IIT JEE 1994 Physics Question Paper with Answer and Solution

(B) Given centripetal acceleration $a_c = k^2 r t^2$.
We know that $a_c = \frac{v^2}{r}$,where $v$ is the speed of the particle.
Equating the two expressions: $\frac{v^2}{r} = k^2 r t^2$.
This simplifies to $v^2 = k^2 r^2 t^2$,so $v = krt$.
The tangential acceleration $a_t$ is given by $a_t = \frac{dv}{dt} = \frac{d}{dt}(krt) = kr$.
The tangential force acting on the particle is $F_t = m a_t = mkr$.
The power $P$ delivered to the particle is the product of the tangential force and the velocity: $P = F_t \cdot v = (mkr) \cdot (krt) = m k^2 r^2 t$.

(B) The normal force $R$ exerted by the wall on the block is equal to the applied horizontal force,so $R = 5 \, N$.
The limiting friction $F_l$ is given by $F_l = \mu R = 0.5 \times 5 = 2.5 \, N$.
The weight of the block acting downwards is $W = mg = 0.1 \times 9.8 = 0.98 \, N$.
Since the downward force (weight) is less than the limiting friction $(0.98 \, N < 2.5 \, N)$,the block remains at rest.
For a block in equilibrium,the static frictional force $F_s$ must balance the downward weight. Therefore,$F_s = W = 0.98 \, N$.

(D) For a uniform sphere of mass $M$ and radius $R$,the gravitational field $g$ at a distance $r$ from the centre is given by:
$1$. Inside the sphere $(r < R)$: $g = \frac{GMr}{R^3}$,which implies $g \propto r$.
$2$. Outside the sphere $(r > R)$: $g = \frac{GM}{r^2}$,which implies $g \propto \frac{1}{r^2}$.
Since the gravitational force $F = mg$,the ratio of forces $\frac{F_1}{F_2}$ is equal to the ratio of gravitational fields $\frac{g_1}{g_2}$.
If $r_1 < R$ and $r_2 < R$,then $\frac{F_1}{F_2} = \frac{r_1}{r_2}$. This matches option $(a)$.
If $r_1 > R$ and $r_2 > R$,then $\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}$. This matches option $(b)$.
Therefore,both $(a)$ and $(b)$ are correct.

(A) Bernoulli's equation states that for an incompressible,non-viscous,and steady fluid flow,the sum of pressure energy,kinetic energy,and potential energy per unit volume remains constant along a streamline.
The dynamic lift of an aeroplane wing is a classic application of this principle.
As the wing moves through the air,the shape of the wing (airfoil) causes the air velocity above the wing to be higher than the air velocity below the wing.
According to Bernoulli's equation,where the velocity of the fluid is higher,the pressure is lower.
Therefore,the pressure below the wing is greater than the pressure above the wing,creating an upward force known as dynamic lift.

(B) Let $M$ be the mass of the ice block.
According to the principle of floatation,the weight of the ice equals the weight of the liquid displaced.
Thus,$M \cdot g = V_D \cdot \sigma_L \cdot g$,where $V_D$ is the volume of liquid displaced and $\sigma_L$ is the density of the liquid.
Therefore,$V_D = \frac{M}{\sigma_L}$.
When the ice melts,it forms water of mass $M$. The volume of this water is $V_F = \frac{M}{\sigma_W}$,where $\sigma_W$ is the density of water $(1 \ g/cm^3)$.
Given $\sigma_L = 1.2 \ g/cm^3$ and $\sigma_W = 1 \ g/cm^3$,we have $\sigma_L > \sigma_W$.
Comparing the volumes: Since $\sigma_L > \sigma_W$,it follows that $\frac{M}{\sigma_L} < \frac{M}{\sigma_W}$,which means $V_D < V_F$.
Since the volume of the water formed $(V_F)$ is greater than the volume of the liquid displaced by the floating ice $(V_D)$,the level of the liquid in the beaker will rise.

(D) According to Stefan-Boltzmann law,the total radiant power $P = e A \sigma T^4$. Since $P_A = P_B$ and $A_A = A_B$,we have $e_A T_A^4 = e_B T_B^4$.
Given $e_A = 0.01$,$e_B = 0.81$,and $T_A = 5802\;K$.
$T_B = T_A \left( \frac{e_A}{e_B} \right)^{1/4} = 5802 \times \left( \frac{0.01}{0.81} \right)^{1/4} = 5802 \times \left( \frac{1}{81} \right)^{1/4} = 5802 \times \frac{1}{3} = 1934\;K$.
From Wien's displacement law,$\lambda_A T_A = \lambda_B T_B = b$ (constant).
$\lambda_A = \frac{b}{T_A}$ and $\lambda_B = \frac{b}{T_B}$.
Given $\lambda_B - \lambda_A = 1.00\;\mu m$.
$\frac{b}{T_B} - \frac{b}{T_A} = 1.00\;\mu m$.
$b \left( \frac{1}{1934} - \frac{1}{5802} \right) = 1.00\;\mu m$.
$b \left( \frac{3 - 1}{5802} \right) = 1.00\;\mu m \Rightarrow b = \frac{5802}{2} = 2901\;\mu m \cdot K$.
Now,$\lambda_B = \frac{b}{T_B} = \frac{2901}{1934} = 1.5\;\mu m$.
Thus,both $(a)$ and $(b)$ are correct.

