IIT JEE 1992 Physics Question Paper with Answer and Solution

(D) Let the dimension of length be $L = G^x c^y h^z$.
The dimensions are:
$G = [M^{-1} L^3 T^{-2}]$
$c = [L T^{-1}]$
$h = [M L^2 T^{-1}]$
Substituting these into the equation:
$[M^0 L^1 T^0] = [M^{-1} L^3 T^{-2}]^x [L T^{-1}]^y [M L^2 T^{-1}]^z$
$[M^0 L^1 T^0] = M^{-x+z} L^{3x+y+2z} T^{-2x-y-z}$
Comparing the powers on both sides:
$1$) $-x + z = 0 \implies x = z$
$2$) $3x + y + 2z = 1$
$3$) $-2x - y - z = 0 \implies y = -2x - z$
Substitute $x=z$ into $(3)$: $y = -2x - x = -3x$.
Substitute $x=z$ and $y=-3x$ into $(2)$:
$3x + (-3x) + 2x = 1$
$2x = 1 \implies x = 1/2$.
Since $x=z$,$z = 1/2$.
Since $y = -3x$,$y = -3/2$.
Thus,$x = 1/2, y = -3/2, z = 1/2$. Both options $(b)$ and $(c)$ are correct.

(D) The block $A$ acts as a mass $M$ and the block $B$ acts as a spring with a spring constant $k$.
For a block of side $L$ and modulus of rigidity $\eta$,the shear force $F$ is related to the displacement $x$ by $F = \eta A \frac{x}{L}$,where $A = L^2$ is the area.
Thus,$F = \eta L^2 \frac{x}{L} = (\eta L)x$.
Comparing this with Hooke's law $F = kx$,we get the spring constant $k = \eta L$.
The time period $T$ of a simple harmonic oscillator is given by $T = 2\pi \sqrt{\frac{M}{k}}$.
Substituting $k = \eta L$,we get $T = 2\pi \sqrt{\frac{M}{\eta L}}$.

(A) $1$. Curie is a unit of radioactivity,defined as the number of disintegrations per second. Its dimension is $[T^{-1}]$,which corresponds to $(G)$.
$2$. Light year is a unit of distance. Its dimension is $[L]$,which corresponds to $(H)$.
$3$. Dielectric strength is the maximum electric field a material can withstand. Electric field $E = F/q$. Dimension: $[MLT^{-2}] / [IT] = [MLT^{-3}I^{-1}]$,which corresponds to $(I)$.
$4$. Atomic weight is a dimensionless ratio of masses. Its dimension is $[M^0L^0T^0]$,but since it represents mass,it is often associated with $[M]$ in specific contexts,which corresponds to $(B)$.
$5$. Decibel is a logarithmic unit used to express the ratio of two values of a physical quantity,such as power or intensity. It is dimensionless,which corresponds to $(C)$.
Thus,the correct match is $(i)-G, (ii)-H, (iii)-I, (iv)-B, (v)-C$.

(A) Let the total distance be $x$. The first half distance is $x/2$ covered with speed $v_1 = 3 \, m/s$. The time taken is $t_1 = \frac{x/2}{3} = \frac{x}{6}$.
The second half distance $x/2$ is covered in two equal time intervals,say $t_2$ each. Let the speeds be $v_2 = 4.5 \, m/s$ and $v_3 = 7.5 \, m/s$.
Distance covered in the second half is $(v_2 \cdot t_2) + (v_3 \cdot t_2) = x/2$.
$(4.5 + 7.5) t_2 = x/2 \Rightarrow 12 t_2 = x/2 \Rightarrow t_2 = x/24$.
The total time taken for the second half is $2 t_2 = 2(x/24) = x/12$.
Total time $T = t_1 + 2 t_2 = \frac{x}{6} + \frac{x}{12} = \frac{2x + x}{12} = \frac{3x}{12} = \frac{x}{4}$.
Average speed $v_{avg} = \frac{\text{Total distance}}{\text{Total time}} = \frac{x}{x/4} = 4 \, m/s$.

