JEE Mains (Apr 10) 2015 Paper

$$P=\dfrac{nM}{3V}v_{rms}^2$$
Thus pressure $$P\propto v_{rms}^2$$
and thus
Force $$\propto v^2_{rms}\propto T$$
1
#306899
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In an ideal gas at temperature T, the average force that a molecules applies on the walls of a closed container depends on T as $$T^q$$. As good estimate for q is.
A
$$\frac {1}{4}$$
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B
$$\frac {1}{2}$$
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C
1
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D
2
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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.
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A 10V battery with internal resistance 1 $$\Omega $$ and a 15V battery with internal resistance $$0.6 \Omega $$ are connected in parallel to a voltmeter. The reading in the voltmeter will be close to.
A
11.9 V
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B
13.1 V
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C
24.5 V
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D
12.5 V
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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:

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

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

3
#306905
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If two glass plates have water between them and are separated by very small distance (see figure), it is very difficult to pull them apart. It is because the water in between forms cylindrical surface on the side that give rise to lower pressure in the water in compression to atmosphere. If the radius of the cylindrical surface is $$R$$ and surface tension of water is $$T$$ then the pressure in water between the plates is lower by.
A
$$\displaystyle \dfrac{T}{4R}$$
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B
$$\displaystyle \dfrac{T}{2R}$$
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C
$$\displaystyle \dfrac{4T}{R}$$
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D
$$\displaystyle \dfrac{2T}{R}$$
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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.$$\dfrac{E}{R}$$.
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In the circuit (a) and (b) switches $$S_1$$ and $$S_2$$ are closed at t-0 and are kept closed for a long time. The variation of currents in the two circuits for $$t \geq 0$$ are roughly shown by(figures are schematic and not drawn to scale).
A
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B
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C
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D
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$$\lambda = \dfrac{h}{p}= \dfrac{h}{\sqrt{2mE}}= \dfrac{h}{\sqrt{2mqV}}$$

Substituting the values,
$$\lambda = 1.7\: \overset{0}{A}$$
5
#306908
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de-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to $$(|e|=1.6 \times 10^{-19} C, m_e=9.1 \times 10^{31}kg, h=6.6 \times 10^{-34} Js).$$ 
A
$$0.5 \mathring {A} $$
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B
$$1.7 \mathring {A} $$
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C
$$2.4 \mathring {A} $$
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D
$$1.2 \mathring {A} $$
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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)

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

$$ \beta = 56.3$$
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#306910
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A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting, the angle formed by the image of the tower is $$\theta$$, then $$\theta$$ is closed to.
A
$$60^o$$
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B
$$1^o$$
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C
$$30^o$$
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D
$$15^o$$
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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

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

$$L=1.67 \;H $$.
7
#306911
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When current is coil changes from 5 A to 2 A in 0.1 s, an average of 50 V is produced. The self-inductance of the coil is.
A
6 H
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B
0.67 H
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C
1.67 H
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D
3 H
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8
#306913
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A 25 cm long solenoid has radius 2 cm and 500 total number of turns. It carries a current of 15 A. If it is equivalent to magnet of the same size and magnetization $$ \overrightarrow{M} $$ Magnetic moment / vloume), then $$\overrightarrow{|M|}$$ is.
A
$$300 Am^{-1}$$
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B
$$30000 Am^{-1}$$
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C
$$30000 \pi Am^{-1}$$
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D
$$3\pi Am^{-1}$$
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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
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#306914
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If it takes $$5$$ minutes to fill a $$15L$$ bucket from a water tap of diameter $$ \dfrac {2}{\sqrt {\pi}} $$ cm then the Reynolds number for the flow is (density of water = $$10^3 kg/m^3$$ and visocsity of water =$$10^{-3}Pa. s$$) close to.
A
$$11,000$$
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B
$$550$$
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C
$$1100$$
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D
$$5500$$
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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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#306915
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In an unbiased n-p junction electrons diffuse from n-region to p- region because
A
electrons travel across the junction due to potential difference.
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B
electrons concentration in n region is more as compared to that in p-region.
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C
holes in p-region attract them.
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D
only electrons move from n- to p- region and not the vice-versa.
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