IIT JEE 1990 Physics Question Paper with Answer and Solution

(D) The direction of the frictional force depends on the state of motion of the bicycle.
$1$. When pedalling: The rear wheel is driven by the chain,pushing the ground backward,so the ground exerts a forward frictional force on the rear wheel. The front wheel is rolling freely,so it experiences a backward frictional force to oppose its motion.
$2$. When coasting (pedalling stopped): Both wheels are rolling freely. In this case,both wheels experience a backward frictional force from the ground to oppose their forward motion.
Since the question asks about the general motion of a bicycle,it covers both scenarios depending on whether the rider is pedalling or coasting. Therefore,both $(a)$ and $(c)$ are valid descriptions of the frictional force acting on the wheels under different conditions.

(D) The fraction of supplied heat energy that increases the internal energy is given by the ratio of the change in internal energy to the heat supplied at constant pressure.
$f = \frac{\Delta U}{(\Delta Q)_P} = \frac{(\Delta Q)_V}{(\Delta Q)_P} = \frac{\mu C_V \Delta T}{\mu C_P \Delta T} = \frac{C_V}{C_P} = \frac{1}{\gamma}$.
For an ideal diatomic gas,the adiabatic index $\gamma = \frac{C_P}{C_V} = \frac{7}{5}$.
Therefore,the fraction $f = \frac{1}{7/5} = \frac{5}{7}$.

(D) The standard wave equation for a wave traveling in the negative $x$-direction is $y = A\sin(kx + \omega t + \phi_0)$.
Comparing the given equation $Y = A\sin(10\pi x + 15\pi t + \frac{\pi}{3})$ with the standard form,we identify the angular frequency $\omega = 15\pi\,rad/s$ and the wave number $k = 10\pi\,rad/m$.
The wave velocity $v$ is given by $v = \frac{\omega}{k} = \frac{15\pi}{10\pi} = 1.5\,m/s$. Since the signs of the $x$ and $t$ terms are both positive,the wave travels in the negative $x$-direction.
The wavelength $\lambda$ is given by $\lambda = \frac{2\pi}{k} = \frac{2\pi}{10\pi} = 0.2\,m$.
Thus,both statements $(b)$ and $(c)$ are correct.

(A) Let the capacity of each capacitor be $C = 2\,\mu F$. We need a total equivalent capacity $C_{eq} = \frac{10}{11}\,\mu F$.
Let $n$ capacitors be connected in parallel and $m$ capacitors be connected in series with this parallel combination.
The equivalent capacity of $n$ capacitors in parallel is $C_p = nC = n(2)\,\mu F$.
The total equivalent capacity $C_{eq}$ is given by the series combination of $C_p$ and $m$ capacitors in series:
$\frac{1}{C_{eq}} = \frac{1}{C_p} + \frac{m}{C} = \frac{1}{2n} + \frac{m}{2} = \frac{1 + nm}{2n}$.
Given $C_{eq} = \frac{10}{11}\,\mu F$,we have $\frac{2n}{1 + nm} = \frac{10}{11}$.
$22n = 10 + 10nm \implies 11n = 5 + 5nm \implies 5nm = 11n - 5$.
Since the total number of capacitors is $n + m = 7$,we have $m = 7 - n$.
Substituting $m$: $5n(7 - n) = 11n - 5 \implies 35n - 5n^2 = 11n - 5 \implies 5n^2 - 24n - 5 = 0$.
Solving the quadratic equation: $5n^2 - 25n + n - 5 = 0 \implies 5n(n - 5) + 1(n - 5) = 0 \implies (5n + 1)(n - 5) = 0$.
Thus,$n = 5$ and $m = 7 - 5 = 2$.
This means $5$ capacitors are in parallel and $2$ capacitors are in series with this parallel group. This matches the arrangement in figure $(a)$.

(B) For dispersion without deviation,the net deviation produced by the combination of two thin prisms must be zero.
The condition for dispersion without deviation is given by: $(\mu - 1)A + (\mu' - 1)A' = 0$.
Here,$A = 4^o$,$\mu = 1.54$,and $\mu' = 1.72$.
Since the prisms are combined to produce dispersion without deviation,the deviation produced by the first prism must be balanced by the second prism in the opposite direction.
Thus,$(\mu - 1)A = -(\mu' - 1)A'$.
Taking the magnitude: $(\mu - 1)A = (\mu' - 1)A'$.
Substituting the values: $(1.54 - 1) \times 4^o = (1.72 - 1) \times A'$.
$0.54 \times 4^o = 0.72 \times A'$.
$A' = \frac{0.54 \times 4}{0.72} = \frac{2.16}{0.72} = 3^o$.
Therefore,the angle of prism $P_2$ is $3^o$.

(B) Let the incident intensity be $I$.
At point $A$,$25\%$ of $I$ is reflected as ray $AB$. So,$I_1 = 0.25I = I/4$.
The intensity of the refracted ray is $0.75I$.
At point $C$,this ray is reflected. $25\%$ of the incident energy $(0.75I)$ is reflected. So,the intensity of the ray reaching $A^1$ is $0.25 \times 0.75I = 0.1875I = (3/16)I$.
At point $A^1$,this ray is refracted out as $A^1B^1$. Since $25\%$ is reflected at $A^1$,$75\%$ of the incident energy is transmitted.
So,$I_2 = 0.75 \times (3/16)I = (3/4) \times (3/16)I = (9/64)I$.
The amplitudes are $a_1 = \sqrt{I_1} = \sqrt{I/4} = (1/2)\sqrt{I}$ and $a_2 = \sqrt{I_2} = \sqrt{9I/64} = (3/8)\sqrt{I}$.
The ratio of amplitudes is $a_1/a_2 = (1/2) / (3/8) = 4/3$.
$I_{\text{max}} / I_{\text{min}} = (a_1 + a_2)^2 / (a_1 - a_2)^2 = ((4+3)/ (4-3))^2 = (7/1)^2 = 49/1$.