(A) Consider a small element of the liquid of length $dx$ at a distance $x$ from the axis of rotation. The mass of this element is $dM = (M/L)dx$.
The centripetal force required for this element to rotate in a circle of radius $x$ is $dF = (dM)\omega^2 x$.
Substituting $dM$,we get $dF = (M/L)dx \cdot \omega^2 x = (M\omega^2/L)x dx$.
The total force $F$ exerted by the liquid at the outer end is the integral of these centripetal forces from $x = 0$ to $x = L$:
$F = \int_0^L \frac{M\omega^2}{L} x dx = \frac{M\omega^2}{L} \left[ \frac{x^2}{2} \right]_0^L = \frac{M\omega^2}{L} \cdot \frac{L^2}{2} = \frac{1}{2}ML\omega^2$.

(C) When the car moves in a circular path,the plumb bob experiences a pseudo-force (centrifugal force) in the frame of the car,directed radially outward,given by $F_c = \frac{mv^2}{r}$.
The gravitational force acting on the bob is $mg$,directed vertically downward.
Let $\theta$ be the angle the rod makes with the vertical. In the equilibrium position,the net force on the bob in the car's frame is zero.
Taking the ratio of the horizontal force to the vertical force,we get:
$\tan \theta = \frac{F_c}{mg} = \frac{mv^2/r}{mg} = \frac{v^2}{rg}$.
Given $v = 10 \, m/s$,$r = 10 \, m$,and taking $g = 10 \, m/s^2$:
$\tan \theta = \frac{10^2}{10 \times 10} = \frac{100}{100} = 1$.
Therefore,$\theta = \tan^{-1}(1) = 45^\circ$.

(C) The stress at any point in the wire is defined as the force acting on the cross-section at that point divided by the area of cross-section $S$.
At a height $3L/4$ from the lower end,the total force acting on the cross-section is the sum of the weight $W_1$ suspended at the bottom and the weight of the portion of the wire below that point.
Since the wire is uniform,the weight of a portion of the wire is proportional to its length.
The length of the wire below the point at height $3L/4$ from the lower end is $3L/4$.
Therefore,the weight of this portion of the wire is $(3/4)W$.
The total force at this cross-section is $F = W_1 + (3W/4)$.
Thus,the stress is $\sigma = \frac{F}{S} = \frac{W_1 + (3W/4)}{S}$.

(B) The elastic potential energy stored per unit volume $(u)$ in a material is given by the formula:
$u = \frac{1}{2} \times \text{stress} \times \text{strain}$
We know that Young's modulus $(Y)$ is defined as the ratio of stress $(S)$ to strain $(\epsilon)$:
$Y = \frac{S}{\epsilon} \implies \epsilon = \frac{S}{Y}$
Substituting the expression for strain into the energy formula:
$u = \frac{1}{2} \times S \times \left( \frac{S}{Y} \right)$
$u = \frac{S^2}{2Y}$
Thus,the correct option is $B$.

(D) thermostat is a device used in electric appliances like refrigerators and irons for automatic temperature control.
It consists of a bimetallic strip made of two different metals joined together.
When the temperature changes, the two metals expand by different amounts because they have different coefficients of linear expansion $(\alpha)$.
This difference in expansion causes the strip to bend, which triggers the switching mechanism to turn the appliance on or off.
Therefore, the two metal strips must necessarily differ in their coefficient of linear expansion.

(B) The power $P$ supplied is used to maintain the substance in the molten state by compensating for the heat lost to the surroundings.
When the power is turned off,the substance solidifies by releasing its latent heat of fusion $L$ over time $t$.
The total heat released during solidification is $Q = mL$.
The rate of heat loss is $\frac{Q}{t} = \frac{mL}{t}$.
Since the power $P$ was required to maintain the molten state,it must be equal to the rate of heat loss: $P = \frac{mL}{t}$.
Solving for $L$,we get $L = \frac{Pt}{m}$.

