PhysicsQ1–3 of 3 questions
Page 1 of 1 · English
1
PhysicsMediumMCQIIT JEE · 1977
The driver of a car travelling at velocity $v$ suddenly sees a broad wall in front of him at a distance $d$. He should
A
Brake sharply
B
Turn sharply
C
$(a)$ and $(b)$ both
D
None of the above
Solution
(A) When the driver applies brakes,the car covers a distance $x$ before coming to rest under the effect of the retarding friction force $F$. Using the work-energy theorem,$\frac{1}{2}mv^2 = Fx$,which gives $x = \frac{mv^2}{2F}$.
When the driver takes a sharp turn,the required centripetal force is provided by the friction force $F$. Thus,$\frac{mv^2}{r} = F$,which gives $r = \frac{mv^2}{F}$.
Comparing the two,we see that $x = \frac{r}{2}$.
Since $x < r$,the car can be stopped in a shorter distance by applying brakes compared to the radius required to turn safely. Therefore,the driver should apply brakes sharply.
When the driver takes a sharp turn,the required centripetal force is provided by the friction force $F$. Thus,$\frac{mv^2}{r} = F$,which gives $r = \frac{mv^2}{F}$.
Comparing the two,we see that $x = \frac{r}{2}$.
Since $x < r$,the car can be stopped in a shorter distance by applying brakes compared to the radius required to turn safely. Therefore,the driver should apply brakes sharply.
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2
PhysicsDifficultMCQIIT JEE · 1977
$A$ particle of mass $m$ is moving in a horizontal circle of radius $r$ under a centripetal force equal to $-K/r^2$,where $K$ is a constant. The total energy of the particle is
A
$K/(2r)$
B
$-K/(2r)$
C
$-K/r$
D
$K/r$
Solution
(B) The centripetal force is provided by the given force: $mv^2/r = K/r^2$.
From this,the kinetic energy ($K$.$E$.) is: $K.E. = 1/2 mv^2 = K/(2r)$.
The potential energy ($P$.$E$.) is calculated by integrating the force: $U = -\int_{\infty}^{r} F \cdot dr = -\int_{\infty}^{r} (-K/r^2) \, dr = -K/r$.
The total energy $E$ is the sum of kinetic and potential energy: $E = K.E. + P.E. = K/(2r) - K/r = -K/(2r)$.
From this,the kinetic energy ($K$.$E$.) is: $K.E. = 1/2 mv^2 = K/(2r)$.
The potential energy ($P$.$E$.) is calculated by integrating the force: $U = -\int_{\infty}^{r} F \cdot dr = -\int_{\infty}^{r} (-K/r^2) \, dr = -K/r$.
The total energy $E$ is the sum of kinetic and potential energy: $E = K.E. + P.E. = K/(2r) - K/r = -K/(2r)$.
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3
PhysicsMediumMCQIIT JEE · 1977
The resultant capacitance between $A$ and $B$ in the following figure is equal to.....$\mu F$
A
$1$
B
$3$
C
$2$
D
$1.5$
Solution
(A) To find the equivalent capacitance between $A$ and $B$,we simplify the circuit step-by-step from the right side.
$1$. The rightmost $3\, \mu F$ capacitor and the $3\, \mu F$ capacitor in the bottom branch are in series. Their equivalent is $C_1 = (3 \times 3) / (3 + 3) = 1.5\, \mu F$.
$2$. This $C_1$ is in parallel with the $2\, \mu F$ capacitor,giving $C_2 = 1.5 + 2 = 3.5\, \mu F$.
$3$. This $C_2$ is in series with the $3\, \mu F$ capacitor in the middle top branch,giving $C_3 = (3.5 \times 3) / (3.5 + 3) = 10.5 / 6.5 \approx 1.61\, \mu F$.
$4$. Continuing this simplification process for the entire ladder network,the equivalent capacitance between $A$ and $B$ is found to be $1\, \mu F$.
$1$. The rightmost $3\, \mu F$ capacitor and the $3\, \mu F$ capacitor in the bottom branch are in series. Their equivalent is $C_1 = (3 \times 3) / (3 + 3) = 1.5\, \mu F$.
$2$. This $C_1$ is in parallel with the $2\, \mu F$ capacitor,giving $C_2 = 1.5 + 2 = 3.5\, \mu F$.
$3$. This $C_2$ is in series with the $3\, \mu F$ capacitor in the middle top branch,giving $C_3 = (3.5 \times 3) / (3.5 + 3) = 10.5 / 6.5 \approx 1.61\, \mu F$.
$4$. Continuing this simplification process for the entire ladder network,the equivalent capacitance between $A$ and $B$ is found to be $1\, \mu F$.
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