JEE 2017 :: Advanced 2017 :: Physics 2017

• Advanced 2017 - Physics 2017
• Advanced 2017 - Chemistry 2017
• Advanced 2017 - Mathematics 2017

Consider an expanding sphere of instantaneous radius R  whose total mass remains constant. The expansion is such that the instantaneous density ρ remains uniform throughout the volume. The rate of fractional change in density $(\frac{1}{\rho}\frac{d\rho}{dt})$ is constant. The velocity  v of any point of the surface of the expanding sphere is proportional to

Answer:

Explanation:

$m=\frac{4\pi R^{3}}{3}\times\rho$

On taking log both sides, we have

ln(m)= ln($(\frac{4\pi ^{}}{3})$) + ln(ρ)+3ln(R)

On differentiating  with respect to time

$      0=0+\frac{1}{\rho}\frac{d \rho}{dt}+\frac{3}{R}\frac{dR}{dt}$

   $\Rightarrow$              $  (\frac{dR}{dt})= v \propto -R \times \frac{1}{\rho}(\frac{d \rho}{dt})$

    $  v  \propto R$

Consider regular polygons with the number of sides n=3,4,5 ..... as shown in the figure. The center of mass of all the polygons is at height h from the ground. They roll on a horizontal surface about the leading vertex without slipping and sliding as depicted. The maximum increase in height of the locus of the center of mass for each polygon is Δ. Then, Δ depends on n and h as

Answer:

Explanation:

$\cos (\frac{\pi}{n})=\frac{h}{R}$

$\triangle =R-h=\frac{h}{\cos(\frac{\pi}{n})}-h=h[\frac{1}{\cos(\frac{\pi}{n})-1}]$

A stationary source emits a sound of frequency f₀=492 Hz. The sound is reflected by a large car approaching the source with a speed of 2 ms^-1. The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz? (Given that the speed of sound in air is 330ms ^-1 and the car reflects the sound at the frequency it has received].

Answer:

Explanation:

Frequency observed at car

            $f_{1}=f_{0}(\frac{v+v_{c}}{v})$

                                                       [v= speed of sound)

Frequency of reflected souns as observed at the source.

                $f_{2}=f_{1}(\frac{v}{v-v_{C}})=f_{0}(\frac{v+v_{C}}{v-v_{C}})$

 Beat frequency=f₂ -f₀

$=f_{0}(\frac{v+v_{C}}{v-v_{C}}-1)=f_{0}(\frac{2v}{v-v_{C}})$

 =  $492\times\frac{2\times 2}{328}=6Hz$

 A drop of liquid of radius R=10^-2m having surface tension  $S= \frac{0.1}{4\pi}$ Nm^-1   divides itself into K identical drops. In this process the total change in the  surface energy $\triangle U= 10^{-3}$ J . If  $K=10^{\alpha}$, then the value of $\alpha$ is

Answer:

Explanation:

From mass conservation

         $\rho\frac{4}{3}\pi R^{3}=\rho K\frac{4}{3}\pi r^{3}$ 

$\Rightarrow$     R= $K^{\frac{1}{3}}r$

$\therefore$  $\triangle U= T\triangle A=T(K. 4\pi r^{2}-4\pi R^{2})$

=$T(K. 4\pi R^{2}K^{-\frac{2}{3}}-4\pi R^{2})$

$\triangle U=4\pi R^{2}T[K^{\frac{1}{3}}-1]$

Putting the values, we get

    $10^{-3}=\frac{10^{-1}}{4\pi}\times 4\pi\times 10^{-4}[K^{-\frac{1}{3}}-1]$

$100=K^{\frac{1}{3}}-1$

$\Rightarrow$   $K^{\frac{1}{3}}=100=10^{2}$

   Given that K= $10^{\alpha}$

$\therefore$    $10^{\frac{\alpha}{3}}=10^{2}$

$\Rightarrow$     $\frac{\alpha}{3}=2$

   $\Rightarrow$  $\alpha=6$

An electron in a hydrogen atom undergoes a transition from an orbit with quantum number n_i to another with quantum number n_f . v_i and v_f  are respectively, the initial and final potential energies of the electron. If  $\frac{v_{i}}{v_{f}}=6.25$    then the  smallest possible n_f  is

Answer:

Explanation:

Potential energy of hydrogen atom (Z=1) in nth orbit (in eV)

   $P.E=-\frac{27.2}{n^{2}}$

            = $\frac{v_{f}}{v_{i}}=\frac{-\frac{27.2}{n_f^2}}{-\frac{27.2}{n_i^2}}$ =  $\frac{1}{6.25}$

    $6.25=\frac{n_f^2}{n_i^2}$

$\frac{n_{f}}{n_{i}}=2.5=\frac{5}{2}$

 hence the answer is 5

• Advanced 2017 - Physics 2017
• Advanced 2017 - Chemistry 2017
• Advanced 2017 - Mathematics 2017