We know,
$\displaystyle v=\dfrac{1}{4l} \sqrt{\dfrac{\gamma RT}{M} }$

$\displaystyle \therefore l=\dfrac{1}{4v} \sqrt{\frac{\gamma RT}{M} }$

therefore solving  we  get

$Neon \Rightarrow 0.459$

$Nitrogen \Rightarrow 0.363$

$Oxygen \Rightarrow 0.340$

$Argon \Rightarrow 0.348$
1
#113709
Solution
A student is performing an experiment using a resonance column and a tuning fork of frequency 244 s$^{-1}$. He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at  which resonance occurs is (0.350 $\pm$ 0.005)m, the gas in the tube is (Useful information : $\sqrt{167RT} = 640 J^{1/2}; \sqrt{140 RT}= 590 J^{1/2} mole^{-\frac{1}{2}}.$ The molar masses M in grams are given in the option. Take the values of $\displaystyle \sqrt{\frac{10}{M}}$ for each gas given)
A
Neon $\displaystyle ( M = 20, \sqrt{\frac{10}{20}} = \frac{7}{10})$
B
Nitrogen $\displaystyle ( M = 28, \sqrt{\frac{10}{28}} = \frac{3}{5} )$
C
Oxygen $\displaystyle ( M= 32, \sqrt{\frac{10}{32}} = \frac{9}{16})$
D
Argon $\displaystyle ( M = 36, \sqrt{\frac{10}{36}} = \frac{17}{32})$
As current leads voltage by $\pi /2$ in the given circuit initially, then ac voltage can be represent as

$V = V_0 sin \omega t$

$\therefore q = CV_0 sin \omega t = Q sin \omega t$

where, $Q = 2 \times 10^{-3}C$

At $t = 7\pi / 6 \omega; I = - \dfrac{\sqrt{3}}{2} I_0$ and hence current is
anticlockwise.

Current 'i' immediately after $t= \dfrac{7 \pi}{6 \omega}$ is $\displaystyle i = \dfrac{V_c + 50}{R} = 10 A$

Charge flow $= Q_{final }-Q_{7\pi/6\omega} = 2 \times 10^{-6} C$

Hence C & D are correct options.
2
#113716
Solution
At time t $=$ 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current $I(t) = I_0 cos (\omega t),$ with $I_0 = 1 A$ and $\omega = 500 rad/s$ starts flowing in it with the initial direction shown in the figure. At $t = \displaystyle \frac{7 \pi}{6 \omega}$, the key is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If C $=$ 20 $\mu$F, R $= 10 \Omega$ and the battery is ideal with emf of 50 V, identify the correct statement(s).
This question has multiple correct options
A
Magnitude of the maximum charge on the capacitor before $t = \displaystyle \frac{7 \pi}{6 \omega}$ is $1 \times 10^{-3}C$.
B
The current in the left part of the circuit just before $\displaystyle t = \frac{7 \pi}{6 \omega}$ is clockwise.
C
Immediately after A is connected to D, the current in R is 10 A.
D
$Q = 2 \times 10^{-3}C$
As E $=$ V/d
$E_1 / E_2=1$ (both parts have common potential difference)
Assume $C_0$ be the capacitance without dielectric for whole capacitor.
$\displaystyle k \frac{C_0}{3} + \frac{2C_0}{3} = C$
$\displaystyle \frac{C}{C_1} = \frac{2+k}{k}$
$\displaystyle \frac{Q_1}{Q_2} = \frac{k}{2}$
3
#113720
Solution
A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the area of its plates, as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is C$_1$. When the capacitor is charged, the plate area covered by the dielectric gets charge Q$_1$ and the rest of the area gets charge Q$_2$. The electric field in the dielectric is E$_1$ and that in the other portion is E$_2$. Choose the correct option/options, ignoring edge effects.
This question has multiple correct options
A
$\displaystyle \frac{E_1}{E_2} = 1$
B
$\displaystyle \frac{E_1}{E_2} = \frac{1}{K}$
C
$\displaystyle \frac{Q_1}{Q_2} = \frac{3}{K}$
D
$\displaystyle \frac{C}{C_1} = \frac{2+K}{K}$
$y=A\sin kx.\cos\omega x$
$x=3, l=3$
at $x=0$       node
at $x=3$      antinode
$v=f\lambda$
$\lambda = 4l = 12$
$\displaystyle \therefore \omega_{fundamental}= \frac{50\pi}{3}$
$\displaystyle Thus, \omega= \dfrac{50\pi}{3} , \dfrac{150\pi}{3} , \dfrac{250\pi}{3} , \dfrac{350\pi}{3} , \dfrac{450\pi}{3} , \dfrac{450\pi}{3} , \dfrac{650\pi}{3} , \dfrac{750\pi}{3}$
4
#113724
Solution
One end of a taut string of length $3 m$ along the x axis is fixed at $x$ $=$ $0$. The speed of the waves in the string is $100 ms^{-1}$. The other end of the string is vibrating in the y direction so that stationary waves are set up in the string. The possible waveform(s) of these stationary waves is (are)
This question has multiple correct options
A
$\displaystyle y (t) = A \sin \dfrac{\pi x}{6} \cos \dfrac{50 \pi t}{3}$
B
$\displaystyle y (t) = A \sin \dfrac{\pi x}{3} \cos \dfrac{100 \pi t}{3}$
C
$\displaystyle y (t) = A \sin \dfrac{5\pi x}{6} \cos \dfrac{250 \pi t}{3}$
D
$\displaystyle y (t) = A \sin \dfrac{5 \pi x}{2} \cos 250 \pi t$
Using lensmakers formula,
$\displaystyle \Rightarrow \dfrac{1}{f} = \big( \dfrac{ \mu _{2} }{ \mu _{1} }-1 \big) \big( \dfrac{ 1}{ R_{1} }-\dfrac{ 1}{ R_{2} } \big)$

