IIT JEE 1983 Physics Question Paper with Answer and Solution

(A) The formula for heat energy $Q$ required to change the state of a substance of mass $m$ is given by $Q = mL$,where $L$ is the latent heat.
Therefore,the latent heat $L$ is given by $L = \frac{Q}{m}$.
Since heat $Q$ is a form of energy,its dimensional formula is $[M^1 L^2 T^{-2}]$.
The dimensional formula for mass $m$ is $[M^1]$.
Substituting these into the formula for $L$:
$L = \frac{[M^1 L^2 T^{-2}]}{[M^1]} = [M^0 L^2 T^{-2}]$.
Thus,the correct option is $A$.

(B) Angular momentum $(L)$ is defined as the product of linear momentum $(p)$ and the perpendicular distance $(r)$ from the axis of rotation.
Mathematically,$L = p \times r$.
Since linear momentum $p = m \times v$,where $m$ is mass and $v$ is velocity.
The dimensional formula for mass $m$ is $[M]$.
The dimensional formula for velocity $v$ is $[L{T^{ - 1}}]$.
The dimensional formula for distance $r$ is $[L]$.
Therefore,the dimensional formula for angular momentum is $[M] \times [L{T^{ - 1}}] \times [L] = [M{L^2}{T^{ - 1}}]$.

(A) The capacitance $C$ is defined as the ratio of charge $Q$ to potential difference $V$,given by $C = \frac{Q}{V}$.
Since $V = \frac{W}{Q}$,where $W$ is work done,we can write $C = \frac{Q^2}{W}$.
The dimensional formula for charge $Q$ is $[A^1T^1]$.
The dimensional formula for work $W$ is $[M^1L^2T^{-2}]$.
Substituting these into the formula: $C = \frac{[A^1T^1]^2}{[M^1L^2T^{-2}]} = \frac{[A^2T^2]}{[M^1L^2T^{-2}]}$.
Simplifying the expression,we get $[C] = [M^{-1}L^{-2}T^4A^2]$.

(A) The energy $E$ stored in an inductor is given by the formula $E = \frac{1}{2} L i^2$.
Rearranging for $L$,we get $L = \frac{2E}{i^2}$.
The dimensional formula for energy $E$ is $[M L^2 T^{-2}]$.
The dimensional formula for current $i$ is $[A]$.
Substituting these into the expression for $L$:
$L = \frac{[M L^2 T^{-2}]}{[A]^2} = [M L^2 T^{-2} A^{-2}]$.
Thus,the correct option is $A$.

(A) Torque $( \tau)$ is defined as the product of force and the perpendicular distance from the axis of rotation.
Mathematically,$\tau = \text{Force} \times \text{Distance}$.
The dimensional formula for force is $[M L T^{-2}]$.
The dimensional formula for distance is $[L]$.
Therefore,the dimensional formula for torque is $[M L T^{-2}] \times [L] = [M L^2 T^{-2}]$.
Thus,the correct option is $A$.

(B) The initial moment of inertia of the ring about its axis is $I = Mr^2$.
The initial angular momentum is $L = I\omega = Mr^2\omega$.
When two objects of mass $m$ are attached to the opposite ends of a diameter,the new moment of inertia $I'$ becomes the sum of the ring's moment of inertia and the moment of inertia of the two point masses: $I' = Mr^2 + m(r)^2 + m(r)^2 = (M + 2m)r^2$.
According to the principle of conservation of angular momentum,the external torque is zero,so $L_{initial} = L_{final}$.
$Mr^2\omega = (M + 2m)r^2\omega'$
Solving for the new angular velocity $\omega'$:
$\omega' = \frac{Mr^2\omega}{(M + 2m)r^2} = \frac{M\omega}{M + 2m}$.

(B) For a charged hollow metal sphere,the electric field inside the sphere is zero $(E = 0)$.
Since the electric field is the negative gradient of the potential $(E = -dV/dr)$,if $E = 0$,then the potential $V$ must be constant throughout the interior of the sphere.
Therefore,the potential at any point inside the sphere,including the centre,is equal to the potential on its surface.
Given that the potential on the surface is $10\, V$,the potential at the centre is also $10\, V$.

(C) Initially,the switch $S$ is closed. Both capacitors $A$ and $B$ are connected in parallel to the battery of potential $V$. The capacitance of each capacitor is $C$.
The initial total energy stored in the system is:
$U_1 = \frac{1}{2}CV^2 + \frac{1}{2}CV^2 = CV^2$ --- $(i)$
When the switch $S$ is opened,capacitor $A$ remains connected to the battery,so its potential difference remains $V$. Capacitor $B$ is disconnected from the battery,so its charge $Q$ remains constant. The initial charge on capacitor $B$ was $Q = CV$.
After filling the space with a dielectric of constant $K = 3$,the new capacitance of each capacitor becomes $C' = KC = 3C$.
For capacitor $A$,the potential difference remains $V$. The new energy is $U_{A}' = \frac{1}{2}C'V^2 = \frac{1}{2}(3C)V^2 = \frac{3}{2}CV^2$.
For capacitor $B$,the charge $Q = CV$ remains constant. The new energy is $U_{B}' = \frac{Q^2}{2C'} = \frac{(CV)^2}{2(3C)} = \frac{1}{6}CV^2$.
The final total energy stored in the system is:
$U_2 = U_{A}' + U_{B}' = \frac{3}{2}CV^2 + \frac{1}{6}CV^2 = \left(\frac{9+1}{6}\right)CV^2 = \frac{10}{6}CV^2 = \frac{5}{3}CV^2$ --- (ii)
The ratio of total electrostatic energy before and after is:
$\frac{U_1}{U_2} = \frac{CV^2}{(5/3)CV^2} = \frac{3}{5}$

