Category Archives: JEE Main

JEE Main

A cube is placed inside an electric field, $\vec E = 150 y^2 \hat j $ ….

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A cube is placed inside an electric field, $\vec E = 150 y^2 \hat j $. The side of the cube is 0.5 m and is placed in the field as shown in the given figure. The charge inside the cube is:

(A) $8.3 \times 10^{-11} C$
(B) $8.3 \times 10^{-12} C$
(C) $3.8 \times 10^{-12} C$
(D) $3.8 \times 10^{-11} C$

Solution

We have, $\phi = \frac {q_{in}}{\epsilon_0}$

Since electric field is upwardly directed, it will only cut the top and bottom surfaces.

Since y = 0 at the bottom surface makes electric field 0 there, no flux is present there. The only flux through the cube is contributed by the flux through the top surface.

Flux through the top surface $=150 \times 0.5^2 \times 0.5^2 $

So, $\phi =150 \times 0.5^2 \times 0.5^2 = \frac {q_{in}}{8.85 \times 10^{-12}}$

$\therefore q_{in} \approx 8.3 \times 10^{-11} C $

Answer: (A)

JEE Main

A glass tumbler having inner depth of 17.5 cm ….

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A glass tumbler having inner depth of 17.5 cm is kept on a table. A student starts pouring water $(\mu = 4/3 )$ into it while looking at the surface of water from the above. When he feels that the tumbler is half filled, he stops pouring water. Up to what height, the tumbler is actually filled?

(A) 11.7 cm
(B) 8.75 cm
(C) 7.5 cm
(D) 10 cm

Solution

$\mu = \frac {Real\, Depth}{Apparent\, Depth} = \frac {17.5-y}{y}$

$\therefore \frac {4}{3} = \frac {17.5-y}{y} $

$\therefore y = 7.5 cm$

Real Depth = 17.5 – y = 10 cm

Answer: (D)

JEE Main

${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} – x} \right) – \cos \left( {\frac{{3\pi }}{4} – x} \right)} \right]$

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The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} – x} \right) – \cos \left( {\frac{{3\pi }}{4} – x} \right)} \right]$ is:

(A) $[ – 2,2]$
(B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$
(C) $(0,\sqrt 5 )$
(D) $[ 0,2]$

Solution

We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 – 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$

$ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x – \sin x)} \right]$

Now, $ – \sqrt 2  \le \cos x – \sin x \le \sqrt 2 $

$\therefore – 2 \le \sqrt 2 (\cos x – \sin x) \le 2$

$\therefore 1 \le 3 + \sqrt 2 (\cos x – \sin x) \le 5$

$\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x – \sin x)] \le {\log _{\sqrt 5 }}5$

$ \Rightarrow 0 \le f(x) \le 2$

Answer: (D)

JEE Main

An engine is attached to a wagon through a shock absorber ….

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An engine is attached to a wagon through a shock absorber of length 1.5 m. The system with a total mass of 40,000 kg is moving with a speed of $72 kmh^{-1}$ when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0 m. If 90% of energy of the wagon is lost due to friction, the spring constant is _ _ _ _ $\times 10^5 Nm^{-1}$.

Solution

10 % of K.E. = $\frac {1}{2} kx^2 $

$\therefore 0.1 \times \frac {1}{2} mv^2 = \frac {1}{2} kx^2 $

$\therefore 0.1 \times 40,000 \times (72 \times \frac {5}{18})^2 = k \times 1.0^2 $

$\Rightarrow k = 4,000 \times 20^2 = 16 \times 10^5 Nm^{-1} $

Answer: 16

JEE Main

The average translational kinetic energy of $N_2 $ ….

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The average translational kinetic energy of $N_2 $ gas molecules at _ _ _ _ $^\circ C $ becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 Volt. (Given $k_b = 1.38 \times 10^{-23} J/K $) (Fill the nearest integer)

Solution

K.E. of electron =  Translational K.E. of $N_2 $ molecules

$\therefore eV = \frac {3}{2} kT $

$\therefore 2 \times 1.6 \times 10^{-19} \times 0.1 = 3 \times 1.38 \times 10^{-23} \times T $

$\therefore T = \frac {1.6 \times 10^3 }{3 \times 0.69 } \approx 773 K = 500 ^\circ C $

Answer: 500

JEE Main

The width of one of the two slits in a YDSE ….

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The width of one of the two slits in a Young’s double slit experiment is three times the other slit. If the amplitude of light coming from a slit is proportional to the slit-width, the ratio of minimum to maximum intensity in the interference pattern is x:4 where x is _ _ _ _ .

Solution

Given, $\frac {A_1}{A_2} = \frac {3}{1} $

Now, $\frac {I_1}{I_2} = \frac {A_1 ^2 }{A_2 ^2} = (\frac {3}{1})^2 = \frac {9}{1} $

$\frac {I_{min}}{I_{max}} = \frac {(\sqrt I_1 – \sqrt I_2 )^2 }{(\sqrt I_1 + \sqrt I_2 )^2} = \frac {(3-1)^2}{(3+1)^2}=1:4 $

Answer: 1.00

JEE Main

Due to cold weather a 1 m water pipe of cross-sectional area ….

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Due to cold weather a 1 m water pipe of cross-sectional area $1 cm^2$ is filled with ice at $-10^\circ C $. Resistive heating is used to melt the ice. Current of 0.5 A is passed through $4 k\Omega $ resistance. Assuming that all the heat produced is used for melting, what is the minimum time required?
(Given latent heat of fusion for water/ice $=3.33 \times 10^5 J Kg^{-1}$, specific heat of ice $=2\times 10^3 J Kg^{-1}$ and density of ice $=10^3 Kg m^{-3} $ )

(A) 3.53 s
(B) 0.353 s
(C) 35.3 s
(D) 70.65 s

Solution

Electrical Energy = Thermal Energy

$\therefore I^2 Rt = Cm\Delta T + mL = m (C\Delta T + L) $

$\Rightarrow 0.5^2 \times 4000 \times t = \rho V (2 \times 10^3 \times 10 + 3.33 \times 10^5 )$

$\Rightarrow 1000 t = 10^3 \times Al (2 \times 10^4 + 33.3 \times 10^4 )$

$\Rightarrow t = 1 \times 10^{-4} \times 1 (35.3 \times 10^4 ) = 35.3 s $

Answer: (C)