(D) Given: $L_1 = 8 \, mH$,$L_2 = 2 \, mH$,and $\frac{di_1}{dt} = \frac{di_2}{dt} = k$ (constant).
$1$. Induced voltage: $V = L \frac{di}{dt}$. Since $\frac{di}{dt}$ is the same for both,$\frac{V_2}{V_1} = \frac{L_2}{L_1} = \frac{2}{8} = \frac{1}{4}$. Thus,option $(b)$ is correct.
$2$. Power supplied: $P = V \cdot i$. Given $P_1 = P_2$,so $V_1 i_1 = V_2 i_2$. Therefore,$\frac{i_1}{i_2} = \frac{V_2}{V_1} = \frac{1}{4}$. Thus,option $(a)$ is correct.
$3$. Energy stored: $W = \frac{1}{2} L i^2$. The ratio is $\frac{W_2}{W_1} = \left( \frac{L_2}{L_1} \right) \left( \frac{i_2}{i_1} \right)^2 = \left( \frac{1}{4} \right) \left( 4 \right)^2 = \frac{1}{4} \times 16 = 4$. Thus,option $(c)$ is correct.
Since $(a)$,$(b)$,and $(c)$ are all correct,the answer is $(d)$.

(D) Using Einstein's photoelectric equation: $K_{max} = E - \Phi$,where $\Phi$ is the work function.
For metal $A$: $T_A = 4.25 - \Phi_A$ ... $(i)$
For metal $B$: $T_B = T_A - 1.50 = 4.70 - \Phi_B$ ... (ii)
From $(i)$,$\Phi_A = 4.25 - T_A$. From (ii),$\Phi_B = 4.70 - (T_A - 1.50) = 6.20 - T_A$.
The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mK}}$,so $\lambda \propto \frac{1}{\sqrt{K}}$.
Given $\lambda_B = 2\lambda_A$,we have $\frac{\lambda_B}{\lambda_A} = \sqrt{\frac{T_A}{T_B}} = 2$.
Squaring both sides: $\frac{T_A}{T_A - 1.50} = 4$.
$T_A = 4T_A - 6.00 \Rightarrow 3T_A = 6.00 \Rightarrow T_A = 2.00\, eV$.
Substituting $T_A = 2.00\, eV$ into $(i)$: $\Phi_A = 4.25 - 2.00 = 2.25\, eV$.
Substituting $T_A = 2.00\, eV$ into (ii): $T_B = 2.00 - 1.50 = 0.50\, eV$. Then $\Phi_B = 4.70 - 0.50 = 4.20\, eV$.
Thus,all statements are correct.

(D) The rest mass of a stable nucleus is always less than the sum of the rest masses of its constituent nucleons. This difference is known as the mass defect,$\Delta m$. The energy equivalent to this mass defect,given by $\Delta m c^2$,is the binding energy that keeps the nucleons together.
Nuclear fission is a process in which a very heavy nucleus splits into two or more lighter nuclei,releasing a significant amount of energy.
Nuclear fusion involves the combining of two light nuclei to form a heavier nucleus,not medium-mass nuclei as stated in option $(c)$.
Therefore,both statements $(a)$ and $(b)$ are correct.

(B) Fast neutrons are slowed down by passing them through a moderator.
Materials containing light nuclei,such as water $(H_2O)$,heavy water $(D_2O)$,or graphite,are effective moderators.
When a fast neutron collides elastically with a light nucleus (like hydrogen or deuterium),it transfers a significant portion of its kinetic energy to the nucleus,thereby reducing its speed.
Therefore,passing neutrons through water is an effective method to slow them down.

(A) The penetrating power of radioactive radiations depends on their mass and charge.
Alpha particles $(\alpha)$ are heavy and carry a $+2e$ charge, making them highly ionizing but having the least penetrating power.
Beta particles $(\beta)$ are high-speed electrons with much smaller mass, giving them greater penetrating power than $\alpha$ particles.
Gamma rays $(\gamma)$ are high-energy electromagnetic waves with no mass and no charge, allowing them to have the highest penetrating power.
Specifically, the penetrating power of $\gamma$ is approximately $100$ times that of $\beta$, and the penetrating power of $\beta$ is approximately $100$ times that of $\alpha$.
Therefore, the increasing order of penetrating power is $\alpha$ < $\beta$ < $\gamma$.

(A) $PN$ junction diode acts as a conductor when it is forward-biased.
In forward bias,the positive terminal of the battery is connected to the $P$-region and the negative terminal to the $N$-region.
This reduces the width of the depletion layer and the height of the potential barrier.
For significant current to flow,the applied forward bias voltage must exceed the barrier potential (which is approximately $0.7 \ V$ for Silicon and $0.3 \ V$ for Germanium).
Therefore,the correct condition is that the forward bias should be greater than the barrier potential.

(D) Given: $\lambda = 600 \, nm = 600 \times 10^{-9} \, m$,slit width $a = 1 \, mm = 10^{-3} \, m$,and distance $D = 2 \, m$.
The distance between the first dark fringes on either side of the central bright fringe is equal to the width of the central maximum.
The formula for the width of the central maximum is given by $w = \frac{2 \lambda D}{a}$.
Substituting the values:
$w = \frac{2 \times (600 \times 10^{-9} \, m) \times 2 \, m}{10^{-3} \, m}$
$w = \frac{2400 \times 10^{-9}}{10^{-3}} \, m$
$w = 2400 \times 10^{-6} \, m = 2.4 \times 10^{-3} \, m = 2.4 \, mm$.