when light travels from air to glass,

$\displaystyle \mu \Rightarrow 1\rightarrow1.4\rightarrow1.5$

$\displaystyle \Rightarrow \dfrac{1.5}{f_{1}} = \big( \dfrac{1.4-1}{R} \big) + \big( \dfrac{1.5-1.4}{R} \big)$

$\displaystyle \therefore f_{1} = 3R$

when light travels from glass to air,

$\displaystyle \mu \Rightarrow 1.5\rightarrow1.4\rightarrow1$

$\displaystyle \Rightarrow \dfrac{1}{f_{2}} = \big( \dfrac{1.4-1.5}{R} \big) + \big( \dfrac{1-1.4}{R} \big)$

$\displaystyle \therefore f_{2} = 2R$
5
#113726
Solution
A transparent thin film of uniform thickness and refractive index $n_1 = 1.4$ is coated on the convex spherical surface of radius R at one end of a long solid glass cylinder of refractive index $n_2 = 1.5$, as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance f$_1$ from the film, while rays of light traversing from glass to air get focused at distance f$_2$ from the film. Then
This question has multiple correct options
A
$|f_1| = 3R$
B
$|f_1| = 2.8R$
C
$|f_2| = 2R$
D
$|f_2| = 1.4R$
Joules law,

$H=I^{2}Rt= \dfrac{V^{2}t}{R}$

Also,

$\displaystyle R=\rho \dfrac{l}{A} \Rightarrow R \propto \dfrac{1}{A} \Rightarrow R \propto \dfrac{1}{r^{2}}$

$\displaystyle\therefore R' =\dfrac{R}{4}$

$\displaystyle parallel \Rightarrow R_{p}=(\dfrac{1}{R/4}+\frac{1}{R/4})^{-1}=\frac{R}{8}$

$\displaystyle series \Rightarrow R_{s}=\dfrac{R}{4}+\dfrac{R}{4}=\dfrac{R}{2}$

$\displaystyle \therefore \frac{V^{2}4}{R} = \frac{V^{2}t_{p}}{R_{p}}= \frac{V^{2}t_{s}}{R_{s}}$

