Physics Test - 35

Force \(=\mathrm{G} \times \mathrm{m}_{1} \times \mathrm{m}_{2} \times\left[\mathrm{r}^{2}\right]^{-1}\)

\(\Rightarrow \mathrm{G}=\) Force \(\times \mathrm{r}^{2} \times\left[\mathrm{m}_{1} \times \mathrm{m}_{2}\right]^{-1} \ldots \ldots(1)\)

Where, \(\mathrm{G}=\) Universal Gravitational Constant

Now, the dimensions of,

Mass \(=\left[M^{1} L^{0} T^{0}\right] \ldots .(2)\)

Radius \(=\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}\right] \ldots(3)\)

Force \(=\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\right] \ldots \ldots(4)\)

On substituting equation \((2),(3)\) and (4) in equation (1), we get

Universal Gravitational Constant = Force \(\times \mathrm{r}^{2} \times\left[\mathrm{m}_{1} \times \mathrm{m}_{2}\right]^{-1}\)

\(\mathrm{G}=\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\right] \times\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}\right]^{2} \times\left[\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]^{-1} \times\left[\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]^{-1}\)

\(=\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\)

Therefore, the Universal Gravitational constant is dimensionally represented as \(\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]\).

In the convex lens, the rays are converted to a point on the other side lens when the parallel rays are incident on the surface of the lens. If the central portion of the lens wrapped by black paper, the full image will be formed but with less brightness.

Thermodynamics is primarily based on a set of four rules that are universally applicable when applied to systems that fall within their respective limitations. They are as follows:

• Zeroth law of thermodynamics
• First law of thermodynamics
• Second law of thermodynamics
• Third law of thermodynamics

Given that,

Kinetic Energy, \((K.E.)=E_{2}=2 E_{1}\)

From de-Broglie wavelength formula, we know that

\(\lambda=\frac {h}{p}=\frac{h}{\sqrt {2m(K.E.)}}\)

Therefore,

\(\lambda_{1}=\frac{h}{\sqrt{2 m E_{1}}} \quad \ldots \ldots\) (i)

\(\lambda_{2}=\frac{h}{\sqrt{2 m E_{2}}}\)\(\quad \ldots \ldots\) (ii)

Dividing equation (i) by equation (ii), we get

\(\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{E_{2}}{E_{1}}}=\sqrt{\frac{2 E_{1}}{E_{1}}}\)

\(\frac{\lambda_{2}}{\lambda_{1}}=\sqrt{\frac{1}{2}}\)

\(\lambda_{2}=\lambda_{1} \left(\frac{1}{\sqrt{2}}\right)\)

Given,

Electrical field \(E=100 \) V/m

Magnetic field \({B}=0.265 \) A/m

The energy flow is given by the Poynting vector 

\(\vec{S}=\vec{E} \times \vec{B}\)

\(\Rightarrow S=E B \sin \phi\)

\(\Rightarrow S=E B\)  (\(\phi=90^{\circ},\) as \(E\) and \(B\) are perpendicular to each other)

\(\Rightarrow S=100  \times 0.265 \)

\(\Rightarrow S=26.5\) W/m\(^2\)

For an earth satellite moving in a circular orbit, centripetal force required for its circular motion is provided by the gravitational force exerted by earth on it. It means resultant force on the astronaut is equal to the gravitational force exerted by earth on him. So, no reaction is exerted by floor of the satellite on him. In other words, his acceleration towards earth centre (centripetal acceleration) is exactly equal to acceleration caused by the gravitational force alone.

The property of viscosity of liquids arises when there is a relative motion between its layers. The viscous force acts in a direction opposite to the direction of motion, i.e. it opposes the motion.