Physics Test - 43

For a closed organ pipe, the frequency of fundamental mode is \(\mathrm{f}_{\mathrm{c}}=\frac{\mathrm{v}}{4 \mathrm{~L}_{\mathrm{c}}}\).

where \( v\) is velocity of sound in air and \(L_{c}\) is length of the closed pipe. For an open pipe, the frequency of fundamental mode is:

\(\mathrm{f}_{0}=\frac{\mathrm{v}}{2 \mathrm{~L}_{0}}\) Where \(\mathrm{L}_{0}\) is length of open pipe

\(\because \mathrm{L}_{\mathrm{c}}=\mathrm{L}_{0}\) (given)

\(\mathrm{f}_{0}=2 \mathrm{f}_{\mathrm{c}} \ldots \ldots \ldots \ldots .(1)\)

\(\mathrm{f}_{0}-\mathrm{f}_{\mathrm{c}}=2 \ldots \ldots \ldots \ldots(2)\)

Solving (1) and (2) we get \(\mathrm{f}_{0}=4,\mathrm{f}_{\mathrm{c}}=2\)

When length of open pipe is halved, its frequency of fundamental mode is:

\(\mathrm{f}_{0}^{\prime}=\frac{\mathrm{v}}{2\left(\frac{\mathrm{L}_{0}}{2}\right)}=2 \mathrm{f}_{0}\)

\(=2 \times 4=8\)

When Length of closed pipe is doubled, its frequency

\(\mathrm{f}_{\mathrm{c}}^{\prime}=\frac{\mathrm{v}}{4\left(2 \mathrm{~L}_{\mathrm{c}}\right)}\)

\(=\frac{1}{2} \mathrm{f}_{\mathrm{c}}=\frac{1}{2} \times 2 =1\)

Number of beats produced per second

\(=\mathrm{f}_{0}^{\prime}-\mathrm{f}_{\mathrm{c}}^{\prime}\)

\(=8-1\)

\(=7\)

Due to heating, number of electron-hole pairs will increase, so overall resistance of diode will change.

Due to which forward biasing and reversed biasing both are changed.

The figure given below represents the system mentioned in the question:

The charge at point \(\mathrm{A (q_{A})=2.5 \times 10^{-7} C}\)

The charge at point \(\mathrm{B (q_{B})=-2.5 \times 10^{-7} C}\)

Then, the net charge \(\mathrm{q=(q_{A}+q_{B})=2.5 \times 10^{-7} C-2.5 \times 10^{-7} C=0}\)

The distance between two charges at \(A\) and \(B\) would be,

\(\mathrm{d=15+15=30 \mathrm{~cm}}\)

\(\mathrm{d=0.3 \mathrm{~m}}\)

The electric dipole moment of the system could be given by,

\(\mathrm{P=q_{A} \times d=q_{B} \times d}\)

\(\mathrm{P=2.5 \times 10^{-7} \times 0.3}\)

\(\mathrm{P=7.5 \times 10^{-8} \mathrm{Cm}}\) along the \(+\mathrm{z}\) axis.

Therefore, the electric dipole moment of the system is found to be \(7.5 \times 10^{-8} \mathrm{Cm}\) and it is directed along the positive \(z\)-axis.

The only acceleration in projectile motion is the acceleration due to gravity. If the maximum height of the projectile motion is not large, and can be neglected with respect to the radius of earth (which usually is the case), the acceleration due to gravity remains constant. Hence, the acceleration in projectile motion remains constant.

In a Young's double slit experiment for interference pattern, the position of bright fringe is given by

\(\mathrm{y}_{\mathrm{n}}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}\)

Here, \(\mathrm{D}=1.5 \mathrm{~m}\),

\(\mathrm{d}=0.3 \times 10^{-3} \times \mathrm{m}\)

and \(\mathrm{n}=1\)

\(\lambda=\frac{\mathrm{y}_{\mathrm{n}} \mathrm{d}}{\mathrm{D}}\)

For first violet, \(\mathrm{y}_{\mathrm{n}}=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}\)

\(\therefore \lambda_{\text {violet }}=2 \times 10^{-3} \frac{\mathrm{d}}{\mathrm{D}}\)

For first red \(\mathrm{y}_{\mathrm{n}}=3.5 \mathrm{~mm}=3.5 \times 10^{-3} \mathrm{~m}\)

\(\lambda_{\text {red }}=3.5 \times 10^{-3} \frac{\mathrm{d}}{\mathrm{D}}\)