(B) The average speed of gas molecules is given by the formula $V_{av} = \sqrt{\frac{8RT}{\pi M}}$.
Since the temperature $T$ is the same for all vessels and the molar mass $M$ of $O_2$ is constant,the average speed of $O_2$ molecules depends only on the temperature and the molar mass of the gas.
In vessel $C$,the $O_2$ and $N_2$ molecules behave independently as an ideal gas mixture.
Therefore,the average speed of $O_2$ molecules in vessel $C$ remains the same as in vessel $A$,which is $V_1$.

(B) Given the displacement equation: $y = 4\cos^2(t/2)\sin(1000t)$.
Using the trigonometric identity $2\cos^2(\theta) = 1 + \cos(2\theta)$,we can rewrite the expression as:
$y = 2(2\cos^2(t/2))\sin(1000t) = 2(1 + \cos(t))\sin(1000t)$.
Expanding this,we get:
$y = 2\sin(1000t) + 2\cos(t)\sin(1000t)$.
Using the product-to-sum identity $2\sin(A)\cos(B) = \sin(A+B) + \sin(A-B)$,where $A = 1000t$ and $B = t$:
$y = 2\sin(1000t) + \sin(1000t + t) + \sin(1000t - t)$.
$y = 2\sin(1000t) + \sin(1001t) + \sin(999t)$.
This expression is the sum of $3$ independent simple harmonic motions ($S$.$H$.$M$.).
Therefore,the correct option is $B$.

(D) The superposition of two waves travelling in opposite directions with the same frequency and amplitude results in a stationary wave.
Using the principle of superposition,the resultant wave is $y = y_1 + y_2 = A\sin[k(x - ct)] + A\sin[k(x + ct)]$.
Using the trigonometric identity $\sin(C) + \sin(D) = 2\sin(\frac{C+D}{2})\cos(\frac{C-D}{2})$,we get:
$y = 2A\sin(kx)\cos(kct)$.
Nodes occur where the displacement is zero for all time $t$,which happens when $\sin(kx) = 0$.
This implies $kx = n\pi$,where $n = 0, 1, 2, \dots$.
The positions of the nodes are $x = \frac{n\pi}{k}$.
The distance between two adjacent nodes is the difference between consecutive positions:
$\Delta x = \frac{(n+1)\pi}{k} - \frac{n\pi}{k} = \frac{\pi}{k}$.
Alternatively,since $k = \frac{2\pi}{\lambda}$,the distance between adjacent nodes is $\frac{\lambda}{2} = \frac{\pi}{k}$.

(D) According to the theorem of perpendicular axes,the moment of inertia $I_z$ of a planar body about an axis perpendicular to its plane and passing through a point $O$ is equal to the sum of the moments of inertia about two mutually perpendicular axes in the plane of the body,both passing through $O$.
In the given figure,axes $1$ and $2$ are the diagonals of the square,which are mutually perpendicular. Thus,$I_0 = I_1 + I_2$.
Axes $3$ and $4$ are lines passing through the centre $O$ and parallel to the sides of the square. These are also mutually perpendicular. Thus,$I_0 = I_3 + I_4$.
Furthermore,due to the symmetry of the square,the moment of inertia about any axis passing through the centre and lying in the plane of the plate is the same if the angle of the axis is the same relative to the sides. Specifically,for a square plate,$I_1 = I_2 = I_3 = I_4 = \frac{ML^2}{12}$.
Therefore,$I_0 = I_1 + I_2 = I_3 + I_4 = I_1 + I_3$ (since $I_1 = I_3$).
Thus,all the given options are correct.

(A) The electric potential is given by $V(x, y, z) = 4x^2 \text{ volt}$.
The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} = -\nabla V = -\left( \hat{i} \frac{\partial V}{\partial x} + \hat{j} \frac{\partial V}{\partial y} + \hat{k} \frac{\partial V}{\partial z} \right)$.
Calculating the partial derivatives:
$\frac{\partial V}{\partial x} = \frac{d}{dx}(4x^2) = 8x$
$\frac{\partial V}{\partial y} = 0$
$\frac{\partial V}{\partial z} = 0$
Substituting these into the expression for $\vec{E}$:
$\vec{E} = -(8x \hat{i} + 0 \hat{j} + 0 \hat{k}) = -8x \hat{i} \text{ V/m}$.
At the point $(1 \text{ m}, 0, 2 \text{ m})$,the $x$-coordinate is $1 \text{ m}$.
Therefore,$\vec{E} = -8(1) \hat{i} = -8 \hat{i} \text{ V/m}$.
The negative sign indicates that the electric field is directed along the negative $X$-axis with a magnitude of $8 \text{ V/m}$.

