Physics Test - 22

In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes four times.

Fringe width in Young's double slit experiment is given by,

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

If the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width,

\(\beta=\frac{\lambda 2 D}{\frac{d}{2}}=\frac{4 \lambda D}{d}=4 \beta\)

Surface tension will be less as temperature increases

\(\mathbf{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho g \mathrm{r}}\)

Height of capillary rise will be smaller in hot water and larger in cold water.

The wavelength of the incident light \(= 4000\) Å \(=4000 \times 10^{-10}m\)

The value of the work function \(=2 \mathrm{eV}\)

As we know,

Work function \(\phi=h f-K E\)

Where \(h f=\) incident energy of the light, \(\mathrm{KE}=\) kinetic energy of the ejected photoelectron.

Therefore, \(\mathrm{KE}=h f-\phi\)

We know that \(f=\frac{c}{\lambda}\)

Thus, \(K E=\frac{h c}{\lambda}-\phi\)........(i)

We know that:

\(c=3 \times 10^{8}\)m/s, \(h=606 \times 10^{-34}\), \(\phi=2\)

Put the given values in equation (i),

\(=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{4000 \times 10^{-10} \times 1.6 \times 10^{-19}}-2\quad\) ( \(1.6 \times 10^{-19}\) is used to get the value in terms of eV)

\(\Rightarrow K E=10.37 \mathrm{eV}\)

Therefore, the value of the kinetic energy of the ejected photoelectron is \(10.37 \mathrm{eV}\).

Given:

\(C =100 \mu F =100 \times 10^{-6} F\), and \(V =100. \sin (10 t )\)

The maximum voltage is given as,

\( V_{\max }=100 V \quad \ldots\)(1)

And angular frequency is given as,

\( \omega=10\) rad/sec \(\quad \ldots\)(2)

We know that the capacitive reactance is given as,

\( X_{C}=\frac{1}{\omega C}\quad \ldots\)(3)

Where \(\omega=\) angular frequency, and \(C =\) capacitance

By equation (2) and equation (3),

\( X_{C}=\frac{1}{10 \times 100 \times 10^{-6}}\)

\(\Rightarrow X _{ C }=1000 \Omega\quad \ldots\)(4)

When an AC voltage is applied to a capacitor, the current in the circuit is given as,

\(I=\frac{V}{X_{C}} \quad \ldots\)(5)

By equation (1), equation (4), and equation (5), the maximum current in the circuit is given as,

\( I_{\max }=\frac{V_{\max }}{X_{C}}\)

\(\Rightarrow I_{\max }=\frac{100}{1000}\)

\(\Rightarrow I_{\max }=0.1 A\)

Given,

Focal length\(=20 \mathrm{~cm}= 0.2 m\)

Relative speed of the first car\(=15 \mathrm{~ms}^{-1}\)

Relative speed of the second car\(=15 \mathrm{~ms}^{-1}\)

For the mirror, \(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\)

Differentiate this equation with respect to \(t,\) we get

\(\Rightarrow-\frac{1}{u^{2}} \frac{d u}{d t}-\frac{1}{v^{2}} \frac{d v}{d t}=0\)

\(\Rightarrow \frac{d v}{d t}=-\frac{v^{2}}{u^{2}}\left(\frac{d u}{d t}\right)\)

But \(\frac{v}{u}=\frac{f}{u-f}\)

\(\therefore \frac{d v}{d t}=-\left(\frac{f}{u-f}\right)^{2}\left(\frac{d u}{d t}\right)\)

\(=\left(\frac{0.2}{-2.8-0.2}\right)^{2} \times 15\)

\(=\frac{1}{15} \mathrm{~ms}^{-1}\)

\(\begin{aligned} & g^{\prime}=\frac{g}{\left(1+\frac{h}{R}\right)^2} \\ & g^{\prime}=g\left\{1-\frac{d}{R}\right\}\end{aligned}\)

Torque is a physical computation with the dimension of force times distance. Its SI unit is a newton meter. It also has a unit as joule per radian.

We know that,

Torque \((T)=\) Moment of Inertia \(\times\) Angular Acceleration

We know that Moment of Inertia (M.O.I) \(=\) Radius of Gyration \(^{2} \times\) Mass

The dimensional formula of Moment of Inertia \(=M^{1} L^{2} T^{0}\)...(i)

And, Angular Acceleration \(=\) Angular velocity \(\times\) Time \(^{-1}\)

\(\therefore\) The dimensional formula of Angular Acceleration \(=M^{0} L^{0} T^{-2}\)......(ii)

On substituting equation (i) and (ii) in the formula of torque, we get

Torque \(=\) Moment of Inertia \(\times\) Angular Acceleration

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

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

Acceleration due to gravity at height \(h\)

\(g^{\prime}=\frac{g}{\left[1+\frac{h}{R}\right]^2}\)

So weight at given height

\(m g^{\prime}=\frac{m g}{\left[1+\frac{h}{R}\right]^2}=\frac{18}{\left[1+\frac{1}{2}\right]^2}=8 \mathrm{~N}\)