IIT JEE 1978 Physics Question Paper with Answer and Solution

(D) Let the car accelerate at a constant rate $\alpha$ for time $t_1$. The maximum velocity $v$ attained is given by $v = u + \alpha t_1$. Since it starts from rest,$u = 0$,so $v = \alpha t_1$.
Next,the car decelerates at a constant rate $\beta$ for the remaining time $(t - t_1)$ until it comes to rest. Using the equation $v = u + at$ for the deceleration phase,we have $0 = v - \beta(t - t_1)$.
Substituting $v = \alpha t_1$ into the equation,we get $0 = \alpha t_1 - \beta t + \beta t_1$.
Rearranging the terms,we get $(\alpha + \beta) t_1 = \beta t$,which gives $t_1 = \frac{\beta}{\alpha + \beta} t$.
Substituting $t_1$ back into the expression for $v$,we get $v = \alpha \left( \frac{\beta}{\alpha + \beta} t \right) = \frac{\alpha \beta}{\alpha + \beta} t$.

(D) The heat produced by the heater per unit time is $P = \frac{V^2}{R} = \frac{200^2}{20} = 2000 \ W = 2000 \ J/s$.
The heat conducted through the glass window per unit time is given by $H = \frac{K A (T_{in} - T_{out})}{d}$.
Here, $K = 0.2 \ W/(m \cdot K)$, $A = 1 \ m^2$, $d = 0.2 \ cm = 0.002 \ m$, and $T_{in} = 20^{\circ}C$.
Equating the heat produced to the heat conducted: $2000 = \frac{0.2 \times 1 \times (20 - T_{out})}{0.002}$.
$2000 = 100 \times (20 - T_{out})$.
$20 = 20 - T_{out}$.
$T_{out} = 0^{\circ}C$. Since this result is not among the options, the correct choice is $D$.