(C) The circuit consists of a $2 \, V$ battery connected to a delta network of three $30 \, \Omega$ resistors.
One node of the triangle is connected to the negative terminal,and the opposite side is connected to the positive terminal.
This results in two $30 \, \Omega$ resistors being in series,and this combination being in parallel with the third $30 \, \Omega$ resistor.
Equivalent resistance $R_{eq} = \frac{(30 + 30) \times 30}{(30 + 30) + 30} = \frac{60 \times 30}{90} = 20 \, \Omega$.
Using Ohm's law,the current $i = \frac{V}{R_{eq}} = \frac{2}{20} = \frac{1}{10} \, A$.

(C) The magnetic force on a small current element $d\overrightarrow{l}$ in a magnetic field $\overrightarrow{B}$ is given by $d\overrightarrow{F} = i(d\overrightarrow{l} \times \overrightarrow{B})$.
For a closed loop,the net magnetic force is $\overrightarrow{F} = \oint i(d\overrightarrow{l} \times \overrightarrow{B})$.
Since the magnetic field $\overrightarrow{B}$ is uniform,it can be taken out of the integral: $\overrightarrow{F} = i(\oint d\overrightarrow{l}) \times \overrightarrow{B}$.
For any closed loop,the vector sum of all infinitesimal length elements $\oint d\overrightarrow{l}$ is zero.
Therefore,the net magnetic force $\overrightarrow{F} = i(0) \times \overrightarrow{B} = 0$.

(A) The maximum number of electrons (and thus elements) that can be accommodated in a shell with principal quantum number $n$ is given by the formula $2n^2$.
For $n = 1$,the number of elements is $2(1)^2 = 2$.
For $n = 2$,the number of elements is $2(2)^2 = 8$.
For $n = 3$,the number of elements is $2(3)^2 = 18$.
For $n = 4$,the number of elements is $2(4)^2 = 32$.
Since the principal quantum number $n$ cannot exceed $4$,the total number of possible elements is the sum of elements in shells $n=1, 2, 3,$ and $4$.
Total elements $= 2 + 8 + 18 + 32 = 60$.

(D) The wavelength $\lambda$ of the spectral line emitted during a transition between energy levels is given by the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For a transition from $n_2 = 2$ to $n_1 = 1$,the term $\left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \frac{3}{4}$ is constant for all hydrogen-like species.
Therefore,$\frac{1}{\lambda} \propto Z^2$,which implies $\lambda \propto \frac{1}{Z^2}$.
To obtain the minimum wavelength $\lambda$,the atomic number $Z$ must be maximum.
- Hydrogen atom $(H)$: $Z = 1$
- Deuterium atom $(D)$: $Z = 1$
- Uni-ionized helium $(He^+)$: $Z = 2$
- Di-ionized lithium $(Li^{2+})$: $Z = 3$
Since $Li^{2+}$ has the highest atomic number $(Z = 3)$,it produces the spectral line with the minimum wavelength.

(C) Radioactive decay involves the transformation of a neutron into a proton and an electron (or vice versa) within the nucleus.
$\beta$-particles are high-energy electrons or positrons emitted from the nucleus during this process.
Since they are electrons or positrons,they carry a net electric charge (negative for $\beta^-$ and positive for $\beta^+$).
Therefore,$\beta$-rays are charged particles emitted by the nucleus.

(B) The intensity of radiation $I$ decreases according to the law of radioactive decay: $I = I_0 \left( \frac{1}{2} \right)^n$, where $n$ is the number of half-lives.
Given that the initial intensity $I_0 = 64 I_{\text{safe}}$, we need the intensity to reach $I_{\text{safe}}$.
Thus, $\frac{I_{\text{safe}}}{64 I_{\text{safe}}} = \left( \frac{1}{2} \right)^n$.
$\frac{1}{64} = \left( \frac{1}{2} \right)^n$.
Since $64 = 2^6$, we have $\left( \frac{1}{2} \right)^6 = \left( \frac{1}{2} \right)^n$, which implies $n = 6$.
The total time $t$ is given by $t = n \times T_{1/2}$, where $T_{1/2} = 2 \text{ hours}$.
$t = 6 \times 2 = 12 \text{ hours}$.

(C) According to the law of reflection,the angle of incidence $i$ is equal to the angle of reflection $r$,so $i = r$.
Given that the reflected ray and the refracted ray are mutually perpendicular,the sum of the angle of reflection $r$,the angle between the reflected ray and the refracted ray $(90^{\circ})$,and the angle of refraction $r'$ is $180^{\circ}$.
Thus,$r + 90^{\circ} + r' = 180^{\circ}$,which implies $r' = 90^{\circ} - r$. Since $i = r$,we have $r' = 90^{\circ} - i$.
Applying Snell's law at the interface,the refractive index of the denser medium with respect to the rarer medium is given by $\mu = \frac{\sin r'}{\sin i}$.
We know that the critical angle $C$ is related to the refractive index by $\sin C = \frac{1}{\mu}$.
Therefore,$\sin C = \frac{\sin i}{\sin r'} = \frac{\sin i}{\sin(90^{\circ} - i)} = \frac{\sin i}{\cos i} = \tan i$.
Hence,the critical angle is $C = \sin^{-1}(\tan i)$.