IIT JEE 1984 Physics Question Paper with Answer and Solution

(A) Due to symmetry,at any instant,the four persons will be at the corners of a square whose side length gradually decreases. They will eventually meet at the center $O$ of the square.
The velocity of each person is $v$. At any instant,the velocity component of a person directed towards the center $O$ is $v \cos(45^{\circ}) = \frac{v}{\sqrt{2}}$.
The initial distance of each person from the center $O$ is half the diagonal of the square,which is $\frac{d\sqrt{2}}{2} = \frac{d}{\sqrt{2}}$.
The time taken to reach the center is the distance divided by the velocity component along the radial direction:
$t = \frac{\text{distance}}{\text{velocity component}} = \frac{d/\sqrt{2}}{v/\sqrt{2}} = \frac{d}{v}$.

(C) Given that power $P$ is constant. We know that $P = Fv = mav = m \left( \frac{dv}{dt} \right) v$.
Integrating this expression: $\frac{P}{m} dt = v dv$.
Integrating both sides: $\int \frac{P}{m} dt = \int v dv \implies \frac{P}{m} t = \frac{v^2}{2}$.
Thus,$v^2 = \frac{2P}{m} t$,which gives $v = \sqrt{\frac{2P}{m}} t^{1/2}$.
Since $v = \frac{ds}{dt}$,we have $ds = \sqrt{\frac{2P}{m}} t^{1/2} dt$.
Integrating with respect to $t$: $s = \int \sqrt{\frac{2P}{m}} t^{1/2} dt = \sqrt{\frac{2P}{m}} \left( \frac{t^{3/2}}{3/2} \right) = \sqrt{\frac{2P}{m}} \left( \frac{2}{3} t^{3/2} \right)$.
Therefore,$s \propto t^{3/2}$.

(A) Let the mass of the shell be $M$. When fired with velocity $v$ at an angle $\theta$,its velocity at the highest point is $v_x = v \cos \theta$ in the horizontal direction.
The momentum of the shell just before the explosion is $P_i = Mv \cos \theta$.
After the explosion,the shell splits into two equal pieces of mass $m = M/2$.
One piece retraces its path,meaning its velocity becomes $-v \cos \theta$ (opposite to the original horizontal direction).
Let the velocity of the other piece be $V$. By the law of conservation of linear momentum:
$P_i = P_f$
$Mv \cos \theta = m(-v \cos \theta) + mV$
Since $m = M/2$,we have:
$Mv \cos \theta = \frac{M}{2}(-v \cos \theta) + \frac{M}{2}V$
Dividing by $M/2$:
$2v \cos \theta = -v \cos \theta + V$
$V = 3v \cos \theta$
Thus,the speed of the other piece is $3v \cos \theta$.

(A) The formula for $r.m.s.$ speed is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Rearranging for molar mass $M$,we get $M = \frac{3RT}{v_{rms}^2}$.
Given: $R = 8.3\, J/(mol \cdot K)$,$T = 300\, K$ (room temperature),and $v_{rms} = 1920\, m/s$.
Substituting the values: $M = \frac{3 \times 8.3 \times 300}{(1920)^2}$.
$M = \frac{7470}{3686400} \approx 0.002026\, kg/mol \approx 2 \times 10^{-3}\, kg/mol = 2\, g/mol$.
The molar mass of $H_2$ is $2\, g/mol$. Therefore,the gas is $H_2$.

(B) Comparing the given equation $Y = Y_0 \sin 2\pi \left( ft - \frac{x}{\lambda} \right)$ with the standard wave equation $y = a \sin(\omega t - kx)$:
We identify the amplitude $a = Y_0$,angular frequency $\omega = 2\pi f$,and wave number $k = \frac{2\pi}{\lambda}$.
The maximum particle velocity is given by $(v_{\max})_{\text{particle}} = a\omega = Y_0 \times 2\pi f$.
The wave velocity (phase velocity) is given by $v_{\text{wave}} = \frac{\omega}{k} = \frac{2\pi f}{2\pi / \lambda} = f\lambda$.
According to the problem,the maximum particle velocity is four times the wave velocity:
$(v_{\max})_{\text{particle}} = 4 v_{\text{wave}}$
$Y_0 \times 2\pi f = 4 f\lambda$
Dividing both sides by $4f$,we get:
$\lambda = \frac{2\pi Y_0}{4} = \frac{\pi Y_0}{2}$.

