Physics Test - 3

Time \(t=8 min =480 sec\)
Speed of light \(c=3 \times 10^{8} ms ^{-1}\)
Distance \(=t \times c=1.44 \times 10^{11} m =1.44 \times 10^{8} km\)
So order of magnitude os distance is \(10^{8} km\)

\begin{equation}\begin{array}{l}
\text { 1 } \text {dyne}=10^{-5} N \text { \& } 1 cm =10^{-2} m \\
\text { So, } S=70 \frac{\text { dyne }}{ cm }=\frac{70 \times 10^{-5} N }{10^{-2} m } \\
=70 \times 10^{-3} N / m \\
\Rightarrow S=7 \times 10^{-2} N / m
\end{array}\end{equation}

Given force, \(F=A \sin C t+B \cos D x\)
We know that the trigonometric functions are dimensionless because they are ratio of length. Thus, \(\sin C t\) and \(\cos D x\) will be dimensionless. So dimension of C should be \(\left[T^{-1}\right]\) and dimension of D should be \(\left[L^{-1}\right]\)
So, \(C / D=\left[T^{-1}\right] /\left[L^{-1}\right]=\left[L T^{-1}\right]=\left[M^{0} L^{1} T^{-1}\right]\)
since the dimension on each side of a equation is same so dimension of A and B should be equal to dimension of force.
Thus, \(A / B=1=\left[M^{0} L^{0} T^{0}\right]\)

Surface energy per unit area is the measure of surface tension. Surface tension \(=\frac{\text {Force}}{\text {length}}=\frac{N}{m}\)
Thus the unit of surface energy per unit area may be expressed as \(N m^{-1}\)

Surface tension \(=\frac{F}{l}=\frac{N}{m}=\frac{N m}{m^{2}}=\frac{J}{m^{2}}\)
Viscosity \(=\frac{\text {Force}}{\text {Area} \times \frac{d v}{d z}}=\frac{N}{m^{2} \frac{m}{s m}}=\frac{N m s}{m^{3}}=\frac{J s}{m^{3}}\)
Strain energy \(=J\)
Power \(=J / s\)

For a perfect black body, the energy radiated per unit area per unit time is given by, \(E =\sigma T^{4}\) where \(\sigma\) is Stefan's constant and \(T\) is temperature in Kelvin scale. We know that energy per unit time is power \((P.)\)
So power radiated per unit area \(=\frac{P}{A}=\sigma T^{4}\)
or \(\sigma=\frac{P}{A T^{4}}\)
Now substituting the unit of \(P, A\) and \(T\) as \(W, m^{2}\) and \(k^{4}\) respectively.
We get the unit of Stefan's constant i.e, \(W m^{-2} k^{-4}\)

From Biot Savart's law
\(B=\frac{\mu_{o}}{4 \pi} \frac{i . d l \sin \theta}{r^{2}}\)
\(\mu_{o}=\frac{4 \pi B r^{2}}{i d l \sin \theta}=\frac{W b m^{-2} m^{2}}{A \cdot m}=W b A^{-1} m^{-1}\)

\(\frac{d y n e}{c m^{2}}=\frac{\left[M^{1} L^{1} T^{-2}\right]}{L^{2}}=\left[M^{1} L^{-1} T^{-2}\right]\)

Pressure \(=\frac{\text {force}}{\text {Area}}=\frac{\left[M^{1} L^{1} T^{-2}\right]}{\left[L^{2}\right]}=\left[M^{1} L^{-1} T^{-2}\right]\)

Longitudinal stress \(=\) Young's Modulus \(=\left[M^{1} L^{-1} T^{-2}\right]\)

Longitudinal strain \(=\frac{\Delta l}{l}=\left[M^{0} L^{0} T^{0}\right]\)

Hence option C is correct.

Lux, unit of illumination (see luminous intensity) in the International System of Units (SI). One lux (Latin for “light”) is the amount of illumination provided when one lumen is evenly distributed over an area of one square metre.

Hence option B is correct.

A Physical quantity is a quantity in physics that can be measured or can be quantified. 

Power is the rate of doing work. It is the amount of energy consumed per unit time. 

Momentum is the product of the mass and velocity of an object.

Latent heat is energy released or absorbed, by a body or a thermodynamic system, during a constant-temperature process. All of these are physical quantities.

Radian is the standard unit of angular measure, used in many areas of mathematics not a physical quantity.

Hence option D is correct.