(A) The structural analysis of crystals relies on the phenomenon of diffraction. For diffraction to occur, the wavelength of the incident radiation must be comparable to the spacing between the scattering centers (atoms) in the crystal lattice. The interatomic spacing in crystals is typically in the range of $0.1 \, nm$ to $0.5 \, nm$ ($1 \, \text{AA}$ to $5 \, \text{AA}$). $X-$ rays possess wavelengths in this same range, making them ideal for probing the atomic structure of crystals through $X-$ ray diffraction $(XRD)$. Therefore, option $A$ is correct.

(B) The cut-off voltage (stopping potential) depends only on the frequency of the incident light,not on its intensity. Therefore,it remains $0.6 \ V$.
The intensity of light from a point source follows the inverse square law,$I \propto \frac{1}{d^2}$.
When the distance increases from $0.2 \ m$ to $0.6 \ m$,the distance becomes $3$ times $(d' = 3d)$.
Thus,the intensity becomes $I' = \frac{I}{3^2} = \frac{I}{9}$.
Since the saturation current is directly proportional to the intensity of incident light,the new saturation current will be $I'_{sat} = \frac{18 \ mA}{9} = 2 \ mA$.
Comparing this with the given options,option $(b)$ is correct regarding the stopping potential.

(A) In a typical nuclear fusion reaction,such as the fusion of two deuterium nuclei $(^2H + ^2H \rightarrow ^3He + n + 3.27 \ MeV)$ or the fusion of deuterium and tritium $(^2H + ^3H \rightarrow ^4He + n + 17.6 \ MeV)$,the energy released per reaction is relatively small compared to fission.
However,when considering the energy released per unit mass,fusion is much more efficient.
For a typical fusion reaction involving light nuclei,the energy released is in the range of a few $MeV$ to about $25 \ MeV$.
Therefore,among the given options,$25 \ MeV$ is the most appropriate value for a typical fusion reaction.

(C) When a donor impurity ($+5$ valence) is added to pure silicon ($+4$ valence),the donor atom replaces a silicon atom in the crystal lattice.
Since the donor atom has one extra valence electron,it creates a net positive charge of $+1e$ at the site.
The four valence electrons form covalent bonds with neighboring silicon atoms.
The fifth electron is loosely bound to the donor site and can be modeled as an electron revolving around a $+1e$ charge in a medium with permittivity $\varepsilon = \varepsilon_0 \varepsilon_r$.
The energy levels are given by $E_n = -\frac{13.6 \times m^* / m_e}{\varepsilon_r^2 n^2} \text{ eV}$.
For silicon,the relative permittivity $\varepsilon_r \approx 12$.
Substituting $n=1$ and accounting for the effective mass of the electron in the crystal,the ionization energy is approximately $E \approx -\frac{13.6}{12^2} \approx -0.094 \text{ eV}$,which is approximately $0.1 \text{ eV}$.

(D) For an astronomical telescope in normal adjustment,the distance between the objective and the eye-piece is $L = f_o + f_e$. Given $f_o = 16 \, m$ and $f_e = 2 \, cm = 0.02 \, m$,so $L = 16 + 0.02 = 16.02 \, m$. Statement $A$ is correct.
The angular magnification $m$ is given by $m = -f_o / f_e = -16 / 0.02 = -800$. The magnitude of magnification is $800$. Statement $B$ is correct.
The negative sign in the magnification indicates that the final image formed by an astronomical telescope is inverted with respect to the object. Statement $C$ is correct.
Since all statements $A, B,$ and $C$ are correct,the correct option is $D$.