JEE Main

An empty LPG cylinder weighs 14.8 kg ….

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An empty LPG cylinder weighs 14.8 kg. When full, it weighs 29.0 kg and shows a pressure of 3.47 atm. In the course of use at ambient temperature, the mass of the cylinder is reduced to 23.0 kg. The final pressure inside the cylinder is _ _ _ _ atm. (Nearest integer)
(Assume LPG to be an ideal gas)

Solution

Weight of gas before use = 29.0 – 14.8 = 14.2 kg

Weight of gas after use = 23.0 – 14.8 = 8.2 kg

Now, $\frac {P_1}{n_1} = \frac {P_2}{n_2} $

$\Rightarrow \frac {P_1}{w_1} = \frac {P_2}{w_2} $

$\therefore \frac {3.47}{14.2} =  \frac {P_2}{8.2} $

$\Rightarrow P_2 = \frac {3.47 \times 8.2}{14.2} \approx 2 atm $

Answer: 2

JEE Main

$f(x) = x + \int\limits_0^{\pi /2} {\sin x.\cos yf(y)dy} $

f(x)=?

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The function f(x), that satisfies the condition $f(x) = x + \int\limits_0^{\pi /2} {\sin x.\cos yf(y)dy} $ is:

(A) $x + \frac{2}{3}(\pi  – 2)\sin x$
(B) $x + (\pi  + 2)\sin x$
(C) $x + \frac{\pi }{2}\sin x$
(D) $x + (\pi  – 2)\sin x$

Solution

We have, $f(x) = x + \int\limits_0^{\pi /2} {\sin x.\cos yf(y)dy}  = x + \sin x\int\limits_0^{\pi /2} {\cos yf(y)dy}  = x + \sin x.k$

Where, $k = \int\limits_0^{\pi /2} {\cos yf(y)dy}  = \int\limits_0^{\pi /2} {\cos y(y + \sin y.k)dy} $

$\therefore k = \int\limits_0^{\pi /2} {y\cos ydy}  + k\int\limits_0^{\pi /2} {\cos y\sin ydy} $

$ \Rightarrow k = \left. {y\sin y} \right|_0^{\pi /2} – \int\limits_0^{\pi /2} {\sin ydy}  + \frac{k}{2}\int\limits_0^{\pi /2} {sin2ydy}  = \frac{\pi }{2} + \left. {\cos y} \right|_0^{\pi /2} – \frac{k}{2}.\frac{1}{2}\left. {\cos 2y} \right|_0^{\pi /2}$

$ \Rightarrow k = \frac{\pi }{2} – 1 – \frac{k}{4}( – 1 – 1) = \frac{\pi }{2} – 1 + \frac{k}{2}$

$ \Rightarrow \frac{k}{2} = \frac{\pi }{2} – 1 = \frac{{\pi  – 2}}{2}$

$\therefore k = \pi  – 2$

Now, $f(x) = x + \sin x.k = x + (\pi  – 2)\sin x$

Answer: (D)

JEE Main

A 2 kg steel rod of length 0.6 m is clamped ….

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A 2 kg steel rod of length 0.6 m is clamped on a table vertically at its lower end and is free to rotate in vertical plane. The upper end is pushed so that the rod falls under gravity. Ignoring the friction due to clamping at its lower end, the speed of the free end of rod when it passes through its lowest position is _ _ _ _ $m s^{-1}$.

Solution

Decrease in gravitational potential energy = Increase in kinetic energy

So, $mgl = \frac {1}{2} I \omega ^2 = \frac {1}{2} .\frac {1}{3} ml^2 \omega ^2 $

$\Rightarrow 6g = l \omega ^2 $

Using $v = l \omega $, we have

$6 g = l. \frac {v^2}{l^2}$

$\Rightarrow v^2 = 6gl $

$\Rightarrow v = \sqrt {6gl} = \sqrt {6\times 10 \times 0.6 } = 6ms^{-1} $

Answer: 6

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)