The difference in wavelengths of red and violet light is

\(\Delta \lambda=\lambda_{\text {red }}-\lambda_{\text {violet }} \)

\(=3.5 \times 10^{-3} \frac{\mathrm{d}}{\mathrm{D}}-2 \times 10^{-3} \frac{\mathrm{d}}{\mathrm{D}} \)

\(=\frac{\mathrm{d}}{\mathrm{D}}(1.5) \times 10^{-3} \)

\(=\frac{0.3 \times 10^{-3}}{1.5} \times 1.5 \times 10^{-3} \)

\(=0.3 \times 10^{-6} \)

\(=0.3 \times 10^{-6} \times 10^3 \times 10^{-3} \)

\(=300 \times 10^{-9} \)

\(=300 \mathrm{~nm}\)

Using Kirchhoff's junction rule.

\(\begin{aligned} & \frac{2-\mathrm{V}_0}{1}+\frac{4-\mathrm{V}_0}{1}+\frac{6-\mathrm{V}_0}{1}=0 \\ & 12-3 \mathrm{~V}_0=0 \\ & \mathrm{~V}_0=4 \mathrm{~V}\end{aligned}\)

As we know that the Einstein's photo-electric equation is given by,

\( K_{\max }=h \nu-\phi_{0}.....(1)\)

Also we know that the frequency of a light wave is given by

\( \nu=\frac{c}{\lambda}.....(2)\)

Substituting the value from (2) to (1), we get:

\( K_{\max }=\frac{h c}{\lambda}-\phi_{0}\)

So the work function of the metal is given by 

\( \phi_{0}=\frac{h c}{\lambda}-K_{\max }.....(3)\)

According to the question,

\(\lambda=400 \mathrm{~nm}, K_{\max }=1.68 \mathrm{eV}\), \(h c=1240 \mathrm{eV}\mathrm{~nm}\)

Putting these values in \((3)\) we get,

\( \phi_{0}=\frac{1240}{400}-1.68\)

\(\Rightarrow \phi_{0}=1.42 \mathrm{eV}\)

Thus the work function of the metal comes out to be equal to \(1.42 \mathrm{eV}\).

The formula to calculate the rate of cooling of the object is given by

\(\frac{d T}{d t}=K\left[\frac{T_f+T_i}{2}-T_0\right]\)

For the first case, it can be written, using equation (1) that

\(\frac{80-60}{5} =K\left[\frac{80+60}{2}-20\right] \)

\(4 =50 K \ldots(2)\)

If \(t\) is the required time for the second case, from equation (1), it can be written that

\(\frac{60-40}{t} =K\left[\frac{60+40}{2}-20\right] \)

\(\frac{20}{t} =30 K \ldots(3)\)

Divide equation (2) by equation (3) and solve to calculate the required time.

\(\frac{4}{\frac{20}{t}} =\frac{50 K}{30 K} \)

\( \Rightarrow \frac{t}{5}=\frac{5}{3} \)

\( \Rightarrow t=\frac{25}{3} \min =500 s\)

The mass of planets is measured using the gravitational method. In this method, the mass is compared to the force of gravity of a known mass with the force of gravity of a known mass. The strength of the gravitational force is directly proportional to the mass.

The frequency of light wave does not change as the light moves from air to glass.

Speed of light wave \(v\) is related to both the frequency \(f\) and the wavelength \(l\) as:

\(v=f \lambda\)

Let \(V_{\mathrm{a}}\) and \(V_{g}\) be the speed of light in air and glass respectively. 

Let \(\lambda_{\mathrm{a}}\) and \(\lambda_{\mathrm{g}}\) be the wavelength of light wave in air and in glass:

i.e. \(\lambda_{{g}}\)- wavelength of incident light wave(in air)

\(\lambda_{g}\)- wavelength of refracted light wave(in glass)

Therefore we can write:

\(v_{a}=f \lambda_{a}\)........(i)

\(v_{g}=f \lambda_{g}\)........(ii)

Dividing equation (i) by (ii) gives:

\(\frac{v_{a}}{v_{g}}=\frac{\lambda_{a}}{\lambda_{g}}\)

But we know that \(\frac{v_{a}}{v_{g}}=\) refractive index \(\mu\)

Therefore:

\(\frac{\lambda_{\mathrm{a}}}{\lambda_{\mathrm{g}}}=\frac{\mu}{1}\)

i.e \(\frac{\lambda_{\mathrm{a}}}{\lambda_{\mathrm{g}}}=\mu: 1\)