Physics Test - 26

Let power of lens \(=P_{L}\)

Power of mirror \(= P_{M}\)

In case of silvering plane surface, refraction will take place through curve part twice, and reflection will take place through plane part once.

\(P=2 P_{L}-P_{M}=2 P_{L}=\frac{2}{f_{L}}=\frac{1}{30} ~cm-1(\) as power of plane mirror \(=0)\)

By lens maker formula,

\(\frac{1}{f_{L}}=(\mu-1) \frac{1}{R}\)

\(\frac{1}{60}=\frac{1}{2R}\)

\(R=30\)cm

Fringe width \(\beta=\frac{D \lambda} d\)

where \(D\) is the distance between slits and screen

and \(d\) is the distance between the slits.

Thus, From question \(D^{\prime}=2 D\) and \(d^{\prime}= \frac {d} 2\)

When D is doubled and \(d\) is reduced to half, then the fringe width becomes

\(\beta^{\prime}=\frac{\lambda 2 {D}} {{(\frac{d} {2})}}\)

\(=\frac{4 \lambda {D}}{d}\)

\(=4 \beta\)

Thus, when the double slit experiment the distance between the slits and the screen is doubled and the separation between the slits is reduced to half, the fringe width becomes four times.

The Einstein's photoelectric equation is:

\(h v=h v_{o}+K_{\max }\)

So,

\(K_{\max }=h v-h v_{0}=h v+\text { constant }\)

On comparing it with

\(y=m x+c\)

This is the equation of straight line having positive slope (h) and negative intercept \(h v_{0}\).

We know that:

Time period of SHM of small vertical oscillations in a liquid is given by:

\(T=2 \pi \sqrt{\frac{l}{g}}\), where \(l\) is the length of cube/cylinder dipped in the water.

So according to law of floatation,

weight of the cube \(=\) weight of the water displaced

\(a b c \times d \times g=b c l \times l \times g\)

\(\Rightarrow l=d a\)

\(\Rightarrow T=2 \pi \sqrt{\frac{d a}{g}}\)

An astronomical unit is defined as the average distance between the center of the Earth and the center of the Sun. The astronomical unit is the unit of length, which is approximately \(150\) million kilometers and is based on the distance from the Earth to the Sun.

1 astronomical unit \((1 \mathrm{AU})=1.496 \times 10^{11} \mathrm{~m}\)

Given:

\(\vec{F}=q \vec{B}|\vec{v} \sin \theta|\)

\([F]=\left[M L T^{-2}\right]\)

\([Q]=[A T]\)

\([V]=\left[L T^{-1}\right]\)

\(\operatorname{Sin} \theta\) = Dimensionless

\([B]=\frac{\mid F]}{\mid q][v]}\)

\(=\frac{\left[M L T^{-2}\right]}{\mid A T]\left[L T^{-1}\right]}\)

\(=\left[M L^{0} T^{-2} A^{-1}\right]\)

\(=\left[M T^{-2} A^{-1}\right]\)

Pressure of a gas increases due to increase of its temperature because at higher temperature kinetic energy of the gas molecules are higher.

Root mean square speed of gas molecules:

\(\left(V_{r m s}\right)=\sqrt{\frac{\gamma R T}{M}}\)

According to the above-given formula for the speed of molecules, when the temperature of the gases increases then the speed of gaseous molecules increases.

The kinetic energy of the molecules increases by an increase in speed.

Due to an increase in the speed of molecules, it moves faster and the collisions between them increase.

Due to an increase in collisions between the molecules, the gaseous pressure increases. So, due to increasing temperature the kinetic energy of molecules increases and finally the pressure increases.

Given, \(L=25 \mathrm{mH}\) \(=25 \times 10^{-3} H\),\(f=50 \mathrm{~Hz}\),\(V_{r m s}=220 \mathrm{~V}\)
The inductive reactance,\(X_{L}=2 \pi f L=2 \times \frac{22}{7} \times 50 \times 25 \times 10^{-3} \Omega\)
The rms current in the circuit is:\(I_{r m s}=\frac{V_{r m s}}{X_{L}}=\frac{220}{2 \times \frac{22}{7} \times 50 \times 25 \times 10^{-3}}=\frac{7 \times 1000}{2 \times 5 \times 25}=28 A\)