Physics Test - 2

Stress has the units of Force/Area \(= N / m ^{2},\) while strain \((=\Delta l / 1)\) is dimensionless.
So option \(A\) is correct.
Surface tension is Force/Length, thus \(N / m ,\) hence \(B\) is also correct.
Energy has the units of \(\left[ ML ^{2} T ^{-2}\right]\), hence the units have to be \(kgm ^{2} / s ^{2}\). Thus \(C\) is
incorrect option.
Pressure is Force/Area \(= N / m ^{2}\)

\begin{equation}\begin{array}{l}
\text { Heat Energy }=I^{2} R t \\
\text { So, } R=\frac{J}{A^{2} s}=K g m^{2} s^{-3} A^{-2}
\end{array}\end{equation}

The ratio of increase in resistance per degree rise in temperature to its resistance at \(0^{0} C\) is defined as the temperature coefficient of resistance \((\alpha).\)
\(\alpha=\frac{R_{t}-R_{0}}{R_{0} t}\)
The unit of \(\alpha\) is \(K^{-1}\)

\begin{equation}\begin{array}{l}
\text { 1 } \text {dyne}=10^{-5} N \\
\text { 1 } cm =10^{-2} m \\
\therefore \frac{d y \text { ne }}{ cm }=\frac{10^{-5} N }{10^{-2} m }=\frac{10^{-3} N }{ m } \\
\therefore \frac{45 dyne }{ cm }=\frac{45 \times 10^{-3} N }{ m }=0.045 N / m
\end{array}\end{equation}

\(1 N=10^{5} D y n e\)
\(1 m ^{2}=10^{4} cm ^{2}\)
\(N / m^{2}=10 D y n e / c m^{2}\)
Modulus \(\propto\) stress So, if SI unit value is \(\beta\) the CGS unit will be \(10 \beta.\)

\(W b / m^{2}\) and \(T\) are the SI units of magnetic induction. Mathematically, 1 \(W b / m^{2}=1 T=10^{4} G\)

Young's modulus \(\quad Y=\frac{\text {Stress}}{\text {Strain}}\)
As we know, strain is a dimensionless quantity whereas stress is defined as force per unit area.
Stress has units as \(N m^{-2}\) (SI unit), Dyne \(cm ^{-2}\) (cgs unit) or Pascal \((P a).\)
As Young's modulus has same unit as that of stress, thus \(N m^{-1}\) is not a unit of Young's modulus.

\begin{equation}\begin{array}{l}
\text {pico}=10^{-12} \\
\text {giga}=10^{9} \\
\Rightarrow \frac{\text {pico}}{\text {giga}}=\frac{10^{-12}}{10^{9}}=10^{-21}
\end{array}\end{equation}

The force experienced by a moving charge in a magnetic field is given by \(F=q(\vec{v} \times \vec{B})\)
Here, \(\vec{B}\) is the magnetic field or magnetic flux density or magnetic flux per unit area.
Thus, dimensions of magnetic flux \([\Phi]=[B] \times[\) Area \(]=\frac{[F][A]}{[q][v]}\)
We know the dimensions of the following as
\([F]=\left[M^{1} L^{1} T^{-2}\right]\)
\([A]=\left[L^{2}\right]\)
\([q]=\left[A^{1} T^{1}\right]\)
\([ v ]=\left[ L ^{1} T ^{- 1 }\right]\)
Thus, \([\Phi]=\frac{\left[M^{1} L^{1} T^{-2}\right]\left[L^{2}\right]}{\left[A^{1} T^{1}\right]\left[L^{1} T^{-1}\right]}=\left[M^{1} L^{2} T^{-2} A^{-1}\right]\)

Dimensional formula of work is \(M L^{2} T^{-2}\)
Multiply and divide by \(T^{2}\)
\(M L^{2} T^{-2} \frac{T^{2}}{T^{2}}=M L^{2} T^{-4} T^{2}\)
The term \(L^{2} T^{-4}=\left(L T^{-2}\right)^{2}=a^{2} \operatorname{can} be\)
written as square of acceleration
Dimensional formula of work is
same as that of Mass\(\times(\text {acceleration} \times T)^{2}\)
Hence, new unit of work
\(\begin{aligned}=100 \times(10 \times 60)^{2} & \\ &=36 \times 10^{6} J \end{aligned}\)