } \\
& \text { (c) – (iv) } \\
& \text { (d) – (i) }
\end{aligned}
\)

Hint

\(
\text { Dimensional formula of magnetic permeability is }\left[M L T^{-2} A^{-2}\right]
\)

Hint

\(
\begin{aligned}
\text { Plane angle } & =\frac{\text { Arc }}{\text { Radius }}=\frac{[\mathrm{L}]}{[\mathrm{L}]} \longrightarrow \text { Unit }=\text { Radian } \\
& =\left[M^0 L^0 T^0\right]
\end{aligned}
\)
\(
\begin{aligned}
\text { Solid angle } & =\frac{\text { Area }}{(\text { Radius })^2} \longrightarrow \text { Unit }=\text { Steradian } \\
& =\frac{L^2}{L^2}=\left[M^0 L^0 T^0\right]
\end{aligned}
\)
\(\therefore\) Both have units but no dimensions.

Hint

(a)
\(
\begin{aligned}
& {[G]=\frac{F r^2}{m_1 m_2}} \\
& {[G]=\frac{F r^2}{m_1 m_2}=\frac{\left[M L T^{-2}\right] L^2}{[\mathrm{MM}]}=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]}
\end{aligned}
\)
(b) Gravitational potential energy \(=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\)
(c) Gravitational potential \(=\frac{\mathrm{PE}}{m}=\left[\mathrm{L}^2 \mathrm{~T}^{-2}\right]\)
(d) Gravitational field intensity \(=\frac{F}{m}=\left[\mathrm{LT}^{-2}\right]\)

Hint

\(
\begin{aligned}
\text { Area } & =\text { Length } \times \text { Breadth } \\
& =55.3 \times 25 \mathrm{~m}^2 \\
& =1382.5 \mathrm{~m}^2 \\
& =14 \times 10^2 \mathrm{~m}^2 \text { (Rounding off of two significant figures) }
\end{aligned}
\)

Hint

The errors which cannot be associated with any systematic or constant cause are called random errors. These errors can arise due to unpredictable fluctuations in experimental conditions. e.g., random change in pressure, temperature, voltage supply, etc.

Hint

\(
\text { We know, } \rho=\frac{\text { Mass }}{\text { Volume }}=\frac{M}{\pi r^2 l}
\)
Using the concept of errors we know,
\(
\begin{aligned}
& \frac{\Delta \rho}{\rho}=\frac{\Delta M}{M}+\frac{2 \Delta r}{r}+\frac{\Delta l}{l}=\left(\frac{0.002}{0.4}+\frac{2 \times 0.001}{0.3}+\frac{0.02}{5}\right) \\
& \frac{\Delta \rho}{\rho}=0.0156 \\
& \frac{\Delta \rho}{\rho} \%=1.56 \% \approx 1.6 \%
\end{aligned}
\)