# JEE Mains (Apr 10) 2015

$$P=\dfrac{nM}{3V}v_{rms}^2$$

Thus pressure $$P\propto v_{rms}^2$$

and thus

Force $$\propto v^2_{rms}\propto T$$

Thus pressure $$P\propto v_{rms}^2$$

and thus

Force $$\propto v^2_{rms}\propto T$$

Applying KVL in the circuit

$$15-10-1*i-.6*i=0$$ The current through voltmeter will be negligible as voltmeter have very high resistance.

$$\Rightarrow i=\frac{5}{1.6}=\frac{25}{8}$$

The voltmeter reading is given as potential difference across points A and B

$$\Delta V= -i*.6+15=15-\frac{25}{8}*.6=15-1.875=13.12 V$$

reading across voltmeter will be 13.2 V, correct option is B.

$$15-10-1*i-.6*i=0$$ The current through voltmeter will be negligible as voltmeter have very high resistance.

$$\Rightarrow i=\frac{5}{1.6}=\frac{25}{8}$$

The voltmeter reading is given as potential difference across points A and B

$$\Delta V= -i*.6+15=15-\frac{25}{8}*.6=15-1.875=13.12 V$$

reading across voltmeter will be 13.2 V, correct option is B.

To determine the excess pressure, we make a cut along the longitudinal axis and construct a surface as illustrated in the diagram.

Consider the equilibrium of the surface, surface ABCDE. Forces acting on this surface are

(i) $$F_1$$, due to the surface tension of the surface ACE in contact,

(ii) $$F_2$$, due to the air outside the surface ACE and

(iii) $$F_3$$, due to the liquid inside the surface ACE.

The free body is in static equilibrium. According to Newton's first law of motion:

$$P_1\pi R^2+2SR=P_2\pi R^2$$

$$P_1-P_2=\dfrac{2S}{\pi R}$$

In capacitor circuit at time t=0 capacitor behaves like zero resistance and the current is maximum i.e. $$\dfrac{E}{R}$$ in circuit, after infinite time the capacitor offers infinite resistance like and current in the circuit is zero.

In inductive circuit at time t=0 inductor behaves like infinite resistance and the current is zero in circuit, after infinite time the capacitor behaves like zero resistance and current in the circuit is maximum i.e.$$\dfrac{E}{R}$$.

In inductive circuit at time t=0 inductor behaves like infinite resistance and the current is zero in circuit, after infinite time the capacitor behaves like zero resistance and current in the circuit is maximum i.e.$$\dfrac{E}{R}$$.

$$\lambda = \dfrac{h}{p}= \dfrac{h}{\sqrt{2mE}}= \dfrac{h}{\sqrt{2mqV}}$$

Substituting the values,

$$\lambda = 1.7\: \overset{0}{A}$$

Substituting the values,

$$\lambda = 1.7\: \overset{0}{A}$$

The tower PQ in figure (19.5) subtends an angle $$\alpha$$ on the objective. As $$u_o$$ is very large, the first image P Q' is formed in the focal plane of the objective.

$$tan \alpha=\alpha=\dfrac{P'Q'}{f\circ{}}$$ ..(i)

The final image P"Q" subtends an angle $$\beta$$ on the eyepiece (and

hence on the eye). We have from the triangle P Q'E

$$tan \beta=\beta= \dfrac{P'Q'}{EP'}$$ ...(ii)

the telescope is set for normal adjustment so that the final image is formed at infinity, the first image P'Q' must be in the focal plane of the eyepiece.Then EP

=$$ {f_e}$$ . Thus, equation (ii) becomes

$$tan \beta=\beta= \dfrac{f\circ{}}{f_e}$$ ...(iii)

dividing equation (iii) by (i)

$$\dfrac{\beta}{\alpha}=\dfrac{f\circ{}}{{f_e}}$$...(iv)

from equation (i) The $$ \alpha =\dfrac{P'Q'}{f\circ{}} $$

$$P'Q'= m h_1$$ ...(v) where m is magnification of objective lens

$$m=\dfrac{v}{u}$$ where v is position of first image P'Q' and u is position of tower $$v= f\circ{}=150 \;cm\;, u=1000\;m $$ substituting value of m in equation (v) m

$$P'Q'= \dfrac{.15}{1000} 50m=.0075 m$$

$$ \alpha =\dfrac{.0075}{.15}=.05$$

value of $$\beta$$ can be obtained from equation (v)

$$ tan \beta =\dfrac{.15}{.05}\times \alpha=30\times.05=1.5 rad{}$$

$$ \beta = 56.3$$

Inudctor opposes change in curren, as the current is decreased, the inductor will convert magnetic energy into electrical by creating a potential difference. The voltage drop across inductor is given by

$$V=L\dfrac{di}{dt}$$

$$ -50=L \dfrac{2-5}{.1}$$ here negative sign is because voltage is increasing instead of drop across inductor

$$ -50=L \dfrac{2-5}{.1}$$

$$L=1.67 \;H $$.

The velocity can be determined from $$Q=Av$$...(i) where $$Q$$ is volume flow rate, $$A$$ and $$V$$ are area and velocity of tube.

Given $$ V=15L=15\times 10^{-3} \; m^3, \;A=\dfrac{\pi D^2}{4} =\dfrac{\pi \dfrac{0.02}{\sqrt{\pi}}^2}{4}=10^{-4} m^2$$

$$V=Qt$$

$$Q=\dfrac{V}{t}=\dfrac{15\times 10^{-3}}{60\times5}= 0.5 \times 10^{-4}m^3/s$$

Using Q in equation (i)

$$v=\dfrac{Q}{A}=\dfrac{0.5 \times 10^{-4}}{10^{-4}}=0.5 m/s$$

Reynolds number for flow in a tube is given as $$ R_n=\dfrac{vD}{\nu}$$ where $$\nu$$ is kinematic viscosity of fluid.

$$ R_n=\dfrac{vD}{\dfrac{\mu}{\rho}}$$ where $$\mu$$ is viscosity of fluid

$$ R_n=\dfrac{0.5 \dfrac{0.02}{\sqrt{\pi}}\times 10^3}{10^{-3}}$$

$$ R_n=5649$$ hence correct answer is option D

In p-type materials holes are majority carriers and electrons are majority carriers in n-type materials. When the two types of semiconductor materials are joined together, the electrons from the n-type material diffuse into p-type material and combines with holes as their concentration is higher in n-type layer. This creates a layer of negative ions near the junction in p-type material. Negative ions are formed because the trivalent impurities (e.g., Aluminum) now has an extra electron from the n-type material. Similarly, the holes from the p-type material diffuse into n-type material resulting in a layer of positive ions in the n-type material.

These negative ions creates an electric field in the direction from n-type to p-type. As more electrons diffuse into p-type material, the electric field strength goes on increasing. The electrons from n-type material now diffusing into p-type material will have to overcome the electric field due to negative ions. At one point, the electric field becomes sufficiently strong to stop further diffusion of electrons.

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