IIT JEE 1985 Physics Question Paper with Answer and Solution

(C) The energy of a photon is given by the equation $E = h \nu$,where $E$ is energy,$h$ is Planck's constant,and $\nu$ is frequency.
Rearranging for $h$,we get $h = \frac{E}{\nu}$.
The dimensional formula for energy $E$ is $[M L^2 T^{-2}]$.
The dimensional formula for frequency $\nu$ is $[T^{-1}]$.
Substituting these into the equation: $[h] = \frac{[M L^2 T^{-2}]}{[T^{-1}]} = [M L^2 T^{-1}]$.
Therefore,the correct option is $C$.

(D) When the block $m$ is immersed in the liquid,it experiences an upward buoyant force (upthrust) $F_B$ according to Archimedes' principle.
For balance $A$,the reading is the tension $T$ in the string. Initially,$T = mg = 2 \, kg \cdot g$. When immersed,$T = mg - F_B$. Since $F_B > 0$,the reading of balance $A$ will be less than $2 \, kg$.
For balance $B$,it measures the normal force exerted by the beaker on the pan. Initially,it reads the weight of the beaker plus the liquid. When the block is immersed,the liquid exerts an upward force $F_B$ on the block. By Newton's third law,the block exerts an equal and opposite downward force $F_B$ on the liquid. Thus,the total downward force on the balance $B$ increases by $F_B$. Therefore,the reading of balance $B$ will be more than $5 \, kg$.
Since both statements $(b)$ and $(c)$ are correct,the correct option is $(d)$.

(D) The mass of the hanging part of the chain is $m = M/3$.
The length of the hanging part is $l = L/3$.
The center of mass of the hanging part is at a distance $h = l/2 = L/6$ below the edge of the table.
The potential energy of the hanging part relative to the table surface is $U = -mgh = -(M/3)g(L/6) = -MgL/18$.
To pull the hanging part onto the table, the work done by an external force must be equal to the change in potential energy, which is $W = \Delta U = U_{final} - U_{initial} = 0 - (-MgL/18) = MgL/18$.

(B) Given: $\mu = 2 \, moles$,$(\Delta Q)_P = 70 \, cal$,$\Delta T = 35^{\circ}C - 30^{\circ}C = 5 \, K$,$R = 2 \, cal/mol \cdot K$.
At constant pressure,the heat supplied is given by $(\Delta Q)_P = \mu C_P \Delta T$.
Substituting the values: $70 = 2 \times C_P \times 5$.
$70 = 10 \times C_P$,which gives $C_P = 7 \, cal/mol \cdot K$.
Using the Mayer's relation: $C_P - C_V = R$.
$C_V = C_P - R = 7 - 2 = 5 \, cal/mol \cdot K$.
Now,at constant volume,the heat required is $(\Delta Q)_V = \mu C_V \Delta T$.
$(\Delta Q)_V = 2 \times 5 \times 5 = 50 \, cal$.

(B) The angular momentum $L$ of a particle about the origin is given by $L = r \times p$,where $r$ is the position vector and $p$ is the linear momentum.
Alternatively,the magnitude of angular momentum is $L = p \times d$,where $d$ is the perpendicular distance from the origin to the line of motion.
Here,the mass $m$ moves with constant velocity $v$,so the linear momentum $p = mv$ is constant.
The perpendicular distance $d$ from the origin to the line of motion is constant and equal to $a$.
Therefore,the angular momentum $L = mv \times a = mva$,which is a constant value.

(D) When a dielectric slab is inserted into a parallel plate capacitor while the battery remains connected,the potential difference $V$ across the plates remains constant,i.e.,$V = {V_0}$.
Since the capacitance $C$ increases by a factor of $K$ (dielectric constant),where $C = KC_0$,the charge $Q$ on the plates increases according to $Q = CV = K C_0 V_0 = K Q_0$. Thus,$Q > {Q_0}$.
The electric field $E$ between the plates is given by $E = V/d$. Since $V$ and $d$ remain constant,$E = {E_0}$.
The energy stored in the capacitor is $U = \frac{1}{2}CV^2$. Since $C$ increases and $V$ remains constant,$U$ increases,i.e.,$U > {U_0}$.
Therefore,both statements $(a)$ and $(b)$ are correct.