$\displaystyle \Rightarrow t_{p}=0.5 min, t_{s}=2 min$
6
#113728
Solution
Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4 minutes to raise the temperature of 0.5 kg water by 40 K. This heater is replaced by a new heater having two wires of the same material, each of length L and diameter 2d. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40 K?
This question has multiple correct options
A
4 if wires are in parallel
B
2 if wires are in series
C
1 if wires are in series
D
0.5 if wires are in parallel
Writing Kirchoff's laws for the top and bottom loops ,
$V_1 -i_1R_1 - ( i_1 - i_2 ) R_2 = 0$
$V_2 -i_2 R_3 + ( i_1 - i_2) R_2 = 0$
If Current through $R_2$ is zero , $i_1 = i_2$
Hence , $\dfrac{V_1}{R_1} = \dfrac{V_2}{R_3}$
Substituting the values given in the options and verifying ,
Options A , B and D satisfy this condition.
7
#113730
Solution
Two ideal batteries of emf $V_1$ and $V_2$ and three resistances $R_1$, $R_2$ and $R_3$ are connected as shown in the figure. The current in resistance $R_2$ would be zero if
This question has multiple correct options
A
$V_1 = V_2$ and $R_1 = R_2 = R_3$
B
$V_1 = V_2$ and $R_1 = 2 R_2 = R_3$
C
$V_1 = 2V_2$ and $2R_1 = 2 R_2 = R_3$
D
$2V_1 = V_2$ and $2R_1 = R_2 = R_3$
$\displaystyle \frac{Q}{4 \pi \varepsilon_0r_0^2}=\frac{\lambda}{2 \pi \varepsilon_0 r_0} = \frac{\sigma}{2 \varepsilon_0}$
$\displaystyle E_1 \left ( \frac{r_0}{2} \right ) = \frac{Q}{\pi \varepsilon_0r_0^2}, E_2 \left ( \frac{r_0}{2} \right ), E_3 \left ( \frac{r_0}{2} \right ) = \frac{\sigma}{2 \varepsilon_0}$
$\displaystyle \therefore E_1 \left ( \frac{r_0}{2} \right ) = 2E_2 \left ( \frac{r_0}{2}\right)$
8
#113731
Solution
Let E$_1$(r), E$_2$(r) and E$_3$(r) be the respective electric fields at a distance r from a point charge Q, an infinitely long wire with constant linear charge density $\lambda$, and an infinite plane with uniform surface charge density $\sigma$. If E$_1(r_0) = E_2 (r_0) = E_3 (r_0)$ at a given distance r$_0$, then
A
$Q = 4 \sigma \pi r_0^2$
B
$r_0 =\frac{\lambda}{2 \pi \sigma}$
C
$E_1 (r_0 / 2) = 2E_2 (r_0/2)$
D
$E_2 (r_0/2) = 4E_3 (r_0/2)$
$\beta = \dfrac{D \lambda}{d}$
$\because \lambda_2 > \lambda_1 \Rightarrow \beta_2 > \beta_1$
Also $m_1\beta_1 = m_2 \beta_2 \Rightarrow m_1 > m_2$
Also $3 \left ( \dfrac{D}{d} \right ) (600 nm) = (2 \times 5 -1) \left ( \dfrac{D}{2d} \right ) 400 nm$
Angular width $\theta = \dfrac{\lambda}{d}$
9
#113733
Solution
A light source, which emits two wavelengths $\lambda_1 = 400$ nm and $\lambda_2 = 600$ nm, is used in a Young's double slit experiment. If recorded fringe widths for $\lambda_1$ and $\lambda_2$ are $\beta_1$ and $\beta_2$ and the number of fringes for them within a distance y on one side of the central maximum are m$_1$ and m$_2$, respectively, then
This question has multiple correct options
A
$\beta_2 > \beta_1$
B
$m_1 > m_2$
C
From the central maximum, 3$^{rd}$ maximum of $\lambda_2$ overlaps with 5$^{th}$ minimum of $\lambda_1$
D
The angular separation of fringes for $\lambda_1$ is greater than $\lambda_2$
Condition of translation equilibrium
$N_1 = \mu_2 N_2$
$N_2 + \mu_1 N_1 = Mg$
Solving $N_2 = \frac{mg}{1 + \mu_1 \mu_2}$
$N_1 = \frac{\mu_2 mg}{1 +\mu_1 \mu_2}$
Applying torque equation about corner (left) point on the floor
$mg \frac{l}{2} cos \theta = N_1 l sin \theta + \mu_1 N_1 l cos \theta$
Solving tan $\theta = \frac{1- \mu_1 \mu_2}{2 \mu_2}$
From the condition of equilibrium of bodies:
$\Sigma F_{x} = 0$
$\therefore - N_{1} + \mu _{2} N_{2} = 0$
$\therefore N_{1} = \mu _{2} N_{2}$
Also $\Sigma F_{y} = 0$
$\therefore N_{2} + \mu _{1} N_{1} -mg = 0$
$\therefore N_{2} + \mu _{1} \mu _{2} N_{2} -mg = 0$
$\therefore N_{2} (1 + \mu _{1} \mu _{2}) = mg$
$\therefore N_{2} = \dfrac{mg}{1 + \mu _{1} \mu _{2}}$
Applying torque equation about corner (left) point on the floor:
$mg \frac{l}{2} cos \theta = N_1 l sin \theta + \mu_1 N_1 l cos \theta$

Solving: $N_1tan\theta =\dfrac{mg}{2}-\mu_1N_1=\dfrac{mg}{2}-\dfrac{\mu_1\mu_2mg}{1+\mu_1\mu_2}$
Solving we get: $N_1tan\theta=\dfrac{mg}{2}$
10
#113736
Hint Solution
In the figure, a ladder of mass $m$ is shown leaning against a wall. It is in static equilibrium making an angle $\theta$ with the horizontal floor. The coefficient of friction between the wall and the ladder is $\mu_1$ and that between the floor and the ladder is $\mu_2$. The normal reaction of the wall on the ladder is $N_1$ and that of the floor is $N_2$. If the ladder is about to slip, then
This question has multiple correct options
A
$\mu_1 = 0 \mu_2 \neq 0$ and $N_2 tan \theta = \frac{mg}{2}$
$\mu_1 \neq 0 \mu_2 = 0$ and $N_1 tan \theta = \frac{mg}{2}$
$\mu_1 \neq 0 \mu_2 \neq 0$ and $N_2 = \frac{mg}{1 + \mu_1 \mu_2}$
$\mu_1 = 0 \mu_2 \neq 0$ and $N_1 tan \theta = \frac{mg}{2}$