(D) The dimensions of the given quantities are:
$L = [M L^2 T^{-2} A^{-2}]$
$C = [M^{-1} L^{-2} T^4 A^2]$
$R = [M L^2 T^{-3} A^{-2}]$
$1$. For $\frac{1}{RC}$: The time constant $\tau = RC$ has dimensions of time $[T]$. Thus,$\frac{1}{RC}$ has dimensions of $[T^{-1}]$,which is frequency.
$2$. For $\frac{R}{L}$: The ratio $\frac{R}{L}$ has dimensions $[M L^2 T^{-3} A^{-2}] / [M L^2 T^{-2} A^{-2}] = [T^{-1}]$,which is frequency.
$3$. For $\frac{1}{\sqrt{LC}}$: The resonant frequency $\omega = \frac{1}{\sqrt{LC}}$ has dimensions $[T^{-1}]$,which is frequency.
$4$. For $\frac{C}{L}$: The dimensions are $[M^{-1} L^{-2} T^4 A^2] / [M L^2 T^{-2} A^{-2}] = [M^{-2} L^{-4} T^6 A^4]$. This does not represent frequency.
Therefore,the correct option is $D$.

(D) By the symmetry of the problem,the components of the force on $Q$ due to the charges at $A$ and $B$ along the $Y$-axis will cancel each other,while the components along the $X$-axis will add up and be directed towards the origin $O$. Under the action of this force,the charge $Q$ will move towards $O$.
If at any time the charge $Q$ is at a distance $x$ from $O$,the net force on charge $Q$ is:
$F_{net} = 2F \cos \theta = 2 \left( \frac{1}{4\pi \varepsilon_0} \frac{qQ}{a^2 + x^2} \right) \left( \frac{x}{\sqrt{a^2 + x^2}} \right)$
$F_{net} = - \frac{1}{4\pi \varepsilon_0} \frac{2qQx}{(a^2 + x^2)^{3/2}}$
Since the restoring force $F_{net}$ is not proportional to the displacement $x$ (it is not linear),the motion will be oscillatory (with an amplitude of $2a$) but not simple harmonic.

(C) The given circuit consists of $4$ capacitors connected in parallel. Each capacitor is formed by two adjacent plates with area $A$ and separation $d$. The capacitance of each capacitor is $C = \frac{\varepsilon_0 A}{d}$.
Plate $1$ is connected to the positive terminal of the battery. It forms one capacitor with plate $2$. The charge on plate $1$ is $q_1 = +CV = +\frac{\varepsilon_0 AV}{d}$.
Plate $4$ is connected to the negative terminal of the battery. It acts as one plate for two capacitors: one with plate $3$ and one with plate $5$. Since plate $4$ is connected to the negative terminal,the charge on both sides of plate $4$ will be negative. The total charge on plate $4$ is $q_4 = -CV - CV = -2CV = -\frac{2\varepsilon_0 AV}{d}$.
Thus,the charges on plates $1$ and $4$ are $\frac{\varepsilon_0 AV}{d}$ and $-\frac{2\varepsilon_0 AV}{d}$ respectively.

(D) Nuclear reactions occur when nuclei interact to form new products. In the carbon-nitrogen cycle,the reactions $_6C^{12} + _1H^1 \to _7N^{13} + 2 \text{ MeV}$ and $_7N^{14} + _1H^1 \to _8O^{15} + 7.3 \text{ MeV}$ are well-known,experimentally verified nuclear fusion processes that occur in stars. The reaction in option $(a)$ is not a standard nuclear fusion reaction in this context. Therefore,both $(b)$ and $(c)$ are possible nuclear reactions.

(D) diode acts as a unidirectional switch,allowing current to flow in one direction,which makes it suitable for rectification. Thus,statement $(A)$ is correct.
$A$ triode can also be used as a rectifier by connecting its grid to the cathode,effectively making it act like a diode. However,in standard electronic circuit theory,a triode is primarily used for amplification. Statement $(B)$ is technically incorrect because a triode can be used as a rectifier.
The amplification of a signal by a triode is achieved by operating it in the linear region of its $I-V$ characteristic curve,which ensures that the output signal is a faithful reproduction of the input signal without distortion. Thus,statement $(C)$ is correct.
Since both $(A)$ and $(C)$ are correct,the correct option is $(D)$.

(D) Let $P$ be the point on the screen directly in front of slit $S_1$. The path difference between the light rays reaching $P$ from $S_1$ and $S_2$ is given by:
$\Delta x = S_2P - S_1P = \sqrt{b^2 + d^2} - d$
Using binomial expansion for $d >> b$:
$\Delta x = d(1 + \frac{b^2}{d^2})^{1/2} - d \approx d(1 + \frac{b^2}{2d^2}) - d = \frac{b^2}{2d}$
For destructive interference (missing wavelengths),the path difference must be an odd multiple of $\frac{\lambda}{2}$:
$\Delta x = (2n - 1)\frac{\lambda}{2}$,where $n = 1, 2, 3, ...$
Equating the two expressions for path difference:
$\frac{b^2}{2d} = (2n - 1)\frac{\lambda}{2}$
$\lambda = \frac{b^2}{(2n - 1)d}$
For $n = 1$,$\lambda = \frac{b^2}{d}$.
For $n = 2$,$\lambda = \frac{b^2}{3d}$.
Thus,both $(a)$ and $(c)$ are correct.