(D) The force on a proton moving with velocity $\overrightarrow{v}$ in the presence of electric field $\overrightarrow{E}$ and magnetic field $\overrightarrow{B}$ is given by the Lorentz force law: $\overrightarrow{F} = q(\overrightarrow{E} + \overrightarrow{v} \times \overrightarrow{B})$.
Since the proton moves with a constant velocity,the net force $\overrightarrow{F}$ must be zero.
Case $1$: If $\overrightarrow{E} = 0$ and $\overrightarrow{B} = 0$,then $\overrightarrow{F} = 0$. The proton continues in a straight line.
Case $2$: If $\overrightarrow{E} = 0$ and $\overrightarrow{B} \neq 0$,the force $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$. If $\overrightarrow{B}$ is parallel or anti-parallel to $\overrightarrow{v}$,then $\overrightarrow{v} \times \overrightarrow{B} = 0$,so $\overrightarrow{F} = 0$.
Case $3$: If $\overrightarrow{E} \neq 0$ and $\overrightarrow{B} \neq 0$,the forces can cancel each other if $q\overrightarrow{E} = -q(\overrightarrow{v} \times \overrightarrow{B})$,i.e.,$\overrightarrow{E} = -(\overrightarrow{v} \times \overrightarrow{B})$. This is the principle of a velocity selector.
Thus,all the given conditions are possible.

(C) The magnetic field produced by the long straight wire at a distance $r$ is given by $B = \frac{\mu_0 i}{2\pi r}$.
For the side of the loop closer to the wire at distance $r_1$,the current flows in the same direction as the wire. According to Fleming's left-hand rule,the force $F_1$ on this side is attractive (towards the wire).
For the side of the loop farther from the wire at distance $r_2$,the current flows in the opposite direction to the wire. The force $F_2$ on this side is repulsive (away from the wire).
Since $r_1 < r_2$,the magnetic field $B_1$ at the closer side is stronger than the magnetic field $B_2$ at the farther side $(B_1 > B_2)$.
Consequently,the attractive force $F_1$ is greater than the repulsive force $F_2$ $(F_1 > F_2)$.
Therefore,the net force $F_{net} = F_1 - F_2$ acts towards the wire,and the loop will move towards the wire.

(C) In an $X-$ray tube,electrons are accelerated by a potential difference $V$ and strike a metal target.
This process produces a continuous spectrum of $X-$rays due to Bremsstrahlung (braking radiation).
As an electron decelerates,it emits a photon with energy $E = hf = hc/\lambda$.
The maximum energy of the emitted photon corresponds to the kinetic energy of the incident electron,$E_{max} = eV = hc/\lambda_{min}$.
Therefore,the emitted $X-$ray spectrum contains all wavelengths greater than or equal to this minimum wavelength $\lambda_{min} = hc/eV$.
Thus,the beam consists of all wavelengths larger than a certain minimum wavelength.

(C) To produce characteristic $X$-rays,the incident electron must have enough energy to knock out an electron from the inner shell of the target atom.
This requires the kinetic energy of the incident electron to be at least equal to the binding energy of the electron in that shell.
The kinetic energy of an electron accelerated through a potential difference $V$ is given by $K = eV$.
Therefore,to eject an electron from the innermost shell (binding energy $E_b = 40 \text{ keV}$),the condition is $eV > E_b$.
Substituting the values,$eV > 40 \text{ keV}$,which implies $V > 40 \text{ kV}$.

(A) The characteristic $X$-rays are produced when an electron from a higher energy shell (e.g.,$L, M, N, \dots$ shells) transitions to a vacancy in the innermost shell (the $K$-shell).
The energy of the emitted characteristic $X$-ray photon is given by the difference in energy between the two shells involved in the transition: $E_{X-ray} = E_{higher} - E_{lower}$.
Since the electron transitions from a higher energy level (closer to $0 \, eV$) to the innermost shell (which has a binding energy of $40 \, keV$,meaning its energy level is $-40 \, keV$),the energy released must be less than the binding energy of the innermost shell.
Therefore,the energy of the characteristic $X$-rays is always less than the binding energy of the innermost electron.

(B) In a triode,the plate current is determined by the number of electrons reaching the plate $P$ from the cathode $K$.
When the grid $G$ is given a positive potential,it exerts an attractive force on the electrons emitted from the cathode,helping them pass through the grid mesh towards the plate.
Since the plate $P$ must also be at a positive potential to attract these electrons,having both the grid and the plate at positive potentials results in the maximum flow of electrons to the plate,thereby maximizing the plate current.
Therefore,the correct option is $B$.

(B) For a parallel beam of light to remain parallel after passing through a combination of two lenses,the second focal point of the first lens must coincide with the first focal point of the second lens.
Let $f_1 = 20 \, cm$ (convex lens) and $f_2 = -5 \, cm$ (concave lens).
The distance $d$ between the lenses is given by the formula $d = f_1 + f_2$.
Substituting the values,we get $d = 20 \, cm + (-5 \, cm) = 15 \, cm$.
Thus,the distance $d$ is $15 \, cm$.